Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 4 Laws of Motion Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 4 Laws of Motion

Question 1.
‘Rest and motion are relative concepts.’ Explain the statement with an example.
Answer:

  1. A body can be described to be at rest or in motion with respect to a system of co¬ordinate axes known as the frame of reference.
  2. A body is in motion if it changes its position with respect to a fixed reference point in a frame of reference. On the other hand, a body is at rest if it does not change its position with respect to a fixed reference point in a frame of reference.
  3. An object can be said to be at rest with respect to a frame of reference while the same object can be said to be in motion with respect to a different frame of reference.
    Example: In a running train, all the travellers in the train are in a state of rest if the train is taken as the frame of reference. On the other hand, all the travellers in the train are in a state of motion if ground (or platform) is taken as the frame of reference.
  4. Thus, motion and rest always need a frame of reference to be described. Hence, rest and motion are relative concepts.

Question 2.
Explain how acceleration and initial velocity decides trajectory of a motion.
Answer:

  1. The resultant motion is linear if:
    • initial velocity \(\overrightarrow{\mathrm{u}}\) = 0 (starting from rest) and acceleration \(\overrightarrow{\mathrm{a}}\) is in any direction.
    • initial velocity \(\overrightarrow{\mathrm{u}}\) ≠ 0 and acceleration a is in line with the initial velocity (same or opposite direction).
  2. The resultant motion is circular if initial velocity \(\overrightarrow{\mathrm{u}}\) ≠ 0 and acceleration \(\overrightarrow{\mathrm{a}}\) is perpendicular to the velocity throughout.
  3. The resultant motion is parabolic if the initial velocity \(\overrightarrow{\mathrm{u}}\) is not in line with the acceleration \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{a}}\) = constant.
    e.g., trajectory of a projectile motion.
  4. Similarly, various other combinations of initial velocity and acceleration will result into more complicated motions.

Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion

Question 3.
State Newton’s first law of motion.
Answer:
Statement: Every inanimate object continues to be in a state of rest or of uniform unaccelerated motion along a straight line, unless it is acted upon by an external, unbalanced force.

Question 4.
An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a constant rate of 100 m s-2. What is the acceleration of the astronaut the instant after he is outside the spaceship? (Assume that there are no nearby stars to exert gravitational force on him.) (NCERT)
Answer:

  1. Assuming absence of stars in the vicinity, the only gravitational force exerted on astronaut is by the spaceship.
  2. But this force is negligible.
  3. Hence, once astronaut is out of the spaceship net external force acting on him can be taken as zero.
  4. From the first law of motion, the acceleration of astronaut is zero.

Question 5.
Give the magnitude and direction of the net force acting on:

  1. a drop of rain falling down with a constant speed.
  2. a cork of mass 10 g floating on water.
  3. a kite skilfully held stationary in the sky.
  4. a car moving with a constant velocity of 30 kmh-1 on a rough road.
  5. a high speed electron in space far from all gravitating objects, and free of electric and magnetic fields. (NCERT)

Answer:

  1. The drop of rain falls down with a constant speed, hence according to the first law of motion, the net force on the drop of rain is zero.
  2. Since the 10 g cork is floating on water, its weight is balanced by the up thrust due to water. Therefore, net force on the cork is zero.
  3. As the kite is skilfully held stationary in the sky, in accordance with first law of motion, the net force on the kite is zero.
  4. As the car is moving with a constant velocity of 30 km/h on a road, the net force on the car is zero.
  5. As the high-speed electron in space is far from all material objects, and free of electric and magnetic fields, it doesn’t accelerate and moves with constant velocity. Hence, net force acting on the electron is zero.

Question 6.
State Newton’s second law of motion and its importance.
Answer:
Statement: The rate of change of linear momentum of a rigid body is directly proportional to the applied (external unbalanced) force and takes place in the direction of force.
\(\overrightarrow{\mathrm{F}}=\frac{\mathrm{d} \overrightarrow{\mathrm{p}}}{\mathrm{dt}}\)
Where, \(\overrightarrow{\mathrm{F}}\) = Force applied
p = m\(\overrightarrow{\mathrm{v}}\) = linear momentum

Importance of Newton’s second law:

  1. It gives mathematical formulation for quantitative measure of force as rate of change of linear momentum.
  2. It defines momentum instead of velocity as the fundamental quantity related to motion.
  3. It takes into consideration the resultant unbalanced force on a body which is used to overcome Aristotle’s fallacy.

Question 7.
Explain why a cricketer moves his hands backwards while holding a catch. (NCERT)
Answer:

  1. In the act of catching the ball, by drawing hands backward, cricketer allows longer time for his hands to stop the ball.
  2. By Newton’s second law of motion, force applied depends on the rate of change of momentum.
  3. Taking longer time to stop the ball ensures smaller rate of change of momentum.
  4. Due to this the cricketer can stop the ball by applying smaller amount of force and thereby not hurting his hands.

Question 8.
Large force always produces large change in momentum on a body than a small force. Is this correct?
Answer:
No. From Newton’s second law, we have.
\(\frac{\mathrm{dP}}{\mathrm{dt}}=\mathrm{F}\) …. (i)
dP = Fdt …. (ii)
From equation (ii), we can infer that a small force acting for a longer time can produce same change in momentum of a body as a large force acting in the same direction for a short time. Hence, the given statement is incorrect.

Question 9.
Newton’s first law is contained in the second law. Prove it.
Answer:

  1. From Newton’s second law of motion, we have, \(\overrightarrow{\mathrm{F}}=\frac{\mathrm{d} \overrightarrow{\mathrm{p}}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{m} \overrightarrow{\mathrm{v}})\)
  2. For a given body, mass m is constant.
    ∴ \(\overrightarrow{\mathrm{F}}=\mathrm{m} \frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}=\mathrm{m} \overrightarrow{\mathrm{a}}\)
  3. If \(\overrightarrow{\mathrm{F}}\) = zero, \(\overrightarrow{\mathrm{v}}\) is constant. Hence if there is no force, velocity will not change. This is nothing but Newton’s first law of motion.

Question 10.
State Newton’s third law of motion
Answer:
Statement: To every action (force) there is always an equal and opposite reaction force).

Question 11.
State the importance of Newton’s third law of motion.
Answer:

  1. Newton’s third law of motion defines action and reaction as a pair of equal and opposite forces acting along the same line.
  2. Action and reaction forces always act on different objects.

Question 12.
State the consequences of Newton’s third law of motion.
Answer:

  1. Two interacting bodies exert forces which are always equal in magnitude, have the same line of action and are opposite in direction, upon each other. Thus, forces always occur in pairs.
  2. If a body A exerts an action force \(\overrightarrow{\mathrm{F}}_{\mathrm{BA}}\) on body B, then body B also exerts an equal and opposite reaction force \(\overrightarrow{\mathrm{F}}_{\mathrm{AB}}\) on body A, simultaneously.
  3. Body A experiences the force \(\overrightarrow{\mathrm{F}}_{\mathrm{AB}}\) only and
    body B experiences the force \overrightarrow{\mathrm{F}}_{\mathrm{BA}} only.
  4. Both the forces, action and reaction act at the same instant.
  5. Both the forces always act on different bodies. Hence, they never cancel each other.
  6. Both the forces do not necessarily arise due to contact i.e., they can be non-contact forces. Example: Repulsive forces between two magnets.

Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion

Question 13.
If a constant force of 800 N produces an acceleration of 5 m/s2 in a body, what is its mass? If the body starts from rest, how much distance will it travel in 10 s?
Solution:
Given: F = 800 N, a = 5 m/s2, u = 0, t = 10 s
To find: mass (m), distance travelled (s)
Formulae:

i. F = ma
ii. s = ut + \(\frac{1}{2} \mathrm{at}^{2}\)

Calculation:
From formula (i),
∴ m = \(\frac{\mathrm{F}}{\mathrm{a}}=\frac{800}{5}\) = 160 kg
From formula (ii),
s = \(\frac{1}{2}\) × 5 × (10)2 [∵ u = 0]
∴ s = 250 m
Answer:
Mass of the body is 160 kg and the distance travelled by the body is 250 m.

Question 14.
A constant force acting on a body of mass 3 kg changes its speed from 2 m s-1 to 3.5 m/s in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force? (NCERT)
Solution:
Given: u = 2 ms-1, m = 3 kg,
v = 3.5 m s-1, t = 25s
To find: Force (F)
Formula: F = ma
Calculation: Since, v = u + at
∴ 3.5 = 2 + a × 25
a = \(\frac{3.5-2}{25}\) = 0.06 m s-2
From formula,
F = 3 × 0.06 = 0.18 N
Since, the applied force increases the speed of the body, it acts in the direction of the motion.
Answer:
The applied force is 0.18 N along the direction of motion.

Question 15.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms-1. How long does the body take to stop? (NCERT)
Solution:
Given: m = 20 kg, u = 15ms-1, v = 0,
F = – 50 N (retarding force)
To find: Time (t)
Formula: v = u + at
Calculation: Since, F = ma
∴ a = \(\frac{\mathrm{F}}{\mathrm{m}}=\frac{-50}{20}\) = -2.5 m s-2
From formula,
0 = 15 + (-2.5) × t
∴ t = 6s
Answer:
Time taken to stop the body is 6 s.

Question 16.
A hose pipe used for gardening is ejecting water horizontally at the rate of 0.5 m/s. Area of the bore of the pipe is 10 cm2. Calculate the force to be applied by the gardener to hold the pipe horizontally stationary.
Solution:
Let ejecting water horizontally be considered as the action force on the water, then the water exerts a backward force (called recoil force) on the pipe as the reaction force.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 1
Where, V = volume of water ejected
A = area of cross section of bore = 10 cm2
ρ = density of water = 1 g/cc
l = length of the water ejected in time t
\(\frac{\mathrm{d} l}{\mathrm{dt}}\) = v = velocity of water ejected
= 0.5 m/s = 50 cm/s
F = \(\frac{\mathrm{dm}}{\mathrm{dt}} \mathrm{v}\)
= (Aρv) v
= Aρv2
= 10 × 1 × 502
∴ F = 25000 dyne = 0.25 N
Answer:
The gardener must apply an equal and opposite force of 0.25 N.

Question 17.
What does the term frame of reference mean?
Answer:
A system of co-ordinate axes with reference to which the position or motion of an object is described is called a frame of reference.

Question 18.
Explain the terms inertial and non-inertial frame of reference.
Answer:

  1. Inertial frame of reference:
    • A frame of reference in which Newton ‘s first law of motion is applicable is called inertia/frame of reference.
    • A body moves with a constant velocity (which can be zero) in the absence of a net force. The body does not accelerate.
    • Example: A rocket in inter-galactic space (gravity free space between galaxies) with all its engine shut.
  2. Non-inertial frame of reference:
    • A frame of reference in which an object suffers acceleration in absence of net force is called non-inertial frame of reference.
    • The body undergoes acceleration.
    • Example: If a car just start its motion from rest, then during the time of acceleration the car will be in a non-inertial frame of reference.

Question 19.
State the limitations of Newton’s laws of motion.
Answer:

  1. Newton’s laws of motion are not applicable in a non-inertial (accelerated) frame of reference.
  2. Newton’s laws are only applicable to point objects.
  3. Newton’s laws are only applicable to rigid bodies.
  4. Results obtained by applying Newton’s laws of motion for objects moving with speeds comparable to that of light do not match with the experimental results and Einstein special theory of relativity has to be used.
  5. Newton’s laws of motion fail to explain the behaviour and interaction of objects having atomic or molecular sizes, and quantum mechanics has to be used.

Question 20.
Name the different types of fundamental forces in nature.
Answer:
Fundamental forces in nature are classified into four types:

  1. Gravitational force
  2. Electromagnetic force
  3. Strong nuclear force
  4. Weak nuclear force

Question 21.
Define gravitational force. Give its examples.
Answer:
Force of attraction between two (point) masses separated by a distance is called as gravitational force.
F = \(\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)
where ‘G’ is constant called the universal gravitational constant = 6.67 × 10-11 Nm2/kg2

Examples:

  1. The motion of moon, artificial satellites around the earth and motion of planets around the sun is due to gravitational force of attraction.
  2. The concept of weight of a body in the earth’s gravitational field is due to gravitational force exerted by the earth on a body.

Question 22.
Write down the main characteristics of gravitational force.
Answer:
Characteristics of gravitational force:

  1. It is always attractive.
  2. It is the weakest of the four basic forces in nature.
  3. Its range is infinite.
  4. Structure of the universe is governed by this force.

Question 23.
Write a note on electromagnetic (EM) force.
Answer:
Electromagnetic force:

  1. The attractive and repulsive force between electrically charged particles is called electromagnetic force.
  2. It can be attractive or repulsive.
  3. It is stronger than the gravitational force.
  4. Example: force of friction, normal reaction, tension in strings, collision forces, elastic forces, fluid friction etc. are electromagnetic in nature.
  5. Reaction forces are a result of the action of electromagnetic forces.
  6. Since majority of forces are electromagnetic in nature, our life is practically governed by these forces.

Question 24.
Write a note on strong and weak nuclear force.
Answer:

  1. Strong nuclear force: The strong force which binds protons and neutrons (nucleons) together in the nucleus of an atom is called strong nuclear force.
    Characteristics of strong nuclear force:

    • It is a very strong attractive force.
    • It is a short range force of the order of 10-14 m.
    • it is charge independent.
  2. Weak nuclear force: The force of interaction between subatomic particles which results in the radioactive decay of atoms is called weak nuclear force.

Characteristics of weak nuclear force:

  • It acts between any two elementary particles (pair of subatomic particles).
  • It is a stronger force than gravitational force.
  • It is much weaker than electromagnetic force or strong nuclear force.
  • It is a short range force of the order of 10-16m.

Question 25.
Three identical point masses are fixed symmetrically on the periphery of a circle. Obtain the resultant gravitational force on any point mass M at the centre of the circle. Extend this idea to more than three identical masses symmetrically located on the periphery. How far can you extend this concept?
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 2
i. Consider three identical points A, B and C of mass m on the periphery of a circle of radius r. Mass M is at the centre of the circle.
Gravitational forces on M due to these masses are attractive and are given as,
In magnitude, \(\mathrm{F}_{\mathrm{MA}}=\mathrm{F}_{\mathrm{MB}}=\mathrm{F}_{\mathrm{MC}}=\frac{\mathrm{GMm}}{\mathrm{r}^{2}}\)

ii. Forces \(\overrightarrow{\mathrm{F}}_{\mathrm{MB}}\) and \(\overrightarrow{\mathrm{F}}_{\mathrm{MC}}\) are resolved along \(\overrightarrow{\mathrm{F}}_{\mathrm{MA}}\) and perpendicular to \(\overrightarrow{\mathrm{F}}_{\mathrm{MA}}\). Components perpendicular to \(\overrightarrow{\mathrm{F}}_{\mathrm{MA}}\) cancel each other. Components along \(\overrightarrow{\mathrm{F}}_{\mathrm{MA}}\) are,
FMB cos 60° = FMC cos 60° = \(\frac{1}{2} F_{M A}\) each.

Magnitude of their resultant is FMA and its direction is opposite to that of FMA. Thus, the
resultant force on mass M is zero. For any even number of equal masses, the force due to any mass m is balanced (cancelled) by diametrically opposite mass. For any odd number of masses, the components perpendicular to one of them cancel each other while the components parallel to one of these add up in such a way that the resultant is zero for any number of identical masses m located symmetrically on the periphery.

As the number of masses tends to infinity, their collective shape approaches circumference of the circle, which is nothing but a ring. Thus, the gravitational force exerted by a ring mass on any other mass at its centre is zero.

iii. This concept can be further extended to three-dimensions by imagining a uniform hollow sphere to be made up of infinite number of such rings with a common diameter. Thus, the gravitational force for any mass kept at the centre of a hollow sphere is zero.

Question 26.
A car of mass 1.5 ton is running at 72 kmph on a straight horizontal road. On turning the engine off, it stops in 20 seconds. While running at the same speed, on the same road, the driver observes an accident 50 m in front of him. He immediately applies the brakes and just manages to stop the car at the accident spot. Calculate the braking force.
Solution:
Given: m = 1.5 ton = 1500 kg,
u = 72 kmph = 72 × \(\frac{5}{18} \mathrm{~m} / \mathrm{s}\)m/s = 20 m
s-1 (on turning engine off),
v = 0, t = 20 s, s = 50 m
To find: Braking force (F)

Formula:

i. v = u + at
ii. v2 – u2 = 2as
iii. F = ma

Calculation:
On turning the engine off,
From formula (i),
a = \(\frac{0-20}{20}\) = -1 m s-2
This is frictional retardation (negative acceleration).
After seeing the accident,
From formula (ii),
a1 = \(\frac{0^{2}-20^{2}}{2(50)}\) = -4 m s-2
This retardation is the combined effect of braking and friction
∴ braking retardation =4 – 1 = 3 m s-2
From formula (iii), the braking force, F = 1500 × 3 = 4500 N
Answer:
The braking force is 4500 N.

Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion

Question 27.
A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by \(\overrightarrow{\mathbf{F}}=(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})\) N where \(\hat{\mathbf{i}}\), \(\hat{\mathbf{j}}\), \(\hat{\mathbf{k}}\) are unit vectors along the x, y and z axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis? (NCERT)
Solution:
Given: \(\overrightarrow{\mathbf{F}}=(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})\) N, \(\overrightarrow{\mathrm{s}}=4 \hat{\mathrm{k}}\)
To find: work done (W)

Formula: W = \(\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{s}}\)
Calculation: From formula,
W = \((-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{k}})\)
= \(12 \hat{\mathrm{k}} \cdot \hat{\mathrm{k}}\) = 12 J
Answer:
The work done by the force is in moving the body 12 J.

Question 28.
Over a given region, a force (in newton) varies as F = 3x2 – 2x + 1. In this region, an object is displaced from x1 = 20 cm to x2 = 40 cm by the given force. Calculate the amount of work done.
Solution:
Given: F = 3x2 – 2x + 1, x1 = 20 cm = 0.2 m,
x2 = 40 cm = 0.4 m.
To find: Work done (W)
Formula: W = \(\int_{A}^{B} \vec{F} \cdot \overrightarrow{d s}\)
Calculation:
From formula,
W = \(\int_{x_{1}}^{x_{2}} F \cdot d x=\int_{0.2}^{0.4}\left(3 x^{2}-2 x+1\right) d x\)
= [x3 – x2 + x]0.4
= [0.43 – 0.42 + 0.4] – [0.23 – 0.22 + 0.2]
= 0.304 – 0.168 = 0.136 J
The work done is 0.136 J.

Question 29.
A position dependent force f = 7 – 2x + 3x2 newton acts on a small body of mass 2 kg and displaces from x = 0 m to x = 5 m, calculate the work done.
Solution:
Given: F = 7 – 2x + 3x2, x = 0 at A and x = 5 at B.
To find: Work done (W)
Formula: W = \(\int_{A}^{B} \vec{F} \cdot \overrightarrow{d s}\)
Calculation: From formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 3
∴ W = 135 J
Answer:
The work done is 135 J.

Question 30.
State the principle of work-energy theorem in case of a conservative force and explain.
OR
Show that work done on a body by a conservative force is equal to the change in its kinetic energy
Answer:
Principle: Decrease in the potential energy due to work done by a conservative force is entirely converted into kinetic energy. Vice versa, for an object moving against a conservative force, its kinetic energy decreases by an amount equal to the work done against the force.

Work-energy theorem in case of a conservative force:

  1. Consider an object of mass m moving with velocity u experiencing a constant opposing force F which slows it down to v during displacement s.
  2. The equation of motion can be written as, v2 – u2 = -2as (negative acceleration for
    opposing force.)
    Multiplying throughout by \(\frac{\mathrm{m}}{2}\), we get,
    \(\frac{\mathrm{1}}{2}\)mu2 – \(\frac{\mathrm{1}}{2}\)mv2 = (ma)s …. (1)
  3. According to Newton’s second law of motion,
    F = ma … (2)
  4. From equations (1) and (2), we get,
    \(\frac{\mathrm{1}}{2}\)mu2 – \(\frac{\mathrm{1}}{2}\)mv2 = F.s
  5. But, \(\frac{\mathrm{1}}{2}\)mv2= Kf = final K.E of the body,
    \(\frac{\mathrm{1}}{2}\)mu2 = Ki = initial K.E of the body. and, work done by the force = F.s
    ∴ work done by the force = kf – ki
    = decrease in K.E of the body.
  6. Thus, work done on a body by a conservative force is equal to the change in its kinetic energy.

Question 31.
Explain the work-energy theorem in case of an accelerating conservative force along with a retarding non-conservative force.
Answer:

  1. Consider an object dropped from some point at height h.
  2. While coming down its potential energy decreases.
    ∴ Work done = decrease in P.E of the body.
  3. But, in this case, some part of the energy is used in overcoming the air resistance. This part of energy appears in some other forms such as heat, sound, etc. Thus, the work is not entirely converted into kinetic energy. In this case, the work-energy theorem can mathematically be written as,
    ∴ ∆ PE = ∆ K.E. + Wair resistance
    ∴ Decrease in the gravitational P.E. = Increase in the kinetic energy + work done against non-conservative forces.

Question 32.
A liquid drop of 1.00 g falls from height of cliff 1.00 km. It hits the ground with a speed of 50 m s-1. What is the work done by the unknown force? (Take g = 9.8 m/s2)
Solution:
Given: m = 1.0 g = 1.0 × 10-3 kg,
h = 1 km = 103 m, v = 50 ms-1
To find: Work done (Wf)
Formula: Wf = ∆ K.E – Wg

Calculation:

i. The change in kinetic energy of the drop
∆ K.E = (K.E.)final (K.E.)initial
∴ ∆ K.E. = \(\frac{1}{2} \mathrm{mv}^{2}-0\)
= \(\frac{1}{2} \times 1.0 \times 10^{-3} \times(50)^{2}\)
∴ ∆ K.E.= 1.25 J

ii. Work done by the gravitational force is,
Wg = mgh = 1.0 × 10-3 × 9.8 × 103 = 9.8 J
∴ Wg = 9.8J
From formula,
Wf = ∆K.E. – Wg = 1.25 – 9.8
Wf = -8.55 J
Answer:
Work done by the unknown force is – 8.55 J.

Question 33.
A body of mass 0.5 kg travels in a straight line with velocity y = ax3/2, where a = 5 m1/2s-1. What is the work done by the net force during its displacement from x = 0 to x = 2m? (NCERT)
Solution:
Given: M = 0.5 kg, y = ax3/2,
where a = 5 m-1/2s-1
Let v1 and v2 be the velocities of the body, when x = 0 and x = 2 m respectively. Then,
v1 = 5 × 03/2 = 0, v2 = 5 × 23/2 = \(10 \sqrt{2}\) m
To find: Work done (W)
Formula: Work done = Increase in kinetic energy
W = \(\frac{1}{2} \mathrm{M}\left(\mathrm{v}_{2}^{2}-\mathrm{v}_{1}^{2}\right)\)
Calculation: From formula,
W = \(\frac{1}{2}\) × 0.5 × [latex](10 \sqrt{2})^{2}-0^{2}[/latex]
∴ W = 50J
Answer:
Work done by the net force on the body is 50 J.

Question 34.
A particle of mass 12 kg is acted upon by a force f = (100 – 2x2) where f is in newton and ‘x’ is in metre. Calculate the work done by this force in moving the particle x = 0 to x = -10 m. What will be the speed at x = 10 m if it starts from rest?
Solution:
Given: F = 100 – 2x2
at A, x = 0 and at B, x = -10 m
To find: Work done (W), speed (v)

Formulae:

i. W = \(\int_{A}^{B} \vec{F} \cdot d s\)
ii. W = K.E. = \(\frac{1}{2} \mathrm{mv}^{2}\)

Calculation:
From formula (i),
W = \(\int_{A}^{B} \vec{F} \cdot \overline{d s}=\int_{x=0}^{x=-10} F d x\)
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 4
Answer:

  1. Work done by the force on the particle is 333.3 J.
  2. The speed of the particle at x = 10 will be 7.45 m/s.

Question 35.
Define free body diagram. In the figure given below, draw the free body diagrams for mass of 2 kg, 4 kg and 5 kg and hence state their force equations.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 5
Answer:
i. The diagram showing the forces acting on only one body at a time along-with its acceleration is called a free body diagram.

ii. The free body diagram for the mass of 2 kg is
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 6
Free body diagram for 2 kg mass
The force equation is given as,
2a = T3 – 2g

iii. The free body diagram for the mass of 4 kg is,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 7
The force equation is given as, 4a = T1 + 4g – T2

iv. The free body diagram for the mass of 5kg is,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 8

The force equation for the mass of 5 kg is given as,
N + F sin 60° = 5g, along the vertical direction.
T1 + 10 = F cos 60°, along the horizontal direction (Considering the mass is in equilibrium).

Question 36.
Figure shows a fixed pulley. A massless inextensible string with masses m1 and m2 > m1 attached to its two ends is passing over the pulley. Such an arrangement is called an Atwood machine. Calculate accelerations of the masses and force due to the tension along the string assuming axle of the pulley to be frictionless.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 9
Solution:
Method I: As m2 > m1, mass m2 is moving downwards and mass m1 is moving upwards.
Net downward force = F = (m2) g – (m1) g
= (m2 – m1)g
the string being inextensible, both the masses travel the same distance in the same time. Thus, their accelerations are equal in magnitude (one upward, other downward). Let it be a.
Total mass in motion, M = m2 + m1
∴ a = \(\frac{F}{M}=\left(\frac{m_{2}-m_{1}}{m_{2}+m_{1}}\right) g\) …. (i)

For mass m1, the upward force is the force due to tension T and downward force is mg. It has upward acceleration a. Thus, T – m1g = m1a
∴ T = m1(g + a)
Using equation (i), we get,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 10
From the free body equation for the first body,
T – m1g = m1a .. (i)

From the free body equation for the second body,
m2g – T = m2a … (ii)
Adding (i) and (ii), we get,
a = \(\left(\frac{\mathrm{m}_{2}-\mathbf{m}_{1}}{\mathrm{~m}_{2}+\mathrm{m}_{1}}\right) \mathbf{g}\) ….(iii)
Solving equations. (ii) and (iii) for T, we get,
T = m2(g – a) = \(\left(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\right) g\)

Question 37.
Write a note on elastic collision.
Answer:

  1. Collision between two bodies in which kinetic energy of the entire system is conserved along with the linear momentum is called as elastic collision.
  2. In an elastic collision,
    \(\mathrm{m}_{1} \overrightarrow{\mathrm{u}_{1}}+\mathrm{m}_{2} \overrightarrow{\mathrm{u}_{2}}=\mathrm{m}_{1} \overrightarrow{\mathrm{v}_{1}}+\mathrm{m}_{2} \overrightarrow{\mathrm{v}}_{2}\)
  3. In an elastic collision,
    \(\sum \mathrm{K} \cdot \mathrm{E}_{\text {‘initial }}=\sum \mathrm{K} \cdot \mathrm{E}_{\text {. final }}\)
  4. An elastic collision is impossible in daily life.
  5. However, in many situations, the interatomic and intermolecular collisions are considered to be elastic.

Question 38.
Write a note on inelastic collision.
Answer:

  1. A collision is said to be inelastic if there is a loss in the kinetic energy during collision, but linear momentum is conserved.
  2. In an inelastic collision, m1u1 + m2u2 = m1v1 + m2v2.
  3. In an inelastic collision,
    \(\sum \mathrm{K} \cdot \mathrm{E}_{\text {.initial }} \neq \sum \mathrm{K} \cdot \mathrm{E}_{\text {.final }}\)
  4. The loss in kinetic energy is either due to internal friction or vibrational motion of atoms causing heating effect.

Question 39.
Define perfectly inelastic collision. Give an example of it.
Answer:

  1. Collision in which the colliding bodies stick together after collision and move with a common velocity is called perfectly inelastic collision.
  2. The loss in kinetic energy is maximum in perfectly elastic collision.
  3. Example: Lump of mud thrown on a wall sticks to the wall due to the loss of kinetic energy.

Question 40.
In case of an elastic head on collision between two bodies, derive an expression for the final velocities of the bodies in terms of their masses and velocities before collision.
Answer:
Head on elastic collision of two spheres:
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 11

i. Consider two rotating smooth bodies A and B of masses m1 and m2 respectively moving
along the same straight line.

ii. Let \(\overrightarrow{\mathrm{u}}_{1}\) = initial velocity of the sphere A before collision.
\(\overrightarrow{\mathrm{u}}_{2}\) = initial velocity of the sphere B before collision.
\(\overrightarrow{\mathrm{v}}_{1}\) = velocity of the sphere A after collision.
\(\overrightarrow{\mathrm{v}}_{2}\) = velocity of the sphere B after collision.

iii. After the elastic collision, the spheres separate and move along the same straight line without rotation.

iv. According to the law of conservation of momentum,
m1\(\overrightarrow{\mathrm{u}}_{1}\) + m2\(\overrightarrow{\mathrm{u}}_{2}\) = m1\(\overrightarrow{\mathrm{v}}_{1}\) + m2\(\overrightarrow{\mathrm{v}}_{2}\) ….(i)
According to the law of conservation of energy (as kinetic energy is conserved during elastic collision),
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 12

v. Since kinetic energy is a scalar quantity, the terms involved in the above equations are scalars.

vi. The equation (1) can be written in scalar form as,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 13

vii. Also the equation (2) can be written as,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 14

viii. Now dividing equation (4) by (3) we get,
(u1 + v1) = (u2 + v2)
∴ u1 + v1 = u2 + v2
∴ v2 = u1 – u2 + v1 … (5)

ix. Comparing equation (3) and (5),
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 15
Equations, (6) and (7), represent the final velocities of two spheres after collision.

Question 41.
Are you aware of elasticity of materials? Is there any connection between elasticity of materials and elastic collisions?
Answer:
(Students should answer the question as per their understanding).

Question 42.
Two bodies undergo one-dimensional, inelastic, head-on collision. State an expression for their final velocities in terms
of their masses, initial velocities and coefficient of restitution.
Answer:
If e is the coefficient of restitution, the final velocities after an inelastic, head on collision are given as,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 16

Question 43.
Two bodies undergo one-dimensional, inelastic, head-on collision. State an expression for the loss in the kinetic energy.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 17
ii. As e < 1, (1 – e2) is always positive. Thus, there is always a loss of kinetic energy in an inelastic collision.

Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion

Question 44.
Two bodies undergo one-dimensional, inelastic, head-on collision. Obtain an expression for the magnitude of impulse.
Answer:
i. When two bodies undergo collision, the linear momentum delivered by the first body to the second body must be equal to the change in momentum or impulse of the second body and vice versa.
∴ Impulse,
|J| = |∆p1| = |∆p2|
= |m1v1 – m1u1| = |m2v2 – m2u2| ….(1)

Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 18
In equation (1) and solving, we get,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 19

Question 45.
Two bodies undergo one-dimensional, perfectly inelastic, head-on collision. Derive an expression for the loss in the kinetic energy.
Answer:
i. Let two bodies A and B of masses m1 and m2 move with initial velocity \(\overrightarrow{\mathrm{u}}_{1}\) and \(\overrightarrow{\mathrm{u}}_{2}\), respectively such that particle A collides head on with particle B i.e., u1 > u2.

ii. If the collision is perfectly inelastic, the particles stick together and move with a common velocity \(\overrightarrow{\mathrm{v}}\) after the collision along the same straight line.
loss in kinetic energy = total initial kinetic energy – total final kinetic energy.

iii. By the law of conservation of momentum,
m1u1 + m2u2 = (m1 + m2)v
∴ v = \(\frac{\mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\)

iv. Loss of Kinetic energy,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 20

iv. Both the masses and the term (u1 – u2)2 are positive. Hence, there is always a loss in a perfectly inelastic collision. For a perfectly inelastic collision, as e = 0, the loss is maximum.

Question 46.
Distinguish between elastic and inelastic collision.
Answer:

No. Elastic Collision Inelastic Collision
i. In an elastic collision, both momentum and kinetic energy are conserved. In an inelastic collision, momentum is conserved but kinetic energy is not conserved.
ii. The total kinetic energy after collision is equal to the total kinetic energy before collision. The total kinetic energy after the collision is not equal to the total kinetic energy before collision.
iii. Coefficient of restitution (e) is equal to one. Coefficient of restitution (e) is less than one. For a perfectly inelastic collision coefficient of restitution is equal to zero.
iv. Bodies do not stick together in elastic collision. Bodies stick together in a perfectly inelastic collision.
v. Sound, heat and light are not produced. Sound or light or heat or all of these may be produced.

Question 47.
Explain elastic collision in two dimensions.
Answer:
i. Suppose a particle of mass mi moving with initial velocity \(\overrightarrow{\mathrm{u}_{1}}\), undergoes a non head-on collide with another particle of mass m2 and initial velocity \(\overrightarrow{\mathrm{u}_{2}}\).
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 21

ii. Let us consider two mutually perpendicular directions; Common tangent at the point of impact, along which there is no force (or no change of momentum).
Line of impact which is perpendicular to the common tangent through the point of impact, in the two-dimensional plane of initial and final velocities.

iii. Applying the law of conservation of linear momentum along the line of impact, we have, m1u1 cos α1 + m2u2 cos α2 = m1v1 cos β1 + m2v2 cos β2
As there is no force along the common tangent,
m1u1 sin α1 = m1u1 sin β1 and m2u2 sin α2 = m2v2 sin β2
iv. Coefficient of restitution (e) along the line of impact is given as
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 22

Question 48.
Two bodies undergo a two-dimensional collision. State an expression for the magnitude of impulse along the line of impact and the loss in kinetic energy.
Answer:
i. For two bodies undergoing a two-dimensional collision, the magnitude of impulse along the line of impact is given as, Magnitude of the impulse, along the line of impact,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 23

ii. The loss in the kinetic energy is given as Loss in the kinetic energy = ∆ (K.E.)
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 24

Question 49.
0ne marble collides head-on with another identical marble at rest. If the collision is partially inelastic, determine the ratio of their final velocities in terms of coefficient of restitution e.
Solution:
According to conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
As m1 = m2, we get, u1 + u2 = v1 + v2
∴ If u2 = 0, we get, v1 + v2 = u1 ….. (i)
Coefficient of restitution,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 25

Question 50.
A 10 kg mass moving at 5 m/s collides head- on with a 4 kg mass moving at 2 m/s in the same direction. If e = \(\frac{1}{2}\), find their velocity after impact.
Solution:
Given: m1 = 10 kg, m2 = 4 kg
u1 = 5 m/s, u2 = 2 m/s, e = \(\frac{1}{2}\)
To find: Velocity after impact (v1 and v2)

Formulae:

i. m1u1 + m2u2 = m1v1 + m2v2
ii. e = \(\left(\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\right)\)

Calculation:

From formula (i),
10 × 5 + 4 × 2 = 10v1 + 4v2
∴ 5v1 + 2v2 = 29 … (1)

From formula (ii),
v2 – v1 = e(u1 – u2) = \(\frac{1}{2}\) (5 – 2) = \(\frac{3}{2}\)
∴ 2v2 – 2v1 = 3 … (2)
Solving (1) and (2), we have
∴ v1 = \(\frac{26}{7}\) m/s and v2 = \(\frac{73}{14}\) m/s
Answer:
The respective velocities of the two masses are \(\frac{26}{7}\) m/s and \(\frac{73}{14}\) m/s.

Question 51.
A metal ball falls from a height 1 m on a steel plate and jumps upto a height of 0.81 m. Find the coefficient of restitution.
Solution:
As the ball falls to the steel plate P.E changes to kinetic energy.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 26
As ground is stationary, both its initial and final velocities are zero.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 27

Question 52.
Two bodies of masses 5 kg and 3 kg moving in the same direction along the same straight line with velocities 5 m s-1 and 3 m s-1 respectively suffer one-dimensional elastic collision. Find their velocities after the collision.
Solution:
Given: m1 = 5kg, u1 = 5ms-1, m2 = 3kg, u2 = 3 m s-1
To find: velocities after collision (v1 and v2)
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 281
Answer:
The velocities of the two bodies after collision are 3.5 m/s and 5.5 m/s.

Question 53.
A 20 g bullet leaves a machine gun with a velocity of 200 m/s. If the mass of the gun is 20 kg, find its recoil velocity. If the gun fires 20 bullets per second, what force is to be applied to the gun to prevent recoil?
Solution:
Given: m1 = 20g = 0.02 kg, m2 = 20 kg, v1 = 200 m/s, t = \(\frac{1}{20}\) s,
To find: Recoil velocity (v2), applied force (F)

Formulae:

i. v2 = \(-\frac{\mathrm{m}_{1} \mathrm{v}_{1}}{\mathrm{~m}_{2}}\)
ii. F = ma

Calculation: From formula (i),
∴ v2 = \(-\frac{0.02}{20} \times 200\)
= -0.2 m/s

Negative sign shows that the machine gun moves in a direction opposite to that of the bullet.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 29
From formula (ii),
∴ F = m2 × a = 20 × 4 = 80N
Answer:
The recoil velocity of gun is 0.2 m/s and the required force to prevent recoil is 80 N.

Question 54.
A shell of mass 3 kg is dropped from some height. After falling freely for 2 seconds, it explodes into two fragments of masses 2 kg and 1 kg. Kinetic energy provided by the explosion is 300 J. Using g = 10 m/s2, calculate velocities of the fragments. Justify your answer if you have more than one options.
Solution:
Total mass = m1 + m2 = 3 kg
Initially, when the shell falls freely for 2 seconds, v = u+ at = 0 + 10(2) = 20 ms-1 = u1 = u2
According to conservation of linear momentum,
m1u1 + m2u2 = m1v1 + m2v2
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 30
There are two possible answers since the positions of two fragments can be different as explained below.
Case 1: v1 = 30 m s-1 and v2 = 0 with the lighter fragment 2 above.
Case 2: v1 = 10 m s-1 and v2 = 40 m s-1 with the lighter fragment 2 below, both moving downwards.

Question 55.
Bullets of mass 40 g each, are fired from a machine gun at a rate of 5 per second towards a firmly fixed hard surface of area 10 cm2. Each bullet hits normal to the surface at 400 m/s and rebounds in such a way that the coefficient of restitution for the collision between bullet and the surface is 0.75. Calculate average force and average pressure experienced by the surface due to this firing.
Solution:
For the collision,
u1 = 400 m s-1, e = 0.75
For the firmly fixed hard surface, u2 = v2 = 0
e = 0.75 = \(\frac{v_{1}-v_{2}}{u_{2}-u_{1}}=\frac{v_{1}-0}{0-400}\)
∴ v1 = -300 m/s.
Negative sign indicates that the bullet rebounds in exactly opposite direction.
Change in momentum of each bullet = m(v1 – u1)
The same momentum is transferred to the surface per collision in opposite direction.
∴ Momentum transferred to the surface, per collision,
p = m (u1 – v1) = 0.04(400 – [-300]) = 28 Ns
The rate of collision is same as rate of firing.
∴ Momentum received by the surface per second, \(\frac{\mathrm{dp}}{\mathrm{dt}}\) = average force experienced by the surface = 28 × 5 = 140 N

This is the average force experienced by the surface of area A = 10 cm2 = 10-3 m2
∴ Average pressure experienced,
P = \(\frac{\mathrm{F}}{\mathrm{A}}=\frac{140}{10^{-3}}\) = 1.4 × 105 N m-2
∴ P ≈ 1.4 times the atmospheric pressure.
Answer:
The average force and average pressure experienced by the surface due to the firing is 140 N and 1.4 × 105 N m-2 respectively.

Question 56.
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s-1, what is the recoil speed of the gun? (NCERT)
Solution:
Given: m1 = 0.02 kg, m2 = 100 kg, v1 = 80 m s-1
To find: Recoil speed (v2)
Formula: m1u1 + m2u2 = m1v1 + m2v2
Calculation: Initially gun and shell are at rest.
∴ m1u1 + m2u2 = 0
Final momentum = m1v1 – m2v2
Using formula,
0 = 0.02 (80) – 100(v2)
∴ v2 = \(\frac{0.02 \times 80}{100}\) = 0.016 ms-1
Answer:
The recoil speed of the gun is 0.016 m s-1.

Question 57.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s-1 collide and rebound with the same speed. What is the impulse imparted to each bail due to the other? (NCERT)
Solution:
Given: m = 0.05 kg. u = 6 m/s, v = -6 m/s
To find: Impulse (J)
Formula: J = m (v – u)
calculation: From formula,
J = 0.05 (-6 – 6) = -0.6 kg m s-1
Answer:
Impulse received by each ball is -0.6 kg m s-1.

Question 58.
A bullet of mass 0.1 kg moving horizontally with a velocity of 20 m/s strikes a target and brought to rest in 0.1 s. Find the impulse and average force of impact.
Solution:
Given: m = 0.1 kg, u = 20 m/s, t = 0.1 s
To find: Impulse (J), Average force (F)

Formulae:

i. J = mv – mu
ii. F = \(m \frac{(v-u)}{t}\)

Calculation:

From formula (i).
J = m(v – u) = 0.1 (0 – 20) = -2 Ns
From formula (ii),
F = \(\frac{m(v-u)}{t}=\frac{2}{0.1}=20 N\)
Answer:
Magnitude of impulse is 2 Ns, average force of impact is 20 N.

Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion

Question 59.
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg) (NCERT)
Answer:
Let the point B represents the position of bat. The ball strikes the bat with velocity v along the path AB and gets deflected with same velocity along BC. such that ∠ABC = 45°
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 31
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 32
Thus, impulse imparted to the ball is 4.157 kg ms-1

Question 60.
A cricket ball of mass 150 g moving with a velocity of 12 m/s is turned back with a velocity of 20 m/s on hitting the bat. The force of the ball lasts for 0.01 s. Find the average force exerted on the ball by the bat.
Solution:
m = 0.150 kg, v = 20 m/s,
u = -12 m/s and t = 0.01 s
To find: Average force (F)
Formula: F = \(\frac{m(v-u)}{t}\)
Calculation: From formula,
F = \(\frac{0.150[20-(-12)]}{0.01}\) = 480 N
Answer:
The average force exerted on the ball by the bat is 480 N.

Question 61.
Mass of an Oxygen molecule is 5.35 × 10-26 kg and that of a Nitrogen molecule is 4.65 × 10-26 kg. During their Brownian motion (random motion) in air, an Oxygen molecule travelling with a velocity of 400 m/s collides elastically with a nitrogen molecule travelling with a velocity of 500 m/s in the exactly opposite direction. Calculate the impulse received by each of them during collision. Assuming that the collision lasts for
1 ms, how much is the average force experienced by each molecule?
Solution:
Let, m1 = mO = 5.35 × 10-26 kg,
m2 = mN = 4.65 × 10-26 kg,
∴ u1 = 400 ms-1 and u2 = -500 ms-1 taking direction of motion of oxygen molecule as the positive direction.
For an elastic collision,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 33
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 34
Hence, the net impulse or net change in momentum is zero.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 35
Answer:
The average force experienced by the nitrogen molecule and the oxygen molecule are
-4.478 × 10-20 N and 4.478 × 10-20 N.

Question 62.
Explain rotational analogue of the force. On what factors does it depend? Represent it in vector form.
Answer:

  1. Rotational analogue of the force is called as moment of force or torque.
  2. It depends on the mass of the object, the point of application of the force and the angle between direction of force and the line joining the axis of rotation with the point of application.
  3. In its mathematical form, torque or moment of a force is given by
    \(\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}\)
    where \(\overrightarrow{\mathrm{F}}\) is the applied force and \(\overrightarrow{\mathrm{r}}\) is the position vector of the point of application of the force from the axis of rotation.

Question 63.
Illustrate with an example how direction of the torque acting on any object is determined.
Answer:
i. Consider a laminar object with axis of rotation perpendicular to it and passing through it as shown in figure (a).
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 36

ii. Figure (b) indicates the top view of the object when the rotation is in anticlockwise direction
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 37

iii. Figure (c) shows the view from the top, if rotation is in clockwise direction.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 38

iv. The applied force \(\vec{F}\) and position vector \(\vec{r}\) of the point of application of the force are in the plane of these figures.

v. Direction of the torque is always perpendicular to the plane containing the vectors \(\vec{r}\) and \(\vec{F}\) and can be obtained from the rule of cross product or by using the right-hand thumb rule.

vi. In Figure (b), it is perpendicular to the plane of the figure and outwards while in the figure (c), it is inwards.

Question 64.
State the equation for magnitude of torque and explain various cases of angle between the direction of \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{F}}\).
Answer:
Magnitude of torque, \(\tau\) = r F sin θ
where θ is the smaller angle between the directions of \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{F}}\).

Special cases:

  1. If θ = 90°, \(\tau\) = \(\tau\)max = rF. Thus, the force should be applied along normal direction for easy rotation.
  2. If θ = 0° or 180°, \(\tau\) = \(\tau\)min = 0. Thus, if the force is applied parallel or anti-parallel to \(\overrightarrow{\mathrm{r}}\), there is no rotation.
  3. Moment of a force depends not only on the magnitude and direction of the force, but also on the point where the force acts with respect to the axis of rotation. Same force can have different torque as per its point of application.

Question 65.
A force \(\overrightarrow{\mathbf{F}}=\mathbf{3} \hat{\mathbf{i}}+\hat{\mathbf{j}}-\mathbf{4} \hat{\mathbf{k}}\) is applied at a point (3, 4, -2). Find its torque about the point (-1, 2, 4).
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 39

Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion

Question 66.
Define couple. Show that moment of couple is independent of the points of application of forces.
Answer:
A pair of forces consisting of two forces of equal magnitude acting in opposite directions along different lines of action is called a couple.

  1. Figure shows a couple consisting of two forces \(\overrightarrow{\mathrm{F}}_{1}\) and \(\overrightarrow{\mathrm{F}}_{2}\) of equal magnitudes and opposite directions acting along different lines of action separated by a distance r.
  2. Position vector of any point on the line of action of force \(\overrightarrow{\mathrm{F}}_{1}\) from the line of action of force \(\overrightarrow{\mathrm{F}}_{2}\) is \(\overrightarrow{\mathrm{r}}_{12}\). Similarly, the position vector of any point on the line of action of force \(\overrightarrow{\mathrm{F}}_{2}\) from the line of action of force \(\overrightarrow{\mathrm{F}}_{1}\) is \(\overrightarrow{\mathrm{r}}_{21}\).
  3. Torque or moment of the couple is then given mathematically as
    Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 40
  4. From the figure, it is clear that r12 sinα = r21 sin β = r.
  5. If \(\left|\overrightarrow{\mathrm{F}}_{1}\right|=\left|\overrightarrow{\mathrm{F}}_{2}\right|\) = F, the magnitude of torque is given by
    \(\tau=\mathrm{r}_{12} \mathrm{~F}_{1} \sin \alpha\) = \(r_{21} F_{2} \sin \beta=r F\)
  6. It clearly shows that the torque corresponding to a given couple, i.e., the moment of a given couple is constant, i.e., it is independent of the points of application of forces.

Question 67.
The figure below shows three situations of a ball at rest under the action of balanced forces. Is the bail in mechanical equilibrium? Explain how the three situations differ.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 41
Answer:

  1. In all these cases, as the ball is at rest under the action of balanced forces i.e, there is no net force acting on it. Hence, it is in mechanical equilibrium.
    However, potential energy-wise, the three situations show the different states of mechanical equilibrium.
  2. Stable equilibrium: In situation (a), potential energy of the system is at its local minimum. If it is disturbed slightly from its equilibrium position and released, it tends to recover its position. In this situation, the ball is most stable and is said to be in stable equilibrium.
  3. Unstable equilibrium: In situation (b), potential energy of the system is at its local maximum. If it is slightly disturbed from its equilibrium position, it moves farther from that position. This happens because initially, if disturbed, it tries to achieve the configuration of minimum potential energy. In this situation, the ball is said to be in unstable equilibrium.
  4. Neutral equilibrium: in situation (e), potential energy of the system is constant over a plane and remains same at any position. Thus, even if the ball is disturbed, it still remains in equilibrium, practically at any position. In this situation, the ball is in neutral equilibrium.

Question 68.
Two weights 5 kg and 8 kg are suspended from a uniform rod of length 10 m and weighing 3 kg. The distances of the weights from one end are 2 m and 7 m. Find the point at which the rod balances.
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 42
Let the rod balance at a point P, x m from the end A of the rod AB. The suspended weight and the centre of gravity of the rod G is shown in the figure.
P = 5 + 3 + 8 = 16kg
Taking moments about A,
16 × x = 5 × 2 + 3 × 5 + 8 × 7
16x = 10 + 15 + 56 = 81
∴ x = \(\frac{81}{16}\) = 5.1 m
Answer:
The rod balances at 5.1 m from end A.

Question69.
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick? (NCERT)
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 43
Consider the metre scale AB. Let its mass be concentrated at C at 50 cm. mark. Upon placing coins, balancing point of the scale and the coins system is shifted to C’ at 45 cm mark.
For equilibrium about C’
10(45 – 12) = m (50 – 45)
m = \(\frac{10 \times 33}{5}\) = 66g
Answer:
The mass of the metre scale is 66 g.

Question 70.
The diagram shows a uniform beam of length 10 m, used as a balance. The beam is pivoted at its centre. A 5.0 N weight is attached to one end of the beam and an empty pan weighing 0.25 N Is attached to the other end of the beam.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 44
i. What is the moment of couple at pivot?
ii. If pivot is shifted 2 ni towards left, then what will be moment of couple at new position?
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 45
Answer:

  1. The moment of couple at center is 23.75 N-m.
  2. The moment of couple at new position is 13.25 N-m.
    ∴ For equilibrium,
    40 × x = 20 × 0.5
    ∴ x = \(\frac{1}{4}\) = 0.25 m

Hence, the total distance walked by the person is 1.25 m.

Ans
The person can walk 1.25 m before the plank topples.

Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion

Question 71.
Define centre of mass.
Answer:
Centre of mass of a body is a point about which the summation of moments of masses in the system is zero.

Question 72.
Derive an expression for the position of centre of mass of a system of n particles and for continuous mass distribution.
Answer:
System of n particles;
i. Consider a system of n particles of masses m1, m2, …, mn having position vectors \(\overrightarrow{\mathrm{r}_{1}}\), \(\overrightarrow{\mathrm{r}_{2}}\),….., \(\overrightarrow{\mathrm{r}_{n}}\) from the origin O.
The total mass of the system is,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 46
Centre of mass for n particles

ii. Position vector \(\vec{r}\) of their centre of mass from the same origin is then given by
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 47

iii. If the origin is at the centre of mass, \(\overrightarrow{\mathbf{r}}\) = 0
∴ \(\sum_{1}^{\mathrm{n}} \mathrm{m}_{\mathrm{i}} \overrightarrow{\mathrm{r}}_{\mathrm{i}}\) = 0,

iv. In this case, \(\sum_{1}^{n} m_{i} \vec{r}_{i}\) gives the moment of masses (similar to moment of force) about the centre of mass.

v. If (x1, x2, …… xn), (y1, y2, …..yn), (z1, z2, …. zn) are the respective x, y and z – coordinates of (r1, r2,…….. rn) then x,y and z – coordinates of the centre of mass are given by,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 48

vi. Continuous mass distribution: For a continuous mass distribution with uniform density, the position vector of the centre of mass is given by,
r = \(\frac{\int \vec{r} d m}{\int d m}=\frac{\int \vec{r} d m}{M}\)
Where \(\int \mathrm{dm}=\mathrm{M}\) is the total mass of the object.

vii. The Cartesian coordinates of centre of mass are
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 49

Question 73.
State the expression for velocity of the centre of mass of a system of n particles and for continuous mass distribution.
Answer:
Let v1, v2,…..vn be the velocities of a system of point masses m1, m2, … mn. Velocity of the centre of mass of the system is given by
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 50
x, y and z components of \(\overrightarrow{\mathbf{v}}\) can be obtained similarly.
For continuous distribution, \(\overrightarrow{\mathrm{v}}_{\mathrm{cm}}\) = \(\frac{\int \vec{v} \mathrm{dm}}{\mathrm{M}}\)

Question 74.
State the expression for acceleration of the centre of mass of a system of n particles and for continuous mass distribution.
Answer:
Let a1, a2,…. an be the accelerations of a system of point masses m1, m2 … mn.
Acceleration of the centre of mass of the system is given by
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 51
x, y and z components of can be obtained similarly.
For continuous distribution, \(\overrightarrow{\mathrm{a}}_{\mathrm{cm}}\) = \(\frac{\int \vec{a} d m}{\mathrm{M}}\)

Question 75.
State the characteristics of centre of mass.
Answer:

  1. Centre of mass is a hypothetical point at which entire mass of the body can be assumed to he concentrated.
  2. Centre of mass is a location, not a physical quantity.
  3. Centre of mass is particle equivalent of a given object for applying laws of motion.
  4. Centre of mass is the point at which applied force causes only linear acceleration and not angular acceleration.
  5. Centre of mass is located at the centroid, for a rigid body of uniform density.
  6. Centre of mass is located at the geometrical centre, for a symmetric rigid body of uniform density.
  7. Location of centre of mass can be changed only by an external unbalanced force.
  8. Internal forces (like during collision or explosion) never change the location of centre of mass.
  9. Position of the centre of mass depends only upon the distribution of mass, however, to describe its location we may use a coordinate system with a suitable origin.
  10. For a system of particles, the centre of mass need not coincide with any of the particles.
  11. While balancing an object on a pivot, the line of action of weight must pass through the centre of mass and the pivot. Quite often, this is an unstable equilibrium.
  12. Centre of mass of a system of only two particles divides the distance between the particles in an inverse ratio of their masses, i.e., it is closer to the heavier mass.
  13. Centre of mass is a point about which the summation of moments of masses in the system is zero.
  14. If there is an axial symmetry for a given object, the centre of mass lies on the axis of symmetry.
  15. If there are multiple axes of symmetry for a given object, the centre of mass is at their point of intersection.
  16. Centre of mass need not be within the body.
    Example: jumper doing fosbury flop.

Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion

Question 76.
The mass of moon is 0.0 123 times the mass of the earth and separation between them is 3.84 × 108 m. Determine the location of C.M as measured from the centre of the earth.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 52
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 53
Answer:
The location of centre of mass as measured from the center of the earth is 4.67 × 106 m.

Question 77.
Three particles of masses 3 g, 5 g and 8 g are situated at point (2, 2, 2), (-3, 1, 4) and (-1, 3, -2) respectively. Find the position vector of their centre of mass.
Solution:
m1 = 3 g, m2 = 5 g, m3 = 8 g, x1 = 2, y1 = 2, z1 = 2
x2 = -3, y2 = 1, z2 = 4,
x3 = -1, y3 = 3, z3 = -2
Let (X, Y, Z) be co-ordinates of C.M., then
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 54
The co-ordinates of the C.M. are \(\left(-\frac{17}{16}, \frac{35}{16}, \frac{5}{8}\right)\)
Answer:
The position vector of the C.M. is \(-\frac{17}{16} \hat{\mathbf{i}}+\frac{35}{16} \hat{\mathbf{j}}+\frac{5}{8} \hat{\mathbf{k}}\)

Question 78.
In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus. (NCERT)
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 55
Claculation: From formula
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 56
Answer:
The location of centre of mass from the nucleus of hydrogen atom is 1.235 A.

Question 79.
Three thin walled uniform hollow spheres of radii 1 cm, 2 cm and 3 cm are so located that their centres are on the three vertices of an equilateral triangle ABC having each side 10 cm. Determine centre of mass of the system.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 57
Solution:
Mass of a thin walled uniform hollow sphere is proportional to its surface area. (as density is constant) hence proportional to r2.

Thus, if mass of the sphere at A is mA = m, then mB = 4m and mC = 9m. By symmetry of the spherical surface, their centres of mass are at their respective centres, i.e., at A, B and C. Let us choose the origin to be at C, where the largest mass 9m is located and the point B with mass 4m on the positive x-axis. With this, the co-ordinates of C are (0, 0) and that of B are (10, 0). If A of mass m is taken in the first quadrant, its co-ordinates will be \([5,5 \sqrt{3}]\)
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 58

Question 80.
Locate the centre of mass of three particles of mass m1 = 1 kg, m2 = 2 kg and m3 = 3 kg at the corner of an equilateral triangle of each side of 1 m.
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 59
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 60
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 61
Answer:
The centre of mass of the system of three particles lies at \(\left(\frac{2}{3} m, \frac{\sqrt{3}}{6} m\right)\) with respect to the particle of mass 1 kg as the origin.

Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion

Question 81.
A letter ‘E’ is prepared from a uniform cardboard with shape and dimensions as shown in the figure. Locate its centre of mass.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 62
Solution:
As the sheet is uniform, each square can be taken to be equivalent to mass m concentrated at its respective centre. These masses will then be at the points labelled with numbers 1 to 10, as shown in figure. Let us select the origin to be at the left central mass m5, as shown and all the co-ordinates to be in cm.

By symmetry, the centre of mass of m1, m2 and m3 will be at m2 (1, 2) having effective mass 3m. Similarly, effective mass 3m due to m8, m9 and m10 will be at m9 (1, -2). Again, by symmetry, the centre of mass of these two (3m each) will have co-ordinates (1, 0). Mass m6 is also having co-ordinates (1, 0). Thus, the
effective mass at (1, 0) is 7m.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 63
Using symmetry for m4, m5 and m7, there will be effective mass 3m at the origin (0, 0). Thus, effectively, 3m and 7m are separated by 1 cm along X-direction. Y-coordinate is not required.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 64

Question 82.
A hole of radius r is cut from a uniform disc of radius 2r. Centre of the hole is at a distance r from centre of the disc. Locate centre of mass of the remaining part of the disc.
Solution:
Before cutting the hole, c.în. of the full disc was at its centre. Let this be our origin O. Centre of mass of the cut portion is at its centre D. Thus, it is at a distance x1 = r form the origin. Let C be the centre of mass of the
remaining disc, which will he on the extension of the line DO at a distance x2 = x from the origin. As the disc is uniform, mass of any of its part is proportional to the area of that part.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 65
Thus, if m is the mass of the cut disc, mass of the entire disc must be 4m and mass of the remaining disc will be 3m.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 66

Alternate method: (Using negative mass):

Let \(\overrightarrow{\mathrm{R}}\) be the position vector of the centre of mass of the uniform disc of mass M. Mass m is with centre of mass at position vector \(\overrightarrow{\mathrm{r}}\) from the centre of the disc be cut out from the complete disc. Position vector of the centre of mass of the remaining disc is then given by
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 67

Question 83.
Define centre of gravity of a body. Under what conditions the centre of gravity and centre of mass coincide?
Answer:

  1. centre of gravity of a body is the point around which the resultant torque due to force of gravity on the body is zero.
  2. The centre of mass coincides with centre of gravity when the body is in a uniform gravitational field.

Question 84.
Explain how to find the location of centre of mass or centre of gravity of a laminar object.
Answer:

  1. A laminar object is suspended from a rigid support at two orientations.
  2. Lines are drawn on the object parallel to the plumb line as shown in the figure.
  3. Plumb line is always vertical, i.e., parallel to the line of action of gravitational force.
  4. Intersection of the lines drawn is then the point through which line of action of the gravitational force passes for any orientation. Thus, it gives the location of the c.g. or c.m.
    Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 68

Question 85.
Why do cricketers wear helmet and pads while playing? Is it related with physics?
Answer:
Helmet and pads used by cricketers protects the head, using principles of physics.

  1. The framing between the interior of the helmet and pads increases the time over which the impulse acts on the head resulting into reduction of force.
    (Impulse = force × time = constant)
  2. The pads spread the force applied by the ball over a wider area reducing pressure at a point.

Question 86.
The diagram shows four objects placed on a flat surface.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 69
The centre of mass of each object is marked M. Which object is likely to fall over?
Answer:
Object C will fall because its centre of mass is not exactly at centre, in turn applying more force on one side of the object resulting in to unbalance force. Whereas, in other objects center of mass is exactly at center resulting into zero rotational or translational motion maintaining their equilibrium.

Question 87.
A boy is about to close a large door by applying force at A and B as shown. State with a reason, which of the two positions, A or B, will enable him to close the door with least force.
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 70
Answer:
Boy has to apply more force at point B as compared to point A, because point B is at least distance from the hinge of the door while point A is at maximum distance from hinge of the door. Opening of door applies a moment, which is. given by,
M = F × perpendicular distance
F ∝ \(\frac{1}{\text { distance }}\)
More is the distance from the axis of rotation less will be the force.

Question 88.
How is a seat belt useful for safety?
Answer:
When car hits another car or an object with high speed it applies a high impulse on the driver and due to inertia driver tends to move in forward direction towards the steering. Seat belts spreads the force over large area of the body and holds the driver and protects him from crashing at the steering.

Question 89.
According to Newton’s third law of motion for every action there is equal and opposite reaction, why two equal and opposite forces don’t cancel each other?
Answer:
Forces of action and reaction always act on different bodies, hence they never cancel each other.

Question 90.
Linear momentum depends on frame of reference, but principle of conservation of linear momentum is independent of frame of reference. Why?
Answer:
Observers in different frame find different values of linear momentum of a system, hence linear momentum depends upon frame of reference, but each would observe that the value of linear momentum does not change with time (provided the system is isolated), hence principle of conservation of linear momentum is independent of frame of reference.

Question 91.
Multiple choice Questions

Question 1.
A body of mass 2 kg moving on a horizontal surface with initial velocity of 4 m s-1 comes to rest after two seconds. If one wants to keep this body moving on the same surface with a velocity of 4 m s-1, the force required is
(A) 2 N
(B) 4 N
(C) 0
(D) 8 N
Answer:
(B) 4 N

Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion

Question 2.
What force will change the velocity of a body of mass 1 kg from 20 m s-1 to 30 m s-1 in two seconds?
(A) 1N
(B) 5 N
(C) 10 N
(D) 25 N
Answer:
(B) 5 N

Question 3.
A force of 5 newton acts on a body of weight 9.80 newton. What is the acceleration produced in m/s2?
(A) 0.51
(B) 1.96
(C) 5.00
(D) 49.00
Answer:
(C) 5.00

Question 4.
A body of mass m strikes a wall with velocity v and rebounds with the same speed. Its change in momentum is
(A) 2 mv
(B) mv/2
(C) – mv
(D) Zero
Answer:
(A) 2 mv

Question 5.
A force of 6 N acts on a body of mass 1 kg initially at rest and during this time, the body attains a velocity of 30 m/s. The time for which the force acts on a body is
(A) 10 second
(B) 8 second
(C) 7 second
(D) 5 second
Answer:
(D) 5 second

Question 6.
A bullet of mass 10 g is fired from a gun of mass 1 kg with recoil velocity of gun = 5 m/s. The muzzle velocity will be
(A) 30 km/min
(B) 60 km/min
(C) 30 m/s
(D) 500 m/s
Answer:
(D) 500 m/s

Question 7.
The velocity of rocket with respect to ground is v1 and velocity of gases ejecting from rocket with respect to ground is v2 Then velocity of gases with respect to rocket is given by
(A) v2
(B) v1 + v2
(C) v1 × v2
(D) v1
Answer:
(B) v1 + v2

Question 8.
Two bodies A and B of masses 1 kg and 2 kg moving towards each other with velocities 4 m/s and 1 m/s suffers a head on collision and stick together. The combined mass will
(A) move in direction of motion of lighter mass.
(B) move in direction of motion of heavier mass.
(C) not move.
(D) move in direction perpendicular to the line of motion of two bodies.
Answer:
(A) move in direction of motion of lighter mass.

Question 9.
Which of the following has maximum momentum?
(A) A 100 kg vehicle moving at 0.02 m s-1.
(B) A 4 g weight moving at 1000 cm s-1’.
(C) A 200 g weight moving with kinetic energy of 10-6 J
(D) A 200 g weight after falling through one kilometre.
Answer:
(D) A 200 g weight after falling through one kilometre.

Question 10.
A bullet hits and gets embedded in a solid block resting on a horizontal frictionless table. What is conserved?
(A) Momentum alone.
(B) K.E. alone.
(C) Momentum and K.E. both.
(D) P.E. alone.
Answer:
(A) Momentum alone.

Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion

Question 11.
The force exerted by the floor of an elevator on the foot of a person standing there, is more than his weight, if the elevator is
(A) going down and slowing down.
(B) going up and speeding up.
(C) going up and slowing down.
(D) either (A) and (B).
Answer:
(D) either (A) and (B).

Question 12.
If E, G and N represents the magnitudes of electromagnetic, gravitational and nuclear forces between two electrons at a given separation, then,
(A) N = E = G
(B) E < N < G (C) N > G < E (D) E > G > N
Answer:
(D) E > G > N

Qestion 13.
For an inelastic collision, the value of e is
(A) greater than 1
(B) less than 1
(C) equal to 1
(D) none of these
Answer:
(B) less than 1

Question 14.
A perfect inelastic body collides head on with a wall with velocity y. The change in momentum is
(A) mv
(B) 2mv
(C) zero
(D) none of these.
Answer:
(A) mv

Question 15.
Two masses ma and rnb moving with velocitics va and vb in opposite direction collide
elastically and after the collision ma and mb move with velocities vb and va respectively. Then the ratio mamb is
(A) \(\frac{v_{a}-v_{b}}{v_{a}+v_{b}}\)
(B) \(\frac{\mathrm{m}_{\mathrm{a}}+\mathrm{m}_{\mathrm{b}}}{\mathrm{m}_{\mathrm{a}}}\)
(C) 1
(D) \(\frac{1}{2}\)
Answer:
(C) 1

Question 16.
The frictional force acts _____.
(A) in direction of motion
(B) against the direction of motion
(C) perpendicular to the direction of motion
(D) at any angle to the direction of motion
Answer:
(B) against the direction of motion

Question 17.
A marble of mass X collides with a block of mass Z, with a velocity Y. and sticks to it. The final velocity of the system is
(A) \(\frac{\mathrm{Y}}{\mathrm{X}+\mathrm{Y}} \mathrm{Y}\)
(B) \(\frac{X}{X+Z} Y\)
(C) \(\frac{X+Y}{Z}\)
(D) \(\frac{X+Z}{X}\)
Answer:
(B) \(\frac{X}{X+Z} Y\)

Question 18.
Two balls lying on the same plane collide. Which of the following will be always conserved?
(A) heat
(B) velocity
(C) kinetic energy
(D) linear momentum.
Answer:
(D) linear momentum.

Question 19.
A body is moving with uniform velocity of 50 km h-1, the force required to keep the body in motion in SI unit is
(A) zero
(B) 10
(C) 25
(D) 50
Answer:
(A) zero

Question 20.
A coolie holding a suitcase on his head of 20 kg and travels on a platform. then work done in joule by the coolie is
(A) 198
(B) 98
(C) 49
(D) zero
Answer:
(D) zero

Question 21.
Out of the following forces, which force is non-conservative?
(A) gravitational
(B) electrostatic
(C) frictional
(D) magnetic
Answer:
(C) frictional

Question 22.
The work done in conservative force is ____.
(A) negative
(B) zero
(C) positive
(D) infinite
Answer:
(B) zero

Question 23.
The angle between the line of action of force and displacement when no work done (in degree) is
(A) zero
(B) 45
(C) 90
(D) 120
Answer:
(C) 90

Question 24.
If the momentum of a body is doubled, its KE. increases by
(A) 50%
(B) 300%
(C) 100%
(L)) 400%
Answer:
(B) 300%

Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion

Question 25.
In perfectly inelastic collision, which is conserved?
(A) P.E. only
(B) K.E. only
(C) momentum only
(D) K.E. and momentum
Answer:
(C) momentum only

Question 26.
In case of elaine collision, which s
(A) Momentum and K.E. is conserved.
(B) Momentum conserved and K.E. not conserved,
(C) Momentum not conserved and K.E. conserved.
(D) Momentum and K.E. both not conserved,
Answer:
(A) Momentum and K.E. is conserved.

Question 27.
Pseudo force is true only in
(A) frame of reference which is at rest
(B) inertial frame of reference.
(C) frame of reference moving with constant velocity.
(D) non-inertial frame of reference
Answer:
(D) non-inertial frame of reference

Question 28.
A men weighing 90kg carries a stone of 20 kg to the top of the building 30m high. The work done by hint is (g = 9.8 m/s2)
(A) 80 J
(B) 100 J
(C) 980 J
(D) 29,400 J
Answer:
(D) 29,400 J

Question 29.
A weight lifter is holding a weight of 100 kg on his shoulders for 45 s, the amount of work done by him in joules is
(A) 4500
(B) 100
(C) 45
(D) zero
Answer:
(D) zero

Question 30.
If m is the mass of a body and E its K.E..then its linear momentum is
(A) \(\mathrm{m} \sqrt{\mathrm{E}}\)
(B) \(2 \sqrt{\mathrm{m}} \mathrm{E}\)
(C) \(\sqrt{m} E\)
(D) \(\sqrt{2 \mathrm{mE}}\)
Answer:
(D) \(\sqrt{2 \mathrm{mE}}\)

Question 31.
Torque applied is masimum when the angle between the directions of \(\overrightarrow{\mathrm{r}}\) and \(\overrightarrow{\mathrm{F}}\) is
(A) 90°
(B) 180°
(C) 0°
(D) 45°
Answer:
(A) 90°

Question 92.
A particle moving with velocity \(\vec{v}\) is acted by three forces shown by the vector triangle PQR. The velocity of the particle will:
(A) remain constant
(B) change according to the smallest force \(\overrightarrow{\mathrm{QR}}\)
(C) increase
(D) decrease
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 71
Hint: As the three forces acting on a particle represents a triangle (i.e., a closed loop)
∴ Fnet = 0
∴ m \(\vec{a}\) = 0
∴ m\(\frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}\)
∴ v remains constant
Answer:
(A) remain constant

Question 93.
A force F = 20 + 10y acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is:
(A) 25J
(B) 20J
(C) 30J
(D) SJ
Hint: Work done by variable force, W = \(\int_{y_{\text {initial }}}^{y_{\text {final }}} \mathrm{Fdy}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 72
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 73
Answer:
(A) 25J

Question 94.
Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is:
(A) \(\frac{4}{9}\)
(B) \(\frac{5}{9}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{8}{9}\)
Hint:
Fractional loss of K.E. of colliding bodies,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 74
Answer:
(D) \(\frac{8}{9}\)

Question 95.
An object of mass 500 g, initially at rest, is acted upon by a variable force whose X-component varies with X in the manner shown. The velocities of the object at the points X = 8 m and X = 12 m, would have the respective values of
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 75
(A) 18m/s and 20.6 m/s
(B) 18 m/s and 24.4 m/s
(C) 23 m/s and 24.4 m/s
(D) 23 m/s and 20.6 m/s
Hint: From work-energy theorem
∆ K.E. = work = area under F-x graph
From x = 0 to x = 8m
\(\frac{1}{2} \mathrm{mv}^{2}\) = (5 × 20) + (3 × 10)
∴ \(\frac{1}{2} \mathrm{mv}^{2}\) = 100 + 30
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 76
Answer:
(D) 23 m/s and 20.6 m/s

Question 96.
The centre of mass of two particles system lies
(A) at the midpoint on the line joining the two particles.
(B) at one end of line joining the two particles.
(C) on the line perpendicular in the line joining two particles.
(D) on the line joining the Iwo particles.
Answer:
(D) on the line joining the two particles.

Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion

Question 97.
A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration ‘a’ towards the right. The relation between a and θ for the block to remain stationary on the wedge is
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 77
Hint:
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 78
The mass of block is given to be m. It will remain stationary if forces acting on it are in equilibrium i.e., ma cos θ = mg sin θ
Here, ma = Pseudo force on block.
∴ a = g tan θ
Answer:
a = g tan θ

Question 98.
A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision, When the initial velocity of the lighter block is v, then the value of coefficient of restitution
(e) will
(A) 0.5
(B) 0.25
(C) 0.8
(D) 0.4
Hint:
Given: m1 = m, m2 = 4m, u1 = v, u2 = 0, v1 = 0 According to law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
mv + 4m × 0 = m × 0 + 4mv2
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 79
Answer:
(B) 0.25

Question 99.
In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is:
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 80
Hint:
According to law of conservation of momentum,
mv0 = mv1 + mv2
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 81
Answer:
(D) \(\sqrt{2} \mathbf{v}_{0}\)

Question 100.
The mass of a hydrogen molecule is 3.32 × 10-27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall of area 2 cm2 at an angle of 45° to the normal and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly:
(A) 2.35 × 102 N/m2
(B) 4.70 × 102 N/m2
(C) 2.35 × 103 N/m2
(D) 4.70 × 103 N/m2
Hint:
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 82
Answer:
(C) 2.35 × 103 N/m2

Question 101.
A bomb at rest explodes into 3 parts of same mass. The momentum of two parts is -3P\(\hat{\mathrm{i}}\) and 2P\(\hat{\mathrm{j}}\) respectively. The magnitude of momentum of the third part is
(A) P
(B) \(\sqrt{5} \mathrm{P}\)
(C) \(\sqrt{11} \mathrm{P}\)
(D) \(\sqrt{13} \mathrm{P}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 83
Answer:
(D) \(\sqrt{13} \mathrm{P}\)

Question 102.
A sphere of mass ‘m’ moving with velocity V collides head-on on another sphere of same mass which is at rest. The ratio of final velocity of second sphere to the initial velocity of the first sphere is (e is coefficient of restitution and collision is inelastic)
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 84
Hint:
Initial momentum = mv
Final momentum = mv1 + mv2
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 85
Answer:
(C) \(\frac{\mathrm{e}+1}{2}\)

Question 103.
Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively:
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 86
Hint:
Tension in spring before cutting the strip,
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 87
Answer:
(B) \(\frac{\mathrm{g}}{3}\), g

Question 104.
A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be
(A) 9 J
(B) 18 J
(C) 4.5 J
(D) 22 J
Hint:
F = 6t
m = 1 kg
∴ a = 6t
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 88
Answer:
(C) 4.5 J

Question 105.
Consider a drop of rain water having mass Ig falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take ‘g’ constant with a value 10 m/s2. The work done by the
(i) gravitational force and the
(ii) resistive force of air is:
(A) (i) -10 J
(ii) -8.25 J
(B) (i) 1.25 J
(ii) -8.25 J
(C) (i) 100 J
(ii) 8.75 J
(D) (i) 10 J
(ii) -8.75 J
Hint: Work done by gravitation force is given by (Wg)
Maharashtra Board Class 11 Physics Important Questions Chapter 4 Laws of Motion 89
Answer:
(D) (i) 10 J
(ii) -8.75 J

Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 2 Mathematical Methods Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 2 Mathematical Methods

Question 1.
Explain representation of a vector graphically and symbolically.
Answer:

  1. Graphical representation:
    A vector is graphically represented by a directed line segment or an arrow.
    eg.: displacement of a body from P to Q is represented as P → Q.
  2. Symbolic representation:
    Symbolically a vector is represented by a single letter with an arrow above it, such as \(\overrightarrow{\mathrm{A}}\). The magnitude of the vector \(\overrightarrow{\mathrm{A}}\) is denoted as |A| or | \(\overrightarrow{\mathrm{A}}\) | or A.

Question 2.
A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector?
Answer:

  1. For a physical quantity, only having magnitude and direction is not a sufficient condition to be a vector.
  2. A physical quantity also has to obey vectors law of addition to be termed as vector.
  3. Hence, anything that has magnitude and direction is not necessarily a vector.
    Example: Though current has definite magnitude and direction, it is not a vector.

Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods

Question 3.
Define and explain the following terms:
i. Zero vector (Null vector)
ii. Resultant vector
iii. Negative vectors
iv. Equal vectors
v. Position vector
Answer:
i. Zero vector (Null vector):
A vector having zero magnitude and arbitrary direction is called zero vector. It is denoted as \(\overrightarrow{0}\).
Example: Velocity vector of stationary particle, acceleration vector of a body moving with uniform velocity.

ii. Resultant vector:
The resultant of two or more vectors is defined as that single vector, which produces the same effect as produced by all the vectors together.

iii. Negative vectors:
A negative vector of a given vector is a vector of the same magnitude but opposite in direction to that of the given vector.
Negative vectors are antiparallel vectors.
In figure, \(\vec{b}\) = – \(\vec{a}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 1

iv. Equal vectors:
Two vectors A and B representing same physical quantity are said to be equal if and only if they have the same magnitude and direction.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 2
In the given figure |\(\overrightarrow{\mathrm{P}}\)| = |\(\overrightarrow{\mathrm{Q}}\)| = |\(\overrightarrow{\mathrm{R}}\)| = |\(\overrightarrow{\mathrm{S}}\)|

v. Position vector:
A vector which gives the position of a particle at a point with respect to the origin of chosen co-ordinate system is called position vector.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 3
In the given figure \(\overrightarrow{\mathrm{OP}}\) represents position vector of \(\vec{P}\) with respect to O.

Question 4.
Whether the resultant of two vectors of unequal magnitude be zero?
Answer:
The resultant of two vectors of different magnitude cannot give zero resultant.

Question 5.
Define unit vector and give its physical significance.
Answer:
Unit vector: A vector having unit magnitude in a given direction is called a unit vector in that direction.
If \(\vec{p}\) is a non zero vector (P ≠ 0) then the unit vector \(\hat{\mathrm{u}}_{\mathrm{p}}\) in the direction of \(\overrightarrow{\mathrm{P}}\) is given by,
\(\hat{\mathrm{u}}_{\mathrm{p}}\) = \(\frac{\overrightarrow{\mathrm{P}}}{\mathrm{P}}\)
∴ \(\overrightarrow{\mathrm{P}}\) = \(\hat{u}_{p} P\)

Significance of unit vector:

i. The unit vector gives the direction of a given vector.

ii. Unit vector along X, Y and Z direction of a rectangular (three dimensional) coordinate is represented by \(\hat{\mathrm{i}}\), \(\hat{\mathrm{j}}\) and \(\hat{\mathrm{k}}\) respectively Such that \(\hat{\mathbf{u}}_{x}\) = \(\hat{\mathrm{i}}\), \(\hat{\mathbf{u}}_{y}\) = \(\hat{\mathrm{j}}\) and \(\hat{\mathbf{u}}_{z}\) = \(\hat{\mathrm{k}}\)
This gives \(\hat{\mathrm{i}}\) = \(\frac{\overrightarrow{\mathrm{X}}}{\mathrm{X}}\), \(\hat{\mathrm{j}}\) = \(\frac{\overrightarrow{\mathrm{Y}}}{\mathrm{X}}\) and \(\hat{\mathrm{k}}\) = \(\frac{\overrightarrow{\mathrm{Z}}}{\mathrm{Z}}\)

Question 6.
Explain multiplication of a vector by a scalar.
Answer:

  1. When a vector \(\overrightarrow{\mathrm{A}}\) is multiplied by a scalar ‘s’, it becomes ‘s\overrightarrow{\mathrm{A}}’ whose magnitude is s times the magnitude of \(\overrightarrow{\mathrm{A}}\).
  2. The unit of \(\overrightarrow{\mathrm{A}}\) is different from the unit of ‘s \(\overrightarrow{\mathrm{A}}\)’.
    For example,
    If \(\overrightarrow{\mathrm{A}}\) = 10 newton and s = 5 second, then s\(\overrightarrow{\mathrm{A}}\) = 10 newton × 5 second = 50 Ns.

Question 7.
Explain addition of vectors.
Answer:

  1. The addition of two or more vectors of same type gives rise to a single vector such that the effect of this single vector is the same as the net effect of the original vectors.
  2. It is important to note that only the vectors of the same type (physical quantity) can be added.
  3. For example, if two vectors, \(\overrightarrow{\mathrm{P}}\) = 3 unit and \(\overrightarrow{\mathrm{Q}}\) = 4 unit are acting along the same line, then they can be added as, |\(\overrightarrow{\mathrm{R}}\)| = |\(\overrightarrow{\mathrm{P}}\)| + |\(\overrightarrow{\mathrm{Q}}\)|
    |\(\overrightarrow{\mathrm{R}}\)| = 3 + 4 = 7
    Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 4
    [Note: When vectors are not in the same direction, then they can be added using triangle law of vector addition.]

Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods

Question 8.
State true or false. If false correct the statement and rewrite.
It is possible to add two vectors representing physical quantities having different dimensions.
Answer:
False.
It is not possible to add two vectors representing physical quantities having different dimensions.

Question 9.
Explain subtraction of vectors.
Answer:

  1. When two vectors are anti-parallel (in the opposite direction) to each other, the magnitude
  2. It is important to note that only vectors of the same type (physical quantity) can be subtracted.
  3. For example, if two vectors \(\overrightarrow{\mathrm{P}}\) = 3 unit and \(\overrightarrow{\mathrm{Q}}\) = 4 unit are acting in opposite direction, they are subtracted as, |\(\overrightarrow{\mathrm{R}}\)| = ||\(\overrightarrow{\mathrm{P}}\)| – |\(\overrightarrow{\mathrm{Q}}\)||
    = |3 – 4| = 1 unit, directed along \(\overrightarrow{\mathrm{Q}}\)
    Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 5

Question 10.
How can resultant of two vectors of a type inclined with each other be determined?
Answer:
When two vectors of a type are inclined with each other, their resultant can be determined by using triangle law of vector addition.

Question 11.
What is triangle law of vector addition?
Answer:
Triangle law of vector addition:
If two vectors describing the same physical quantity are represented in magnitude and direction, by the two sides of a triangle taken in order, then their resultant is represented in magnitude and direction by the third side of the triangle drawn in the opposite sense, i.e., from the starting point (tail) of the first vector to the end point (head) of the second vector.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 6
Let \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) be the two vectors of same type taken in same order as shown in figure.
∴ Resultant vector will be given by third side taken in opposite order, i.e., \(\overline{\mathrm{OA}}\) + \(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{OB}}\)
∴ \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) = \(\overrightarrow{\mathrm{R}}\)

Question 12.
Using triangle law of vector addition, explain the process of adding two vectors which are not lying in a straight line.
Answer:
i. Two vectors in magnitude and direction are drawn in a plane as shown in figure (a)
Let these vectors be \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 7

ii. Join the tail of \(\overrightarrow{\mathrm{Q}}\) to head of \(\overrightarrow{\mathrm{P}}\) in the given direction. The resultant vector will be the line which is obtained by joining tail of \(\overrightarrow{\mathrm{P}}\) to head of \(\overrightarrow{\mathrm{Q}}\) as shown in figure (b).
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 8

iii. If \(\overrightarrow{\mathrm{R}}\) is the resultant vector of \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) then using triangle law of vector addition, we have, \(\overrightarrow{\mathrm{R}}\) = \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\)

Question 13.
Is it possible to add two velocities using triangle law?
Answer:
Yes, it is possible to add two velocities using triangle law.

Question 14.
Explain, how two vectors are subtracted. Find their resultant by using triangle law of vector addition.
Answer:

  1. Let \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) be the two vectors in a plane as shown in figure (a).
    Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 10
  2. To subtract \(\overrightarrow{\mathrm{Q}}\) from \(\overrightarrow{\mathrm{P}}\), vector \(\overrightarrow{\mathrm{Q}}\) is reversed so that we get the vector –\(\overrightarrow{\mathrm{Q}}\) as shown in figure (b).
  3. The resultant vector is obtained by –\(\overrightarrow{\mathrm{R}}\) joining tail of \(\overrightarrow{\mathrm{P}}\) to head of – \(\overrightarrow{\mathrm{Q}}\) as shown in figure (c).
  4. From triangle law of vector addition, \(\overrightarrow{\mathrm{R}}\) = \(\overrightarrow{\mathrm{P}}\) + (-\(\overrightarrow{\mathrm{Q}}\)) = \(\overrightarrow{\mathrm{P}}\) – \(\overrightarrow{\mathrm{Q}}\)

Question 15.
Prove that: Vector addition is commutative.
Answer:
Commutative property of vector addition:
According to commutative property, for two
vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\), \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) = \(\overrightarrow{\mathrm{Q}}\) + \(\overrightarrow{\mathrm{p}}\)

Proof:

i. Let two vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) be represented in magnitude and direction by two sides \(\overrightarrow{\mathrm{OA}}\) and \(\overrightarrow{\mathrm{AB}}\) respectively.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 11

ii. Complete a parallelogramOABC such that
\(\overrightarrow{\mathrm{OA}}\) = \(\overrightarrow{\mathrm{CB}}\) = \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{OC}}\) = \(\overrightarrow{\mathrm{Q}}\) then join OB.

iii. In △OAB, \(\overrightarrow{\mathrm{OA}}\) + \(\overrightarrow{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{OB}}\)
(By triangle law of vector addition)
∴ \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) = \(\overrightarrow{\mathrm{R}}\) … (1)
In △OCB, \(\overrightarrow{\mathrm{OC}}\) + \(\overrightarrow{\mathrm{CB}}\) = \(\overrightarrow{\mathrm{OB}}\)
(By triangle law of vector addition)
∴ \(\overrightarrow{\mathrm{Q}}\) + \(\overrightarrow{\mathrm{P}}\) = \(\overrightarrow{\mathrm{R}}\) … (2)

iv. From equation (1) and (2),
\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) = \(\overrightarrow{\mathrm{Q}}\) + \(\overrightarrow{\mathrm{P}}\)
Hence, addition of two vectors obeys commutative law.

Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods

Question 16.
Prove that: Vector addition is associative.
Answer:
Associative property of vector addition:
According to associative property, for three vectors \(\overrightarrow{\mathrm{P}}\), \(\overrightarrow{\mathrm{Q}}\) and \(\overrightarrow{\mathrm{R}}\),
(\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\)) + \(\overrightarrow{\mathrm{R}}\) = \(\overrightarrow{\mathrm{P}}\) + (\(\overrightarrow{\mathrm{Q}}\) + \(\overrightarrow{\mathrm{R}}\))
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 12
Proof:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 13
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 14
On comparing, equation (2) and (4), we get,
(\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\)) + \(\overrightarrow{\mathrm{R}}\) = \(\overrightarrow{\mathrm{P}}\) + (\(\overrightarrow{\mathrm{Q}}\) + \(\overrightarrow{\mathrm{R}}\))
Hence, associative law is proved.

Question 17.
State true or false. If false correct the statement and rewrite.
The subtraction of given vectors is neither commutative nor associative.
Answer:
True.

Question 18.
State and prove parallelogram law of vector addition and determine magnitude and direction of resultant vector.
Answer:

i. Parallelogram law of vector add addition;
If two vectors of same type starting from the same point (tails cit the same point), are represented in magnitude and direction by the two adjacent sides of a parallelogram then, their resultant vector is given in magnitude and direction, by the diagonal of the parallelogram starting from the same point.

ii. Proof:

a. Consider two vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) of the same type, with their tails at the point O’ and θ’ is the angle between \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) as shown in the figure below.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 15
b. Join BC and AC to complete the parallelogram OACB, with \(\overline{\mathrm{OA}}\) = \(\overrightarrow{\mathrm{P}}\) and \(\overline{\mathrm{AC}}\) = \(\overrightarrow{\mathrm{Q}}\) as the adjacent sides. We have to prove that diagonal \(\overline{\mathrm{OC}}\) = \(\overrightarrow{\mathrm{R}}\), the resultant of sum of the two given vectors.

c. By the triangle law of vector addition, we have,
\(\overrightarrow{\mathrm{OA}}\) + \(\overrightarrow{\mathrm{AC}}\) = \(\overrightarrow{\mathrm{OC}}\) … (1)
As \(\overrightarrow{\mathrm{AC}}\) is parallel to \(\overrightarrow{\mathrm{OB}}\),
\(\overrightarrow{\mathrm{AC}}\) = \(\overrightarrow{\mathrm{OB}}\) = \(\overrightarrow{\mathrm{Q}}\)
Substituting \(\overrightarrow{\mathrm{OA}}\) and \(\overrightarrow{\mathrm{OC}}\) in equation (1) we have,
\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) = \(\overrightarrow{\mathrm{R}}\)
Hence proved.

iii. Magnitude of resultant vector:

a. To find the magnitude of resultant vector \(\overrightarrow{\mathrm{R}}\) = \(\overrightarrow{\mathrm{OC}}\), draw a perpendicular from C to meet OA extended at S.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 16

c. Using Pythagoras theorem in right angled triangle, OSC
(OC)2 = (OS)2 + (SC)2
= (OA + AS)2 + (SC)2
∴ (OC)2 = (OA)2 + 2(OA).(AS) + (AS2) + (SC)2 . . . .(4)

d. From right angle trianle ASC,
(AS)2 + (SC)2 = (AC)2 …. (5)

e. From equation (4) and (5), we get
(OC)2 = (OA)2 + 2(OA) (AS) + (AC)2
… .(6)

f. Using (2) and (6), we get
(OC)2 = (OA)2 + (AC)2 + 2(OA)(AC) cos θ
∴ R2 = P2 + Q2 + 2 PQ cos θ
∴ R = \(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cos \theta}\) ….(7)
Equation (7) gives the magnitude of resultant vector \(\overrightarrow{\mathrm{R}}\).

iv. Direction of resultant vector:
To find the direction of resultant vector \(\overrightarrow{\mathrm{R}}\), let \(\overrightarrow{\mathrm{R}}\) make an angle α with \(\overrightarrow{\mathrm{P}}\).
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 17
Equation (9) represents direction of resultant vector.
[Note: If β is the angle between \(\overrightarrow{\mathrm{R}}\) and \(\overrightarrow{\mathrm{Q}}\), it can be similarly derived that
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 18

Question 19.
Complete the table for two vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) inclined at angle θ.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 19
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 20

Question 20.
The diagonal of the parallelogram made by two vectors as adjacent sides is not passing through common point of two vectors. What does it represent?
Answer:
The diagonal of the parallelogram made by two vectors as adjacent sides not passing through common point of two vectors represents triangle law of vector addition.

Question 21.
If | \(\overrightarrow{\mathbf{A}}\) + \(\overrightarrow{\mathbf{B}}\) | = | \(\overrightarrow{\mathbf{A}}\) – \(\overrightarrow{\mathbf{B}}\) | then what can be the
angle between \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\) ?
Answer:
Let θ be the angle between \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\), then
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 21

Thus, if |\(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)| = |\(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\) |, then vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) must be at right angles to each other.

Question 22.
Express vector \(\overrightarrow{\mathbf{A C}}\) in terms of vectors \(\overrightarrow{\mathbf{A B}}\) and \(\overrightarrow{\mathbf{C B}}\) shown in the following figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 22
Solution:
Using the triangle law of addition of vectors,
\(\overrightarrow{\mathbf{A C}}\) + \(\overrightarrow{\mathbf{C B}}\) = \(\overrightarrow{\mathbf{A B}}\)
∴\(\overrightarrow{\mathbf{A C}}\) = \(\overrightarrow{\mathbf{A B}}\) – \(\overrightarrow{\mathbf{C B}}\)

Question 23.
From the following figure, determine the resultant of four forces \(\overrightarrow{\mathbf{A}}_{1}\), \(\overrightarrow{\mathbf{A}}_{2}\), \(\overrightarrow{\mathbf{A}}_{3}\), \(\overrightarrow{\mathbf{A}}_{4}\).
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 23
Solution:
Join \(\overrightarrow{\mathrm{OB}}\) to complete ∆OAB as shown in figure below
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 24
Now, using triangle law of vector addition,
\(\overrightarrow{\mathrm{OB}}\) = \(\overrightarrow{\mathrm{OA}}\) + \(\overrightarrow{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{A}}_{1}\) + \(\overrightarrow{\mathrm{A}}_{2}\)
Join \(\overrightarrow{\mathrm{OC}}\) to complete triangle OBC as shown figure below
Similarly, \(\overrightarrow{\mathrm{OC}}\) = \(\overrightarrow{\mathrm{OB}}\) + \(\overrightarrow{\mathrm{BC}}\) = \(\overrightarrow{\mathrm{A}}_{1}\) + \(\overrightarrow{\mathrm{A}}_{2}\) + \(\overrightarrow{\mathrm{A}}_{3}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 25
Answer:
\(\overrightarrow{O D}\) is the resultant of the four vectors.

Question 24.
Find the vector that should be added to the sum of (2\(\hat{\mathbf{i}}\) – 5\(\hat{\mathbf{j}}\) + 3\(\hat{\mathbf{k}}\)) and (4\(\hat{\mathbf{i}}\) + 7\(\hat{\mathbf{j}}\) – 4\(\hat{\mathbf{k}}\)) to give a unit vector along the X-axis.
Solution:
Let vector \(\overrightarrow{\mathrm{p}}\) be added to get unit vector (\(\hat{\mathbf{i}}\)) along X-axis.
Sum of given vectors is given as,
(2\(\hat{\mathbf{i}}\) – 5\(\hat{\mathbf{j}}\) + 3\(\hat{\mathbf{k}}\) ) + (4\(\hat{\mathbf{i}}\) + 7\(\hat{\mathbf{j}}\) – 4\(\hat{\mathbf{k}}\)) = 6\(\hat{\mathbf{i}}\) + 2\(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\)
According to given condition, (6\(\hat{\mathbf{i}}\) + 2\(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\)) + \(\hat{\mathbf{P}}\) = \(\hat{\mathbf{i}}\)
∴ \(\overrightarrow{\mathrm{P}}\) = \(\hat{\mathbf{i}}\) – (6\(\hat{\mathbf{i}}\) + 2\(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\)) = \(\hat{\mathbf{i}}\) – 6\(\hat{\mathbf{i}}\) – 2\(\hat{\mathbf{j}}\) + \(\hat{\mathbf{k}}\) = -5\(\hat{\mathbf{i}}\) – 2\(\hat{\mathbf{j}}\) + \(\hat{\mathbf{k}}\)
Answer:
The required vector is -5\(\hat{\mathbf{i}}\) – 2\(\hat{\mathbf{j}}\) + \(\hat{\mathbf{k}}\).

Question 25.
If \(\overrightarrow{\mathbf{P}}\) = 2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{Q}}\) = 2\(\hat{\mathbf{i}}\) – 5\(\hat{\mathbf{j}}\) + 2\(\hat{\mathbf{k}}\).Find
i. \(\overrightarrow{\mathbf{P}}\) + \(\overrightarrow{\mathbf{Q}}\)
ii. 3\(\overrightarrow{\mathbf{P}}\) – 2\(\overrightarrow{\mathbf{Q}}\)
Solution:
Given \(\overrightarrow{\mathbf{P}}\) = 2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\), \(\overrightarrow{\mathbf{Q}}\) = 2\(\hat{\mathbf{i}}\) – 5\(\hat{\mathbf{j}}\) + 2\(\hat{\mathbf{k}}\)
To find:

i. \(\overrightarrow{\mathbf{P}}\) + \(\overrightarrow{\mathbf{Q}}\)
ii. 3\(\overrightarrow{\mathbf{P}}\) – 2\(\overrightarrow{\mathbf{Q}}\)

Calculation:

i. \(\overrightarrow{\mathbf{P}}\) + \(\overrightarrow{\mathbf{Q}}\) = (2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\) – k) + (2\(\hat{\mathbf{i}}\) – 5\(\hat{\mathbf{j}}\) + 2k)
= (2 + 2)\(\hat{\mathbf{i}}\) + (3 – 5)\(\hat{\mathbf{j}}\) + (-1 + 2)\(\hat{\mathbf{k}}\)
= 4\(\hat{\mathbf{i}}\) – 2\(\hat{\mathbf{j}}\) + \(\hat{\mathbf{k}}\)

ii. 3\(\overrightarrow{\mathbf{P}}\) = 3(2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\)) = 6\(\hat{\mathbf{i}}\) + 9\(\hat{\mathbf{j}}\) – 3\(\hat{\mathbf{k}}\)
2\(\overrightarrow{\mathbf{Q}}\) = 2(2\(\hat{\mathbf{i}}\) – 5\(\hat{\mathbf{j}}\) + 2\(\hat{\mathbf{k}}\)) = 4\(\hat{\mathbf{i}}\) – 10\(\hat{\mathbf{j}}\) + 4\(\hat{\mathbf{k}}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 26

Question 26.
Find unit vector parallel to the resultant of the vectors \(\overrightarrow{\mathbf{A}}\) = \(\hat{\mathbf{i}}\) + 4\(\hat{\mathbf{j}}\) – 2\(\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}\) = 3\(\hat{\mathbf{i}}\) – 5\(\hat{\mathbf{j}}\) + \(\hat{\mathbf{k}}\).
Solution:
The resultant of \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) is,
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 27
Answer:
The required unit vector is \(\frac{1}{3 \sqrt{2}}\)(4\(\hat{\mathbf{i}}\) – \(\hat{\mathbf{j}}\) – \(\hat{\mathbf{k}}\))

Question 27.
Two forces, F1 and F2, each of magnitude 5 N are inclined to each other at 60°. Find the magnitude and direction of their resultant force.
Solution:
Given: F1 = 5 N, F2 = 5 N, θ = 60°
To find: Magnitude of resultant force (R),
Direction of resultant force (α)
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 28
Answer:
i. The magnitude of resultant force is 8.662 N.

ii. The direction of resultant force is 30° w.r.t. \(\overrightarrow{\mathrm{F}_{1}}\).

Question 28.
Water is flowing in a stream with velocity 5 km/hr in an easterly direction relative to the shore. Speed of a boat relative to still water is 20 km/hr. If the boat enters the stream heading north, with what velocity will the boat actually travel?
Answer:
The resultant velocity \(\overrightarrow{\mathrm{R}}\) of the boat can be obtained by adding the two velocities using ∆ OAB shown in the figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 29
The direction ot the resultant velocity is
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 30
Answer: The velocity of the boat is 20.616 km/hr in a direction 14.04° east of north. .
[Note: tan-1 (0.25) ≈ 14.04° which equals 14°2]

Question 29.
Rain is falling vertically with a speed of 35 m/s. Wind starts blowing at a speed of 12 m/s in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella? (NCERT)
Solution:
Let the velocity of rain and wind be \(\overrightarrow{\mathbf{V}_{\mathrm{R}}}\) and \(\overrightarrow{\mathbf{V}_{\mathrm{W}}}\), then resultant velocity \(\overrightarrow{\mathrm{v}}\) has magnitude of
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 31
If \(\overrightarrow{\mathrm{v}}\) makes an angle θ with vertical then, from the figure
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 32
Answer: The boy should hold his umbrella in vertical plane at an angle of about 19° with vertical towards the east.

Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods

Question 30.
What are components of a vector?
Answer:

  1. The given vector can be written as sum of two or more vectors along certain fixed directions. The vectors into which the given single vector is splitted are called components of the vector.
  2. Let \(\overrightarrow{\mathrm{A}}\) = \(\mathrm{A}_{1} \hat{\alpha}\) + \(\mathrm{A}_{2} \hat{\beta}\) + \(\mathrm{A}_{3} \hat{\gamma}\) where, \(\hat{\alpha}\), \(\hat{\beta}\) and \(\hat{\gamma}\) are unit vectors along chosen directions. Then, A1, A2 and A3 are known as components of \(\overrightarrow{\mathrm{A}}\) along three directions \(\hat{\alpha}\), \(\hat{\beta}\) and \(\hat{\gamma}\).
  3. It two vectors are equal then, their corresponding components are also equal and vice-versa.
    Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 33

[Note: The magnitude of a vector is a scalar while each component of a vector is always a vector.]

Question 31.
What is meant by resolution of vector?
Answer:

  1. The process of splitting a given vector into its components is called resolution of the vector.
  2. Resolution of vector is equal to replacing the original vector with the sum of the component vectors.

Question 32.
That are rectangular components of vectors? Explain their uses.
Answer:
i. Rectangular components of a vector:
If components of a given vector are mutually perpendicular to each other then they are called rectangular components of that vector.

ii. Consider a vector \(
\overrightarrow{\mathrm{R}}
\) = \(
\overrightarrow{\mathrm{OC}}
\) originating from the origin O’ of a rectangular co-ordinate system as shown in figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 34

iii. Draw CA ⊥ OX and CB ⊥ OY.
Let component of \(
\overrightarrow{\mathrm{R}}
\) along X-axis \(
\overrightarrow{\mathrm{R}}_{\mathrm{x}}
\) and component of \(
\overrightarrow{\mathrm{R}}
\) along Y-axis = \(
\overrightarrow{\mathrm{R}}_{\mathrm{y}}
\)
By parallelogram law of vectors,
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 35
where, \(
\hat{i}
\) and \(
\hat{j}
\) are unit vectors along positive direction of X and Y axes respectively.

iv. If θ is angle made by \(
\overrightarrow{\mathrm{R}}
\) with X-axis, then
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 36

v. Squaring and adding equation (1) and (2) we get,
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 37
Equation (3) gives the magnitude of \(
\overrightarrow{\mathrm{R}}
\).

vi. Direction of \(
\overrightarrow{\mathrm{R}}
\) can be found out by dividing equation (2) by (1),
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 38
Equation (4) gives direction of \(
\overrightarrow{\mathrm{R}}
\)

vii. When vectors are noncoplanar, it becomes necessary to use the third dimension. If \(
\overrightarrow{\mathrm{R}}_{\mathrm{x}}
\), \(
\overrightarrow{\mathrm{R}}_{\mathrm{y}}
\) and \(
\overrightarrow{\mathrm{R}}_{\mathrm{z}}
\) are three rectangular components of \(
\overrightarrow{\mathrm{R}}
\) along X, Y and Z axes of a three dimensional rectangular cartesian co-ordinate system then.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 39
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 40

Question 33.
Find a unit vector in the direction of the vector 3\(
\hat{i}
\) + 4\(
\hat{j}
\).
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 41

Question 34.
Given \(
\overrightarrow{\mathbf{a}}
\) = \(
\hat{\mathbf{i}}
\) + 2\(
\hat{\mathbf{j}}
\) and \(
\overrightarrow{\mathbf{b}}
\) = 2\(
\hat{\mathbf{i}}
\) + \(
\hat{\mathbf{j}}
\), what are the magnitudes of the two vectors? Are these two vectors equal?
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 42
The magnitudes of \(
\vec{a}
\) and \(
\vec{b}
\) are equal. However, their corresponding components are not equal, i.e., ax ≠ bx and ay ≠ by. Hence, the two vectors are not equal.
Answer:
Magnitudes of two vectors are equal, but vectors are unequal.

Question 35.
Find the vector drawn from the point (-4, 10, 7) to the point (3, -2, 1). Also find its magnitude.
Solution:
If \(
\overrightarrow{\mathrm{A}}
\) is a vector drawn from the point (x1, y1, z1) to the point (x2, y2, z2), then
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 43

Question 36.
In a cartesian co-ordinate system, the co-ordinates of two points P and Q are (2, 4, 4) and (-2, -3, 7) respectively, find \(
\overrightarrow{\mathbf{P Q}}
\) and its magnitude.
Solution:
Given: Position vector of P = (2,4,4)
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 44
∴ |\(
\overrightarrow{\mathrm{PQ}}
\)| = 8.6 units
Answer: Vector \(
\overrightarrow{\mathrm{PQ}}
\) is -4\(
\hat{\mathbf{i}}
\) – 7\(
\hat{\mathbf{j}}
\) + 3\(
\hat{\mathbf{k}}
\) and its magnitude is 8.6 units.

Question 37.
If \(
\overrightarrow{\mathbf{A}}
\) = 3\(
\hat{i}
\) + 4[/latex] = 3\(
\hat{j}
\) and \(
\overrightarrow{\mathbf{B}}
\) = 7\(
\hat{i}
\) + 24\(
\hat{j}
\), find a vector having the same magnitude as \(
\overrightarrow{\mathbf{B}}
\) and parallel to \(
\overrightarrow{\mathbf{A}}
\).
Solution:
The magnitude of vector \(
\overrightarrow{\mathrm{A}}
\) is | \(
\overrightarrow{\mathrm{A}}
\) |
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 45
Answer: The required vector is 15\(
\hat{\mathbf{i}}
\) + 20\(
\hat{\mathbf{j}}
\).

Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods

Question 38.
Complete the table.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 46
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 47

Question 39.
State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful.
i. Adding any two scalars,
ii. Adding a scalar to a vector of the same dimensions,
iii. Multiplying any vector by any scalar,
iv. Multiplying any two scalars,
v. Adding any two vectors. (NCERT)
Answer:

  1. Not any two scalars can be added. To add two scalars it is essential that they represent same physical quantity.
  2. This operation is meaningless. Only a vector can be added to another vector.
  3. This operation is possible. When a vector is multiplied with a dimensional scalar, the resultant vector will have different dimensions.
    eg.: acceleration vector is multiplied with mass (a dimensional scalar), the resultant vector has the dimensions of force.
    When a vector is multiplied with non – dimensional scalar, it will be a vector having dimensions as that of the given vector.
    eg.: \(
    \overrightarrow{\mathrm{A}}
    \) × 3 = 3\(
    \overrightarrow{\mathrm{A}}
    \)
  4. This operation is possible. Multiplication of non-dimensional scalars is simply algebraic multiplication. Multiplication of non dimensional scalars will result into scalar with different dimensions.
    eg.: Volume × density = mass.
  5. Not any two vectors can be added. To add two vectors it is essential that they represent same physical quantity.

Question 40.
Explain scalar product of two vectors with the help of suitable examples.
Answer:
Scalar product of two vectors:

  1. The scalar product of two non-zero vectors is defined as the product of the magnitude of the two vectors and cosine of the angle θ between the two vectors.
  2. The dot sign is used between the two vectors to be multiplied therefore scalar product is also called dot product.
  3. The scalar product of two vectors \(
    \overrightarrow{\mathrm{P}}
    \) and \(
    \overrightarrow{\mathrm{Q}}
    \) is given by, \(
    \overrightarrow{\mathrm{P}}
    \) . \(
    \overrightarrow{\mathrm{Q}}
    \) = PQ cos θ
    where, p = magnitude of \(
    \overrightarrow{\mathrm{P}}
    \), Q = magnitude of \(
    \overrightarrow{\mathrm{Q}}
    \)
    θ = angle between \(
    \overrightarrow{\mathrm{P}}
    \) and \(
    \overrightarrow{\mathrm{Q}}
    \)
  4. Examples of scalar product:
    1. Power (P) is a scalar product of force (\(
      \overrightarrow{\mathrm{F}}
      \)) and velocity (\(
      \overrightarrow{\mathrm{v}}
      \))
      ∴ P = \(
      \overrightarrow{\mathrm{F}}
      \) . \(
      \overrightarrow{\mathrm{v}}
      \)
    2. Work is a scalar product of force (\(
      \overrightarrow{\mathrm{F}}
      \)) and displacement (\(
      \overrightarrow{\mathrm{s}}
      \)).
      ∴ W = \(
      \overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{s}}
      \)

Question 41.
Discuss characteristics of scalar product of two vectors.
Answer:
Characteristics of the scalar product of two vectors:
i. The scalar product of two vectors is equivalent to the product of magnitude of one vector with component of the other in the direction of the first.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 48
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 49
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 50
vi. Scalar product of two vectors is expressed in terms of rectangular components as
\(
\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}
\) = Ax + Bx + AyBy + AzBz

vii. For \(
\vec{a} \neq 0, \vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}
\) does not necessarily mean \(
\vec{b}
\) = \(
\vec{c}
\)

Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods

Question 42.
Complete the table vector given below:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 51
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 52

Question 43.
Define and explain vector product of two vectors with suitable examples.
Answer:
i. The vector product of two vectors is a third vector whose magnitude is equal to the product of magnitude of the two vectors and sine of the smaller angle θ between the two vectors.

ii. Vector product is also called cross product of vectors because cross sign is used to represent vector product.

iii. Explanation:

a. The vector product of two vectors \(
\overrightarrow{\mathrm{A}}
\) and \(
\overrightarrow{\mathrm{B}}
\), is a third vector \(
\overrightarrow{\mathrm{R}}
\) and is written as, \(
\overrightarrow{\mathrm{R}}
\) = \(
\overrightarrow{\mathrm{A}}
\) × \(
\overrightarrow{\mathrm{B}}
\) = AB sin θ \(
\hat{\mathrm{u}}_{\mathrm{r}}
\) where, \(
\hat{\mathrm{u}}_{\mathrm{r}}
\) is unit vector in direction of \(
\overrightarrow{\mathrm{R}}
\), i.e., perpendicular to plane containing two vectors. It is given by right handed screw rule.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 53
c. Examples of vector product:

1. Force experienced by a charge q moving with velocity \(\overrightarrow{\mathrm{V}}\) in uniform magnetic field of induction (strength) \(\overrightarrow{\mathrm{B}}\) is given as \(\overrightarrow{\mathrm{F}}\) = q\(\overrightarrow{\mathrm{V}}\) × \(\overrightarrow{\mathrm{B}}\)

2. Moment of a force or torque (\(\begin{aligned}
&\rightarrow\\
&\tau
\end{aligned}\)) is the vector product of the position vector (\(\vec{r}\)) and the force (\(\overrightarrow{\mathrm{F}}\)).
i.e., \(\begin{aligned}
&\rightarrow\\
&\tau
\end{aligned}\) = \(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}\)

3. The instantaneous velocity (\(\overrightarrow{\mathrm{v}}\)) of a rotating particle is equal to the cross product of its angular velocity (\(\vec{\omega}\)) and its position (\(\overrightarrow{\mathrm{r}}\)) from axis of rotation.
\(\overrightarrow{\mathrm{v}}\) = \(\overrightarrow{\mathrm{r}}\) × \(\vec{\omega}\)

Question 44.
State right handed screw rule.
Answer:
Statement of Right handed screw rule: Hold a right handed screw with its axis perpendicular to the plane containing vectors and the screw rotated from first vector to second vector through a small angle, the direction in which the screw tip would advance is the direction of the vector product of two vectors.

Question 45.
State the characteristics of the vector product (cross product) of two vectors.
Answer:
Characteristics of the vector product (cross product):
i. The vector product of two vectors does not obey the commutative law of multiplication.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 54
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 55
vi. The magnitude of cross product of two vectors is numerically equal to the area of a parallelogram whose adjacent sides represent the two vectors.

Question 46.
Derive an expression for cross product of two vectors and express it in determinant form.
Answer:
Expression for cross product of two vectors:
i. Let two vectors \(\overrightarrow{\mathrm{R}}\) and \(\overrightarrow{\mathrm{Q}}\) be represented in magnitude and direction by,
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 56
ii.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 57
iii. Determinant form of cross product of two vectors \(\overrightarrow{\mathrm{R}}\) and \(\overrightarrow{\mathrm{Q}}\) is given by,
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 58

Question 47.
Show that magnitude of vector product of two vectors is numerically equal to the area of a parallelogram formed by the two
vectors.
Answer:
Suppose OACB is a parallelogram of adjacent sides, \(\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{Q}}\).
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 59
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 60

Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods

Question 48.
Distinguish between scalar product (dot product) and vector product (cross product).
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 61

Question 49.
Given \(\overrightarrow{\mathbf{P}}\) = 4\(\hat{\mathbf{i}}\) – \(\hat{\mathbf{j}}\) + 8\(\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{Q}}\) = 2\(\hat{\mathbf{i}}\) – m\(\hat{\mathbf{j}}\) + 4\(\hat{\mathbf{k}}\) find m if \(\overrightarrow{\mathbf{P}}\) and \(\overrightarrow{\mathbf{Q}}\) have the same direction. Solution:
Since \(\overrightarrow{\mathbf{P}}\) and \(\overrightarrow{\mathbf{Q}}\) have the same direction, their corresponding components must be in the same proportion, i.e.,
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 62

Question 50.
Find the scalar product of the two vectors \(\overrightarrow{\mathbf{v}}_{1}\) = \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\mathbf{3} \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{v}}_{2}\) = \(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\)
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 63
Answer: Scalar product of two given vectors is – 4.

Question 51.
A force \(\overrightarrow{\mathbf{F}}\) = \(4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) acting on a particle produces a displacement of \(\overrightarrow{\mathbf{S}}\) = \(\overrightarrow{\mathrm{s}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\mathbf{5} \hat{\mathbf{k}}\) where F is expressed in newton and s in metre. Find the work done by the force.
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 64
Answer: The work done by the force is 41 J.

Question 52.
Find ‘a’ if \(\overrightarrow{\mathbf{A}}\) = \(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}\) = \(\mathbf{a} \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\) are perpendicular to one another.
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 65

Question 53.
If \(\overrightarrow{\mathbf{A}}\) = \(5 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}\) = \(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) determine the angle between \(\) and \(\). Solution:
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 66

Question 54.
Find the angle between the vectors
\(\overrightarrow{\mathbf{A}}\) = \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}\) = \(-\hat{\mathbf{i}}+\hat{\mathbf{j}}-\mathbf{2} \hat{\mathbf{k}}\).
Solution:
Let angle between the vectors be θ
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 67
Answer: The angle between the vectors is 60°.

Question 55.
If \(\overrightarrow{\mathbf{A}}\) = \(2 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) and \(\vec{B}\) = \(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\mathbf{5} \hat{\mathbf{k}}\), find the component of \(\overrightarrow{\mathbf{A}}\) along \(\overrightarrow{\mathbf{B}}\).
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 68

Question 56.
\(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) are unit vectors along X-axis and Y-axis respectively. What is the magnitude and direction of the vector \(\hat{\mathbf{i}}+\hat{\mathbf{j}}\) and \(\hat{\mathbf{i}}-\hat{\mathbf{j}}\)? What are the components of a vector
\(\overrightarrow{\mathbf{A}}=2 \hat{\mathbf{i}}+\mathbf{3} \hat{\mathbf{j}}\) along the directions of \((\hat{\mathbf{i}}+\hat{\mathbf{j}})\) and \((\hat{\mathbf{i}}-\hat{\mathbf{j}})\)? (NCERT)
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 69
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 70

Question 57.
The angular momentum \(\overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}\), where \(\overrightarrow{\mathbf{r}}\) is a position vector and \(\overrightarrow{\mathrm{p}}\) is linear momentum of a body.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 71
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 72

Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods

Question 58.
If \(\overrightarrow{\mathbf{A}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\) are two vectors, find \(|\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}|\)
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 73

Question 59.
Find unit vectors perpendicular to the plane of the vectors, \(\overrightarrow{\mathbf{A}}\) = \(\) and
\(\overrightarrow{\mathbf{B}}\) = \(2 \hat{\mathbf{i}}-\hat{\mathbf{k}}\)
Solution:
Let required unit vector be \(\hat{\mathrm{u}}\).
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 74
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 75

Question 60.
\(\overrightarrow{\mathbf{P}}\) = \(\hat{\mathbf{i}}+\mathbf{2} \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{Q}}\) = \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}\) are two vectors, find the unit vector parallel to \(\overrightarrow{\mathbf{P}} \times \overrightarrow{\mathbf{Q}}\). Also find the vector perpendicular to P and Q of magnitude 6 units.
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 76

Question 61.
Find the area of a triangle formed by \(\overrightarrow{\mathbf{A}}\) = \(\hat{3} \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}\) = \(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\boldsymbol{2} \hat{\mathbf{k}}\) as adjacent sides measure in metre. Solution:
Given: Two adjacent sides of triangle,
\(\overrightarrow{\mathrm{A}}\) = \(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\), \(\overrightarrow{\mathrm{B}}\) = \(\hat{i}+\hat{j}-2 \hat{k}\)
To find: Area of triangle
Formula: Area of triangle =
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 77
Answer:
Area of the triangle is 6.1 m2.

Question 62.
Find the derivatives of the functions,
i. f(x) = x8
ii. f(x) = x3 + sin x
Solution:
i. Using \(\frac{\mathrm{dx}^{\mathrm{n}}}{\mathrm{dx}}\) = nxn-1,
\(\frac{d\left(x^{8}\right)}{d x}\) = 8x7

ii.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 78

Question 63.
Find derivatives of e2x – tan x
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 79

Question 64.
Find the derivatives of the functions.
f(x) = x3 sin x
Solution:
Using,
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 80

Question 65.
Find derivatives of \(\frac{d}{d x}(x \times \ln x)\)
Solution:
Using,
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 81

Question 66.
Evaluate the following integrals.

i. \(\int x^{8} d x\)
Solution:
Using formula \(\int x^{n} d x\) = \(\frac{x^{n+1}}{n+1}\),
\(\int x^{8} d x\) = \(\frac{x^{9}}{9}\)

ii. \(\int_{2}^{5} x^{2} d x\)
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 82

iii) \(\int(x+\sin x) d x\)
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 83

iv) \(\int\left(\frac{10}{x}+e^{x}\right) d x\)
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 84

v) \(\int_{1}^{4}\left(x^{3}-x\right) d x\)
Answer:
Using,
f1(x) – f2(x) = \(\int f_{1}(x)-\int f_{2}(x)\)
Here,
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 85

Question 67.
A man applies a force of 10 N on a garbage crate. If another man applies a force of 8 N on the same crate at an angle of 60° with respect to previous, then what will be the resultant force and direction of the crate, if crate is stationary.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 86
Answer:
A resultant force of 15.62 N is applied on a crate at an angle of 26.56°.

Question 68.
A lady dropped her wallet in the parking lot of a super market. A boy picked the wallet up and ran towards the lady. He set off at 60° to the verge, heading towards the lady with a speed of 10 m s-1, as shown in the diagram.
Find the component of velocity of boy directly across the parking strip.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 87
Answer:
The angle between velocity vector and the direction of path is 60°.
∴ Component of velocity across the parking strip
= v × cos 60°
= 10-1 × cos 60°
= 5 m s-1

Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods

Question 69.
On an open ground, a biker follows a track that turns to his left by an angle of 60° after every 600 m. Starting from a given turn, specify the displacement of the biker at the third and sixth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
Answer:
The path followed by the biker will be a closed hexagonal path. Suppose the motorist starts his journey from the point O.
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 88
= 1200 m
= 1.2 km
∴ Total path length = \(|\overrightarrow{\mathrm{OA}}|+|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{BC}}|\)
= 600 + 600 + 600
= 1800 m
= 1.8 km
The ratio of the magnitude of displacement to the total path-length = \(\frac{1.2}{1.8}\) = \(\frac{2}{3}\) = 0.67

ii. The motorist will take the sixth turn at O.
Displacement is zero.
path-length is = 3600 m or 3.6 km.
Ration of magnitude of displacement and path-length is zero.

Question 70.
What is the resultant of vectors shown in the figure below?
Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods 89
Answer:
If number of vectors are represented by the various sides of a closed polygon taken in one order then, their resultant is always zero.

Question 71.
If \(\overrightarrow{\mathbf{P}}\) is moving away from a point and \(\overrightarrow{\mathbf{Q}}\) is moving towards a point then, can their resultant be found using parallelogram law of vector addition?
Answer:
No. Resultant cannot be found by parallelogram law of vector addition because to apply law of parallelogram of vectors the two vectors and should either act towards a point or away from a point.

Question 72.
Which of the throwing is a vector?
(A) speed
(B) displacement
(C) mass
(D) time
Answer:
(B) displacement

Question 73.
The equation \(\vec{a}+\vec{a}=\vec{a}\) is
(A) meaningless
(B) always truc
(C) may he possible for limited values of a’
(D) true only when \(\overrightarrow{\mathrm{a}}=0\)
Answer:
(D) true only when \(\overrightarrow{\mathrm{a}}=0\)

Question 74.
The minimum number of numerically equal vectors whose vector sum can be zero is
(A) 4
(B) 3
(C) 2
(D) 1
Answer:
(C) 2

Question 75.
If \(\vec{A}+\vec{B}=\vec{A}-\vec{B}\) then vector \(\overrightarrow{\mathrm{B}}\) must be
(A) zero vector
(B) unit vector
(C) Non zero vector
(D) equal to \(\overrightarrow{\mathrm{A}}\)
Answer:
(A) zero vector

Question 76.
If \(\hat{\mathrm{n}}\) is the unit vector in the direction of \(\overrightarrow{\mathrm{A}}\), then,
(A) \(\hat{n}=\frac{\vec{A}}{|\vec{A}|}\)
(B) \(\hat{\mathrm{n}}=\overrightarrow{\mathrm{A}}|\overrightarrow{\mathrm{A}}|\)
(C) \(\hat{\mathrm{n}}=\frac{|\overrightarrow{\mathrm{A}}|}{\overrightarrow{\mathrm{A}}}\)
(D) \(\hat{\mathrm{n}}=\hat{\mathrm{n}} \times \overrightarrow{\mathrm{A}}\)
Answer:
(A) \(\hat{n}=\frac{\vec{A}}{|\vec{A}|}\)

Question 77.
Two quantities of 5 and 12 unit when added gives a quantity 13 unit. This quantity is
(A) time
(B) mass
(C) linear momentum
(D) speed
Answer:
(C) linear momentum

Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods

Question 78.
A force of 60 N acting perpendicular to a force of 80 N, magnitude of resultant force is
(A) 20N
(B) 70N
(C) 100 N
(D) 140 N
Answer:
(C) 100 N

Question 79.
A river is flowing at the rate of 6 km h-1. A man swims across it with a velocity of 9 km h-1. The resultant velocity of the man will be
(A) \(\sqrt{15} \mathrm{~km} \mathrm{~h}^{-1}\)
(B) \(\sqrt{45} \mathrm{~km} \mathrm{~h}^{-1}\)
(C) \(\sqrt{117} \mathrm{~km} \mathrm{~h}^{-1}\)
(D) \(\sqrt{225} \mathrm{~km} \mathrm{~h}^{-1}\)
Answer:
(C) \(\sqrt{117} \mathrm{~km} \mathrm{~h}^{-1}\)

Question 80.
If \(\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}\) and magnitudes of \(\overrightarrow{\mathrm{A}}\), \(\overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{C}}\) are 5, 4 and 3 unit respectively, then angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) is
(A) sin-1 (3/4)
(B) cos-1 (4/5)
(C) tan-1 (5/3)
(D) cos-1 (3/5)
Answer:
(B) cos-1 (4/5)

Question 81.
If \(\vec{A}=\hat{i}+2 \hat{j}+3 \hat{k}\) and \(\overrightarrow{\mathrm{B}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\), then the area of parallelogram formed from these vectors as the adjacent sides will be
(A) 2\(\sqrt{3}\) square units
(B) 4\(\sqrt{3}\) square units
(C) 6\(\sqrt{3}\) square units
(D) 8\(\sqrt{3}\) square units
Answer:
(D) 8\(\sqrt{3}\) square units

Question 82.
A person moves from a point S and walks along the path which is a square of each side 50 m. He runs east, south, then west and finally north. Then the total displacement covered is
(A) 200m
(B) 100 m
(C) 50\(\sqrt{2}\) m
(D) zero
Answer:
(D) zero

Question 83.
The maximum value of magnitude of \((\vec{A}-\vec{B})\) is
(A) A – B
(B) A
(C) A + B
(D) \(\sqrt{\left(A^{2}+B^{2}\right)}\)
Answer:
(C) A + B

Question 84.
The magnitude of the X and Y components of \(\overrightarrow{\mathrm{A}}\) are 7 and 6. Also the magnitudes of the X and Y components of \(\vec{A}+\vec{B}\) are 11 and 9 respectively. What is the magnitude of
(A) 5
(B) 6
(C) 8
(D)
Answer:
(A) 5

Question 85.
What is the maximum n Limber of components into which a force can be resolved?
(A) Two
(B) Three
(C) Four
(D) Any number
Answer:
(D) Any number

Question 86.
The resultant of two vectors of magnitude \(|\overrightarrow{\mathrm{P}}|\) is also \(|\overrightarrow{\mathrm{P}}|\). They act at an angle
(A) 60°
(B) 90°
(C) 120°
(D) 180°
Answer:
(C) 120°

Question 87.
The vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are such that \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\) = \(\overrightarrow{\mathrm{C}}\) and A2 + B2 = C2. Angle θ between positive directions of \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) is
(A) \(\frac{\pi}{2}\)
(B) 0
(C) π
(D) \(\frac{2 \pi}{3}\)
Answer:
(A) \(\frac{\pi}{2}\)

Maharashtra Board Class 11 Physics Important Questions Chapter 2 Mathematical Methods

Question 88.
The expression \(\frac{1}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})\) is a
(A) unit vector
(B) null vector
(C) vector of magnitude \(\sqrt{2}\)
(D) scalar
Answer:
(A) unit vector

Question 89.
What is the angle between \(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{\mathrm{i}}\)?
(A) 0°
(B) \(\frac{\pi}{6}\)
(C) \(\frac{\pi}{3}\)
(D) None of the above
Answer:
(D) None of the above

Question 90.
\((\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}})\) is a unit vector along X-axis. If \(\overrightarrow{\mathrm{P}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\) then \(\overrightarrow{\mathrm{Q}}\) is
(A) \(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\)
(B) \(\hat{\mathrm{j}}-\hat{\mathrm{k}}\)
(C) \(\hat{i}+\hat{j}+\hat{k}\)
(D) \(\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
Answer:
(B) \(\hat{\mathrm{j}}-\hat{\mathrm{k}}\)

Question 91.
The magnitude of scalar product of the vectors \(\overrightarrow{\mathrm{A}}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{B}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}\) is
(A) 20
(B) 22
(C) 26
(D) 29
Answer:
(C) 26

Question 92.
Three vectors \(\overrightarrow{\mathrm{A}}\), \(\overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{C}}\) satisfy the relation \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{B}}\) = 0 and \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{C}}\) = 0, then \(\overrightarrow{\mathrm{A}}\) is parallel to
(A) \(\overrightarrow{\mathrm{B}}\)
(B) \(\overrightarrow{\mathrm{C}}\)
(C) \(\overrightarrow{\mathrm{B}}\) × \(\overrightarrow{\mathrm{C}}\)
(D) \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{C}}\)
Answer:
(C) \(\overrightarrow{\mathrm{B}}\) × \(\overrightarrow{\mathrm{C}}\)

Question 93.
What vector must be added to the sum of two vectors \(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}\) and \(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) so that the resultant is a unit vector along Z axis?
(A) \(5 \hat{\hat{i}}+\hat{\mathrm{k}}\)
(B) \(-5 \hat{i}+3 \hat{j}\)
(C) \(3 \hat{j}+5 \hat{k}\)
(D) \(-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
Answer:
(B) \(-5 \hat{i}+3 \hat{j}\)

Question 94.
\(\overrightarrow{\mathrm{A}}=5 \overrightarrow{\mathrm{i}}-2 \overrightarrow{\mathrm{j}}+3 \overrightarrow{\mathrm{k}}\) and \(\overrightarrow{\mathrm{B}}=2 \overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{j}}+2 \overrightarrow{\mathrm{k}}\), then component of \(\overrightarrow{\mathrm{B}}\) along \(\overrightarrow{\mathrm{A}}\) is
(A) \(\frac{\sqrt{28}}{38}\)
(B) \(\frac{28}{\sqrt{38}}\)
(C) \(\frac{\sqrt{28}}{48}\)
(D) \(\frac{14}{\sqrt{38}}\)
Answer:
(D) \(\frac{14}{\sqrt{38}}\)

Question 95.
Choose the WRONG statement
(A) The division of vector by scalar is valid.
(B) The multiplication of vector by scalar is valid.
(C) The multiplication of vector by another vector is valid by using vector algebra.
(D) The division of a vector by another vector is valid by using vector algebra.
Answer:
(D) The division of a vector by another vector is valid by using vector algebra.

Question 96.
The resultant of two forces of 3 N and 4 N is 5 N, the angle between the forces is
(A) 30°
(B) 60°
(C) 90°
(D) 120°
Answer:
(C) 90°

Question 97.
The unit vector along \(\hat{\mathrm{i}}+\hat{\mathrm{j}}\) is
(A) \(\hat{\mathrm{k}}\)
(B) \(\hat{\mathrm{i}}+\hat{\mathrm{j}}\)
(C) \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\)
(D) \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{2}\)
Answer:
(C) \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\)

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 3 Motion in a Plane Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 3 Motion in a Plane

Question 1.
Explain the term: Displacement.
Answer:
Displacement:

  1. Displacement of a particle for a time interval is the difference between the position vectors of the object in that time interval.
  2. Let \(\overrightarrow{\mathrm{x}_{1}}\) and \(\overrightarrow{\mathrm{x}_{2}}\) be the position vectors of a particle at time t1 and t2 respectively. Then the displacement \(\overrightarrow{\mathrm{S}}\) in time ∆t = (t2 – t\overrightarrow{\mathrm{s}}=\Delta \overrightarrow{\mathrm{x}}=\overrightarrow{\mathrm{x}}_{2}-\overrightarrow{\mathrm{x}}_{1}) is given by \(\overrightarrow{\mathrm{s}}=\Delta \overrightarrow{\mathrm{x}}=\overrightarrow{\mathrm{x}}_{2}-\overrightarrow{\mathrm{x}}_{1}\)
  3. Dimensions of displacement are equal to that of length i.e.. [L1M0T0].
  4. Displacement is a vector quantity.
  5. Example:
    • For an object has travelled through 1 m from time t1 to t2 along the positive X-direction, the magnitude of its displacement is I m and its direction is along the positive X-axis.
    • On the other hand, for an object has travelled through I m from time t1 to t1 along the positive Y-direction, the magnitude of its displacement remains same i.e., I m but the direction of the displacement is along the positive Y-axis.

Question 2.
Explain the term: Path length.
Answer:

  1. Path length is the actual distance travelled by the particle during its motion.
  2. It is a scalar quantity.
  3. Dimensions of path length are equal to that of length i.e.. [L1M0T0]
  4. Example:
    • If an object travels along the X-axis from x = 3 m to x = 6 m then the distance travelled is 3 m. In this case the displacement is also 3 m and its direction is along the positive X-axis.
    • However, if the object now comes back to x 5, then the distance through which the object has moved increases to 3 + I = 4 m. Its initial position was x 3 m and the final position is now x = 5 m and thus, its displacement is ∆x = 5 – 3 = 2 m, i.e., the magnitude of the displacement is 2 m and its direction is along the positive X-axis.
    • If the object now moves to x =1, then the distance travelled, i.e., the path length increases to 4 + 4 = 8 m while the magnitude of displacement becomes 3 – 1 = 2 m and its direction is along the negative X-axis.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 3.
Explain the terms:
i. Average velocity
ii. Instantaneous velocity
iii. Average speed
iv. Instantaneous speed
Answer:
i) Average velocity:

  1. Average velocity (\(\overrightarrow{\mathrm{v}}_{\mathrm{av}}\)) of an object is the displacement (\(\Delta \overrightarrow{\mathrm{x}}\)) of the object during the time interval (∆t) over which average velocity is being calculated, divided by that time interval.
  2. Average velocity = (\(\frac{\text { Displacement }}{\text { Time interval }}\))
    \(\overrightarrow{\mathrm{V}_{\mathrm{av}}}=\frac{\overrightarrow{\mathrm{x}}_{2}-\overrightarrow{\mathrm{x}}_{1}}{\mathrm{t}_{2}-\mathrm{t}_{1}}=\frac{\Delta \overrightarrow{\mathrm{x}}}{\Delta \mathrm{t}}\)
  3. Average velocity is a vector quantity.
  4. Its SI unit is m/s and dimensions are [M0L1T-1]
  5. For example, if the positions of an object are x +4 m and x = +6 m at times t = O and t = 1 minute respectively, the magnitude of its average velocity during that time is Vav = (6 – 4)1(1 – 0) = 2 m per minute and its direction will be along the positive X-axis.
    ∴ \(\overrightarrow{\mathrm{v}}_{\mathrm{av}}\) = 2 i m/min
    Where, i = unit vector along X-axis.

ii) Instantaneous velocity:

  1. The instantaneous velocity (\(\overrightarrow{\mathrm{V}}\)) is the limiting value of ¡he average velocity of the object over a small time interval (∆t) around t when the value of lime interval goes to zero.
  2. It is the velocity of an object at a given instant of time.
  3. \(\overrightarrow{\mathrm{v}}=\lim _{\Delta t \rightarrow 0} \frac{\Delta \overrightarrow{\mathrm{x}}}{\Delta \mathrm{t}}=\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\)
    where \(\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\) derivative of \(\overrightarrow{\mathrm{x}}\) with respect to t.

iii) Average speed:

  1. Average speed of an object is the total path length (distance) travelled by the object during the time interval over which average speed is being calculated, divided by that time interval.
  2. Average speed = \(\frac{\text { Total path length }}{\text { Total time interval }}\)
  3. Average speed is a scalar quantity.
  4. Its S.I. unit is m/s and dimensions are [M0V1T-1].
  5. In rectilinear motion;
    • If the motion of the object is only in one direction, then the magnitude of displacement will be equal to the path length and hence the magnitude of average velocity will be equal to the average speed.
    • If the motion of the object reverses its direction, then the magnitude of displacement will be less then the path length and hence the magnitude of average velocity will be less than the average speed.

iv) Instantaneous speed:
The instantaneous speed is the limiting value of the average speed of the object over a small time interval ‘∆t’ around t when the value of time interval goes to zero.

Question 4.
Distinguish between uniform rectilinear motion and non-uniform rectilinear motion.
Answer:

No. Uniformly rectilinear motion Non-uniform rectilinear motion
i. The object is moving with constant velocity. The object is moving with variable velocity.
ii. The average and instantaneous velocities are same. The average and instantaneous velocities are different.
iii. The average and instantaneous speeds are the same. The average and instantaneous speeds are different.
iv. The average and instantaneous speeds are equal to the magnitude of the velocity. The average speed will be different from the magnitude of average velocity.

Question 5.
Explain the terms:

  1. Acceleration
  2. Average acceleration
  3. Instantaneous acceleration

Answer:

  1. Acceleration:
    • Acceleration is the rate of change of velocity with respect to time.
    • It is a vector quantity.
    • Dimension: [M0L1T-2]
    • If a particle moves with constant velocity, its acceleration is zero.
  2. Average acceleration:
    • Average acceleration is the change in velocity divided by the total time required for the change.
    • If \(\overrightarrow{\mathrm{v}_{\mathrm{1}}}\) and \(\overrightarrow{\mathrm{v}_{\mathrm{2}}}\) are the velocities of the T particle at time t1 and t2 respectively, then the change in velocity is \(\) and time required for this change is ∆t = t2 – t1
      ∴ \(\vec{a}_{a v}=\frac{\vec{v}_{2}-\vec{v}_{1}}{t_{2}-t_{1}}=\frac{\Delta \vec{v}}{\Delta t}\)
  3. Instantaneous acceleration:
    • The instantaneous acceleration a is the limiting value of the average acceleration of the object over a small time interval ‘∆t’ around t when the value of time interval goes to zero.
      \(\overrightarrow{\mathrm{a}}_{\text {inst }}=\lim _{\Delta t \rightarrow 0} \frac{\Delta \overrightarrow{\mathrm{v}}}{\Delta \mathrm{t}}=\frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{d} \mathrm{t}}\)
    • Instantaneous acceleration is the slope of the tangent to the velocity-time graph at a position corresponding to given instant of time.
      [Note: Generally, when the term acceleration is used, it is an instantaneous acceleration.]

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 6.
Draw and explain the position-time graph of:

  1. An object at rest.
  2. An object moving with uniform velocity along positive x-axis.
  3. An object moving with uniform velocity along negative x-axis.
  4. An object moving with non-uniform velocity.
  5. An object performing oscillatory motion with constant speed.

Answer:

  1. The position-time graph of an object at rest:
    • For an object at rest, the position-time graph is a horizontal straight line parallel to time axis.
    • The displacement of the object is zero as there is no change in the object’s position.
    • Slope of the graph is zero, which indicates that velocity of the particle is zero.
      Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 1
  2. The position-time graph of an object moving with uniform velocity along positive x-axis:
    • When an object moves, the position of the particle changes with respect to time.
    • Since velocity is constant, displacement is proportional to elapsed time.
    • The graph is a straight line with positive slope, showing that the velocity is along the positive x-axis.
    • In this case, as the motion is uniform, the average velocity and instantaneous velocity are equal at all times.
    • Speed is equal to the magnitude of the velocity.
      Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 2
  3. Position-time graph of an object moving with uniform velocity along negative x- axis:
    • The graph is a straight line with negative slope, showing that the velocity is along the negative x-axis.
    • Displacement decreases with increase in time.
      Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 3
  4. Position-time graph of a particle moving with non-uniform velocity;
    • When the velocity of an object changes with time, slope of the graph is different at different points. Therefore, the average and instantaneous velocities are different.
    • Average velocity over time interval from t1 to t4 around time t0 = slope of line AB.
    • Average velocity over time interval from t2 to t3 = slope of line CD
    • On further reducing the time interval around t0, it can be deduced that, instantaneous interval at t0 = the slope of the tangent PQ at t0.
      Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 4
  5. Position-time graph of an object performing oscillatory motion with constant speed:
    For an object performing oscillatory motion with constant speed, the direction of velocity changes from positive to negative and vice versa over fixed intervals of time.
    Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 5

Question 7.
Explain the velocity-tune graphs of an object:
i) Moving with zero acceleration.
ii) Moving with constant positive acceleration.
iii) Moving with constant negative acceleration.
iv) Moving with non-uniform acceleration.
Answer:
i) Object is moving with zero acceleration:

  1. Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 6
  2. As the acceleration is zero, the graph will be a straight line parallel to time axis.
  3. Velocity of the particle is constant as the acceleration is zero.
  4. Magnitude of displacement of object from t1 to t2 = v0 × (t2 – t1) shaded area under velocity-time graph.

ii) Object is moving with constant positive acceleration:

  1. The velocity-time graph is linear.
  2. Velocity increases with increase in time. as acceleration is positive (along the direction of velocity).
  3. The area under the velocity-time graph between two instants of time t1 and t2 gives the displacement of the object during that time interval.
  4. Slope of the graph is \(\frac{\Delta \mathrm{v}}{\Delta \mathrm{t}}\) = positive acceleration
    Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 7

iii) Object is moving with constant negative acceleration:

  1. The velocity-time graph is linear.
  2. Velocity decreases with increase in time as acceleration is negative (opposite to the direction of velocities).
  3. The area tinder the velocity-time graph between two instants of time t1 and t2 gives the displacement 0f the object during that time interval.
  4. Slope of the graph is \(\frac{\Delta \mathrm{v}}{\Delta \mathrm{t}}\) negative acceleration
    Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 8

iv. Object is moving with non-uniform acceleration:

  1. Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 9
  2. Velocity-time graph is non-linear.
  3. The area under the velocity-time graph between two instants of time t1 and t2 gives the displacement of the object during that time interval area under the velocity-time curve =
    Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 10
    = x(t2) – x(t1)
    = displacement of the object from t1 to t2.

Question 8.
A ball thrown vertically upwards from a point P on earth reaches a point Q and returns back to earth striking at a point R. Draw speed-time graph to depict the motion of the ball (Neglect air resistance).
Answer:

  • Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 11
  • A ball which is thrown up with a certain initial speed goes up with decreasing speed to a certain height where its speed becomes zero.
  • Now, during its downward motion, the speed goes on increasing from zero and reaches its initial value when it strikes the ground.
  • The speed-time graph for the motion of a ball is as shown in the figure.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 9.
Figure shows velocity-time graph for various situations. What does each graph indicate?
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 12
Answer:

  1. Initial velocity, u > 0. Also, velocity is constant with time. Hence, acceleration is zero.
  2. As finite initial velocity is increasing with time, acceleration, a > 0 and is constant.
  3. Initial velocity, u = 0. Velocity is increasing with time so, acceleration a is positive. But it is decreasing in magnitude with time.
  4. Initial velocity, u = 0. Velocity is linearly increasing with time. Hence, starting from rest acceleration is constant.
  5. Initial velocity, u = 0. Acceleration and velocity is increasing with time.
  6. Initial velocity u > 0. Velocity decreases and ultimately comes to rest. Hence, acceleration a < 0.

Question 10.
‘The distances travelled by an object starting from rest and having a positive uniform acceleration in successive seconds are in the ratio 1:3:5:7….’ Prove it.
Answer:

  1. Consider an object under free fall, Initial velocity u = 0, acceleration a = g
  2. The distance travelled by the object in equal time intervals t0 can be given by the second law of motion as,
    s = ut0 + \(\frac{1}{2}\) gt02
  3. Distance travelled in the first time interval to,
    s1 = 0 + \(\frac{1}{2}\)gt02 = \(\frac{1}{2}\) gt02
    Substituting \(\frac{\mathrm{g}}{2}\) = A, we have s1 = At02
  4. Distance travelled in the time interval 2t0 = A (2t0)2
    ∴ The distance travelled in the second t0 interval, s2 = A(4t02 – t02) = 3At0\frac{\mathrm{g}}{2} = 3s1
  5. Distance travelled in the time interval 3t0 = A(3t0)2
    ∴ The distance travelled in the third to interval,
    s3 = A (9t02 – 4t02) = 5 At02 = 5s1
  6. On continuing, it can be seen that the distances travelled, (s1 : s2 : s3 ….) are in the ratio (1 : 3 : 5 :….)

Question 11.
Explain the concept of relative velocity along a straight line with the help of an example.
Answer:

  1. Consider two trains A and B moving on two parallel tracks in the same direction.
  2. Case 1: Train B overtakes train A.
    For a passenger in train A, train B appears to be moving slower than train A. This happens because the passenger in train A perceives the velocity of train B with respect to him/her i.e., the difference in the velocities of the two trains which is much smaller than the velocity of train A.
  3. Case 2: Train A overtakes train B.
    For a passenger in train A, train B appears to be moving faster than train A. This happens because the passenger in train A perceives the velocity of the train B w.r.t. to him/her i.e., the difference in the velocities of the two trains which is larger than the velocity of train A.
  4. If \(\vec{v}_{\mathrm{A}}\) and \(\vec{v}_{\mathrm{B}}\) be the velocities of two bodies then relative velocity of A with respect to B is given by \(\vec{v}_{A B}=\vec{v}_{A}-\vec{v}_{B}\).
  5. Similarly the velocity of B with respect to A is given by, \(\vec{v}_{A B}=\vec{v}_{B}-\vec{v}_{A}\).
    Thus, relative velocity of an object w.r.t. another object is the difference in their velocities
  6. If two objects start form the same point at t = 0, with different velocities, distance between them increases with time in direct proportion to the relative velocity between them.

Solved Problems

Question 12.
A person walks from point P to point Q along a straight road ¡n 10 minutes, then turns back and returns to point R which ¡s midway between P and Q after further 4 minutes. If PQ is 1 km, find the average speed and velocity of the person in going from P to R.
Solution:
Given: time taken (t) = 10 + 4 = 14 minutes,
distance (s) = PQ + QR = 1 + 0.5 = 1.5 km,
displacement = PQ – QR = 1 – 0.5 = 0.5km
To find: Average speed, average velocity (v)
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 13
The average speed and average velocity of the person is 6.42 km/hr and 2.142 km/hr respectively.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 13.
A car moves at a constant speed of 60 km/hr for 1 km and 40 km/hr for next 1 km. What ¡s the average speed of the car?
Solution:
Given. v1 = 60 km/hr, x1 = 1 km,
v2 = 40 km/hr, x2 = 1 km
To find: Average speed of car (Vav)
Formula: vav = \(\frac{\text { total path length }}{\text { total time interval }}\)
Calculation: From given data,
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 14
∴ Average speed of car = 48 km/hr
The average speed of the car is 48 km/hr

Question 14.
A stone is thrown vertically upwards from the ground with a velocity 15 m/s. At the same instant a ball is dropped from a point directly above the stone from a height of 30 m. At what height from the ground will the stone and the ball meet and after how much time? (Use g = 10 m/s2 for ease of calculation).
Solution:
Let the stone and the ball meet after time t0. From second equation of motion, the distances travelled by the stone and the ball in that time is given as,
Sstone = 15 t0 – \(\frac{1}{2}\) gt02
Sball = \(\frac{1}{2}\) gt02
When they meet. Sstone + Sball = 30
∴ 15t0 – \(\frac{1}{2}\) gt02 + \(\frac{1}{2}\) gt02 = 30
t0 = \(\frac{30}{15}\) = 2 s
∴ Sstone = 15 (2) – \(\frac{1}{2}\) (10) (2)2 = 30 – 20 = 10 m
The stone and the ball meet at a height of 10 m after time 2s.

Question 15.
A ball is dropped from the top of a building 122.5 m high. How long will it take to reach the ground? What wilt be its velocity when it strikes the ground?
Solution:
Given: s = h = 122.5 m, u = 0,
a = g = 9.8 ms2
To find: i) Time taken to reach the ground (t)
ii) Velocity of ball when it strikes ground (v)

Formulae: i) s = ut + \(\frac{1}{2}\) at2
ii) v = u + gt
Calculation: From formula (i),
122.5 = 0 + \(\frac{1}{2}\) × 9.8 t2
t2 = \(\frac{122.5}{4.9}\) = 25
t = \(\sqrt {25}\) = 5 second
From formula (ii),
v = u + gt
v = 0 + 9.8 × 5 = 49 m/s

i) Time taken to reach the ground is 5 s.
ii) Velocity of the ball when it strikes the ground is 49 m/s.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 16.
The position vectors of three particles are given by
\(\overrightarrow{\mathrm{x}}_{1}=(5 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}) \mathrm{m}\), \(\overrightarrow{\mathrm{x}}_{2}=(5 \mathrm{t} \hat{\mathrm{i}}+5 \mathrm{t} \hat{\mathrm{j}}) \mathrm{m}\) and \(\overrightarrow{\mathrm{x}}_{3}=\left(5 \mathrm{t} \hat{\mathrm{i}}+10 \mathrm{t}^{2} \hat{\mathrm{j}}\right) \mathrm{m}\) as a function of time t.
Determine the velocity and acceleration for each, in SI units.
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 15
v2 = \(\sqrt{5^{2}+5^{2}}\) = 5\(\sqrt{2}\) m/s
tan θ = \(\frac{5}{5}\) = 1
∴ θ = 45°
Direction of v2 makes an angle of 45° to the horizontal.
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 16
Thus, third particle is getting accelerated along the y-axis at 20 m/s2.

Question 17.
The initial velocity of an object is \(\overrightarrow{\mathrm{u}}=5 \hat{\mathrm{i}}+10 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}\). Its constant acceleration is \(\vec{a}=2 \hat{i}+3 \hat{j} \mathrm{~m} / \mathrm{s}^{2}\). Determine the velocity and the displacement after 5 s.
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 17
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 18

Question 18.
An aeroplane A, is travelling in a straight line with a velocity of 300 km/hr with respect to Earth. Another aeroplane B, is travelling in the opposite direction with a velocity of 350 km/hr with respect to Earth. What is the relative velocity of A with respect to B? What should be the velocity of a third aeroplane C moving parallel to A, relative to the Earth if it has a relative velocity of 100 km/hr with respect to A?
Solution:
Given: vA = 300 km/hr, vB = 350 km/hr,
vCA = 100 km/hr
To find: i) Velocity of plane A relative to B (vA – vB)
ii) Velocity of aeroplane C (vC)

Formula: i) vAB = vA – vB
ii) vCA = vC – vA

Calculations: From formula (i),
vAB = vA – vB = 300 – (-350)
∴ vAB = 650 km/hr
From formula (ii),
vC = vCA + vA = 100 + 300 = 400 km/hr

i) The relative velocity of A with respect to B is 650 km/hr.
ii) The velocity of plane C relative to Earth is 400 km/hr.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 19.
A car moving at a speed 10 m/s on a straight road is ahead of car B moving in the same direction at 6 m/s. Find the velocity of A relative to B and vice-versa.
Solution:
Given: vA = 10 m/s, vB = 6 m/s,
To find: i) Velocity of A relative to B (vA – vB)
ii) Velocity of B relative to A (vB – vA)

Formulae: i) vAB = vA – vB
ii) vBA = vB – vA

Calculation: From formula (i),
vAB = 10 – 6 = 4 m/s
From formula (ii),
vBA = 6 – 10 = -4 m/s
-ve sign indicates that driver of car A sees the car B lagging behind at the rate of 4 m/s.
∴ vAB = 4 m/s, vBA = -4 m/s

i) Velocity of A relative to B is 4 m/s.
ii) Velocity of B relative to A is -4 m/s.

Question 20.
Two trains 120 m and 80 m in length are running in opposite directions with velocities 42 km/h and 30 km/h respectively. In what time will they completely cross each other?
Solution:
Given: l1 = 120 m, l2 = 80 m,
vA = 42 km/h = 42 × \(\frac{5}{18}\) = \(\frac{35}{3}\) m/s,
vB = -30km/h= -30 × \(\frac{5}{18}\) = \(\frac{-25}{3}\) m/s
To find: Time taken by trains to cross each other (t)
Formula: Time = \(\frac{\text { Distance }}{\text { speed }}\)

Calculation :
Total distance to be travelled
= sum of lengths of two trains
= 120 + 80 = 200m
Relative velocity of A with respect to B is vAB,
vAB = vA – vB
= \(\frac{35}{3}\) – (\(\frac{-25}{3}\))
= \(\frac{60}{3}\)
∴ vAB = 20m/S
From formula,
∴ Time taken to cross each other (t) = \(\frac{\text { Distance }}{\text { speed }}\)
= \(\frac{200}{20}\)
= 10 s
Time taken by the two trains to cross each other is 10 s.

Question 21.
A jet aeroplane travelling at the speed of 500 km/hr ejects its products of combustion at speed of 1500 km/hr relative to jet plane. What is the relative velocity of the latter with respect to an observer on the ground?
Solution:
Let us consider the positive direction of motion towards the observer on the ground.
Suppose \(\vec{v}_{\mathrm{a}}\) and \(\vec{v}_{\mathrm{cj}}\) be the velocities of the aeroplane and relative velocity of combustion products w.r.t. aeroplane respectively.

∴ \(\vec{v}_{\mathrm{cj}}\) = 1,500 km/hr (towards the observer on the ground) and \(\vec{v}_{\mathrm{a}}\) = 500 km/hr (away from the observer on the ground)
∴ – \(\vec{v}_{\mathrm{a}}\) = -500 km/ hr (towards the observer on the ground)

Let \(\vec{v}_{\mathrm{c}}\) be the velocity of the combustion products towards the observer on ground then,
\(\vec{v}_{\mathrm{c} j}=\vec{v}_{\mathrm{c}}-\vec{v}_{\mathrm{a}}\)
∴ \(\vec{v}_{\mathrm{c} }=\vec{v}_{\mathrm{cj}}-\vec{v}_{\mathrm{a}}\)
= 1500 + (-500)
= 1000 km/hr
∴ \(\vec{v}_{\mathrm{c}}\) = 1000 km/hr
The relative velocity of the combustion products w.r.t. the observer is 1000 km/hr.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 22.
Derive the expression for average velocity and instantaneous velocity for the motion of an object in x-y plane.
Answer:
i) Consider an object to be at point A at time t1 in an x—y plane.
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 19
ii) At time t1, the position vector of the object is given as,
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 20
vii) The instantaneous velocity of the object at point A along the trajectory is along the tangent to the curve at A. This is shown by the vector AB.
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 21
Equation (3) is the slope of the tangent to the curve at the point at which we are calculating the instantaneous velocity.

Question 23.
Derive the expression for average acceleration and instantaneous acceleration for the motion of an object in x-y plane.
Answer:
i) Consider an object moving in an x-y plane.
Let the velocity of the particle be \(\vec{v}_{1}\) and \(\vec{v}_{2}\) at time t1 and t2 respectively.
ii) The average acceleration (\(\overrightarrow{\mathrm{a}}_{\mathrm{av}}\)) of the particle between t1 and t2 is given as,
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 22
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 23
Equation (1) is the slope of the tangent to the curve at the point at which we are calculating the instantaneous acceleration.

Question 24.
Explain relative velocity between two objects moving in a plane.
Answer:

  1. If \(\vec{v}_{A}\) and \(\vec{v}_{B}\) be the velocities of two bodies then relative velocity of A with respect to B is given by, \(\vec{v}_{A B}=\vec{v}_{A}-\vec{v}_{B}\)
  2. Similarly, the velocity of B with respect to A is given by, \(\vec{v}_{\text {BA }}=\vec{v}_{\text {B }}-\overrightarrow{v_{A}}\)
  3. Thus, the magnitudes of the two relative velocities are equal and their directions are opposite.
  4. For a number of objects A, B, C, D—Y, Z, moving with respect to the other. The velocity of A relative to Z can be given as, \(\overrightarrow{\mathrm{v}}_{\mathrm{AZ}}=\overrightarrow{\mathrm{v}}_{\mathrm{AB}}+\overrightarrow{\mathrm{v}}_{\mathrm{BC}}+\overrightarrow{\mathrm{v}}_{\mathrm{CD}}+\ldots+\overrightarrow{\mathrm{v}}_{\mathrm{XY}}+\overrightarrow{\mathrm{v}}_{\mathrm{YZ}}\)
    The order of subscripts is: A → B → C → D → Z

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 25.
Write a note on projectile motion.
Answer:

  1. An object in flight after being thrown with some velocity is called a projectile and its motion is called projectile motion.
  2. Example: A bullet fired from a gun, football kicked in air, a stone thrown obliquely in air etc.
  3. In projectile motion, the object is moving freely under the influence of Earth’s gravitational field.
  4. The projectile has two components of velocity, one in the horizontal i.e., along the x- direction and the other in the vertical i.e., along the y-direction.
  5. As acceleration due to gravity acts only along the vertically downward direction, the vertical component changes in accordance with the laws of motion with ax = 0 and ay = -g.
  6. As no force is acting in the horizontal direction, the horizontal component of velocity remains unchanged.
    [Note: Retarding forces like air resistance etc. are neglected in projectile motion unless otherwise stated.]

Question 26.
Obtain an expression for the time of flight of a projectile.
Answer:
Expression for time of flight:

  1. Consider a body projected with velocity \(\vec{u}\), at an angle θ of projection from point O in the co-ordinate system of the X-Y plane, as shown in figure.
    Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 24
  2. The initial velocity \(\vec{u}\) can be resolved into two rectangular components:
    ux = u cos θ (Horizontal component)
    uy = u sin θ (Vertical component)
  3. Thus, the horizontal component remains constant throughout the motion due to the absence of any force acting in that direction, while the vertical component changes according to, vy = uy + ay t with ay = – g and uy = u sin θ
  4. The components of velocity of the projectile at time t are given by, vx = ux = u cos θ
    vy = uy – gt = u sin θ – gt ………….. (1)
  5. At maximum height.
    vy = 0, t = tA = time of ascent = time taken to reach maximum height.
    ∴ 0 = u sin θ – gtA ……..[From(l)]
    u sin θ = gtA
    tA = \(\frac{\mathrm{u} \sin \theta}{\mathrm{g}}\) ………….. (2)
    This is time of ascent of projectile.
  6. The total time in air i.e., time of flight T is given as,
    T = 2tA ………… [From(2)]
    = \(\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\) ………… (3)
    Equation (3) represents time of flight of projectile.

Question 27.
Define Horizontal range of projectile:
Answer:
The maximum horizontal distance travelled by the projectile is called the horizontal range (R) of the projectile.

Solved Examples

Question 28.
An aeroplane Is travelling northward with a velocity of 300 km/hr with respect to the Earth. Wind is blowing from east to west at a speed of 100 km/hr. What is the velocity of the aeroplane with respect to the wind?
Solution:
Given:
velocity of aeroplane w.r.t Earth,
\(\vec{v}_{A E}=300 \hat{j}\)
velocity of wind w.r.t Earth,
\(\vec{v}_{w E}=-100 \hat{i}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 25

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 29.
A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 ms-1. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground. Take = 9.8 m/s2.
Solution:
Given: h = 490m, ux = 15 ms-1, ay = 9.8 ms-1,
ax = 0
To find: i) Time taken (t)
ii) Velocity (v)

Formulae: i) t = \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\)
ii) v = \(\sqrt{v_{x}^{2}+v_{y}^{2}}\)

Calculation: t = \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 490}{9.8}}\) = 10 s
vx = ux + axt= 15 + 0 × 10 = 15 m/s
uy = uy + ayt = 0 + 9.8 × 10 = 98 m/s
∴ v = \(\sqrt{\mathrm{v}_{\mathrm{x}}^{2}+\mathrm{v}_{\mathrm{y}}^{2}}=\sqrt{15^{2}+98^{2}}\)
= 99.1 m/s
The stone taken 10 s to reach the ground and hits the ground with 99.1 m/s.

Question 30.
A body is projected with a velocity of 40 ms-1. After 2 s it crosses a vertical pole of height 20.4 m. Find the angle of projection and horizontal range of projectile, (g = 9.8 ms-2).
Solution:
Given: u = 40 ms-1, t = 2 s, y = 20.4 m,
ay = -9.8 m/s2

To find: i) Angle of projection (θ)
ii) Horizontal range of projectile (R)

Formulae: i) y = uy t + \(\frac{1}{2}\) ay t2
ii) R = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\)

Calculation: Taking vertical upward motion of the projectile from point of projection up to the top of vertical pole we have
uy = 40 sinθ,
From formula (i),
∴ 20.4 = 40 sinθ × 2 + \(\frac{1}{2}\) (-9.8) × 22
∴ 20.4 = 80 sinθ – 19.6
or sinθ = \(\frac{(20.4+19.6)}{80}=\frac{1}{2}\)
or θ = 30°.
From formula (ii),
Horizontal range = \(\frac{40^{2}}{9.8}\) sin 2 × 30°
= 141.4 m
The angle of projection is 30°. The horizontal range of projection is 141.4m

Question 31.
A stone is thrown with an initial velocity components of 20 m/s along the vertical, and 15 m/s along the horizontal direction. Determine the position and velocity of the stone after 3 s. Determine the maximum height that it will reach and the total distance travelled along the horizontal on reaching the ground. (Assume g = 10 m/s2)
Solution:
The initial velocity of the stone in x-direction = u cos θ = 15 m/s and in y-direction = u sin θ = 20 m/s
After 3 s, vx = u cos θ = 15 m/s
vy = u sin θ – gt
= 20 – 10(3)
= -10 m/s
10 m/s downwards.
∴ v = \(\sqrt{\mathrm{v}_{x}^{2}+\mathrm{v}_{\mathrm{y}}^{2}}=\sqrt{15^{2}+10^{2}}=\sqrt{225+100}=\sqrt{325}\)
∴ v = 18.03m/s
tan α = vy/vx = 10/15 = 2/3
∴ α = tan-1 (2/3) = 33° 41’ with the horizontal.
Sx = (u cos θ)t = 15 × 3 = 45m,
Sy = (u sin θ)t – \(\frac{1}{2}\)gt2 = 2o × 3 – 5(3)2
∴ Sy = 15m
The maximum vertical distance travelled is given by,
H = \(\frac{(\mathrm{u} \sin \theta)^{2}}{(2 \mathrm{~g})}=\frac{20^{2}}{(2 \times 10)}\)
∴ H = 20m
Maximum horizontal distance travelled
R = \(\frac{2 \cdot u_{x} \cdot u_{y}}{g}=\frac{2(15)(20)}{10}\) = 60 m

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 32.
A body is projected with a velocity of 30 ms-1 at an angle of 300 with the vertical.
Find
i) the maximum height
ii) time of flight and
iii) the horizontal range
Solution:
Given:
30 ms-1, θ = 90° – 30° = 60°

To find: i) The maximum height reached (H)
ii) Time of flight (T)
iii) The horizontal range (R)
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 26
i) The maximum height reached by the body is 34.44 m.
ii) The time of flight of the body is 5.3 s.
iii) The horizontal range of the body is 79.53 m.

Question 33.
A projectile has a range of 50 m and reaches a maximum height of 10 m. What is the e1eation of the projectile?
Solution:
Given: R = 50m, H = 10 m
To find: Elevation of the projectile (θ)
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 27
∴ θ = tan-1 (0.8)
∴ θ = 38.66°
The elevation of the projectile is 38.66°

Question 34.
A bullet fired at an angle of 300 with the horizontal hits the ground 3 km away. By adjusting the angle of projection, can one hope to hit a target 5 km away? Assume the muzzle speed to be fixed and neglect air resistance.
Solution:
R = 3km = 3000m, θ = 30°,
Distance of target R’ = 5km
Horizontal range, R = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\)
∴ 3000 = \(\frac{\mathrm{u}^{2} \sin 60^{\circ}}{\mathrm{g}}\)
∴ \(\frac{\mathrm{u}^{2}}{\mathrm{~g}}=\frac{3000}{\sin 60^{\circ}}=\frac{3000 \times 2}{\sqrt{3}}\) = 2000\(\sqrt {3}\)
Maximum horizontal range,
Rmax = \(\frac{\mathrm{u}^{2}}{\mathrm{~g}}\) = 2000 \(\sqrt {3}\) m
= 2000 × 1.732 = 3464m = 3.46km
Since R’ > Rmax, Target cannot be hit.

Question 35.
Q.54. A ball is thrown at an angle θ and another ball ¡s thrown at an angle (90° – θ) with the horizontal direction from the same point with velocity 39.2 ms-1. The second ball reaches 50 m higher than the first balL find their individual heights. [Take g = 9.8 ms-2]
Solution:
For the first ball: Angle of projection = θ,
u = 39.2 ms-1
H = \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\)
H = \(\frac{(39.2)^{2} \sin ^{2} \theta}{2 \times 9.8}\) …………… (i)
For the second ball: Angle of projection
= 90° – θ,
u = 39.2 ms-1,
maximum height reached = (H + 50) m
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 28
or 2H = 78.4 – 50 = 28.4
∴ H = 14.2 m
∴ Height of first ball = H = 14.2 m
Height of second ball = H + 50 = 14.2 + 50 = 64.2 m

i) Height reached by the first ball is 14.2 m.
ii) Height reached by the second ball is 64.2m.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 36.
A body is thrown with a velocity of 49 m/s at an angle of 30° with the horizontal. Find
i) the maximum height attained by it
ii) the time of flight and
iii) the horizontal range.
Solution:
Given: u = 49 m/s. θ = 30°
To find: i) The maximum height attained (H)
ii) The time of flight (T)
iii) The horizontal range (R)
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 29
i) The maximum height attained by the body is 30.625 m
ii) The time of flight of the body is 5 s.
iii) The horizontal range of the body is 212.2 m.

Question 37.
A fighter plane flying horizontally at a altitude of 1.5 km with a speed of 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with a muzzle velocity of 600 m/s to hit the plane? At what minimum altitude should the pilot fly to avoid being hit? [Take g = 10 m /s2]
Solution:
Given: h = 1.5 km = 1500 m,
u = 600 m/s,
y = 720 km/h = 720 × \(\frac{5}{18}\) = 200 m/s

To find: i) Angle of firing (θ)
ii) Minimum altitude (H)

Formula: H = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)

Calculation:
Let the shell hit the plane t seconds after firing,
∴ 600 cos(90 – θ) × t = 200 t
∴ cos(90 – θ) = \(\frac{1}{6}\)
∴ 90° – θ = cos-1(\(\frac{1}{3}\))
cos -1(\(\frac{1}{3}\)) = 90° – θ
∴ 70°28’ = 90° – θ
∴ θ = 90° – cos-1 (\(\frac{1}{3}\))
∴ θ = 19°47’ with vertical
To avoid being hit, the plane must be at a minimum height, i.e., maximum height reached by the shell.
From formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 30
∴ H = 15.9 km

i) Angle made by gun with the vertical is 19°47′.
ii) Minimum altitude at which the pilot should fly is 15.9 km.

Question 38.
A both is thrown with a velocity of 40 m/s in a direction making an angle of 30° with the horizontal. Calculate
i) Horizontal range
ii) Maximum height and
iii) Time taken to reach the maximum height.
Solution:
Given: u = 40 m/s, θ = 30°
To find: i) Horizontal range (R)
ii) Maximum height (Hmax)
iii) Time to reach max. height (tA)
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 31
i) Horizontal range of the body is 141.4 m.
ii) Maximum height reached by the body is 20.41 m.
iii) Time taken by the body to reach the maximum height is 2.041 s.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 39.
If a child launches paper plane with a velocity of 6 m/s2 at an angle θ with vertical.
i) What will be the maximum range of the projectile?
ii) What will be the maximum height of the projectile?
iii) Will the plane hit a lady standing at a distance of 6m?
Solution:
i) Range of projectile to given by.
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 32
Hmax = 1.63 m
iii) As maximum range of projectile is 6.53 m and lady is standing 6m away, plane will hit the lady.

Question 40.
Explain the term uniform circular motion.
Answer:

  1. The motion of a body along the circumference of a circle with constant speed is called uniform circular motion.
  2. The magnitude of velocity remains constant and its direction is tangential to its circular path.
  3. The acceleration is of constant magnitude and it is perpendicular to the tangential velocity. It is always directed towards the centre of the circular path. This acceleration is called centripetal acceleration.
  4. The centripetal force provides the necessary centripetal acceleration.
  5. Examples of U.C.M:
    • Motion of the earth around the sun.
    • Motion of the moon around the earth.
    • Revolution of electron around the nucleus of atom.

Question 41.
What is meant by period of revolution of U.C.M. Obtain an expression for the period of revolution of a particle performing uniform circular motion.
Answer:
The time taken by a particle performing uniform circular motion to complete one revolution is called as period of revolution. It is denoted by T.

Expression for time period:
During period T, particle covers a distance equal to circumference 2πr of circle with uniform speed v.
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 33

Question 42.
For a particle performing uniform circular motion, derive an expression for angular speed and state its unit.
Answer:

  1. Consider an object of mass m, moving with a uniform speed v, along a circle of radius r. Let T be the time period of revolution of the object, i.e., the time taken by the object to complete one revolution or to travel a distance of 2πr.
    Thus, T = 2πr/v
    ∴ Speed, v = \(\frac{\text { Distance }}{\text { Time }}=\frac{2 \pi \mathrm{r}}{\mathrm{T}}\) …………….. (1)
  2. During circular motion of a point object, the position vector of the object from centre of the circle is the radius vector r. Its magnitude is radius r and it is directed away from the centre to the particle, i.e., away from the centre of the circle.
  3. As the particle performs UCM, this radius vector describes equal angles in equal intervals of time.
  4. The angular speed gives the angle described by the radius vector.
  5. During one complete revolution, the angle described is 2π and the time taken is period T. Hence, the angular speed ω is given as, ….[From (1)]
    ω = \(\frac{\text { Angle }}{\text { time }}=\frac{2 \pi}{\mathrm{T}}=\frac{\left(\frac{2 \pi \mathrm{r}}{\mathrm{T}}\right)}{\mathrm{r}}\) …………….. [From (1)]
    = \(\frac{\mathrm{v}}{\mathrm{r}}\)
  6. The unit of angular speed is radian/second.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 43.
Derive an expression for centripetal acceleration of a particle performing uniform circular motion.
Answer:
Expression for centripetal acceleration by calculus method:
i) Suppose a particle is performing U.C.M in anticlockwise direction.
The co-ordinate axes are chosen as shown in the figure.
Let,
A = initial position of the particle which lies on positive X-axis
P = instantaneous position after time t
θ = angle made by radius vector
ω = constant angular speed
\(\overrightarrow{\mathrm{r}}\) = instantaneous position vector at time t

ii) From the figure,
\(\vec{r}=\hat{i} x+\hat{j} y\)
where, \(\hat{i}\) and \(\hat{j}\) are unit vectors along X-axis and Y-axis respectively.
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 34
iii) Also, x = r cos θ and y = r sin θ
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 35

iv) Velocity of the particle is given as rate of change of position vector.
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 36

vi) From equation (1) and (2),
\(\overrightarrow{\mathrm{a}}=-\omega^{2} \overrightarrow{\mathrm{r}}\) ……….. (3)
Negative sign shows that direction of acceleration is opposite to the direction of position vector. Equation (3) is the centripetal acceleration.

vii) Magnitude of centripetal acceleration is given by,
a = ω2r

viii) The force providing this acceleration should also be along the same direction, hence centripetal.
∴ \(\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{a}}=-\mathrm{m} \omega^{2} \overrightarrow{\mathrm{r}}\)
This is the expression for the centripetal force on a particle undergoing uniform circular motion.

ix) Magnitude of F = mω2r = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) = mωv

Question 44.
Discuss the factors on which time period of conical pendulum depends.
Answer:
Time period of conical pendulum is given by,
T = 2π \(\sqrt{\frac{\cos \theta}{\mathrm{g}}}\) …………. (i)
where, l = length of the string
g = acceleration due to gravity
θ = angle of inclination
From equation (i), it is observed that period of conical pendulum depends on following factors.
i) Length of pendulum (l): Time period of conical pendulum increases with increase in length of pendulum, i.e., T ∝ \(\sqrt {l}\)
ii) Acceleration due to gravity (g): Time period of conical pendulum decreases with increase in g. i.e., T ∝ \(\frac{1}{\sqrt{g}}\)
iii) Angle of inclination (θ): As θ increases, cos θ decreases, hence, time period of conical pendulum decreases with increase in θ. (For 0 < θ < π) i.e., T ∝ \(\sqrt{\cos \theta}\)

Question 45.
Is there any limitation on semi vertical angle in conical pendulum? Give reason.
Answer:
Yes.

  1. For a conical pendulum, Period T ∝ \(\sqrt{\cos \theta}\)
    ∴ Tension F ∝ \(\frac{1}{\cos \theta}\)
    Speed v ∝ \(\sqrt{\tan \theta}\)
    With increase in angle θ, cos θ decreases and tan θ increases. For θ = 90°, T = 0, F = ∞ and v = ∞ which cannot be possible.
  2. Also, θ depends upon breaking tension of string, and a body tied to a string cannot be resolved in a horizontal circle such that the string is horizontal. Hence, there is limitation of semi vertical angle in conical pendulum.

Solved Examples

Question 46.
An object of mass 50 g moves uniformly along a circular orbit with an angular speed of 5 rad/s. If the linear speed of the particle is 25 m/s, ¡s the radius of the circle? Calculate the centripetal force acting on the particle.
Solution:
Given: ω = 5 rad/s, v = 25 m/s,
m = 50 g = 0.05 kg
To find: radius (r), centipetal force (F)
Formula: i) v = ωr
ii) F = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\)

Calculation: From formula (i),
r = v/ω = 25/5 m = 5 m.
From formula (ii),
F = \(\frac{0.05 \times 25^{2}}{5}\) = 6.25 N.

i) Radius of the circle is 5 m.
ii) Centripetal force acting on the particle is 6.25 N.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 47.
An object is travelling in a horizontal circle with uniform speed. At t = 0, the velocity is given by \(\overrightarrow{\mathbf{u}}=20 \hat{\mathbf{i}}+35 \hat{\mathbf{j}}\) km/s. After one minute the velocity becomes \(\overrightarrow{\mathbf{v}}=-20 \hat{\mathbf{i}}-35 \hat{\mathbf{j}}\). What is the magnitude of the acceleration?
Solution:
Magnitude of initial and final velocities,
u= \(\sqrt{(20)^{2}+(35)^{2}}\) m/s
∴ u = \(\sqrt{1625}\) m/s
∴ u = 40.3 m/s
As the velocity reverses in 1 minute, the time period of revolution is 2 minutes.
T = \(\frac{2 \pi \mathrm{r}}{\mathrm{u}}\), giving r = \(\frac{\text { uT }}{2 \pi}\)
Now,
a = \(\frac{\mathrm{u}^{2}}{\mathrm{r}}=\frac{\mathrm{u}^{2} 2 \pi}{\mathrm{uT}}=\frac{2 \pi \mathrm{u}}{\mathrm{T}}=\frac{2 \times 3.142 \times 40.3}{2 \times 60}\)
= {antilog[log(3.142) + log(40.3) – log(60)]}
= {antilog(0.4972 + 1.6053 – 1.7782)}
= {antilog(0.3243)}
= 2.110.
∴ a = 2.11 m/s2
The magnitude of acceleration is 2.11 m/s2.

Question 48.
A racing car completes 5 rounds of a circular track in 2 minutes. Find the radius of the track if the car has uniform centripetal acceleration of π2/s2.
Solution:
Given: 5 rounds = 2πr(5),
t = 2minutes = 120s
To find: Radius (r)
Formula: acp = ω2r
Calculation: From formula,
acp = ω2r
∴ π2 = \(\frac{\mathrm{v}^{2}}{\mathrm{r}}\)
But v = \(\frac{2 \pi r(5)}{t}=\frac{10 \pi r}{t}\)
∴ π2 = \(\frac{100 \pi^{2} \mathrm{r}^{2}}{\mathrm{rt}^{2}}\)
∴ r = \(\frac{120 \times 120}{100}\) =144m
The radius of the track is 144 m.

Question 49.
A car of mass 1500 kg rounds a curve of radius 250m at 90 km/hour. Calculate the centripetal force acting on it.
Solution:
Given: m= 1500 kg, r = 250m,
v = 90 km/h = 90 × \(\frac{5}{18}\) = 25m/s
To find: Centripetal force (FCP)
Formula: FCP = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\)
Calculation: From formula,
FCP = \(\frac{1500 \times(25)^{2}}{250}\)
∴ FCP = 3750 N
The centripetal force acting on the car is 3750 N.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 50.
A one kg mass tied at the end of the string 0.5 m long is whirled ¡n a horizontal circle, the other end of the string being fixed. The breaking tension in the string is 50 N. Find the greatest speed that can be given to the mass.
Solution:
Given: Breaking tension, F = 50 N,
m = 1 kg, r = 0.5m
To find: Maximum speed (vmax)
Formula: B.T (F) = max. C.F \(\frac{m v_{\max }^{2}}{r}\)
Calculation: From formula,
v2max = \(\frac{F \times r}{m}\)
∴ v2max = \(\frac{50 \times 0.5}{1}\)
∴ vmax = \(\sqrt{50 \times 0.5}\) = 5 m/s
The greatest speed that can be given to the mass is 5 m/s.

Question 51.
A mass of 5 kg is tied at the end of a string 1.2 m long revolving in a horizontal circle. If the breaking tension in the string is 300 N, find the maximum number of revolutions per minute the mass can make.
Solution:
Given: Length of the string, r = 1.2 m,
Mass attached. m = 5 kg,
Breaking tension, T = 300 N
To find: Maximum number of revolutions per minute (nmax)
Formula: Tmax = Fmax = mrω2max
Calculation: From formula,
∴ 5 × 1.2 × (2πn)2 = 300
∴ 5 × 1.2 × 4π2n2 = 300
∴ n2max = \(\frac{300}{4 \times(3.142)^{2} \times 6.0}\) = 1.26618
∴nmax = \(\sqrt{1.26618}\) = 1.125 rev/s
∴ nmax = 1.125 × 60
∴ nmax = 67.5 rev/min
The maximum number of revolutions per minute made by the mass is 67.5 rev /min.

Question 52.
A coin placed on a revolving disc, with its centre at a distance of 6 cm from the axis of rotation just slips off when the speed of the revolving disc exceeds 45 r.p.m. What should be the maximum angular speed of the disc, so that when the coin is at a distance of 12 cm from the axis of rotation, it does not slip?
Solution:
Given. r1 = 6cm, r2 = 12cm, n1 = 45 r.p.m
To Find: Maximum angular speed (n2)
Formula: Max. C.F = mrω2
Calculation: Since, mr1ω12 mr2ω22 [As mass is constant]
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 37
The maximum angular speed of the disc should be 31.8 r.p.m.

Question 53.
A stone of mass 0.25 kg tied to the end of a string is whirled in a circle of radius 1.5 m with a speed of 40 revolutions/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Solution:
Given: m = 0.25 kg, r.= 1.5 m, Tmax = 200 N,
n = 40 rev. min-1 = \(\frac{40}{60}\) rev s-1
To find: i) Tension (T)
ii) Maximum speed (vmax)
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 38
i) The tension in the string is 6.55 N.
ii) The maximum speed with which the stone can be whirled around is 34.64 m s-1.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 54.
In a conical pendulum, a string of length 120 cm is fixed at rigid support and carries a mass of 150 g at its free end. If the mass is revolved in a horizontal circle of radius 0.2 m around a vertical axis, calculate tension in the string. (g = 9.8 m/s2)
Solution:
Given: l = 120 cm = 1.2rn, r = 0.2m,
m = 150 g = 0.15 kg
To find: Tension in the string (T)
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 39
∴ Substituting in formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 40
Tension in the string is 1.491 N.

Question 55.
A conical pendulum has length 50 cm. Its bob of mass 100 g performs uniform circular motion in horizontal plane, so as to have radius of path 30 cm. Find
i) The angle made by the string with vertical
ii) The tension in the supporting thread and
iii) The speed of bob.
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 41
Given: l = 150 cm = 0.5 m, r = 30 cm = 0.3 m,
m = 100 g = 100 × 10-3 kg = 0.1 kg
To find: i) Angle made by the string with vertical (θ)
ii) Tension in the supporting thread (T)
iii) Speed of bob (y)

Formulae: i) tan θ = –\(\frac{r}{\mathrm{~h}}\)
ii) tan θ = \(\frac{\mathrm{v}^{2}}{\mathrm{rg}}\)

Calculation: By Pythagoras theorem, l2 = r2 + h2
h2 = l2 – r2
h2 = 0.25 – 0.09 = 0.16
h = 0.4m
i) From formula (1),
tan θ = \(\frac{0.3}{0.4}\) = 0.75
∴ θ = tan-1 (0.75)
θ = 36°52’

ii) The weight of bob is balanced by vertical component of tension T
∴ T cos θ = mg
cos θ = \(\frac{\mathrm{h}}{l}=\frac{0.4}{0.5}\) = 0.8
∴ T = \(\frac{\mathrm{mg}}{\cos \theta}=\frac{0.1 \times 9.8}{0.8}\)
∴ T = 1.225 N

iii) From formula (2),
v2 = rg tan θ
∴ v2 = 0.3 × 9.8 × 0.75 = 2.205
∴ v = 1.485 m/s

i) Angle made by the string with vertical is 36°52′. ‘
ii) Tension in the supporting thread is 1.225 N.
iii) Speed of the bob is 1.485 m/s

Apply Your Knowledge

Question 56.
Explain the variation of acceleration, velocity and distance with time for an object under free fall.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 42
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 43

  1. For a free falling object, considering the downward direction as negative, the object is released from rest.
    ∴ initial velocity u = 0 and a = -g = -9.8 m/s2
    ∴ The kinematical equations become,
    v = u + at = 0 – gt = -gt = -9.8t
    s = ut + \(\frac{1}{2}\)at2 = o + \(\frac{1}{2}\)(-g)t2 = –\(\frac{1}{2}\) 9.8t2
    = -4.9t2
    v2 = u2 + 2as = 0 + 2(-g)s
    = -2gs = -2 × 9.8s
    = -19.6s
  2. These equations give the velocity and the distance travelled as a function of time and also the variation of velocity with distance.
  3. The variation of acceleration, velocity and distance with the time is as shown in figure a, b and c respectively.

Question 57.
The position-time (x – t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in figure. Choose the correct entries in the brackets below.
i) (A/B) lives closer to the school than (B/A)
ii) (A/B) starts from the school earlier than (B/A)
iii) (A/B) walks faster than (B/A)
iv) A and B reach home at the (same/different) time
v) (A/B) overtakes (B/A) on the road (once/twice)
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 44
Answer:

  1. A lives closer to the school than B. This is because, OQ > OP, hence B has to cover larger distance than A.
  2. A starts from the school earlier than B. This is because, A starts at t = 0 whereas B starts at some finite time greater than zero.
  3. As slope of B is greater than that of A, hence B walks faster than A. iv. A and B reach home at different times.
  4. This is because the value of ‘t’ corresponding to P and Q for A and B respectively is different.
  5. B overtakes A on the road once. This is because A and B meet each other only once on their way back home. As B starts from school later than A and walks faster than A, hence B overtakes A once on his way home.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 58.
A bowler throws the ball up to correct distance by controlling his velocity and angle of throw, as shown in the figure given below
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 45
i) What will be the range of the projectile?
ii) What will be the height of the projectile from ground?
Answer:

  1. Range of projectile is given by,
    R = \(\frac{u^{2} \sin 2 \theta}{g}=\frac{6^{2} \times \sin (2 \times 30)}{9.8}\)
    R = 3.18 m
  2. Height of projectile is given by,
    H = \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}=\frac{6^{2} \times \sin ^{2} 30}{2 \times 9.8}\) = 0.46m
    Height achieved by ball from ground is
    H = 0.46 + 1 = 1.46m

i) The range of the projectile is 3.18 m.
ii) The height of the projectile is 1.46 m.

Question 59.
A child takes reading of two cars running on highway, for his school project. He draws a position-time graph of the two cars as shown in the figure
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 46
i) What is the velocity of two cars when they meet together?
ii) What is the difference in velocities of the two cars when they cover their maximum distance?
iii) What will be acceleration of the two cars in first 20 s?
Solution:
i) According to graph, the velocity of two cars when they meet each other are,
x = 70m
t = 308
v = \(\frac{x}{t}=\frac{70}{30}\) = 2.33 m/s

ii) According to graph, for maximum distance.
For 1st car,
x1 = 120m
t1 = 50 s
v1 = \(\frac{x_{1}}{t_{1}}=\frac{120}{50}\)
v1 = 2.4 m/s

For 2nd car,
x2 = 90 m
t2 = 60 s
v2 = \(\frac{x_{2}}{t_{2}}=\frac{90}{60}\)
v2 = 1.5 m/s
Difference in velocities is given by,
v1 – v2 = 2.4 – 1.5 = 0.9 m/s

iii) According to graph,
Acceleration of 1st car in first 20 s
v1 = \(\frac{\mathrm{x}_{1}}{\mathrm{t}}\)
v1 = \(\frac{60}{20}\)
v1 = 3 m/s
a1 = \(\frac{\mathrm{v}_{1}}{\mathrm{t}}=\frac{3}{20}\)
a1 = 0.15 m/s2
Acceleration of 2nd car in first 20 s
v2 = \(\frac{\mathrm{x}_{2}}{\mathrm{t}}\)
v2 = \(\frac{40}{20}\)
v2 = 2 m/s
a2 = \(\frac{\mathrm{v}_{2}}{\mathrm{t}}=\frac{2}{20}\)
a2 = 0.1 m/s2
Now,
a1 – a2 = 0.15 – 0.1
= 0.05 m/s2

i) The velocity of two cars when they meet together is 2.33 m/s.
ii) The difference in velocities of two cars when they cover maximum distance is 0.9 m/s.
iii) The accelerator of two cars in 20 s is 0.05 m/s2.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 60.
The speed of a projectile u reduces by 50% on reaching maximum height. What is the range on the horizontal plane?
Solution:
If θ is the angle of projection, then velocity of projectile at height point = u cos θ
u cos θ = \(\frac{50}{100}\) u = \(\frac{1}{2}\) u
or cos θ = \(\frac{1}{2}\) cos 60°
or θ = 60°
Horizontal range,
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 47

Question 61.
In projectile motion, vertical motion and horizontal motion are dependent of each other. Yes or No? Justify your answer.
Answer:
No. In projectile motion, the horizontal and vertical motion are independent of each other because both motions don’t affect each other.

Question 62.
In angular projection of a projectile, at highest point, what will be the components of horizontal and vertical velocities?
Answer:
At highest point of angular projection of a projectile, the horizontal component of its velocity is non zero and the vertical component of its velocity is momentarily zero.

Question 63.
What angle will be described between velocity and acceleration at highest point of projectile path?
Answer:
At highest point of projectile path, the angle between velocity and acceleration is 90°.

Quick Review

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 48
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 49
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 50

Multiple Choice Questions

Question 1.
The velocity-time relation of a particle starting from rest is given by v = kt where k = 2 m/s2. The distance travelled in 3 sec is
(A) 9 m
(B) 16 m
(C) 27 m
(D) 36 m
Answer:
(A) 9 m

Question 2.
If the particle is at rest, then the x – t graph can be only
(A) parallel to position – axis
(B) parallel to time – axis
(C) inclined with acute angle
(D) inclined with obtuse angle
Answer:
(B) parallel to time – axis

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 3.
A body is thrown vertically upwards, maximum height is reached, then it will have
(A) zero velocity and zero acceleration.
(B) zero velocity and finite acceleration.
(C) finite velocity and zero acceleration.
(D) finite velocity and finite acceleration.
Answer:
(B) zero velocity and finite acceleration.

Question 4.
Which of the following is NOT a projectile?
(A) A bullet fired from gun.
(B) A shell fired from cannon.
(C) A hammer thrown by athlete.
(D) An aeroplane in flight.
Answer:
(D) An aeroplane in flight.

Question 5.
The range of projectile is 1 .5 km when it is projected at an angle of 15° with horizontal. What will be its range when it is projected at an angle of 45° with the horizontal?
(A) 0.75 km
(B) 1.5 km
(C) 3 km
(D) 6 km
Answer:
(C) 3 km

Question 6.
Which of the following remains constant for a projectile fired from the earth?
(A) Momentum
(B) Kinetic energy
(C) Vertical component of velocity
(D) Horizontal component of velocity
Answer:
(D) Horizontal component of velocity

Question 7.
In case of a projectile, what is the angle between the instantaneous velocity and acceleration at the highest point?
(A) 45°
(B) 1800
(C) 90°
(D) 0°
Answer:
(C) 90°

Question 8.
A player kicks up a ball at an angle θ with the horizontal. The horizontal range is maximum when θ is equal to
(A) 30°
(B) 45°
(C) 60°
(D) 90°
Answer:
(B) 45°

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 9.
The greatest height to which a man can throw a stone is h. The greatest distance to which he can throw it will be
(A) h/2
(B) 2h
(C) h
(D) 3h
Answer:
(B) 2h

Question 10.
Two balls are projected at an angle θ and (90° – θ) to the horizontal with the same speed. The ratio of their maximum vertical
heights is
(A) 1 : 1
(B) tan θ : 1
(C) 1 : tan θ
(D) tan2 θ : 1
Answer:
(D) tan2 θ : 1

Question 11.
When air resistance is taken into account while dealing with the motion of the projectile, to achieve maximum horizontal range, the angle of projection should be,
(A) equal to 45°
(B) less than 45°
(C) greater than 90°
(D) greater than 45°
Answer:
(D) greater than 45°

Question 12.
The maximum height attained by projectile is found to be equal to 0.433 of the horizontal range. The angle of projection of this projectile is
(A) 30°
(B) 45°
(C) 60°
(D) 75°
Answer:
(C) 60°

Question 13.
A projectile is thrown with an initial velocity of 50 m/s. The maximum horizontal distance which this projectile can travel is
(A) 64m
(B) 128m
(C) 5m
(D) 255m
Answer:
(D) 255m

Question 14.
A jet airplane travelling at the speed of 500 kmh-1 ejects the burnt gases at the speed of 1400 kmh-1 relative to the jet airplane. The speed of burnt gases relative to stationary observer on the earth is
(A) 2.8 kmh-1
(B) 190 kmh-1
(C) 700 kmh-1
(D) 900 kmh-1
Answer:
(D) 900 kmh-1

Question 15.
A projectile projected with certain angle reaches ground with
(A) double angle
(B) same angle
(C) greater than 90°
(D) angle between 90° and 180°
Answer:
(B) same angle

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 16.
The time period of conical pendulum is _________.
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 51
Answer:
(C) \(2 \pi \sqrt{\frac{l \cos \theta}{\mathrm{g}}}\)

Question 17.
A projectile projected with certain velocity reaches ground with (magnitude)
(A) zero velocity
(B) smaller velocity
(C) same velocity
(D) greater velocity
Answer:
(C) same velocity

Question 18.
The period of a conical pendulum in terms of its length (l), semivertical angle (θ) and acceleration due to gravity (g) is:
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 52
Answer:
(C) \(4 \pi \sqrt{\frac{l \cos \theta}{4 \mathrm{~g}}}\)

Question 19.
Consider a simple pendulum of length 1 m. Its bob performs a circular motion in horizontal plane with its string making an angle 600 with the vertical. The period of rotation of the bob is(Take g = 10 m/s2)
(A) 2s
(B) 1.4s
(C) 1.98 s
(D) none of these
Answer:
(B) 1.4s

Question 20.
The period of a conical pendulum is
(A) equal to that of a simple pendulum of same length l.
(B) more than that of a simple pendulum of same length l.
(C) less than that of a simple pendulum of same length l.
(D) independent of length of pendulum.
Answer:
(C) less than that of a simple pendulum of same length l.

Competitive Corner

Question 1.
Two particles A and B are moving in uniform circular motion in concentric circles of radii rA and rB with speed vA and vB respectively. Their time period of rotation is the same. The ratio of angular speed of A to that of B will be:
(A) rB : rA
(B) 1 : 1
(C) rA : rB
(D) vA : vB
Answer:
(B) 1 : 1
Hint:
Time period of rotation (A and B) is same
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 53

Question 2.
When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60° with horizontal, it can travel a distance x1 along the plane. But when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel x2 distance. Then x1 : x2 will be:
(A) 1 : \(\sqrt {3}\)
(B) 1 : 2\(\sqrt {3}\)
(C) 1 : \(\sqrt {2}\)
(D) \(\sqrt {3}\) : 1
Answer:
(A) 1 : \(\sqrt {3}\)
Hint:
v2 = u2 + 2as
∴ v2 = u2 + 2 g sin θ x
sin θ. x = constant
∴ x ∝ \(\frac{1}{\sin \theta}\)
∴ \(\frac{x_{1}}{x_{2}}=\frac{\sin \theta_{2}}{\sin \theta_{1}}=\frac{1 / 2}{\sqrt{3} / 2}\) = 1 : \(\sqrt {3}\)

Question 3.
A person travelling in a straight line moves with a constant velocity v1 for certain distance x’ and with a constant velocity v2 for next equal distance. The average velocity y is given by the relation
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 54
Answer:
(C) \(\frac{2}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}\)
Hint:
Let, t’ be the time taken to travel distance ‘x’ with constant velocity ‘v1
∴ t1 = \(\frac{\mathrm{x}}{\mathrm{v}_{2}}\)
Let ‘t2’ be the time taken to travel equal distance ‘x’ with constant velocity ‘v2
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 55

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 4.
Two bullets are fired horizontally and simultaneously towards each other from roof tops of two buildings 100 m apart and of same height of 200 m, with the same velocity of 25 m/s. When and where will the two bullets collide? (g = 10 m/s2)
(A) They will not collide
(B) After 2 s at a height of 180 m
(C) After 2 s at a height of 20 m
(D) After 4 s at a height of 120 m
Answer:
(B) After 2 s at a height of 180 m
Hint:
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 56
Let the bullets collide at time t
The horizontal displacement x1 and x2 is given by the equation
x1 = ut and x2 = ut
∴ x1 + x2 = 100
∴ 25t + 25t = 100
∴ t = 2s
Vertical displacement ‘y’ is given by
y = \(\frac{1}{2}\) gt2 = \(\frac{1}{2}\) × 10 × 22 = 20m
∴ h = 200 – 20= 180m

Question 5.
A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field \(\vec{E}\). Due to the force q\(\vec{E}\), its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively
(A) 2 m/s, 4 m/s
(B) 1 m/s, 3 m/s
(C) 1 m/s, 3.5 m/s
(D) 1.5 m/s, 3 m/s
Answer:
(B) 1 m/s, 3 m/s
Hint:
Car at rest attains velocity of 6 m/s in t1 = 1 s.
Now as direction of field is reversed, velocity of car will reduce to 0 m/s in next 1 s. i.e., at t2 = 2 s. But, it continues to move for next one second. This will give velocity of -6 m/s to car at t3 = 3 s.
Using this data, plot of velocity versus time will be
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 57

Question 6.
All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 58
Answer:
(D)
Hint:
The graphs (A), (B) and (C) represent the uniformly retarded motion, i.e., velocity decreases uniformly. However, the slope of the curve in graph (D), indicates increasing velocity. Hence, graph (D) is incorrect.

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 7.
A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then:
(A) T ∝ R(n+1)/2
(B) T ∝ Rn/2
(C) T ∝ R3/2 for any n
(D) T ∝ R\(\frac{n}{2}\)+1
Answer:
(A) T ∝ R(n+1)/2
Hint:
The centripetal force acting on the particle is provided by the central force,
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 59

Question 8.
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t1. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t2. The time taken by her to walk up on the moving escalator will be:
(A) \(\frac{\mathrm{t}_{1}+\mathrm{t}_{2}}{2}\)
(B) \(\frac{t_{1} t_{2}}{t_{2}-t_{1}}\)
(C) \(\frac{\mathbf{t}_{1} t_{2}}{\mathbf{t}_{2}+t_{1}}\)
(D) t1 – t2
Answer:
(C) \(\frac{\mathbf{t}_{1} t_{2}}{\mathbf{t}_{2}+t_{1}}\)
Hint:
Let velocity of Preeti be v1, velocity of escalator be v2 and distance travelled be L.
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 60

Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane

Question 9.
A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?
Maharashtra Board Class 11 Physics Important Questions Chapter 3 Motion in a Plane 61
Answer:
(A)
Hint:
If a body is projected in vertically upward direction, then its acceleration is constant and negative. If direction of motion is positive i.e.. vertically up) and initial position of body is taken as origin, then the velocity decreases uniformly. At highest point its velocity is equal to zero and then it accelerates uniformly downwards returning to its reference position.

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 1 Units and Measurements Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 1 Units and Measurements

Question 1.
What is a measurement? How is measured quantity expressed?
Answer:

  1. A measurement is a comparison with internationally accepted standard measuring unit.
  2. The measured quantity (M) is expressed in terms of a number (n) followed by a corresponding unit (u) i.e., M = nu.

Example:
Length of a wire when expressed as 2 m, it means value of length is 2 in the unit of m (metre).

Question 2.
State true or false. If false correct the statement and rewrite. Different quantities are measured in different units.
Answer: True.
[Note: Choice of unit depends upon its suitability for measuring the magnitude of a physical quantity under consideration. Hence, we choose different scales for same physical quantity.]

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 3.
Describe briefly different types of systems of units.
Answer:
System of units are classified mainly into four types:

  1. C.G.S. system:
    It stands for Centimetre-Gram-Second system. In this system, length, mass and time are measured in centimetre, gram and second respectively.
  2. M.K.S. system:
    It stands for Metre-Kilogram-Second system. In this system, length, mass and time are measured in metre, kilogram and second respectively.
  3. F.P.S. system:
    It stands for Foot-Pound-Second system. In this system, length, mass and time are measured in foot, pound and second respectively.
  4. S.I. system:
    It stands for System International. This system has replaced all other systems mentioned above. It has been internationally accepted and is being used all over world. As the SI units use decimal system, conversion within the system is very simple and convenient.

Question 4.
What are fundamental quantities? State two examples of fundamental quantities. Write their S.J. and C.G.S. units.
Answer:
Fundamental quantities:
The physical quantities which do not depend on any other physical quantity for their measurements i.e., they can be directly measured are called fundamental quantities.
Examples: mass, length etc.
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 1

Question 5.
What are fundamental units? State the S.l. units of seven fundamental quantities.
Answer:
Fundamental units:
The units used to measure fundamental quantities are called fundamental units.
S.I. Units of fundamental quantities:
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 2

Question 6.
State and describe the two supplementary units.
Answer:
The two supplementary units are:
i) Plane angle (dθ):
a. The ratio of kngth of arc (ds) of an circle to the radius (r) of the circle is called as Plane angle (dθ)
i.e., dθ = \(\frac{\mathrm{ds}}{\mathrm{r}}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 3
b. Thus, dθ is angle subtended by the arc at the centre of the circle.
c. Unit: radian (rad)
d. Denoted as θc
e. Length of arc of circle = Circumference of circle = 2πr.
∴ plane angle subtended by entire circle at its centre is θ = \(\frac{2 \pi \mathrm{r}}{\mathrm{r}}\) = 2πc

ii) Solid angle (dΩ):
a. solid angle is 3-dimensional analogue of plane angle.
b. Solid angle is defined as area of a portion of surface of a sphere to the square of radius of the sphere.
i.e., dΩ = \(\frac{\mathrm{d} \mathrm{A}}{\mathrm{r}^{2}}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 4
c. Unit: Steradian (sr)
d. Denoted as (Ω)
e. Surface area of sphere = 4πr2
∴ solid angle subtended by entire sphere at its centre is Ω = \(\frac{4 \pi r^{2}}{r^{2}}\) = 4π sr

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 7.
Derive the relation between radian and degree. Also find out 1” and 1’ in terms of their respective values in radian. (Take π = 3.1416)
Answer:
We know that, 2 πc = 360°
∴ πc = 180°
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 5

Question 8.
What are derived quantities and derived units? State two examples. State the corresponding S.L. and C.G.S. units of the examples.
Answer:

  1. Derived quantities: Physical qUantities other than fundamental quantities which depend on one or more fundamental quantities for their measurements are called derived quantities.
  2. Derived units: The units of derived quantities which are expressed in terms of fundamental units for their measurements are called derived units.
  3. Examples and units:
    Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 6

Question 9.
Classify the following quantities into fundamental and derived quantities: Length, Velocity, Area, Electric current, Acceleration, Time, Force, Momentum, Energy, Temperature, Mass, Pressure, Magnetic induction, Density.
Answer:
Fundamental Quantities: Length, Electric current, Time, Temperature, Mass.

Derived Quantities: Velocity, Area, Acceleration, Force, Momentum, Energy. Pressure, Magnetic induction, Density

Question 10.
Classify the following units into fundamental, supplementary and derived units:
newton, metre, candela, radian, hertz. square metre, tesla, ampere, kelvin, volt, mol, coulomb, farad, steradian.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 7

Question 11.
List the conventions followed while using SI units.
Answer:
Following conventions should be followed while writing S.I. units of physical quantities:

  1. Unit of every physical quantity should be represented by its symbol.
  2. Full name of a unit always starts with smaller letter even if it is named after a person, eg.: 1 newton, 1 joule, etc. But symbol for unit named after a person should be in capital letter, eg.: N after scientist Newton, J after scientist Joule, etc.
  3. Symbols for units do not take plural form.
  4. Symbols for units do not contain any full stops at the end of recommended letter.
  5. The units of physical quantities in numerator and denominator should be written as one ratio. For example the SI unit of acceleration is m/s2 or m s-2 but not m/s/s.
  6. Use of combination of units and symbols for units is avoided when physical quantity is expressed by combination of two. For example, The unit J/kg K is correct while joule/kg K is not correct.
  7. A prefix symbol is used before the symbol of the unit.
    • a. Prefix symbol and symbol of unit constitute a new symbol for the unit which can be raised to a positive or negative power of 10.
      For example,
      1 ms = 1 millisecond = 10-3 s
      1 μs = 1 microsecond = 10-6 s
      1 ns = 1 nanosecond = 10-9 s
    • b. Use of double prefixes is avoided when single prefix is available
      10-6 s = 1 μs and not 1 mms
      10-9 s = 1 ns and not 1 mμs
  8. Space or hyphen must be introduced while indicating multiplication of two units e.g., m/s should be written as m s-1 or m-s-1.

Solved Examples

Question 12.
What is the solid angle subtended by the moon at any point of the Earth, given the diameter of the moon is 3474 km and its distance from the Earth 3.84 × 108 m?
Solution:
Given: Diameter (D) = 3474 km
∴ Radius of moon (R) = 1737 km
= 1.737 × 106 m
Distance from Earth r = 3.84 × 108 m
To find: Solid angle (dΩ)
Formula: dΩ = \(\frac{\mathrm{d} \mathrm{A}}{\mathrm{r}^{2}}\)

Calculation:
From formula,
dΩ = \(\frac{\pi \mathrm{R}^{2}}{\mathrm{r}^{2}}\) ……..( cross-sectional area of disc of moon = πR2)
dΩ = \(\frac{\pi \times\left(1.737 \times 10^{5}\right)^{2}}{\left(3.84 \times 10^{8}\right)^{2}}\)
= \(\frac{3.412 \times(1.737)^{2} \times 10^{10}}{(3.84)^{2} \times 10^{16}}\)
= antilog{log(3.142) + 2log(1.737) – 2log(3.84)} × 10-6
= antilog {0.4972 + 2(0.2397) – 2(0.5843)} × 10-6
= antilog{0.4972 + 0.4794 – 1.1686} × 10-6
= antilog{\(\overline{1}\) .8080} × 10-6
= 6.428 × 10-1 × 10-6
= 6.43 × 10-5 sr
Solid angle subtended by moon at Earth is 6.43 × 10-5 sr
[Note: Above answer is obtained substituting value of r as 3.142]

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 13.
Pluto has mean diameter of 2,300 km and very eccentric orbit (oval shaped) around the Sun, with a perihelion (nearest) distance of 4.4 × 109 km and an aphelion (farthest) distance of 7.3 × 109 km. What are the respective solid angles subtended by Pluto from Earth’s perspective? Assume that distance from the Sun can be neglected.
Solution:
Given: Radius of Pluto. R = \(\frac{2300}{2}\) km
= 1150km
Perihelion distance rp = 4.4 × 109 km
Aphelion distance ra = 7.3 × 109 km
To find: Solid angles (dΩp and dΩa)
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 8
Solid angle at perihelion distance is 2.146 × 10-13 sr and at aphelion distance is 7.798 × 10-14 sr.

Question 14.
Define a metre.
Answer:
The metre is the length of the path travelled by light in vacuum during a time interval of 1/299, 792, 458 of a second.
Answer:

Question 15.
What ¡s parallax?
Answer:

  1. Parallax is defined as the apparent change in position of an object due to a change in position of an observer.
  2. Explanation: When a pencil is held in front of our eyes and we look at it once with our left eye closed and then with our right eye closed, pencil appears to move against the background. This effect is called parallax effect.

Question 16.
What is parallax angle?
Answer:
i) Angle between the two directions along which a star or planet is viewed at the two points of observation is called parallax angle (parallactic angle).
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 9
ii) It is given by θ = \(\frac{b}{D}\)
where, b = Separation between two points of observation.
D = Distance of source from any point of observation.

Question 17.
Explain the method to determine distance of a planet from the Earth.
Answer:

  1. Parallax method is used to determine distance of different planets from the Earth.
  2. To measure the distance ‘D’ of a far distant planet S, select two different observatories (E1 and E2).
  3. The planet should be visible from E1 and E2 observatories simultaneously i.e. at the same time.
  4. E1 and E2 are separated by distance ‘b’ shown in figure.
    ∴ E1E3 = b
    Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 10
  5. The angle between the two directions along which the planet is viewed, can be measured. It is parallax angle, which in this case is L ∠E1E2 = θ
  6. The planet is far away from the (Earth) observers, hence
    b < <D
    ∴ \(\frac{b}{D}\) < < 1 and ‘θ’ is also very small.
    Hence, E1E2 can be considered as arc of length b of circle with S as centre and D as radius.
    :. E1S = E2S = D
    ∴ θ = \(\frac{b}{D}\) . . . .(θ is taken in radian)
    ∴ D = \(\frac{b}{\theta}\)
    Thus, the distance ‘D’ of a far away planet ‘S’ can be determined using the parallax method.

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 18.
Explain how parallax method is used to measure distance of a star from Earth.
Answer:

  1. The parallax measured from two farthest distance points on Earth for stars will be too small and hence cannot be measured.
  2. Instead, parallax between two farthest points (i.e., 2 ΔU apart) along the orbit of Earth around the Sun (s) is measured.
    Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 11

Question 19.
Explain how size of a planet or star is measured.
Answer:

  1. To determine the diameter (d) of a planet or star, two diametrically opposite points of the planet are viewed from the same observatory.
  2. If d is diameter of planet or star, angle subtended by it at any single point on the Earth is called angular diameter of planet.
  3. Let angle α be angle between these two directions.
  4. If distance between the Earth and planet or star (D) is known, α = \(\frac{\mathrm{d}}{\mathrm{D}}\)
  5. This relation gives, d = α D
    Thus, diameter (d) of planet or star can be determined.

Question 20.
Name the devices used to measure very small distances such as atomic size.
Answer:
Devices used are:
Electron microscope, tunnelling electron microscope.

Question 21.
Just as large distances are measured in AU, parsec or light year, atomic or nuclear distances are measured with the help of microscopic units. Match the units given in column A with their corresponding SI unit given in column B.
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 39
Answer:
i. – (b)
ii. – (a)

Solved Examples

Question 22.
A star is 5.5 light years away from the Earth. How much parallax in arcsec will it subtend when viewed from two opposite points along the orbit of the Earth?
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 12
Solution:
Two opposite points-A and B along the orbit of the Earth are 2 AU apart. The angle subtended by AB at the position of the star is
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 13
= antilog{log(2.992) – log(5.5) – log(9.46)} × 10-4
= antilog {0.4761 – 0.7404 – 0.9759} × 10-4
= antilog {\(\overline{2}\).7598} × 10-4
= 5.751 × 10-2 × 10-4
= 5.75 × 10-6
= 5.75 × 10-6 rad
= 5.75 × 10-6 × 57.297 × 60 × 60 arcsec
…. (converting radian into arcsecond)
= 1.186 arcsec
Parallax is 1.186 arcsec

Question 23.
The moon is at a distance of 3.84 × 108 m from the Earth. If viewed from two diametrically opposite points on the Earth, the angle subtended at the moon is 1° 54′. What is the diameter of the Earth?
Solution:
Given
Distance (D) = 3.84 × 108 m
Subtended angle (α)
= 1° 54′ = (60’+ 54′)= 114′
= 114 × 2.91 × 10-4 rad
= 3.317 × 10-2 rad
To find: Diameter of Earth (d)
Formula: d = αD
Calculation: From formula,
d = 3.317 × 10-2 × 3.84 × 108
= 1.274 × 107 m
Diameter of Earth is 1.274 × 107 m.

Question 24.
Explain the method to measure mass.
Answer:
Method for measurement of mass:

  1. Mass, until recently, was measured with a standard mass of the international prototype of the kilogram (a platinum-iridium alloy cylinder) kept at international Bureau of Weights and Measures, at Serves, near Paris, France.
  2. As platinum – iridium piece was seen to pick up microparticles and found to be affected by atmosphere, its mass could no longer be treated as constant.
  3. Hence, a new definition of mass was introduced in terms of electric current on 20th May 2019.
  4. Now, one kilogram mass is described in terms of amount of current which has to be passed through electromagnet to pull one side of extremely sensitive balance to balance the other side which holds one standard kg mass.
  5. To measure mass of small entities such as atoms and nucleus, atomic mass unit (amu) is used.
    It is defined as (\(\frac{1}{12}\))th mass of an unexcited atom of carbon -12(C12).
    1 amu ≈ 1.66 × 10-27 kg.

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 25.
That can he the reason for choosing Carbon-12 to define atomic mass unit?
Answer:

  1. Unlike oxygen and hydrogen, which exhibit various isotopes in higher proportions, carbon- 12 is the single most abundant (98% of available carbon) isotope of carbon.
  2. it is also very stable.
    Hence, it makes more accurate unit of measuring mass and is used to define atomic mass unit.

Question 26.
Define mean solar day. Explain the method for measurement of time.
Answer:

  1. A mean solar day is the average time interval from one noon to the next noon.
    Method for measurement of time:
  2. The unit of time, the second, was considered to be \(\frac{1}{86400}\) of the mean solar day, where a mean solar day = 24 hours
    = 24 × 60 × 60
    = 86400 s
  3. However, this definition proved to be unsatisfactory to define the unit of time precisely because solar day varies gradually due to gradual slowing down of the Earth’s rotation. Hence, the definition of second was replaced by one based on atomic standard of time.
  4. Atomic standard of time is now used for the measurement of time. In atomic standard of time, periodic vibrations of caesium atom is used.
  5. One second is time required for 9,192.631,770 vibrations of the radiation corresponding to transition between two hyperfine energy states of caesium-133 (Cs- 133) atom.

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 27.
Define dimensions and dimensional formula of physical quantities. Give few examples of dimensional formula.
Answer:

  1. Dimensions:
    The dimensions of a physical quantity are the powers to which the fundamental units must be raised in order to obtain the unit of a given physical quantity.
  2. Dimensional formula:
    When any derived quantity is represented with appropriate powers of symbols of the fundamental quantities, SUCh an expression is called dimensional formula.
    It is expressed by square bracket with no comma in between the symbols.
  3. Examples of dimensional formula:
    a. Speed = \(\frac{\text { Distance }}{\text { time }}\)
    ∴ Dimensions of speed = \(\frac{[\mathrm{L}]}{[\mathrm{T}]}\) = [L1M0T-1]
    [Note: As power of M is zero, it can be omitted from dimensional formula. Therefore, dimensions of speed can be written as [L1T1]
    Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 14

Question 28.
A book with many printing errors contains four different formulae for the displacement y of a particle undergoing a certain periodic function:
i) y = a sin \(\frac{2 \pi t}{T}\)
ii) y = a sin v t
iii) y = \(\frac{\mathrm{a}}{\mathbf{T}} \sin \frac{\mathrm{t}}{\mathrm{a}}\)
iv) y = \(\frac{a}{\sqrt{2}}\left[\sin \frac{2 \pi t}{T}+\cos \frac{2 \pi t}{T}\right]\)
Here, a is maximum displacement of particle, y ¡s speed of particle, T is time period of motion. Rule out the wrong formulae on dimensional grounds.
Answer:
The argument of trigonometrical function, i.e., angle is dimensionless. Now,
i) The argument, \(\left[\frac{2 \pi \mathrm{t}}{\mathrm{T}}\right]=\frac{[\mathrm{T}]}{[\mathrm{T}]}\) = 1 = [L0M0T0]
which is a dimensionless quantity.
Hence, formula (i) is correct.

ii) The argument,
[vt] = [LT-1] [T] = [L] = [L1M0T0]
which is not a dimensionless quantity.
Hence, formula (ii) is incorrect.

iii) The argument,
\(\left[\frac{\mathrm{t}}{\mathrm{a}}\right]=\frac{[\mathrm{T}]}{[\mathrm{L}]}\) = [L-1M0T1]
which is not a dimensionless quantity.
Hence, formula (iii) is incorrect.

iv) The argument,
\(\left[\frac{2 \pi \mathrm{t}}{\mathrm{T}}\right]=\frac{[\mathrm{T}]}{[\mathrm{T}]}\) = 1 = [L0M0T0]
which is a dimensionless quantity.
Hence, formula (iv) is correct.

Question 29.
State principle of homogeneity of dimensions.
Answer:
Principle of homogeneity of dimensions: The dimensions of all the terms on the two sides of a physical equation relating different physical quantities must be same.

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 30.
State the uses of dimensional analysis.
Answer:
Uses of dimensional analysis:

To check the correctness of a physical equation.
Correctness of a physical equation by dimensional analysis:

  1. A physical equation is correct only if the dimensions of all the terms on both sides of that equations are the same.
  2. For example, consider the equation of motion.
    v = u + at ……………. (1)
  3. Writing the dimensional formula of every term, we get
    Dimensions of LH.S. [v] [L1M0T-1],
    Dimensions of R.H.S. = [u] + [at]
    = [L1M0T-1] + [L1M0T-2] [L1M0T-1]
    = [L1M0T-1] + [L1M0T-1]
    ⇒ [L.HS.] = [R.H.S.]
  4. As dimensions of both side of equation is same, physical equation is dimensionally correct.

To derive the relationship between related physical quantities.
Expression for time period of a simple pendulum by dimensional analysis:

  1. Time period (T) of a simple pendulum depends upon length (l) and acceleration due to gravity (g) as follows:
    T ∝ la gb
    i.e., T = k la gb ………… (1)
    where, k = proportionality constant, which is dimensionless.
  2. The dimensions of T = [L0M0T1)
    The dimensions of l = [L1M0T0]
    The dimensions of g = [L1M0T2]
    Taking dimensions on both sides of equation (1),
    [L0M0T1] = [L1M0T0]a [L1M0T-2]b
    [L0M0T1] = [La+bM0T-2b]
  3. Equating corresponding power of L, M and T on both sides, we get
    a + b = 0 …………. (2)
    and -2b = 1
    ∴ b = –\(\frac{1}{2}\)
  4. Substituting ‘b’ in equation (2), we get
    a = \(\frac{1}{2}\)
  5. Substituting values of a and b in equation (1),
    we have,
    Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 15
  6. Experimentally, it ¡s found that k = 2π
    ∴ T = 2π \(\sqrt{\frac{l}{\mathrm{~g}}}\)
    This is the required expression for time period of a simple pendulum.

To find the conversion factor between the units of the same physical quantity in two different systems of units.
Conversion factor between units of same physical quantity:

  1. let ‘n’ be the conversion factor between the units of work.
    ∴ 1 J = n erg ………….. (1)
  2. Dimensions of work in S.l. system are \(\left[\mathrm{L}_{1}^{2} \mathrm{M}_{1}^{\prime} \mathrm{T}_{1}^{-2}\right]\) and in CGS system are \(\left[\mathrm{L}_{2}^{2} \mathrm{M}_{2}^{1} \mathrm{~T}_{2}^{-2}\right]\)
  3. From (1),
    \(1\left[\mathrm{~L}_{1}^{2} \mathrm{M}_{1}^{1} \mathrm{~T}_{1}^{-2}\right]=\mathrm{n}\left[\mathrm{L}_{2}^{2} \mathrm{M}_{2}^{1} \mathrm{~T}_{2}^{-2}\right]\)
    Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 16
    n= 104 × 103 × 1 = 107
    Hence, the conversion factor, n = 107
    There fore, from equation (1), we have,
    ∴ 1 J = 107 erg.

Question 31.
Explain the use of dimensional analysis to check the correctness of a physical equation.
Answer:
Correctness of a physical equation by dimensional analysis:

  1. A physical equation is correct only if the dimensions of all the terms on both sides of that equations are the same.
  2. For example, consider the equation of motion.
    v = u + at ……………. (1)
  3. Writing the dimensional formula of every term, we get
    Dimensions of LH.S. [v] [L1M0T-1],
    Dimensions of R.H.S. = [u] + [at]
    = [L1M0T-1] + [L1M0T-2] [L1M0T-1]
    = [L1M0T-1] + [L1M0T-1]
    ⇒ [L.HS.] = [R.H.S.]
  4. As dimensions of both side of equation is same, physical equation is dimensionally correct.

Question 32.
Time period of a simple pendulum depends upon the length of pendulum (l) and acceleration due to gravity (g). Using dimensional analysis, obtain an expression for time period of a simple pendulum.
Answer:
Expression for time period of a simple pendulum by dimensional analysis:
i) Time period (T) of a simple pendulum depends upon length (l) and acceleration due to gravity (g) as follows:
T ∝ la gb
i.e., T = k la gb ………… (1)
where, k = proportionality constant, which is dimensionless.

ii) The dimensions of T = [L0M0T1)
The dimensions of l = [L1M0T0]
The dimensions of g = [L1M0T2]
Taking dimensions on both sides of equation (1),
[L0M0T1] = [L1M0T0]a [L1M0T-2]b
[L0M0T1] = [La+bM0T-2b]

iii) Equating corresponding power of L, M and T
on both sides, we get
a + b = 0 …………. (2)
and -2b = 1
∴ b = –\(\frac{1}{2}\)

iv) Substituting ‘b’ in equation (2), we get
a = \(\frac{1}{2}\)

v) Substituting values of a and b in equation (1),
we have,
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 15

vi) Experimentally, it ¡s found that k = 2π
∴ T = 2π \(\sqrt{\frac{l}{\mathrm{~g}}}\)
This is the required expression for time period of a simple pendulum.

Question 33.
Find the conversion factor between the S.I. and the C.OES. units of work using dimensional analysis.
Answer:
Conversion factor between units of same physical quantity:

  1. let ‘n’ be the conversion factor between the units of work.
    ∴ 1 J = n erg ………….. (1)
  2. Dimensions of work in S.l. system are \(\left[\mathrm{L}_{1}^{2} \mathrm{M}_{1}^{\prime} \mathrm{T}_{1}^{-2}\right]\) and in CGS system are \(\left[\mathrm{L}_{2}^{2} \mathrm{M}_{2}^{1} \mathrm{~T}_{2}^{-2}\right]\)
  3. From (1),
    \(1\left[\mathrm{~L}_{1}^{2} \mathrm{M}_{1}^{1} \mathrm{~T}_{1}^{-2}\right]=\mathrm{n}\left[\mathrm{L}_{2}^{2} \mathrm{M}_{2}^{1} \mathrm{~T}_{2}^{-2}\right]\)
    Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 16
    n= 104 × 103 × 1 = 107
    Hence, the conversion factor, n = 107
    There fore, from equation (1), we have,
    ∴ 1 J = 107 erg.

Question 34.
State the limitations of dimensional analysis.
Answer:
Limitations of dimensional analysis:

  • The value of dimensionless constant can be obtained with the help of experiments only.
  • Dimensional analysis cannot be used to derive relations involving trigonometric (sin θ, cos θ, etc.), exponential (ex, ex2, etc.), and logarithmic functions (log x, log x3, etc) as these quantities are dimensionless.
  • This method is not useful if constant of proportionality is not a dimensionless quantity.
  • If the correct equation contains some more terms of the same dimension, it is not possible to know about their presence using dimensional equation.

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 35.
If two quantities have same dimensions, do they always represent the same physical content?
Answer:
When dimensions of two quantities are same, they do not always represent the same physical content.
Example:
Force and momentum both have same dimensions but they represent different physical content.

Question 36.
A dimensionally correct equation need not actually be a correct equation but dimensionally incorrect equation is necessarily wrong. Justify.
Answer:
i) To justify a dimensionally correct equation need not be actually a correct equation, consider equation, v2 = 2as
Dimensions of L.H.S. = [v2] = [L2M0T2]
Dimensions of R.H.S. = [as]= [L2M0T2]
⇒ [L.H.S.] = [R.H.S.]
This implies equation v2 = 2as is dimensionally correct.
But actual equation is, v2 = u2 + 2as
This confirms a dimensionally correct equation need not be actually a correct equation.

ii) To justify dimensionally incorrect equation is necessarily wrong, consider the formula,
\(\frac{1}{2}\) mv = mgh
Dimensions of L.H.S. = [mv] = [L1M1T-1]
Dimensions of R.H.S. = [mgh] = [L2M1T-2]
Since the dimensions of R.H.S. and L.H.S. are not equal, the formula given by equation must be incorrect.
This confirms dimensionally incorrect equation is necessarily wrong.

Question 37.
State, whether all constants are dimensionless or unitless.
Answer:
All constants need not be dimensionless or unitless.
Planck’s constant, gravitational constant etc., possess dimensions and units. They are dimensional constants.

Solved Examples

Question 38.
If length ‘L’, force ‘F’ and time ‘T’ are taken as fundamental quantities, what would be the dimensional equation of mass and density?
Solution:
i) Force = Mass × Acceleration Force
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 17
∴ Dimensional equation of mass
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 18
= [F1L-4T2]

i) The dimensional equation of mass is [F1L-1T2].
ii) The dimensional equation of density is [F1L-4T2].

Question 39.
A calorie is a unit of heat and it equals 4.2 J, where 1 J = kg m2 s-2. A distant civilisation employs a system of units in which the units of mass, length and time are α kg, β m and δ s. Also J’ is their unit of energy. What will be the magnitude of calorie in their units?
Solution:
1 cal = 4.2 kg m2 s-2
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 19
New unit of energy is J’
Dimensional formula of energy is [L2M1T-2] According to the question,
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 20

Question 40.
Assume that the speed (v) of sound in air depends upon the pressure (P) and density (ρ) of air, then use dimensional analysis to obtain an expression for the speed of sound.
Solution:
It is given that speed (v) of sound in air depends upon the pressure (P) and density (ρ) of the air.
Hence, we can write, v = k Pa ρb ……….. (1)
where, k is a dimensionless constant and a and b are powers to be determined.
Dimensions of y = [L1M0T-1]
Dimensions of P = [L-1M1T-2]
Dimensions of ρ = [L-3M1T0]
Substituting the dimensions of the quantities on both sides of equation (1),
∴ [L1M0T-1] = [L-1M1T-2]a [L-3M1T0]b
∴ [L1M0T-1] = [L-aMaT-2a] [L-3bMbT0]
∴ [L1M0T-1] = [L-a-3bMa+bT-2a]
Comparing the powers of L, M and T on both sides, we get,
-2a = -1
∴ a = \(\frac{1}{2}\)
Also, a + b = O
∴ \(\frac{1}{2}\) + b = 0 b = – \(\frac{1}{2}\)
Substituting values of a and b in equation (1), we get
y = k P\(\frac{1}{2}\) ρ–\(\frac{1}{2}\)
∴ v = k \(\sqrt{\frac{\mathrm{p}}{\rho}}\)

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 41.
Density of oil is 0.8 g cm3 in C.G.S. unit. Find its value in S.I. units.
Solution:
Dimensions of density is [L-3M1T0]
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 21
= 0.8 [10-3] [10-2]-3
= 0.8 [10-3] [10]6
n = 0.8 × 103
Substituting the value of ‘n’ in equation (1).
we get, 0.8 g cm3 = 0.8 × 103 kg m-3.
Density of oil in S.l unit is 0.8 × 103 kg m-3.

Question 42.
The value of G in C.G.S system is 6.67 × 10-8 dyne cm2 g-2. Calculate its value in S.l. system.
Solution:
Dimensional formula of gravitational constant
[L3M-1T-2]
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 22
n = 6.67 × 10-8 × 10-6 × 103
n = 6.67 × 10-11
From equation (1),
6.67 × 10-8 dyne cm2 g-2
= 6.67 × 10-11 N-m2 kg-2
Value ofG in S.l. system is 6.67 × 10-11 N-m2 kg-2.

Question 43.
What is accuracy?
Answer:
Accuracy is how close a measurement is to the actual value of that quantity.

Question 44.
What is precision?
Answer:
Precision is a measure of how consistently a device records nearly identical values i.e., reproducible results.

Question 45.
A scale in a lab measures the mass of object consistently more by 500 g than their actual mass. How would you describe the scale in terms of accuracy and precision?
Answer:
The scale is precise but not accurate.
Explanation: Precision measures how consistently a device records the same answer; even though it displays the wrong value. Hence, the scale is precise.

Accuracy is how well a device measures something against its accepted value. As scale in the lab is always off by 500 g, it is not accurate.
[Note: The goal of the observer should be to get accurate as well as precise measurements.]

Question 46.
List reasons that may introduce possible uncertainties in an observation.
Answer:
Possible uncertainties in an observation may arise due to following reasons:

  1. Quality of instrument used,
  2. Skill of the person doing the experiment,
  3. The method used for measurement,
  4. External or internal factors affecting the result of the experiment.

Question 47.
What is systematic error? Classify errors into different categories.
Answer:

  1. Systematic errors are errors that are not determined by chance but are introduced by an inaccuracy (involving either the observation or measurement process) inherent to the system.
  2. Classification of errors:
    Errors are classified into following two groups:
  3. Systematic errors:
    • Instrumental error (constant error),
    • Error due to imperfection in experimental technique,
    • Personal error (human error).
  4. Random error (accidental error)

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 48.
What is instrumental (constant) error?
Answer:
Instrumental error:

  1. It arises due to defective calibration of an instrument.
  2. Example: If a thermometer is not graduated properly, i.e., one degree on the thermometer actually corresponds to 0.99°, the temperature measured by such a thermometer will differ from its value by a constant amount.

Question 49.
What is error due to imperfection in experimental technique?
Answer:
Error due to imperfection in experimental technique:

  • The errors which occur due to defective setting of an instrument is called error due to imperfection in experimental technique.
  • For example the measured volume of a liquid in a graduated tube will be inaccurate if the tube is not held vertical.

Question 50.
What is personal error?
Answer:
Personal error (Human error):

  • The errors introduced due to fault of an observer taking readings are called personal errors.
  • For example, while measuring the length of an object with a ruler, it is necessary to look at the ruler from directly above. If the observer looks at it from an angle, the measured length will be wrong due to parallax.

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 51.
What is random error (accidental)?
Answer:

  1. Random error (accidental):
    The errors which are caused due to minute change in experimental conditions like temperature, pressure change in gas or fluctuation in voltage, while the experiment is being performed are called random errors.
  2. They can be positive or negative.
  3. Random error cannot be eliminated completely but can be minimized by taking multiple observations and calculating their mean.

Question 52.
State general methods to minimise effect of systematic errors.
Answer:
Methods to minimise effect of systematic errors:

  1. By using correct instrument.
  2. Following proper experimental procedure.
  3. Removing personal error.

Question 53.
Define the term:
Arithmetic mean
Answer:
Arithmetic mean:
a. The most probable value of a large number of readings of a quantity is called the arithmetic mean value of the quantity. This value can be considered to be true value of the quantity.

b. If a1, a2, a3, …………… an are ‘n’ number of readings taken for measurement of a quantity, then their mean value is given by,
amean = \(\frac{a_{1}+a_{2}+\ldots \ldots .+a_{n}}{n}\)
∴ amean = \(\frac{1}{n} \sum_{i=1}^{n} a_{i}\)

Question 54.
What does a = amean ± ∆amean signify?
Answer:
a = amean ± ∆ amean signifies that the actual value of a lies between (amean – ∆ amean) and (amean + ∆ amean).

Question 55.
What is meant by the term combination of errors?
Answer:
Derived quantities may get errors due to individual errors of fundamental quantities, such type of errors are called as combined errors.

Question 56.
Explain errors in sum and in difference of measured quantity.
Answer:
Errors in sum and in difference:
i) Suppose two physical quantities A and B have measured values A ± ∆A and B ± ∆B. respectively, where ∆A and ∆B are their mean absolute errors.

ii) Then, the absolute error ∆Z in their sum.
Z = A + B
Z ± ∆Z = (A ± ∆A) + (B ± ∆B)
= (A + B) ± ∆A ± ∆B
∴ ± ∆Z = ± ∆A ± ∆B.

iii) For difference. i.e.. if Z = A – B.
Z ± ∆Z = A ± ∆A) – (B ± ∆B)
= (A – B) ± ∆A ∓ ∆B
∴ ± ∆Z = ± ∆A ∓ ∆B,

iv) There are four possible values for ∆Z. namely (+∆A – ∆B), (+∆A + ∆B), (-∆A -∆B), (-∆A + ∆B). Hence, maximLim value of absolute error is ∆Z = (∆A+ ∆B) in both the cases.

v) Thus. when two quantities are added or subtracted, the absolute error in the final result is the sum of the absolute errors in the individual quantities.

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 57.
Explain errors in product of measured quantity.
Answer:
Errors in product:
i) Suppose Z = AB and measured values of A and B are (A ± ∆A) and (B ± ∆B) then,
Z ± ∆Z = (A ± ∆A) (B ± ∆B)
= AB ± A∆B ± B∆A ± ∆A∆B
Dividing L.H.S by Z and R.H.S. by AB we get
\(\left(1 \pm \frac{\Delta Z}{Z}\right)=\left[1 \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \pm\left(\frac{\Delta A}{A}\right)\left(\frac{\Delta B}{B}\right)\right]\)
Since ∆A/A and ∆B/B are very small, product is neglected. Hence, maximum relative error in Z is \(\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}\)

ii) Thus, when two quantities are multiplied, the maximum relative error in the result is the sum of relative errors in each quantity.

Question 58.
Explain errors due to power (index) of measured quantity.
Answer:
Errors due to the power (index) of measured quantity:

  1. Suppose
    Z = A3 = A × A × A
    then, \(\)
  2. Hence the relative error in Z = A3 is three times the relative error in A.
  3. This means if Z = An
    Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 23
  4. This implies, the quantity in the formula which has large power is responsible for maximum error.

Question 59.
The radius of a sphere measured repeatedly yields values 5.63 m, 5.54 m, 5.44 m, 5.40 m and 5.35 m. Determine the most probable value of radius and the mean absolute, relative and percentage errors.
Solution:
Given: a1 = 5.63 m, a2 = 5.54 m, a3 = 5.44 m
a4 = 5.40 m, a5 = 5.35 m,
To find:
i) Most probable value (Mean value)
ii) Mean absolute error
iii) Relative error
iv) Percentage error
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 24
From formula (ii),
Absolute errors:
∆a1 = |amean – a1| = |5.472 – 5.63| = 0.158
∆a2 = |amean – a2| = |5.472 – 5.54| = 0.068
∆a3 = |amean – a3| = |5.472 – 5.44| = 0.032
∆a4 = |amean – a4| = |5.472 – 5.40| = 0.072
∆a5 = |amean – a5| = |5.472 – 5.35| = 0.122

From formula (ii),
∆amean = \(\frac{0.158+0.068+0.032+0.072+0.122}{5}\)
= \(\frac{0.452}{5}\)
= 0.0904 m
From formula (iii),
Relative error = \(\frac{0.0904}{5.472}\)
= 1.652 × 10-2
(after rounding off to correct significant digits)
= 1.66 × 10-2
= 0.0166
∴ Percentage error = 1.66 × 10-2 × 100 = 1.66%
i) The mean value is 5.472 m.
ii) The mean absolute error is 0.0904 m.
iii) The relative error is 0.0166.
iv) The percentage error is 1.66%
[Note: Answer to relative error is rounded off using rules of significant figures and of rounding off]

Question 60.
Lin an experiment to determine the volume of an object, mass and density are recorded as m = (5 ± 0.15) kg and p = (5 ± 0.2) kg m3 respectively. Calculate percentage error in the measurement of volume.
Solulion:
Given: M = 5kg, ∆M = 0.15 kg, ρ = 5 kg/m3,
∆ρ = 0.2 kg/m3
To find: Percentage error in volume (V)
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 25
The percentage error in the determination of volume is 7%.

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 61.
The acceleration due to gravity is determined by using a simple pendulum of length l = (100 ± 0.1) cm. If its time period is T = (2 ± 0.01) s, find the maximum percentage error in the measurement of g.
Solution:
Given: ∆l = 0.1 cm, l = 100 cm, ∆T = 0.01 s,
T = 2s
To find: Percentage error
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 26
Percentage error in measurement of g is 1.1 %.

Question 62.
Find the number of significant figures in the following numbers,
i. 25.42
ii. 0.004567
iii. 35.320
iv. 91.000
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 27

Solved Examples

Question 63.
Add 7.21, 12.141 and 0.0028 and express the result to an appropriate number of significant figures.
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 28
In the given problem, minimum number of digits after decimal is 2.
∴ Result will be rounded off upto two places of decimal.
Corrected rounded off sum is 19.35.

Question 64.
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (i) the total mass of the box? (ii) the difference in the masses of the pieces to correct significant figures?
Solution:
i) Total mass of the box
= (2.3+ 0.02017 + 0.02015) kg
= 2.34032 kg
Since, the last number of significant figure is 2, therefore, the total mass of the box = 2.3 kg

ii) Difference of mass = (20.17 – 20.15) = 0.02g Since, there are two significant figures so the difference in masses to the correct significant figures is 0.02 g.

i) The total mass of the box to correct significant figures is 2.3 kg.
ii) The difference in the masses to correct significant figures is 0.02 g.

Apply Your Knowledge

Question 65.
Write the dimensions of a and b in the relation
E = \(\frac{b-x^{2}}{a}\)
Where E is energy, x ¡s distance and t is time.
Answer:
The given relation is E = \(\frac{b-x^{2}}{a}\)
As x is subtracted from b,
∴ dimensions of b are x2;
i.e., b = [L2]
∴ We can write equation as E = \(\frac{\mathrm{L}^{2}}{\mathrm{a}}\)
Or a = \(\frac{\mathrm{L}^{2}}{\mathrm{E}}=\frac{\mathrm{L}^{2}}{\left[\mathrm{~L}^{2} \mathrm{MT}^{-2}\right]}\) = [L0M-1T2]

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 66.
What is the difference between 6.0 and 6.00? which Is more accurate?
Answer:
6.0 indicates the measurement is correct up to first decimal place, whereas 6.00 indicates that the measurement is correct up to second decimal place. Thus, 6.00 is a more accurate value than 6.0.

Question 67.
A child walking on a footpath notices that the width of the footpath is uneven. He reported this to his school principal and the complaint was forwarded to the municipal officer.
i. What is the possible error encountered?
ii. What is the relative error in width of footpath if width of footpath in 10 m length are noted as 5 m, 5.5 m, 5 m, 6 m and 4.5 m?
Answer:
i) The error encountered is personal error.

ii) Mean value of widths
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 29
The relative error in width of footpath is 0.084.

Question 68.
A factory owner kept five identical spheres between two wooden blocks on a ruler as shown in figure. He called all his workers and told them to take reading, to check their efficiency and knowledge.
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 30
i. What is the area of central sphere?
ii. What is the absolute error in reading of diameter of second sphere?
Answer:
i) From above diagram radius of central sphere is
r = 1 cm
∴ Area = πr2 = 3.142 × (1)2= 3.142 cm2
The area of central sphere is 3.142 cm2.

ii) Mean value of all reading of diameters
dmean = \(\frac{\mathrm{d}_{1}+\mathrm{d}_{2}+\mathrm{d}_{3}+\mathrm{d}_{4}+\mathrm{d}_{5}}{5}=\frac{2+2+2+2+2}{5}\)
= \(\frac{10}{5}\) = 2 cm
Absolute error in reading of second sphere.
∆d2 = |dmean – d2| = 2 – 2 = 0
The absolute error in reading of diameter of second sphere is zero.

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 69.
A potential difference of V = 100 ± 2 volt, when applied across a resistance R gives a current of 10 ± 0.5 ampere. Calculate percentage error in R given by R V/I.
Answer:
Here. V = 100 ± 2 volt and I = 10 ± 0.5 ampere
Expressing limits of error as percentage error,
We have
V = 100 volt ± \(\frac{2}{100}\) × 100% = 10 volt ± 2%
and I = 10 ampere ± \(\frac{0.5}{10}\) × 100%
= 10 ampere ± 5%
∴ R = \(\frac{V}{I}\)
∴ %error in R = %error in V + %error in I
= 2% + 5% = 7%

Quick Review

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 31
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 32

Multiple Choice Questions

Question 1.
A physical quantity may be defined as
(A) the one having dimension.
(B) that which is immeasurable.
(C) that which has weight.
(D) that which has mass.
Answer:
(A) the one having dimension.

Question 2.
Which of the following is the fundamental unit?
(A) Length, force, time
(B) Length, mass, time
(C) Mass, volume, height
(D) Mass, velocity, pressure
Answer:
(B) Length, mass, time

Question 3.
Which of the following is NOT a fundamental quantity?
(A) Temperature
(B) Electric charge
(C) Mass
(D) Electric current
Answer:
(B) Electric charge

Question 4.
The distance of the planet from the earth is measured by __________.
(A) direct method
(B) directly by metre scale
(C) spherometer method
(D) parallax method
Answer:
(D) parallax method

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 5.
The two stars S1 and S2 are located at distances d1 and d2 respectively. Also if d1 > d22 then following statement is true.
(A) The parallax of S1 and S2 are same.
(B) The parallax of S1 is twice as that of S2
(C) The parallax of S1 is greater than parallax of S2
(D) The parallax of S2 is greater than parallax of S1
Answer:
(D) The parallax of S2 is greater than parallax of S1

Question 6.
Which of the following is NOT a unit of time?
(A) Hour
(B) Nano second
(C) Microsecond
(D) parsec
Answer:
(D) parsec

Question 7.
An atomic clock makes use of _________.
(A) cesium-133 atom
(B) cesium-132 atom
(C) cesium-123 atom
(D) cesium-131 atom
Answer:
(A) cesium-133 atom

Question 8.
S.I. unit of energy is joule and it is equivalent to
(A) 106 erg
(B) 10-7 erg
(C) 107 erg
(D) 105 erg
Answer:
(C) 107 erg

Question 9.
[L1M1T-1] is an expression for __________.
(A) force
(B) energy
(C) pressure
(D) momentum
Answer:
(D) momentum

Question 10.
Dimensions of sin θ is
(A) [L2]
(B) [M]
(C) [ML]
(D) [M0L0T0]
Answer:
(D) [M0L0T0]

Question 11.
Accuracy of measurement is determined by
(A) absolute error
(B) percentage error
(C) human error
(D) personal error
Answer:
(B) percentage error

Question 12.
Zero error of an instrument introduces .
(A) systematic error
(B) random error
(C) personal error
(D) decimal error
Answer:
(A) systematic error

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 13.
The diameter of the paper pin is measured accurately by using ________.
(A) Vernier callipers
(B) micrometer screw gauge
(C) metre scale
(D) a measuring tape
Answer:
(B) micrometer screw gauge

Question 14.
The number of significant figures in 11.118 × 10-6 is
(A) 3
(B) 4
(C) 5
(D) 6
Answer:
(C) 5

Question 15.
0.00849 contains ___________ significant figures.
(A) 6
(B) 5
(C) 3
(D) 2
Answer:
(C) 3

Question 16.
3.310 × 102 has ___________ significant figures.
(A) 6
(B) 4
(C) 2
(D) 1
Answer:
(B) 4

Question 17.
The Earth’s radius is 6371 km. The order of magnitude of the Earth’s radius is
(A) 103 m
(B) 109 m
(C) 107 m
(D) 102 m
Answer:
(C) 107 m

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 18.
__________ is the smallest measurement that can be made using the given instrument
(A) Significant number
(B) Least count
(C) Order of magnitude
(D) Relative error
Answer:
(B) Least count

Competitive Corner

Question 1.
In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement X,
where X = \(\frac{A^{2} \frac{1}{B^{2}}}{C^{\frac{1}{3}} D^{3}}\), will be:
(A) -10 %
(B) 10 %
(C) \(\left(\frac{3}{13}\right) \%\)
(D) 16 %
Answer:
(D) 16 %
Hint:
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 33
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 34
∴ Percentage error in x is given as,
\(\frac{\Delta x}{x}\) × 100 – (error contributed by A) – (error contributed by B) + (error contributed by C) + (error contributed by D)
= 2% + 1% + 1% + 12%
= 16%

Question 2.
The main scale of a vernier callipers has n divisions/cm. n divisions of the vernier scale coincide with (n – 1) divisions of main scale. The least count of the vernier callipers is,
(A) \(\frac{1}{n(n+1)}\) cm
(B) \(\frac{1}{(n+1)(n-1)}\) cm
(C) \(\frac{1}{n}\) cm
(D) \(\frac{1}{n^{2}}\) cm
Answer:
(D) \(\frac{1}{n^{2}}\) cm
Hint:
1 V.S.D. = \(\frac{(n-1)}{n}\) M.S.D.
LC. = 1 M.S.D. – 1 V.S.D.
= 1 M.S.D. – \(\frac{(n-1)}{n}\) M.S.D.
= \(\frac{1}{n}\) M.S.D.
= \(\frac{1}{n}\) × \(\frac{1}{n}\) cm
∴ L.C. = \(\frac{1}{n^{2}}\) cm

Question 3.
A student measures time for 20 oscillations of a simple pendulum as 30 s. 32 s, 35 s and 31 s. 1f the minimum division in the measuring clock is I s, then correct mean time in second is
(A) 32 ± 3
(B) 32 ± 1
(C) 32 ± 2
(D) 32 ± 5
Answer:
(C) 32 ± 2
Hint:
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 35
Hence rounding off,
∆t = ± 2 s
∴ t ± ∆t = 32 ± 2 s

Question 4.
A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference leveL If screw gauge has a zero error of— 0.004 cm, the correct diameter of the ball is
(A) 0.521 cm
(B) 0.525 cm
(C) 0.053 cm
(D) 0.29 cm
Answer:
(D) 0.29 cm

Hint:
Least count of screw gauge = 0.001 cm = 0.01mm
Main scale reading = 5 mm.
Zero error = – 0.004 cm = -0.04 mm
Zero correction = +0.04 mm
Observed reading = Mainscale reading + (Division × least count)
Observed reading = 5 + (25 × 0.01) = 5.25 mm
Corrected reading = Observed reading + Zero correction
Corrected reading = 5.25 + 0.04
= 5.29 mm = 0.529 cm

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 5.
The density of the material in the shape of a cube is determined by measuring three sides of the cube and its mass. 1f the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is:
(A) 4.5%
(B) 6%
(C) 2.5°
(D) 3.5%
Answer:
(A) 4.5%
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 36

Question 6.
Let x = \(\left[\frac{a^{2} b^{2}}{c}\right]\) be the physical quantity. If the percentage error in the measurement of physical quantities a, b and c is 2, 3 and 4 percent respectively then percentage en-or in the measurement of x is
(A) 7%
(B) 14%
(C) 21%
(D) 28%
Answer:
(B) 14%
Hint:
Given: x = \(\frac{a^{2} b^{2}}{c}\)
Percentage error is given by.
\(\frac{\Delta x}{x}=\frac{2 \Delta a}{a}+\frac{2 \Delta b}{b}+\frac{\Delta c}{c}\)
= (2 × 2) + (2 × 3) + 4
= 4 + 6 + 4 = 14
∴ \(\frac{\Delta \mathrm{x}}{\mathrm{x}} \%\) = 14%

Question 7.
A physical quantity of the dimensions of length that can be formed out of c, G and \(\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\) is [c is velocity of light, G is universal constant of gravitation and e is charge]:
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 37
Answer:
(A) \(\frac{1}{\mathrm{c}^{2}}\left[\mathrm{G} \frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\right]^{1 / 2}\)

Hint:
Let the physical quantity formed of the dimensions of length be given as.
[L] = [c]x [G]y \(\left[\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\right]^{z}\) …………….. (i)
Now,
Dimensions of velocity of light [c]x = [LT-1]x
Dimensions of universal gravitational constant
[G]y = [L3T2M-1]y
Dimensions of \(\left[\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\right]^{z}\) = [ML3T-2]z
Substitrning these in equation (i)
[L] [LT-1]x [M-1L3T-2]y [ML3T-2]z
= Lx+3y+3z M-y+z T-x-2y-2z
Solving for x, y, z
x + 3y + 3z = 1
-y + z = 0
x + 2y + 2z = O
Solving the above equation,
x = -2, y = \(\frac{1}{2}\), z = \(\frac{1}{2}\)
∴ L = \(\frac{1}{c^{2}}\left[\mathrm{G} \frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\right]^{1 / 2}\)

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

Question 8.
The following observations were taken for determining surface tension T of water by capillary method:
diameter of capillary, D = 1.25 × 10-2 m
rise of water, h = 1.45 × 10-2 m
Using g = 9.80 m/s2 and the simplified relation
T = \(\frac{\mathrm{rhg}}{2}\) × 103 N/m, the possible error in surface tension is closest to:
(A) 0.15%
(B) 1.5%
(C) 2.4%
(D) 10%
Answer:
(B) 1.5%
Hint:
D = 1.25 × 10-2 m; h = 1.45 × 10-2 m
The maximum permissible error in D
= ∆D = 0.01 × 10-2 m
The maximum permissible error in h
= ∆h = 0.01 × 10-2 m
g is given as a constant and is errorless.
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 38

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 8 Respiration and Circulation Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 8 Respiration and Circulation

Multiple-choice Questions

Question 1.
The nasal cavity is divided into right and left nasal chambers by a …………………..
(a) sphenoid
(b) palatine
(c) mesethmoid
(d) zygomatic
Answer:
(c) mesethmoid

Question 2.
The right lung is divided into …………………..
(a) 3 lobes
(b) 2 lobes
(c) 4 lobes
(d) 6 lobes
Answer:
(a) 3 lobes

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 3.
Carbon dioxide is carried in the blood mainly as …………………..
(a) sodium carbonate
(b) sodium bicarbonate
(c) carbaminohaemoglobin
(d) carbonic acid
Answer:
(b) sodium bicarbonate

Question 4.
Transport of oxygen is carried out by …………………..
(a) plasma
(b) lungs
(c) RBCs
(d) nostrils
Answer:
(c) RBCs

Question 5.
Respiration taking place at the alveoli of lungs is called …………………..
(a) internal respiration
(b) external respiration
(c) cellular respiration
(d) tissue respiration
Answer:
(b) external respiration

Question 6.
The volume of air inspired or expired during normal breathing is …………………..
(a) ERV
(b) IRV
(c) TV
(d) VC
Answer:
(c) TV

Question 7.
What is the partial pressure of oxygen and carbon dioxide respectively in the atmospheric air?
(a) PPO2 159 mmHg, PPCO22 0.3 mmHg
(b) PPO2 104 mmHg, PPCO2 40 mmHg
(c) PPO2 40 mmHg, PPCO2 45 mmHg
(d) PPO2 95 mmHg, PPCO2 40 mmHg
Answer:
(b) PPO2 104 mmHg, PPCO2 40 mmHg

Question 8.
The vital capacity of human lung is equal to …………………..
(a) 3500 ml
(b) 4600 ml
(c) 500 ml
(d) 1200 ml
Answer:
(b) 4600 ml

Question 9.
The exchange of gases between alveolar air and alveolar capillaries occurs by …………………..
(a) osmosis
(b) active transport
(c) absorption
(d) diffusion
Answer:
(d) diffusion

Question 10.
Oxygen dissociation curve will shift to right on the decrease of …………………..
(a) acidity
(b) carbon dioxide concentration
(c) temperature
(d) pH
Answer:
(d) pH

Question 11.
Respiratory organs in scorpion are …………………..
(a) gills
(b) book lungs
(c) skin
(d) book gills
Answer:
(b) book lungs

Question 12.
Breakdown of alveoli of lungs resulting in reducing surface area for gas exchange is known as …………………..
(a) emphysema
(b) sneezing
(c) pneumonia
(d) tuberculosis
Answer:
(a) emphysema

Question 13.
During inspiration, the diaphragm …………………..
(a) relaxes
(b) contracts
(c) expands
(d) shows no change
Answer:
(b) contracts

Question 14.
Over inflation of the lungs is prevented due to …………………..
(a) Bohr’s effect
(b) Conditioned reflex
(c) Hering-Breuer reflex
(d) Haldane effect
Answer:
(c) Hering-Breuer reflex

Question 15.
Which of the following prevents collapsing of trachea?
(a) Muscles
(b) Diaphragm
(c) Ribs
(d) Cartilaginous rings
Answer:
(d) cartilaginous rings

Question 16.
Which one of the following produces antibodies ?
(a) Monocytes
(b) Erythrocytes
(c) Lymphocytes
(d) Monocytes
Answer:
(c) Lymphocytes

Question 17.
Plasma protein which initiate blood coagulation is …………………..
(a) prothrombin
(b) fibrinogen
(c) thrombin
(d) fibrin
Answer:
(a) prothrombin

Question 18.
The covering of heart is …………………..
(a) perichondrium
(b) pericardium
(c) periosteum
(d) peritoneum
Answer:
(b) pericardium

Question 19.
Left atrioventricular aperture is guarded by …………………..
(a) tricuspid valve
(b) Eustachian valve
(c) bicuspid valve
(d) semilunar valve
Answer:
(c) bicuspid valve

Question 20.
The pulmonary trunk and systemic aorta are joined by …………………..
(a) chordae tendinae
(b) columnae carnae
(c) ligamentum arteriosum
(d) Purkinje fibres
Answer:
(c) ligamentum arteriosum

Question 21.
Atrioventricular node is located in …………………..
(a) left atrium
(b) right atrium
(c) left ventricle
(d) right ventricle
Answer:
(b) right atrium

Question 22.
…………………. is most commonly used to feel pulse.
(a) Radial vein
(b) Brachial artery
(c) Brachial vein
(d) Radial artery
Answer:
(d) Radial artery

Question 23.
QRS is related to …………………..
(a) atrial contraction
(b) ventricular contraction
(c) atrial relaxation
(d) ventricular relaxation
Answer:
(b) ventricular contraction

Question 24.
Blood is a fluid connective tissue derived from …………………..
(a) ectoderm
(b) mesoderm
(c) endoderm
(d) epithelium
Answer:
(b) mesoderm

Question 25.
What is the increase in number of RBCs called?
(a) Erythropoiesis
(b) Polycythaemia
(c) Erythrocytopenia
(d) Erythroblastosis
Answer:
(b) Polycythaemia

Question 26.
What is the increase in the number of WBCs called?
(a) Leucopoiesis
(b) Leukopenia
(c) Leucocytosis
(d) Leukaemia
Answer:
(c) Leucocytosis

Question 27.
In which of the following diseases there is uncontrolled increase in number of WBCs ?
(a) Leucopoiesis
(b) Leukopenia
(c) Leucocytosis
(d) Leukaemia
Answer:
(d) Leukaemia

Question 28.
What is the decrease in the number of WBCs called?
(a) Leucopoiesis
(b) Leukopenia
(c) Leucocytosis
(d) Leukaemia
Answer:
(b) Leukopenia

Question 29.
Which is the correct arrangement of types of WBCs with respect to their number in blood?
(Consider Neutrophil = N, Eosinophil = E, Basophil = B, Monocyte = M and Lymphocyte = L)
(a) NLMEB
(b) BEMLN
(c) NEBLM
(d) MEBLN
Answer:
(a) NLMEB

Question 30.
Which is the correct order in which the proteins participate in clotting of blood?
(a) Prothrombinase → Prothrombin → Thromboplastin → Thrombin
(b) Thromboplastin → Prothrombinase → Prothrombin → Thrombin
(c) Prothrombin → Thromboplastin → Thrombin → Prothrombinase
(d) Thrombin → Prothrombin → Thromboplastin → Prothrombinase
Answer:
(b) Thromboplastin → Prothrombinase → Prothrombin → Thrombin

Question 31.
Decrease in platelet count is called …………………..
(a) thrombocytopenia
(b) thrombocytosis
(c) thrombokinase
(d) thromboplastin
Answer:
(a) thrombocytopenia

Question 32.
Atrioventricular groove is also called a …………………..
(a) foramen ovale
(b) ligamentum arteriosum
(c) coronary sulcus
(d) ductus arteriosus
Answer:
(c) coronary sulcus

Question 33.
The coronary sinus opens into the …………………..
(a) left atrium
(b) right atrium
(c) left ventricle
(d) right ventricle
Answer:
(b) right atrium

Question 34.
Name the valve from the following that guards the opening of inferior vena cava.
(a) Tricuspid valve
(b) Semilunar valve
(c) Eustachian valve
(d) Thebesian valve
Answer:
(c) Eustachian valve

Question 35.
Name the valve from the following guarding the opening of coronary sinus …………………..
(a) Thebesian valve
(b) Eustachian valve
(c) Tricuspid valve
(d) Semilunar valve
Answer:
(a) Thebesian valve

Question 36.
What is an oval aperture in the interatrial septum of the foetus called?
(a) Fossa ovalis
(b) Foramen ovalis
(c) Ligamentum arteriosum
(d) Ductus arteriosus
Answer:
(b) Foramen ovalis

Question 37.
What is the meaning of stroke volume?
(a) Amount of blood in the body
(b) Pressure of contraction of heart
(c) Amount of blood put out of the ventricles in one minute
(d) Amount of blood put out of the ventricles in one beat
Answer:
(d) Amount of blood put out of the ventricles in one beat

Question 38.
How much amount of blood is put out of the heart during one minute?
(a) Equal to cardiac output
(b) Equal to stroke volume
(c) Equal to half of blood volume
(d) Equal to quarter of blood volume
Answer:
(a) Equal to cardiac output

Question 39.
What is the time taken for one cardiac cycle of normal human being?
(a) 0.1 second
(b) 0.3 second
(c) 0.4 second
(d) 0.8 second
Answer:
(d) 0 .8 second

Question 40.
Deposition of fatty substances in the lining of arteries results in …………………..
(a) arteriosclerosis
(b) atherosclerosis
(c) hyperglycemia
(d) hypotension
Answer:
(b) atherosclerosis

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 41.
Largest number of white blood corpuscles are …………………..
(a) eosinophils
(b) basophils
(c) neutrophils
(d) monocytes
Answer:
(c) neutrophils

Question 42.
Which of the following animal have open circulatory system?
(a) Earthworm
(b) Cockroach
(c) Frog
(d) Rabbit
Answer:
(b) Cockroach

Question 43.
Which of the following leucocytes have unlobed nucleus?
(a) lymphocyte
(b) eosinophils
(c) neutrophils
(d) basophils
Answer:
(a) lymphocyte

Question 44.
Carbonic anhydrase is found in …………………..
(a) WBC
(b) RBCs
(c) thrombocytes
(d) blood plasma
Answer:
(b) RBCs

Question 45.
The typical Lubb – Dup sounds heard in the heart of a healthy person are due to …………………..
(a) closing of cuspid valves followed by the closing of the semilunar valves
(b) closing of semilunar valves
(c) closing of tricuspid valves
(d) closing of bicuspid valves
Answer:
(a) closing of cuspid valves followed by the closing of the semilunar valves

Match the columns

Question 1.

Animal Respiratory organ
(1) Fishes (a) Trachea
(2) Birds/Reptiles (b) Moist cuticle
(3) Insects (c) Gills
(4) Earthworm (d) Lungs

Answer:

Animal Respiratory organ
(1) Fishes (c) Gills
(2) Birds/Reptiles (d) Lungs
(3) Insects (a) Trachea
(4) Earthworm (b) Moist cuticle

Question 2.

Respiratory organs Alternative name
(1) Larynx (a) Lid of larynx
(2) Trachea (b) Air sacs
(3) Alveoli (c) Sound box
(4) Epiglottis (d) Windpipe

Answer:

Respiratory organs Alternative name
(1) Larynx (c) Sound box
(2) Trachea (d) Windpipe
(3) Alveoli (b) Air sacs
(4) Epiglottis (a) Lid of larynx

Question 3.

Respiratory capacities Respiratory volumes
(1) Residual volume (RV) (a) 500 ml
(2) Vital capacity (VC) (b) 2000 – 3000 ml
(3) Tidal volume (TV) (c) 1100 – 1200 ml
(4) Inspiratory reserve volume (IRV) (d) 4100 – 4600 ml

Answer:

Respiratory capacities Respiratory volumes
(1) Residual volume (RV) (c) 1100 – 1200 ml
(2) Vital capacity (VC) (d) 4100 – 4600 ml
(3) Tidal volume (TV) (a) 500 ml
(4) Inspiratory reserve volume (IRV) (b) 2000 – 3000 ml

Question 4.

Disease Symptoms
(1) Asthma (a) Fully blown out alveoli
(2) Bronchitis (b) Inflammation of lungs with cough and fever
(3) Emphysema (c) Spasm of Bronchial muscles
(4) Pneumonia (d) Inflammation of bronchi

Answer:

Disease Symptoms
(1) Asthma (c) Spasm of Bronchial muscles
(2) Bronchitis (d) Inflammation of bronchi
(3) Emphysema (a) Fully blown out alveoli
(4) Pneumonia (b) Inflammation of lungs with cough and fever

Question 5.

Valves in heart Location
(1) Bicuspid/Mitral valve (a) Opening of inferior vena cava
(2) Tricuspid valve (b) Opening of coronary sinus
(3) Eustachian valve (c) Left atrioventricular aperture
(4) Thebesian valve (d) Right atrioventricular aperture

Answer:

Valves in heart Location
(1) Bicuspid/Mitral valve (c) Left atrioventricular aperture
(2) Tricuspid valve (d) Right atrioventricular aperture
(3) Eustachian valve (a) Opening of inferior vena cava
(4) Thebesian valve (b) Opening of coronary sinus

Question 6.

Blood vessel Functions
(1) Pulmonary aorta (a) Carries oxygenated blood to left atrium
(2) Superior vena cava (b) Carries oxygenated blood to all body parts
(3) Pulmonary vein (c) Carries deoxygenated blood from upper parts of body to right atrium
(4) Aorta (d) Carries deoxygenated blood to lungs

Answer:

Blood vessel Functions
(1) Pulmonary aorta (d) Carries deoxygenated blood to lungs
(2) Superior vena cava (c) Carries deoxygenated blood from upper parts of body to right atrium
(3) Pulmonary vein (a) Carries oxygenated blood to left atrium
(4) Aorta (b) Carries oxygenated blood to all body parts

Question 7.

Cells Functions
(1) T-lymphocytes (a) Phagocytic in function
(2) Neutrophils (b) Responsible for Humoral immunity
(3) Eosinophils/Acidophils (c) Responsible for cell-medicated immunity
(4) B-lymphocytes (d) Anti-allergic [Antihistamine] in function

Answer:

Cells Functions
(1) T-lymphocytes (c) Responsible for cell-medicated immunity
(2) Neutrophils (a) Phagocytic in function
(3) Eosinophils/Acidophils (d) Anti-allergic [Antihistamine] in function
(4) B-lymphocytes (b) Responsible for Humoral immunity

Question 8.

Waves recorded in ECG Heart activity
(1) P wave (a) Ventricular repolarization
(2) QRS complex (b) Atrial depolarization
(3) T wave (c) Isoelectric segment
(4) ST segment (d) Ventricular depolarization

Answer:

Waves recorded in ECG Heart activity
(1) P wave (b) Atrial depolarization
(2) QRS complex (d) Ventricular depolarization
(3) T wave (a) Ventricular repolarization
(4) ST segment (c) Isoelectric segment

Question 9.

Events in cardiac cycle Time duration
(1) Atrial systole (a) 0.3 second
(2) Atrial diastole (b) 0.5 second
(3) Ventricular systole (c) 0.1 second
(4) Ventricular diastole (d) 0.7 second

Answer:

Events in cardiac cycle Time duration
(1) Atrial systole (c) 0.1 second
(2) Atrial diastole (d) 0.7 second
(3) Ventricular systole (a) 0.3 second
(4) Ventricular diastole (b) 0.5 second

Classify the following to form Column B as per the category given in Column A

Question 1.
Classify the following composition of blood plasma given below as per Column ‘A’ and complete Column ‘B’. Select from the given options
(i) Serum albumin
(ii) Bicarbonates
(iii) Urea
(iv) Sulphates of sodium
(v) Fibrinogen
(vi) Uric acid

Column A Column B
(1) Plasma proteins ————
(2) Nitrogenous waste ————
(3) Inorganic salts ————

Answer:

Column A Column B
(1) Plasma proteins Serum albumin Fibrinogen
(2) Nitrogenous waste Urea, Uric acid
(3) Inorganic salts Bicarbonates, Sulphates of sodium

Question 2.
Classify the following animals having different respiratory organs given below as per Column ‘A’ and complete Column ‘B’.
Select from the given options:
(i) Scorpion
(ii) Reptiles
(iii) Amphibian tadpoles of frog
(iv) Spiders
(vi) Salamanders
(v) Birds

Column A Column B
(1) External gills ————
(2) Book lungs ————
(3) Lungs ————

Answer:

Column A Column B
(1) External gills Amphibian tadpoles of frog, Salamanders
(2) Book lungs Scorpion, Spiders
(3) Lungs Reptiles, Birds

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 3.
Classify the following disorders of respiratory system given below as per Column A and complete Column ‘B’. Select from the given options:
(i) Pneumonia
(ii) Asbestosis
(iii) Emphysema
(iv) Laryngitis
(v) Chronic bronchitis
(vi) Silicosis

Column A Column B
(1) Occupational disorders ————
(2) Disorders due to smoking and air pollution ————
(3) Disorders due to viruses and bacteria ————

Answer:

Column A Column B
(1) Occupational disorders Asbestosis, Silicosis
(2) Disorders due to smoking and air pollution Emphysema, Chronic bronchitis
(3) Disorders due to viruses and bacteria Pneumonia, Laryngitis

Question 4.
Classify the following white blood corpuscles given below as per Column A and complete Column ‘B’. Select from the given options:
(i) Eosinophils
(ii) T-lymphocytes
(iii) Neutrophils
(iv) Basophils
(v) B-lymphocytes
(vi) Monocytes

Column A Column B
(1) Phagocytic cells ————
(2) Cells involved in giving immune response ————
(3) Cells that increase during allergic and anti-allergic responses ————

Answer:

Column A Column B
(1) Phagocytic cells Neutrophils Monocytes
(2) Cells involved in giving immune response T-lymphocytes-B-lymphocytes
(3) Cells that increase during allergic and anti-allergic responses Eosinophils, Basophils

Very Short Answer Questions

Question 1.
How many molecules of ATP are formed when one molecule of glucose is oxidized?
Answer:
38 molecules of ATP are formed when one molecule of glucose is oxidized.

Question 2.
What are the three regions of nasal chamber?
Answer:
Vestibule, respiratory part and sensory part are the three regions of nasal chamber.

Question 3.
What is meant by respiratory cycle?
Answer:
Alternate inspiration and expiration together make one respiratory cycle.

Question 4.
Why is it dangerous to sleep in a garage where automobiles have running engines?
Answer:
It is dangerous to sleep in a garage where automobiles have running engines because it may cause carbon monoxide poisoning.

Question 5.
In which form major part of CO2 is transported in the blood?
Answer:
CO2 is transported in the blood in the form of sodium and potassium bicarbonates.

Question 6.
Which are the parts of plant that help in the process of gaseous exchange?
Answer:
The parts of plants that help in the process of gaseous exchange are stomata, lenticels, etc.

Question 7.
Which respiratory membranes help in gaseous exchange between the alveolar air and the blood?
Answer:
The layer of squamous epithelium lining the alveolus, basement membrane and a layer of squamous epithelium lining the capillary wall help in gaseous exchange between the alveolar air and the blood.

Question 8.
When will the oxygen dissociation curve shift towards the right?
Answer:
The oxygen dissociation curve will shift towards the right due to increase in H+ concentration, increase in PPCO2 rise in temperature and rise in DPG (2, 3 diphosphoglycerate), formed in RBCs during glycolysis.

Question 9.
What is the action of carbonic anhydrase in the RBCs of blood?
Answer:
In the RBCs, CO2 combines with water in the presence of a Zn containing enzyme, carbonic anhydrase to form carbonic acid. In the presence of carbonic anhydrase carbonic acid immediately dissociates into HCO3and H+ ions leading to large accumulation of HCO3 inside the RBCs.

Question 10.
How much energy is required for the formation of single molecule of ATP ?
Answer:
For the formation of a single molecule of ATP about 7.3 Kcal of energy is required.

Question 11.
What is Hamburger’s phenomena?
Answer:
The diffusion of Chloride ions into the RBCs to main the ionic balance between RBCs and the plasma is called Hamburger’s phenomena or chloride shift.

Question 12.
What is the role of Hering-Breuer reflex in respiration?
Answer:
The Hering-Breuer reflex controls the depth and rhythm of respiration. It also prevents the lungs from inflating to the point of bursting.

Question 13.
How much blood is present in the human body and from which embryonic germ layer is it derived?
Answer:
An average adult has about 4 to 6 litres of blood, which is red coloured fluid connective tissue derived from embryonic mesoderm.

Question 14.
What is the percentage of plasma in the blood and how much water does it contain?
Answer:
There is 55% of plasma in the blood and it contains 90 to 92% water.

Question 15.
What is the average life span of RBCs?
Answer:
RBCs have a life span of about 120 days.

Question 16.
What is normal RBC count and total WBC count?
Answer:
Average RBC count in adult human is 5.1 to 5.8 million per cubic mm and average total WBC count in adult human is 5000 to 9000 per cubic mm.

Question 17.
What is erythropoiesis?
Answer:
The process of formation of Red Blood Cells is called erythropoiesis.

Question 18.
What is increase in the RBC number called?
Answer:
The increase in the number of RBCs is called polycythemia.

Question 19.
What is leucopenia and erythrocytopenia ?
Answer:
The decrease in the number of white blood cells is called leucopenia whereas decrease in the number of red blood cells is called erythrocytopenia.

Question 20.
Where are Eustachian valve and Thebesian valve located?
Answer:
Eustachian valve is present at the opening of inferior vena cava while Thebesian valve is present near the opening of coronary sinus.

Question 21.
What is foramen ovale and how is it related to fossa ovalis?
Answer:
Foramen ovale is an oval opening in the interatrial septum of the foetal heart representing the fossa ovalis which lies as a depression on the right side of interatrial septum.

Question 22.
When is a person described as having hypertension?
Answer:
When the blood pressure values Eire more than 140 mm Hg systolic pressure and more than 90 mm Hg diastolic pressure, then the person is described as having hypertension.

Question 23.
What are the effects of excessive hypertension?
Answer:
Excessive hypertension of values about 220/120 mm Hg can cause blindness, nephritis, stroke or paralysis.

Question 24.
What is the difference between anemia and leukemia?
Answer:
Anemia is disorder caused due to the deficiency of heaemoglobin while leukemia is blood cancer in which there is abnormal increase in the number of white blood cells.

Question 25.
What is the difference between tachycardia and bradycardia?
Answer:
The faster heart rate over 100 beats per minute is called tachycardia, while the slower heart rate below 60 beats per minute is called bradycardia.

Question 26.
What is the difference between chordae tendinae and columnae carnae?
Answer:
Chordae tendinae are chords that connect bicuspid and tricuspid valves with the papillary muscles in ventricles while columnae carnae are series of irregular muscular ridges present on the inner surface of the ventricles.

Question 27.
Which valves prevent the backward flow of blood at the time of ventricular systole?
Answer:
Semilunar valves located at the base of pulmonary artery and systemic aorta prevent the backward flow of blood at the time of ventricular systole.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 28.
What are the time intervals for atrial systole, ventricular systole and joint diastole?
Answer:
Atrial systole is for 0.1 second, ventricular systole is for 0.3 second and joint diastole is for 0.4 second.

Question 29.
In the electrocardiogram shown below, which wave represents ventricular diastole?
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 1
Answer:
‘T’ wave represents ventricular diastole.

Question 30.
Mention the role of pacemaker in human heart.
Answer:
Pacemaker can generate wave of contraction or cardiac impulse for rhythmic contraction of heart.

Question 31.
Which structure in the heart is called pacemaker?
Answer:
Sinuatrial node [S. A. node] in the heart wall is called a pacemaker.

Question 32.
What is electrocardiograph?
Answer:
The instrument which is used to record action potentials generated in the heart muscles is called an electrocardiograph or ECG machine.

Question 33.
What is angina pectoris?
Answer:
Angina pectoris is the pain in the chest caused due to reduction in blood supply to cardiac muscle caused due to narrowed and hardened coronary arteries.

Question 34.
What is pulse pressure?
Answer:
Difference between systolic and diastolic pressure is called pulse pressure which is normally 40 mm Hg.

Question 38.
What would happen if respiration takes place in one single step?
Answer:
If respiration takes place in one single step, then the chemical energy released at once during that step might result in a brief blast of light and heat and may lead to death of the cell. Hence respiration is a step-wise process.

Question 39.
Why do the veins have valves?
Answer:
The veins have valves at regular intervals to prevent backflow of blood as blood flows through veins with low pressure.

Question 40.
What is Bohr effect?
Answer:
Bohr effect is the shift of oxyhaemoglobin dissociation curve due to change in partial pressure of CO in blood.

Question 41.
What is Haldane effect?
Answer:
Decrease of pH of blood, due to increase in the number of H+ ions, HCO3 changes into H2O and CO2 by the presence of oxyhaemoglobin is called Haldane effect.

Give definitions of the following

Question 1.
Respiration
Answer:
It is a biochemical process of oxidation of organic compounds in an orderly manner for the liberation of chemical energy in the form of ATP.

Question 2.
Breathing
Answer:
It is a physical process by which gaseous exchange takes place between the atmosphere and the lungs. It involves inspiration and expiration.

Question 3.
Tidal Volume (TV)
Answer:
It is the volume of un¬ inspired or expired during normal breathing. It is 500 ml.

Question 4.
Inspiratory reserve volume (IRV)
Answer:
The maximum or the extra volume of air that is inspired during forced breathing in addition to TV (2000 to 3000 ml).

Question 5.
Expiratory reserve volume (ERV)
Answer:
The maximum volume of air that is expired during forced breathing after normal expiration. (1000 to 1100 ml).

Question 6.
Dead space (DS)
Answer:
The volume of air that is present in the respiratory tract (from nose to the terminal bronchioles), but not involved in gaseous exchange (150 ml).

Question 7.
Residual volume (RV)
Answer:
The volume of air that remains in the lungs and the dead space even after maximum expiration (1100 to 1200 ml).

Question 8.
Total lung capacity
Answer:
The maximum amount of air that the lungs can hold after a maximum forceful inspiration (5200 to 5900 ml).

Question 9.
Vital capacity (VC)
Answer:
The maximum amount of air that can be breathed out after of maximum inspiration. It is the sum total of TV, IRV and ERV and is 4100 to 4600 ml.

Question 10.
Oxygen dissociation curve
Answer:
The relationship between HbO2 saturation and oxygen tension (PPO2) is called oxygen dissociation curve.

Question 11.
Phosphorylation
Answer:
The process that involves trapping the heat energy in the form of high energy bond of ATP molecule is called phosphorylation.

Question 12.
Artificial ventilation
Answer:
It is the method of inducing breathing in a person when natural respiration has ceased or is faltering.

Question 13.
Ventilator
Answer:
A ventilator is a machine that supports breathing and is used during surgery, treatment for serious lung diseases or other conditions when normal breathing fails.

Question 14.
Cyclosis
Answer:
Cyclosis is the streaming movement of the cytoplasm shown by almost all living organisms. E.g. Paramoecium, Amoeba, etc.

Question 15.
Single circulation
Answer:
The movement of blood once through the heart during each circulation cycle is called single circulation.

Question 16.
Double circulation
Answer:
The movement of blood twice through the heart during one circulation cycle is called double circulation.

Question 17.
Erythropoiesis
Answer:
The process of formation of RBCs is called erythropoiesis.

Question 18.
Polycythemia
Answer:
The increase in the number of RBCs is called polycythemia.

Question 19.
Erythrocytopenia
Answer:
The decrease in the number of RBCs is called Erythrocytopenia.

Question 20.
Hematocrit
Answer:
The hematocrit is ratio of the volume of RBCs to total blood volume of blood.

Question 21.
Diapedesis
Answer:
Leucocytes perform amoeboid movement. Due to this kind of movement they can move out of the capillary walls. This is called diapedesis.

Question 22.
Leucocytosis
Answer:
Increase in the number of leucocytes or WBCs is called leucocytosis.

Question 23.
Leucopenia
Answer:
The decrease in the number of white blood cells is called leucopenia.

Question 24.
Leukaemia
Answer:
Pathological Increase in the number WBCs is called leukaemia or blood cancer.

Question 25.
Thrombocytopenia
Answer:
Decrease in the number of blood platelets is called thrombocytopenia.

Question 26.
Blood Coagulation
Answer:
Conversion of liquid blood into semisolid jelly is called blood coagulation or blood clotting.

Question 27.
Pericardium
Answer:
Double layered peritoneum that covers the heart from outside is called pericardium.

Question 28.
Pacemaker
Answer:
Pacemaker is the region that has power of generation of wave of contraction. In heart, sinoatrial node is called pacemaker.

Question 29.
Heartbeat
Answer:
The rhythmic contraction and relaxation of the heart is called heartbeat.

Question 30.
Pulse
Answer:
A pressure wave that travels through the arteries after each ventricular systole is called pulse.

Question 31.
Heart rate
Answer:
The rate with which the heart beats per minute is called the heart rate.

Question 32.
Stroke volume
Answer:
The amount of blood thrown out of the ventricles during one systole is called the stroke volume.

Question 33.
Cardiac output
Answer:
The amount of blood thrown out of the ventricles during one minute is called cardiac output.

Question 34.
Tachycardia
Answer:
Higher heart rate over 100 beats per minute is called tachycardia.

Question 35.
Bradycardia
Answer:
Lower heart rate which is lesser than 60 per minute is called bradycardia.

Question 36.
Myogenic
Answer:
When the initiation and further regulation of heartbeats take place in the muscles then such a heart is called myogenic.

Question 37.
Cardiac cycle
Answer:
Consecutive systole and diastole constitutes a single heartbeat or cardiac cycle.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 38.
Arterial blood pressure
Answer:
The pressure exerted by blood on the wall of artery is called arterial blood pressure.

Question 39.
Angiology
Answer:
Study of blood vessels is called angiology.

Question 40.
Angiography
Answer:
X-ray or imaging of the cardiac blood vessels to locate the position of blockages is called angiography.

Question 41.
Heart transplant
Answer:
Replacement of severely damaged heart by normal heart from brain- dead or recently dead donor is called heart transplant.

Question 42.
Silent Heart Attack
Answer:
Silent heart attack, also known as silent myocardial infarction, is a type of heart attack that lacks the general symptoms of classic heart attack like extreme chest pain, hypertension, shortness of breath, sweating and dizziness.

Question 43.
Electrocardiogram
Answer:
Graphical recording of electrical variations detected at the surface of body during their propagation through the wall of heart is electrocardiogram (ECG).

Question 44.
Lymph
Answer:
It is a fluid connective tissue with almost similar composition to the blood except RBCs, platelets and some proteins.

Give functions of the following

Question 1.
Epiglottis.
Answer:
The epiglottis prevents the entry of food into the trachea by closing the glottis temporarily.

Question 2.
Carbonic anhydrase.
Answer:
Carbonic anhydrase enzyme is found inside the RBCs only to accelerate the rate of formation of carbonic acid from CO2 and H2O.

Question 3.
Ventilators.
Answer:
Ventilators used in hospitals are part of life supporting system, which help in breathing by

  1. Pushing oxygen into the lungs
  2. Removing carbon dioxide from the lungs

Question 4.
Erythrocytes.
Answer:
Erythrocytes carry oxygen to all cells of the body from the lungs and bringing carbon dioxide from all the cells back to lungs.

Question 5.
Neutrophils.
Answer:
Neutrophils are responsible for destroying pathogens by the process of phagocytosis.

Question 6.
Thrombocytes/Platelets.
Answer:
Platelets secrete platelet factors which are essential in blood clotting. They also seal v the ruptured blood vessels by formation of platelet plug/thrombus. They secrete serotonin, a local vasoconstrictor.

Question 7.
Pericardial fluid.
Answer:
Pericardial fluid acts as a shock absorber and protects the heart from mechanical injuries. It also keeps the heart moist and acts as lubricant.

Question 8.
Heart walls.
Answer:
The epicardium and endocardium are protective in function whereas myocardium is responsible for contraction and relaxation of heart.

Question 9.
Valves in heart.
Answer:
Valves in the heart prevent the backflow of the blood at the time of systole and help in maintaining a unidirectional flow of blood.

Question 10.
Chordae tendinae.
Answer:
Chordae tendinae attach the bicuspid and tricuspid valves to the ventricular wall (papillary muscles) and regulate their opening and closing.

Question 11.
Semilunar valves.
Answer:
Semilunar valves prevent the backward flow of blood from pulmonary aorta and the aorta into the respective ventricles.

Question 12.
Sinoatrial node [SA] or Pacemaker.
Answer:
SA node acts as pacemaker of heart because it has the power of generating a new wave of contraction and making the pace of contraction.

Question 13.
Electrocardiogram (ECG).
Answer:
ECG helps to diagnose the abnormality in conducting pathway, enlargement of heart chambers, damage to cardiac muscles, reduced blood supply to cardiac muscles and causes of chest pain.

Question 14.
Blood.
Answer:
Functions of blood:

  1. Transport of oxygen and carbon dioxide
  2. Transport of food
  3. Transport of waste product
  4. Transport of hormones
  5. Maintenance of pH
  6. Water balance
  7. Transport of heat
  8. Defence against infection
  9. Temperature regulation
  10. Blood clotting/coagulation
  11. Helps in healing

Name the following

Question 1.
Name two animals in which moist skin acts as a respiratory surface.
Answer:
Earthworm, Frog

Question 2.
Name the respiratory organs in insects and fish.
Answer:
Insects – Tracheal tubes and spiracles
Fish – Internal gills

Question 3.
Name any two disorders of respiratory system.
Answer:
Asthma and pneumonia are the two disorders of respiratory system.

Question 4.
Name the structural and functional unit of lungs.
Answer:
Alveolus is the structural and functional unit of lungs.

Question 5.
Name the energy currency of cell.
Answer:
ATP is the energy currency of cell.

Question 6.
Name the site where actual exchange of O2 and CO2 takes place between air and blood in the body of man.
Answer:
Alveolus of lung.

Question 7.
Name any two respiratory centres required for regulation of breathing.
Answer:
Inspiratory centre, Expiratory centre, Pneumotaxic centre and Apneustic centre.

Question 8.
Name the muscles which move ribs up and down.
Answer:
External intercostal muscles.

Question 9.
Name two phyla where haemocoel is present.
Answer:
Phylum-Arthropoda and Phylum-Mollusca.

Question 10.
Name the animal-group which show single circulation.
Answer:
Fishes

Question 11.
Name the cells which produce thrombocytes.
Answer:
Megakaryocytes produce thrombocytes.

Question 12.
Name the process of formation of red blood corpuscles.
Answer:
Erythropoiesis

Question 13.
Name the space in which human heart is located.
Answer:
Mediastinum is the space in which human heart is located.

Question 14.
Name the types of lymphocytes depending upon functions.
Answer:
B-lymphocytes and T-lymphocytes.

Question 15.
Name the layers of peritoneum that surrounds the heart sequentially from outside to inside.
Answer:
Fibrous pericardium, parietal layer of serous pericardium and visceral layer of serous pericardium.

Question 16.
Name the connection between the pulmonary trunk and systemic aorta.
Answer:
Ligamentum arteriosum that represents remnant of ductus arteriosus of foetus.

Question 17.
Name the valve between left atrium and left ventricle and give its significance.
Answer:
Between left atrium and left ventricle is mitral or bicuspid valve which maintains the unidirectional flow of blood by preventing hs backflow.

Question 18.
Name the walls of an artery.
Answer:
Outer tunica externa, middle tunica media and inner tunica interna.

Question 19.
Name the instrument used to measure blood pressure.
Answer:
Sphygmomanometer is used to measure blood pressure.

Question 20.
Name the plasma proteins involved in the process of blood clotting.
Answer:
Prothrombin and fibrinogen.

Question 21.
Name the various components of conducting system of the heart.
Answer:
Conducting system of the heart consists of SA node, AV node, bundle of His and Purkinje fibers.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 22.
Name the neurotransmitters that decrease and increase the heart rate in human beings respectively.
Answer:
Acetylcholine decreases heart rate and adrenaline or epinephrine increases the heart rate in human.

Question 23.
Who discovered ECG?
Answer:
Willem Einthoven discovered ECG.

Distinguish between the following

Question 1.
Pharynx and Larynx.
Answer:

Pharynx Laryix
1. Pharynx is a short, vertical tube. 1. Larynx is a sound producing organ located at the end of pharynx.
2. Mouth leads to the pharynx. 2. Larynx leads to the oesophagus.
3. Vocal cords are absent. 3. Vocal cords are present.
4. Pharynx does not increase in size at the time of puberty. 4. Larynx increases in size at the time of puberty.
5. Pharynx does not show Adam’s apple. 5. Larynx shows Adam’s apple in adult males.

Question 2.
Inspiration and Expiration.
Answer:

Inspiration Expiration
1. Inspiration is an active process. 1. Expiration is a passive process.
2. During inspiration diaphragm contracts and becomes flattened. 2. During expiration diaphragm relaxes and becomes dome shaped.
3. During inspiration intercostal muscles contract. 3. During expiration intercostal muscles relax.
4. During inspiration ribs are pulled outwards and sternum is raised. 4. During expiration ribs are pulled inwards and sternum is lowered.
5. During inspiration the space in the thoracic cavity increases. 5. During expiration the space in the thoracic cavity decreases.
6. During inspiration pressure in the lungs decreases. 6. During expiration pressure in the lungs increases.
7. During inspiration the volume of the lungs increase. 7. During expiration the volume of the lungs decreases.
8. During inspiration air comes inside the body. 8. During expiration air goes out of the body.

Question 3.
External respiration and Internal respiration.
Answer:

External respiration Internal respiration
1. The respiratory processes occurring in lungs is called external respiration. 1. The respiratory processes that occur in tissues is called internal respiration.
2. During external respiration O2 from the lungs enters into the lung capillaries by diffusion. 2. During internal respiration O2 from the blood enters the tissue cells.
3. During external respiration CO2 from the lung capillaries diffuse into the lungs. 3. During internal respiration CO2 from the tissues enters into the blood.
4. During external respiration exchange of gases takes place between the air and the lungs. 4. During internal respiration exchange of gases take place between the blood and the tissue.
5. Formation of oxyhaemoglobin takes place during external respiration. 5. Oxyhaemoglobin dissociates into oxygen and haemoglobin during internal respiration.
6. During external respiration CO2 is released. 6. During internal respiration carbamino haemoglobin is formed which is carried to the lungs.

Question 4.
Transport of O2 and Transport of CO2.
Answer:

Transport of O2 Transport of CO2
1. Transport of O2 takes place from lungs to the tissues and cells. 1. Transport of CO2 takes place from tissues and cells to the lungs.
2. Oxygen is carried as oxyhaemoglobin to the tissues with the help of RBCs. 2. Carbon dioxide is carried as carbaminohaemoglobin from the tissues with the help of plasma and RBCs.
3. Oxygen does not form oxides or other products during its transport. 3. CO2 forms bicarbonates with sodium and potassium during its transport.
4. O2 does not form acids during its transport. 4. CO2 dissolves in water to form carbonic acid.

Question 5.
Vital Capacity of Lung and Total Lung Capacity.
Answer:

Vital Capacity of Lung Total Lung Capacity
1. It is the maximum amount of air a person can expire and inspire to their maximum extent. 1. It is the maximum amount of air that the lungs can hold after a maximum forceful inspiration.
2. It is the sum total of inspiratory reserve volume, tidal volume and expiratory reserve volume. 2. It is the sum total of vital capacity and residual volume.
3. It ranges from 4100 to 4600 ml. 3. It ranges from 5200 to 5800 ml.

Question 6.
Inspiratory Reserve Volume (IRV) and Expiratory Reserve Volume (ERV).
Answer:

Inspiratory Reserve Volume (IRV) Expiratory Reserve Volume (ERV).
1. It is the maximum volume of air, or the extra volume of air, that is inspired during forced breathing. 1. It is the maximum volume of air that is expired during forced breathing.
2. Its value is 2000/3000 ml. 2. Its value is 1000/1100 ml.

Question 7.
T. S. of artery and T.S. of vein.
Answer:

T. S. of artery T.S. of vein
1. Histologically in transverse section of artery there are three walls, tunica externa, tunica media and tunica interna or endothelium. 1. Histologically in transverse section of vein there are three walls, tunica externa, tunica media and tunica interna or endothelium.
2. Tunica media is thick and muscular. 2. Tunica media is thinner as compared to artery.
3. Lumen of artery is narrow. 3. Lumen of vein is broad.

Question 8.
Erythrocytes and Leucocytes.
Answer:

Erythrocytes Leucocytes
1. Erythrocytes have a definite shape which is elliptical or oval. 1. Leucocytes do not have definite shape as they are amoeboid.
2. They are enucleated. 2. They are nucleated.
3. Erythrocytes contain haemoglobin and hence appear red. 3. Leucocytes are devoid of any respiratory pigment and hence appear colourless.
4. The normal erythrocyte count is 4.3 to 5.8 million per cubic mm of blood. 4. The normal leucocyte count is 4000 to 11000 per cubic mm of blood.
5. The life span of erythrocytes is 100 to 120 days. 5. The life span of leucocytes is 3 to 4 days.
6. The diameter of erythrocytes is 7.2 m and thickness is about 2 to 2.2 m. 6. The size of leucocytes varies with its subtypes and is of average size of 8 to 15 m.
7. Erythrocytes are formed by the process of erythropoiesis in red bone marrow. 7. Leucocytes are formed by the process of leucopoiesis in bone marrow, tonsils, lymph nodes, spleen, thymus, etc.
8. Erythrocytes transport the respiratory gases. 8. Leucocytes help in the formation of antibodies besides fighting against foreign antigens by phagocytic activity.

Question 9.
Eosinophils and Basophils.
Answer:

Eosinophils Basophils
1. Cytoplasmic granules present in eosinophils are stained with acidic stains. 1. Cytoplasmic granules present in basophils are stained with basic stains.
2. Nucleus is bilobed. 2. Nucleus is twisted.
3. Eosinophils constitute 3% of total WBCs. 3. Basophils constitute 0.5% of total WBCs.

Question 10.
Neutrophils and Eosionophils.
Answer:

Neutrophils Eosinophils
1. Cytoplasmic granules present in neutrophils are stained with neutral stains. 1. Cytoplasmic granules present in eosinophils are stained with acidic stains.
2. Nucleus is three to five lobes showing polymorphic form. 2. Nucleus is bilobed.
3. Neutrophils constitute 62% of total WBCs. 3. Eosinophils constitute 3% of total WBCs.

Question 11.
Lymphocytes and Monocytes.
Answer:

Lymphocytes Monocytes
1. Large round nucleus but size of the cell is smaller. 1. Large kidney shaped nucleus and largest size among WBCs.
2. Lymphocytes form 25-33% of WBCs. 2. Monocytes form 3-9% of WBCs.

Question 12.
Granulocytes and Agranulocytes.
Answer:

Granulocytes Agranulocytes
1. WBCs with granular cytoplasm are called granulocytes. Thus, cytoplasmic granules are present. 1. WBCs with agranular cytoplasm are called agranulocytes. Thus, cytoplasmic granules are absent.
2. Nuclei of granulocytes are variously lobed. 2. Nuclei of agranulocytes are not lobed.

Question 13.
Single circulation and Double circulation.
Answer:

Single circulation Double circulation
1. Blood flows only once through the heart in a complete cycle. 1. Blood flows twice through the heart during one complete circulation. Systemic – to and fro ‘ from heart to body and pulmonary – to and fro from heart to lungs.
2. Heart pumps deoxygenated blood only. 2. Heart pumps both deoxygenated and oxygenated blood to lungs and body respectively.
3. Blood is oxygenated in gills. 3. Blood is oxygenated in lungs.
4. Occurs only in fishes. 4. Occurs in amphibians, reptiles, birds and mammals.

Question 14.
Systolic blood circulation and Diastolic blood circulation.
Answer:

Systolic blood circulation Diastolic blood circulation
1. Blood is passed from right ventricle to lungs by pulmonary artery during systolic circulation. Similarly, from left ventricle the oxygenated blood is given to the entire body through systemic aorta during systolic circulation. 1. Blood is passed to left atrium from lungs by pulmonary veins during diastolic circulation. Similarly, deoxygenated blood from entire body is brought back to heart through vena cava during diastolic circulation.
2. Systolic blood circulation is under maximum pressure as heart is forcing the blood to come out of heart. 2. Diastolic blood circulation is under minimum blood pressure as heart is relaxed during diastole.

Question 15.
Atria and Ventricles.
Answer:

Atria Ventricles
1. Atria are upper chambers of the heart. 1. Ventricles are lower chambers of the heart.
2. Atria are thin walled. 2. Ventricles are thick walled.
3. Atria are receiving chambers. 3. Ventricles are distributing chambers.
4. Interatrial septum divides the two auricles (atria). 4. Interventricular septum divides the two ventricles.
5. Right atrium is larger in size than left atrium. 5. Left ventricle is larger in size than the right ventricle.

Question 16.
S.A. Node and A.V. Node.
Answer:

S.A. Node A.V. Node
1. Sinoatrial node is present in the right ventricle near the opening near the opening of the superior vena cava. 1. Atrioventricular node is present in the right ventricle near the opening of the coronary sinus.
2. S.A. node is the pacemaker of the heart and it starts atrial systole. 2. A.V. node starts ventricular systole through bundles of His and Purkinje’s fibre system.

Question 17.
Pulmonary circulation and Systemic circulation.
Answer:

Pulmonary circulation Systemic circulation
1. The course of blood from the right ventricle to the left atrium of the heart through the lungs is called pulmonary circulation. 1. The course of blood from the left ventricle to the right atrium of the heart through the body is called systemic circulation.
2. Pulmonary circulation is mainly for sending the blood for oxygenation in the lungs from the heart and bringing it back to the heart after oxygenation. 2. Systemic circulation is for sending the deoxygenated blood from the body to the heart and sending oxygenated blood from the heart to the body.

Question 18.
Atrio ventricular valves and Semilunar valves.
Answer:

Atrio ventricular valves Semilunar valves
1. Atrio ventricular valves Eire present between the atria and ventricles. On the right side there is tricuspid valve whereas on the left side there is bicuspid valve. 1. Semilunar valves are present at the opening of pulmonary artery and systemic aorta.
2. Atrio ventricular valves prevent the back flow of blood from ventricles to atria at the time of systole. 2. Semilunar valves prevent the back flow of blood from pulmonary artery and systemic aorta back to the heart.

Question 19.
Hypertension and Hypotension.
Answer:

Hypertension Hypotension
1. Blood pressure values more than 140 mm Hg SP and 90 mm HG DP is called hypertension. 1. Blood pressure values less them 120 mm Hg SP and 70 mm HG DP is called hypotension.
2. Excessive hypertension can result into lethal complications such as stroke or paralysis. 2. Hypotension may not be lethal if immediate measures are taken to raise the blood pressure.

Give reasons

Question 1.
ATP is called energy currency of the cell.
Answer:

  1. During cellular respiration, the oxidation of food (glucose) takes place.
  2. This happens in the mitochondria using the oxygen present in the blood.
  3. ATP molecules are formed during this oxidation.
  4. ATP is used for various vital body processes and also for maintaining the body temperature to constancy.
  5. Since energy is stored in the form of ATP it is called an energy currency of the cell.

Question 2.
The vestibule of nasal chamber has fine hair.
Answer:

  1. Vestibule is the anterior most part of the nasal chamber.
  2. The hairs present in this region trap the dust particles and prevent them from entering into the interiors of the respiratory passage.
  3. Therefore, the vestibule of nasal chambers has fine hair.

Question 3.
Glottis is guarded by a flap called epiglottis.
Answer:

  1. The oesophagus and trachea lie side by side.
  2. There is possibility that food particles may enter respiratory passage at the time of gulping.
  3. However, the epiglottis prevents the entry of food into the respiratory passage by closing it temporarily.
  4. Thus, for preventing the entry of food particles into respiratory passage, the glottis is guarded by a flap called epiglottis.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 4.
Alveoli are very flexible.
Answer:

  1. Alveoli are made up of collagen and elastin fibres.
  2. They are very thin (0.0001 mm) and lined by non-ciliated squamous epithelium.
  3. All the above structural components make the alveoli very flexible.

Question 5.
Expiration is called a passive process.
Answer:

  1. During expiration, intercostal muscles relax. This results in pulling the ribs inwards.
  2. Diaphragm also relaxes and returns to its normal dome shape.
  3. The collective contraction of ribs and diaphragm results in the reduction of
    thoracic cavity and hence automatically air is pushed out of the lungs.
  4. Since the pressure on the lungs increase rushing the air to outside, expiration is called a passive process.

Question 6.
Pericardium acts as a defence wall for the heart.
Answer:

  1. Pericardium protects the heart. It is double layered peritoneum, having outer fibrous and inner serous pericardium layers.
  2. Fibrous pericardium being tough gives protection to the heart.
  3. Serous pericardium has two layers, parietal and visceral layer or epicardium.
  4. In between these two layers, there is pericardial fluid, which helps to absorb shocks and provide nourishment.
  5. In this way pericardium acts as a defence wall.

Question 7.
Valves are present in veins.
Answer:

  1. Veins carry blood to the heart.
  2. At that time the backward flow of the blood should be prevented.
  3. Therefore, valves are present in veins.

Question 8.
Atria are thin walled than ventricles.
Answer:

  1. Atria are receiving chambers, while ventricles are distributing chambers.
  2. The blood is driven out from ventricles.
  3. Ventricles are therefore, strong and with thicker walls.
  4. Atria are thin walled as compared to ventricles.

Question 9.
Heart is called a pump.
Answer:

  1. The heart acts as a pumping organ. It shows continuous pumping action.
  2. The rhythmic contraction or systole and relaxation or diastole of heart forms one heartbeat.
  3. Such heartbeats occur about 72 times per minute.
  4. The heart efficiently pumps about 5 litres of blood per minute. Therefore, the heart is called a pump.

Question 10.
In normal human heart, there is no mixing of oxygenated and deoxygenated blood.
Answer:

  1. In normal human heart, there is completely formed atrioventricular septum.
  2. This septum keeps the deoxygenated and oxygenated blood separate.
  3. Hence there is no mixing of the two types of blood.

Question 11.
Blood pressure is inversely related to the elasticity of the blood vessels.
Answer:

  1. When the blood gushes through the blood vessels, the walls of blood vessels -can expand a little due to their elasticity.
  2. But as the age advances, the elasticity is reduced and then the blood vessels do not expand.
  3. Hence the flowing blood gets more resistance and the blood pressure can rise.
  4. Lesser the elasticity more will be the blood pressure, whereas more the elasticity of the vessel wall, then the pressure will not rise.
  5. In this way, the blood pressure is inversely related to the elasticity of the blood vessels.

Write short notes

Question 1.
Chloride shift or Hamburger’s phenomenon.
Answer:

  1. About 70% of CO2 is transported in the form of sodium bicarbonates/potassium bicarbonates from tissue cells to lungs.
  2. In the RBCs, CO2 combines with water in the presence of a Zn containing enzyme, carbonic anhydrase to form carbonic acid. This action is rapid in RBCs as compared to that in the plasma.
  3. Carbonic acid being unstable, immediately dissociates into HCO3 and H+in the presence of same enzyme, leading to large accumulation of HCO3 inside the RBCs. It thus moves out of RBCs. This can bring about imbalance of the charge inside the RBCs.
  4. To maintain the ionic balance between the RBCs and the plasma, Cl diffuses into the RBCs. This movement of chloride ions is known as chloride shift or Hamburger’s phenomenon.
  5. HCO3 that comes in the plasma joins to Na+/K+ forming NaHCO3/KHCO3 which can maintain pH of blood. The remaining H+ ions in the RBCs are buffered by haemoglobin by the formation of oxyhaemoglobin.
  6. At the level of lungs, due to the low partial pressure of carbon dioxide of the alveolar air, hydrogen ion and bicarbonate ions combine to form carbonic acid and under the influence of carbonic anhydrase again yields carbon dioxide and water.

Question 2.
Regulation of breathing.
Answer:
(1) Respiration is under dual control, i.e. nervous and chemical. Normal breathing is an involuntary process. Steady state of respiration is controlled by neurons located in the pons and medulla and are known as the respiratory centres. They regulate the rate and depth of breathing.

(2) These centres are divided into three groups : dorsal group of neurons in the medulla (inspiratory centre), ventro-lateral group of neurons in medulla (inspiratory and expiratory centre) and pneumotaxic centre located in the pons and apneustic centre which is antagonistic in action to pneumotaxic centre.

(3) During inspiration, when the lungs expand to a critical point, the stretch receptors are stimulated and impulses are sent along the vagus nerves to the expiratory centre. It then sends out inhibitory impulses to the inspiratory centre.

(4) The inspiratory muscles relax and expiration follows. As the air leaves but, the lungs are deflated and the stretch receptors are no longer stimulated. Thus, the inspiratory centre is no longer inhibited and a new respiration begins. These events are called the Hering – Breuer reflex. The Hering – Breuer reflex controls the depth and rhythm of respiration. It also prevents the lungs from inflating to the point of bursting.

(5) The respiratory centre has connections with the cerebral cortex that means we can voluntarily change our pattern of breathing. Voluntary control is protective because it enables us to prevent water or irritating gases from entering the lungs.

Question 3.
Carbon monoxide poisoning.
Answer:

  1. Carbon monoxide poisoning is caused when carbon monoxide is combined with haemoglobin.
  2. Haemoglobin is said to have 250 times more affinity for carbon monoxide than that for the oxygen.
  3. Therefore, haemoglobin with carbon monoxide forms a stable compound, the carboxyhemoglobin.
  4. Due to the formation of carboxyhaemoglobin, the haemoglobin no longer carries oxygen to the cells and tissues. Tissues then suffer from oxygen starvation. This leads to asphyxiation and in extreme cases it leads to death.
  5. Carbon monoxide poisoning occurs in closed rooms with incompletely burning substances such as stove burners or furnaces and garages having running automobile engines.
  6. Person suffering from carbon monoxide poisoning has to be administered with oxygen-carbon dioxide mixture, so that high levels of CO2 makes carbon monoxide dissociated from haemoglobin.

Question 4.
Artificial ventilation.
Answer:
(1) Artificial ventilation is the artificial respiration. It is the method of inducing breathing in a person when natural respiration has ceased or is faltering. If used properly and quickly, it can prevent death due to drowning, choking, suffocation, electric shock, etc.

(2) The process involves two main steps:
a. Establishing and maintaining an open air passage from the upper respiratory tract to the lungs.
b. Force inspiration and expiration as in mouth to mouth respiration or by mechanical means like ventilator.

(3) A ventilator is a machine that supports breathing and is used during surgery, treatment for serious lung diseases or other conditions when normal breathing fails.

Question 5.
Erythrocytes.
Answer:

  1. Erythrocytes or red blood corpuscles. They are circular, biconcave, enucleated cells.
  2. The RBC size : 7 pm in diameter and 2.5 pm in thickness.
  3. The RBC count : 5.1 to 5.8 million RBCs/ cu mm of blood in an adult male and 4.3 to 5.2 million/cu mm in an adult female.
  4. The average life span of RBC : 120 days.
  5. RBCs are formed by the process of erythropoiesis. In foetus, RBC formation takes place in liver and spleen whereas in adults it occurs in red bone marrow.
  6. The old and worn out RBCs are destroyed in liver and spleen.
  7. Polycythemia is an increase in number of RBCs while erythrocytopenia is decrease in their (RBCs) number.

Functions of RBCs:

  1. Transport of oxygen from lungs to tissues and carbon dioxide from tissues to lungs with the help of haemoglobin.
  2. Maintenance of blood pH as haemoglobin acts as a buffer.
  3. Maintenance of the viscosity of blood.

Question 6.
Heartbeat.
Answer:

  1. The rhythmic contraction and relaxation of the heart is called heartbeat.
  2. Each heartbeat includes one systole and one diastole. During systole the heart contracts and during diastole it relaxes.
  3. The rate with which the heart beats is called heart rate. The normal heart rate is 72 beats per minute.
  4. Tachycardia means faster heart rate of about more than 100 beats per minute.
  5. Bradycardia means slower heart rate that is below 60 beats per minute.

Question 7.
Pulse.
Answer:

  1. A pressure wave that travels through the arteries after each ventricular systole is called a pulse.
  2. The pulse can be felt in any artery that lies near the surface of the body.
  3. The radial artery at the wrist is most commonly used to feel the pulse.
  4. The pulse rate per minute indicates the heart rate. Since each heartbeat generates one pulse in the arteries, the pulse rate is same as that of heart rate, i.e. 72 times per minute.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 8.
Peacemaker.
Answer:

  1. Pacemaker is the region in tile heart which initiates the beating.
  2. The natural pacemaker of the heart is sinoatrial node (SA node).
  3. The pacemaker is autorhythmic, it is able to repeatedly and rhythmically generate impulses.
  4. SA node is responsible for initiation of cardiac excitation. Therefore, it is called a pacemaker.

Question 9.
Blood pressure.
Answer:

  1. Blood pressure is the pressure exerted by the flowing blood on the walls of arteries.
  2. Blood pressure described in two terms viz. systolic blood pressure and diastolic blood pressure. Systolic blood pressure is the maximum pressure of blood when heart undergoes ventricular systole. It is responsible for flow of blood in the arteries. Normal systolic pressure is 120 mm Hg.
  3. Diastolic blood pressure is the minimum pressure of blood when heart undergoes diastole. Normal diastolic pressure is 80 mm Hg.
  4. Blood pressure is represented as 120/80 mm Hg for a normal human being.

Question 10.
Hypertension.
Answer:

  1. In a normal healthy person the blood pressure values are 120 mm Hg (systolic)/ 80 mm Hg (diastolic).
  2. When the blood pressure is persistently more than 140 mm Hg systolic pressure and 90 mm Hg diastolic arterial blood pressure then it is said to be hypertension or high blood pressure.
  3. Excessively high blood pressure is very dangerous as high blood pressure of about 220/120 mm Hg may cause rupturing of blood vessels.
  4. Rupture of eye blood vessels can lead to blindness.
  5. If blood vessels of kidney are affected then nephritis is caused.
  6. Hemorrhage occurring in the brain can lead to stroke or paralysis. Therefore, hypertension is commonly called silent killer. It may be present for years with no distinct symptoms.
  7. The factors causing hypertension are arteriosclerosis (reduction of elasticity of blood vessels), atherosclerosis (deposition, of cholesterol inside the blood vessels wall), obesity, physical or emotional stress, alcoholism, smoking, cholesterol rich diet, increased secretion of renin, epinephrine or aldosterone, etc.

Question 11.
Coronary artery disease (CAD).
Answer:

  1. Coronary artery disease is a condition caused due to problems like atherosclerosis.
  2. In this disease, coronary arteries are narrowed due to deposition of fatty substances.
  3. Due to this the blood flow to the heart is reduced.
  4. In coronary heart disease, the heart muscle is damaged because of an inadequate amount of blood due to obstruction of its blood supply.
  5. The symptoms of CAD depend upon the degree of obstruction.
  6. Symptoms are mild chest pain or angina pectoris.
  7. In severe cases it results in heart attack known as myocardial infarction.

Question 12.
Angina pectoris.
Answer:

  1. Angina pectoris is the pain in the chest. It results from a reduction in blood supply to cardiac muscle due to narrowed and hardened coronary arteries.
  2. Atherosclerosis and arteriosclerosis can cause this problem. Basically, the coronary arteries are affected during angina pectoris.
  3. It causes heaviness and severe pain in the chest. The pain can also be felt at the neck, lower jaw, left arm and left shoulder.
  4. Angina pectoris often occurs during exertion, when the heart demands more oxygen and narrowed blood vessels cannot supply. It disappears with rest.

Question 13.
Heart failure.
Answer:

  1. Heart failure is caused due to progressive weakening of the heart muscle. This results in the failure of the heart to pump the blood effectively.
  2. Hypertension increase the after load on the heart leading to significant enlargement of the heart.
  3. This finally results in heart failure.
  4. Factors responsible for heart failure are advanced age, malnutrition, chronic infections, toxins, severe anaemia or hyperthyroidism, etc.
  5. Any problem leading to degeneration of heart muscle, may result in heart failure.

Question 14.
Atherosclerosis.
Answer:

  1. Atherosclerosis is the deposition of fatty substances and cholesterol on the inner lining of eateries.
  2. This deposition results in the formation of atherosclerotic plaque.
  3. It results in the decrease of the lumen of the blood vessels causing increasing resistance for the blood to flow which in turn results in the hypertension.
  4. Atherosclerosis of the coronary arteries results in decrease in the blood flow to the heart muscles.
  5. Due to such condition, coronary heart disease is caused.

Question 15.
ECG.
Answer:

  1. Electrocardiogram or ECG is the graphic v record of electrical variations produced by the heart during one heartbeat or cardiac cycle.
  2. ECG is taken with the help of an instrument called electrocardiograph or ECG machine. Electrocardiograph records the action potentials generated by the heart muscles.
  3. The electrical activity of heart is represented in the form of a graph plotted with time on X-axis against voltage displacement on Y-axis.
  4. A normal ECG is a graph having series of ridges and furrows. There are waves such as P-wave, QRS complex and T-wave.
  5. P-wave is a small upwards wave representing impulse generated by SA node. P-wave is caused by atrial depolarization that results in atrial contraction.
  6. QRS-complex wave begins as a downward deflection, continues as a large upright triangular wave and ends’ as a downward wave.
  7. QRS-complex wave represents spreading of impulse from SA node to AV node, then to bundle of His and Purkinje fibres. It causes ventricular depolarization resulting in ventricular contraction.
  8. T-wave is a broad upward wave which represents ventricular repolarization resulting in ventricular relaxation.
  9. Functions of ECG are mainly for diagnosis and also for prognosis. It is useful to detect abnormal functioning of heart as in coronary artery diseases, heart block, angina pectoris, tachycardia, ischemic heart disease, myocardial infarction, cardiac arrest, etc.

Question 16.
Angiography.
Answer:

  1. Angiography is an X-ray imaging of the cardiac blood vessels to locate the position of blockages.
  2. Depending upon the degree of blockage, remedial procedures like angioplasty or by¬pass surgery are performed.
  3. In angioplasty a stent is inserted at the site of blockage to restore the blood supply while in by-pass surgery, the atherosclerotic region is by-passed with part of vein or artery taken from any other suitable part of the body, like hands or legs.

Question 17.
Silent Heart Attack or silent myocardial infarction.
Answer:

  1. Silent heart attack is a type of heart attack that lacks the general symptoms of classic heart attack like extreme chest pain, hypertension, shortness of breath, sweating and dizziness.
  2. Symptoms of silent heart attack are so mild that a person often confuses it for regular « discomfort and thereby ignores it.
  3. Men are more affected by silent heart attack than women.

Question 18.
Heart Transplant.
Answer:

  1. Heart transplant is the replacement of severely damaged heart by normal heart from brain-dead or recently dead donor,
  2. Heart transplant is necessary in case of patients with end-stage heart disease and severe coronary arterial disease.

Short Answer Questions

Question 1.
What is meant by respiration? How is it useful in the production of energy?
Answer:

  1. Respiration is the biochemical process in which organic compound such as glucose are oxidized to liberate chemical energy.
  2. During respiration energy is released in gradual and step wise process. The released energy is in the form of bonds of ATP (Adenosine Tri Phosphate) molecules are shown below:
    C6H12O6 + 6O2 → 6CO2 + 6H2O + 38 ATP
  3. ATP is the biologically useful energy. ATP drives most of the life process.
  4. When cell requires the energy, ATP is hydrolyzed and is converted into ADP with subsequent release of energy.
  5. The respiratory system, blood and the body cells play an important role in the process of respiration.

Question 2.
How does exchange of gases take place at the alveolar level?
Answer:
1. Exchange of gases between the alveolar air and the blood is known as external respiration.

2. Simple squamous epithelial layer of alveolus is intimately associated with a similar layer lining the capillary wall. Both of these layers are thin walled and together they make up the respiratory membrane through which gaseous exchange occurs between the alveolar air and the blood.

3. Diffusion of gases will take place from an area of higher partial pressure to an area of lower partial pressure until the partial pressure in the two regions reaches equilibrium.

4. The partial pressure of carbon dioxide of blood entering the pulmonary capillaries is 45 mmHg while partial pressure of carbon dioxide in alveolar air is 40 mmHg. Due to this difference, carbon dioxide diffuses from the capillaries into the alveolus.

5. Similarly, partial pressure of oxygen of blood in pulmonary capillaries is 40 mmHg while in alveolar blood it is 104 mmHg. Due to this difference oxygen diffuses from alveoli to the capillaries.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 2

Question 3.
What is the role of haemoglobin in the transport of oxygen in the blood?
Answer:

  1. Haemoglobin is a respiratory pigment present in cytoplasm of RBCs. About 97% of oxygen is transported by these haemoglobin molecules from lungs to tissues.
  2. Haemoglobin has a high affinity for Oa and combines with it to form oxyhaemoglobin. One molecule of Hb has four FeT, each of which can pick up a molecule of oxygen (O2). Hb + 4O2 → Hb (4O2)
  3. Oxyhaemoglobin is transported from lungs to the tissues where it readily dissociates to release O2.
    Hb (4O2) → Hb + 4O2
  4. In the alveoli where PPOa is high and PPCO2 is low, oxygen binds with haemoglobin, but in tissues, where PPO2 is lower and PPCO2 is high, Oxyhaemoglobin dissociates and releases O2 for diffusion into the tissue cells.

Question 4.
What is blood? What is the normal quantity of blood in an adult human being?
Answer:

  1. Blood is the fluid connective tissue that circulates in the body.
  2. Blood is derived from mesoderm.
  3. It is bright red, slightly alkaline fluid having pH about 7.4. It is salty, viscous fluid heavier than water.
  4. The average sized adult has about 5 litres of blood in his/her body which constitutes about 8% of the total body weight.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 5.
Describe the structure and the function of thrombocytes.
Answer:

  1. Thrombocytes or platelets are non- nucleated, round and biconvex blood corpuscles.
  2. They are smallest corpuscles measuring about 2.5 to 5 mm in diameter with a count of about 2.5 lakhs/cu mm of blood.
  3. Their life span is about 5 to 10 days.
  4. Thrombocytes are formed from megakaryocytes of bone marrow. They break from these cells as fragments during the process of thrombopoiesis.
  5. Thrombocytosis is the increase in platelet count while thrombocytopenia is decrease in platelet count.
  6. Thrombocytes possess thromboplastin which helps in clotting of blood.
  7. Therefore, at the site of injury platelets aggregate and form a platelet plug. Here they release thromboplastin due to which further blood clotting reactions take place.

Question 6.
Describe the structure of the heart wall.
Answer:

  1. The heart wall is composed of three layers, viz. outer epicardium, middle myocardium and inner endocardium.
  2. Epicardium is composed of single layer of mesothelium having flat epithelial cells.
  3. Myocardium is composed of cardiac muscle fibres. These muscle fibres perform the function of systole and diastole by showing contraction and relaxation of muscle wall of the heart.
  4. Endocardium is composed of single layer of flat epithelial cells called endothelium.

Question 7.
Name the two heart sounds. How and when are they produced?
Answer:

  1. In one normal heartbeat the heart sounds like lubb and dup are produced once each.
  2. The rhythmic contraction (Systole) and relaxation (diastole) forms are heartbeat. The heart sounds are due to closure of valves.
  3. Lubb sound is produced during ventricular systole when the cuspid valves close both the atrioventricular apertures preventing blood flow into atria.
  4. Dub sound is produced during ventricular diastole when semilunar valves are closed, preventing backflow of blood from pulmonary trunk and systemic aorta into ventricles.

Question 8.
What is double circulation? What is its significance?
Answer:
(1) Double circulation : Movement of blood twice through the heart during one circulation cycle is called double circulation. Body → heart → lungs → heart → body is the course of double circulation.

(2) Significance of double circulation:
a. Double circulation is more effective type of circulation in which oxygenated and deoxygenated type of blood do not intermix.
b. The systemic circulation i.e. from body to heart and back to body while the pulmonary circulation i.e. from heart to lungs and back to heart circulate the blood uniformly.

(3) Coronary and hepatic portal circulation is also achieved due to double circulation.

Question 9.
Describe pulmonary and systemic circulation.
Answer:

  1. In human beings, there is double circulation because blood passes twice through the heart during one cardiac cycle.
  2. The blood follows two routes viz. pulmonary and systemic.
  3. Pulmonary circulation is circulation between heart and lungs. Systemic circulation is the circulation between the heart and body organs (except lungs).
  4. During pulmonary circulation, the blood passes from the right ventricle to the left atrium of the heart through lungs.
  5. The right ventricle pumps deoxygenated blood into the pulmonary trunk which carries it to lungs for oxygenation. The oxygenated blood from the lungs is brought to left atrium by two pairs of pulmonary veins.
  6. During systemic circulation, the blood from the left ventricle passes to the right atrium of heart through body organs.
  7. The left ventricle pumps oxygenated blood into the systemic aorta which carries it to all body organs except lungs. The deoxygenated blood from the body organs is brought to right atrium by superior and inferior venae cavae.

Question 10.
How is cardiac activity regulated?
Answer:

  1. Normal activities of the heart are auto regulated. The specialized muscles help in this regulation.
  2. The heart is said to be myogenic due to this ability.
  3. In the medulla oblongata of brain, there is cardiovascular centre,
  4. From this centre, sympathetic and parasympathetic nerves innervate the sinoatrial node.
  5. Sympathetic nerves secrete adrenaline and it stimulates and increases the heartbeat.
  6. Parasympathetic nerves secrete acetylcholine and it decreases the heart rate.
  7. Rate of heartbeat is controlled in response to inputs from various receptors like proprio-receptors.
  8. These receptors monitor the position of limbs and muscles. There are chemoreceptors which monitor chemical change in blood and baroreceptors that monitor the stretching of main arteries and veins.

Chemical control on heart rate:

  1. Hypoxia, acidosis, alkalosis cause decrease in cardiac activity.
  2. Hormones like epinephrine and nor epinephrine enhance the cardiac activity.
  3. Elevated blood level of K+ and Na+ decreases the cardiac activity.

Question 11.
What are the main features of respiratory surface?
Answer:
The respiratory surface, for the efficient gaseous exchange should have the following features:

  1. It should have a large surface area.
  2. It should be thin, highly vascular and permeable to allow exchange of gases.
  3. It should be moist.

Question 12.
What is the co-relationship between activeness of organism and complexity of transport system?
Answer:

  1. As the size of an organism increases, its surface area to volume ratio decreases. This means it has relatively less surface area available for substances to diffuse through.
  2. Large multicellular organisms therefore cannot rely on diffusion alone to supply their cells with Substances such as food and oxygen and to remove waste products. Large multicellular organisms require specialized transport system.
  3. In short for the organisms to become active, they must be having complex transport system to bring about their vital functions rapidly.

Question 13.
What are the granules in granulocytes?
Answer:

  1. Neutrophils : Granules . of neutrophils contain cationic proteins and other proteins that are used to kill bacteria, some enzymes to breakdown bacterial proteins, lysozymes to breakdown bacterial cell wall. etc.
  2. Eosinophils : Granules of eosinophil contains a unique toxic basic protein receptors that bind to IgE used to help in killing parasites.
  3. Basophils : Granules of basophils contain abundant histamine, heparin and platelet activating factor.

Question 14.
What is haemoglobin count in normal human beings? What is the function of haemoglobin?
Answer:

  1. The normal haemoglobin in adult male is 13-18 mg/100 ml of blood.
  2. In a normal adult female, it is about 11.5-16.5 mg/100 ml of blood.
  3. In anaemic individuals there is lesser amount of haemoglobin.
  4. Functions of haemoglobin is to transport oxygen from lungs to tissues and carbon dioxide from tissues of lungs.
  5. Haemoglobin acts as a buffer and maintains the blood pH.

Question 15.
Why has the heart-recipient to rely upon life-time supply of immunosuppressants?
Answer:
Person who has undergone heart transplant needs lifetime supply of immunosuppressants because in these persons organ rejection is a constant threat. Keeping away the immune system from attacking the transplanted organ requires constant supply of immunosuppressant drugs. These drugs help prevent immune system from attacking (“rejecting”) the donor organ. Typically, these drugs are taken for the life-time for maintaining transplanted organ.

Question 16.
Why is it difficult to hold one’s breath beyond a limit?
Answer:

  1. It is difficult to hold one’s breath beyond a limit because the pressure of oxygen and carbon dioxide in blood changes as one holds his breath.
  2. When the breath is held beyond a limit, the urge to breath becomes irresistible.
  3.  When the breath is held forever, body becomes starved of oxygen and person may fall unconscious and the instinct to breath would take over.

Question 17.
Why and when do the leucocytes perform diapedesis?
Answer:

  1. Diapedesis is the movement of leucocytes through the wall of blood capillaries into the tissue space.
  2. Leucocytes perform diapedesis as an important part of their reaction to tissue injury or infection.
  3. This process forms the part of the innate immune response, involving recruitment of non-specific leucocytes.
  4. Monocytes also use this process during their development into macrophages.
  5. Diapedesis helps leucocytes to perform their functions like phagocytosis, production of antibodies, secretion of inflammatory response chemicals, etc.

Question 18.
Why are obese persons prone to hypertension?
Answer:
(1) Being overweight or obese is a major cause of hypertension, accounting for 65% to 75% of the risk for human primary hypertension.

(2) Following factors play an important role in initiating obesity hypertension:

  1. Physical compression of the kidneys by fat.
  2. Activation of the renin-angiotensin – aldosterone system
  3. Increased sympathetic nervous system activity.
  4. Obesity means more body-surface area. In order to supply blood to these parts. Heart and blood vessels work more resulting into hypertensions.

(3) Blood pressure rises as body weight increases and therefore obese persons are prone to hypertension.

Question 19.
Why does the transplanted heart beats at higher rate than normal?
Answer:

  1. The transplanted heart beats at higher rate than normal (about 100 to 110 beats per minute) because the nerves leading to the heart are cut during the operation. These nerves stimulate the pacemaker i.e. Sinoatrial node.
  2. The new heart also responds more slowly to exercise and does not increase its rate as quickly as before.

Question 20.
Why do large animals cannot carry out respiration without the help of circulatory system?
Answer:

  1. Large animals have various organ systems which always work in a coordinated manner.
  2. These animals provide large respiratory surfaces (numerous alveoli) for the exchange of gases. But these respiratory gases must be carried to the cells of tissues which are away from the respiratory surfaces.
  3. To carry these gases to tissues, there is need of transport system. These gasses are transported from respiratory surfaces to the cells of tissues through blood as a transporting medium.
  4. Therefore, large animals cannot carry out respiration without the help of circulatory system.

Question 21.
What is immunity? Name its types.
Answer:

  1. Immunity is the general ability of a body to recognize, neutralize or destroy and eliminate foreign substances or resist a particular infection or disease.
  2. There are two basic types of immunity, viz. innate immunity and acquired immunity, Acquired immunity is further divided into four types, i.e. Natural acquired active immunity, Natural acquired passive immunity, Artificial acquired active passive immunity and Artificial acquired passive immunity.

Question 22.
Why does the platelet count decrease in dengue patient?
Answer:

  1. The causative pathogen of dengue is dengue virus which induces bone marrow suppression. Since in bone marrow blood cells are formed its suppression causes the deficiency of blood cells leading to low platelet count.
  2. The dengue virus also links with platelets in the blood when there is a virus-specific antibody present in the human body.
  3. When vascular endothelial cell which are infected with dengue virus gets combined with platelets, they tend to destroy platelets. This is one of the major causes of low platelet count in dengue fever.
  4. Even the antibodies that are produced after infection of the dengue virus also cause the destruction of platelets, thus lowering the platelet count.

Question 23.
Why does our immune system fail against pathogens like Trypanosoma cruzi and Plasmodium?
Answer:

  1. Microbes have evolved a diverse range of strategies to destroy the host immune system. The protozoan parasite Trypanosoma cruzi and Plasmodium show similar such adaptations to disturb host defence mechanism.
  2. This parasite attacks host tissues including both peripheral and central lymphoid tissues.
  3. This causes systemic acute response in host body which the parasite tries to overcome. The parasite in fact weakens both innate and acquired immunity.
  4. It interferes with the antigen presenting function of dendritic cells via an action on hosts like lectin receptors. These receptors also induce suppression of CD4+ T cells responses. Therefore, our immune system fail against such pathogens.

Question 24.
What is the relation between immunity and organ transplantation?
Answer:

  1. Those who undergo an organ transplant face the possibility that their immune system will reject their new organ and that they will always be at a higher risk for infections.
  2. The immune system is able to recognize the difference between cells that belong to our body and those that do not by learning to identity protein markers (antigens) that are found on cell and infection surfaces.
  3. In people, the antigens or markers that identity their immune system are referred to as the human leukocyte antigen (HLA).
  4. Antigens that are recognized as unfriendly invaders stimulate an immune response to destroy them.
  5. Therefore, when organ transplantation is done, the immune responses are temporarily stalled. This helps in acceptance of the graft in the recipient’s body.

Question 25.
How do monocytes perform amoeboid movement and phagocytosis?
Answer:
Monocytes can perform phagocytosis. They do this by using intermediary or opsonising proteins such as antibodies or complement that coat the pathogen. They also bind to the microbe directly via pattern-recognition receptors that recognize pathogens. In this way they perform amoeboid movement and indulge in phagocytosis.

Question 26.
How do monocytes modify into macrophages?
Answer:
Monocytes upon having inflammation, selectively travel to the sites of inflammation. Here they produce inflammatory cytokines and contribute to local and systemic inflammation. They are highly infiltrative. They differentiate into inflammatory macrophages, which then remove PAMPs or pathogen-associated molecular patterns and cell debris.

Chart based/Table based questions

Question 1.
Complete the following:

Organism Habitat Respiratory surface/organ
1. Insects Terrestrial —————
2. Amphibian tadpoles of frog, salamanders —————- —————-
3. Fish Aquatic ————–
4. Reptiles, Birds and Mammals —————- —————-

Answer:

Organism Habitat Respiratory surface/organ
1. Insects Terrestrial Tracheal tubes and spiracles
2. Amphibian tadpoles of frog, salamanders Aquatic External gills
3. Fish Aquatic Internal gills
4. Reptiles, Birds and Mammals Terrestrial Lungs

Question 2.

Partial pressure of gases Alveolar air Pulmonary, capillaries
PPO2 ————— —————
PPCO2 —————- —————-

Answer:

Partial pressure of gases Alveolar air Pulmonary, capillaries
PPO2 104 mm Hg 40 mm Hg
PPCO2 40 mm Hg 45 mm Hg

Question 3.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 3
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 4

Question 4.
Complete the following flow chart
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 5
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 6

Question 5.
Complete the following Table

Waves on ECG Heart Activity Caused due to
P wave Atrial contraction —————-
QRS wave —————– Ventricular depolarization
T wave —————– Ventricular repolarization

Answer:

Waves on ECG Heart Activity Caused due to
P wave Atrial contraction Atrial depolarization
QRS wave Ventricular contraction Ventricular depolarization
T wave Ventricular contraction Ventricular repolarization

Question 6.

Cardiovascular disorders Symptom
Coronary Artery Diseases (CAD) Deposition of calcium, fat, cholesterol and ———————
—————- Pain in chest resulting from reduction in the blood supply to the cardiac muscles.
Silent Heart Attack Myocardial infarction without ———————-

Answer:

Cardiovascular disorders Symptom
Coronary Artery Diseases (CAD) Deposition of calcium, fat, cholesterol and fibrous tissues in blood vessels.
Angina pectoris Pain in chest resulting from reduction in the blood supply to the cardiac muscles.
Silent Heart Attack Myocardial infarction without showing symptoms of classical heart attack.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 7.

Instrument / Technique Purpose of use
Sphygmomanometer ———————
—————- X-ray imaging of the cardiac blood vessels to locate the position of blockages.
————— To measure ECG.

Answer:

Instrument / Technique Purpose of use
Sphygmomanometer To measure blood pressure.
Angiography X-ray imaging of the cardiac blood vessels to locate the position of blockages.
Electrocardiograph To measure ECG.

Diagram based questions

Question 1.
Give the name and function of A and ‘B’ from the diagram given below
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 7
Answer:

Name Function
(A) Epiglottis Epiglottis prevents the entry of food into trachea.
(B) Tracheal cartilage Tracheal cartilage prevents collapse of trachea and always keeps it open.

Question 2.
Label parts A’ and ‘B’ from the following diagram and answer the following questions
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 8
(a) What is the partial pressure of O2 in part ‘B’?
(b) What is the partial pressure of CO2 in part A’?
(c) How many alveoli are present in the lungs?
Answer:
Label A → Blood from pulmonary artery Label B → Alveolus
(a) The partial pressure of O2 in alveolar air is 104 mm Hg.
(b) The partial pressure of CO2 in pulmonary capillaries is 45 mmHg.
(c) There are about 700 million alveoli in the lungs.

Question 3.
Label parts A’ and ‘B’ from the given diagram and give their functions.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 9
Answer: Part A → Sinoatrial [SA] node
Function : SA node acts as pacemaker of heart because it has the power of generating a new wave of contraction and making the pace of contraction.
Part B → Atrioventricular [AV] node
Function : Atrioventricular [AV] node acts as pace-setter of heart.

Question 4.
Sketch and label the dorsal (posterior) view of human heart.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 10

Question 5.
Sketch and label the ventral (anterior) view of human heart.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 11

Question 6.
Sketch and label – Electrocardiogram or ECG.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 12

Question 7.
Sketch and label – T.S. of Artery, Vein and Capillary.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 13

Question 8.
Observe the diagrams of blood cells and answer the following questions
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 14
(a) Which of the above is agranulocyte ?
(b) Describe its origin and structure.
(c) Mention its types.
(d) Explain its function.
Answer:
(a) The figure ‘D’ is agranulocyte.
(b) Structure : Agranulocytes do not show cytoplasmic granules and their nucleus is not lobed.
(c) Agranulocytes are of two types, viz. lymphocytes and monocytes.
(d) Functions of agranulocytes : Agranulocytes are responsible for immune response of body by producing antibodies and monocytes are phagocytic in function.

Question 9.
Observe the diagrammatic representation of double circulation and answer the given questions.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 15

  1. Why the circulation shown in the above diagram is called double circulation?
  2. What are the two main routes of double circulation?
  3. Which blood vessels carry oxygented blood to heart and deoxygenated blood to lungs?
  4. Which blood vessels carry deoxygented blood to heart and oxygenated blood to body organs?

Answer:

  1. During circulation, blood passes twice through the heart, therefore it is called double circulation.
  2. (1) Pulmonary circulation which is from heart to lungs and back from lungs to heart.
    (2) Systemic circulation which is from heart to body and back from all body organs to the heart.
  3. Oxygenated blood is carried to the heart by pulmonary veins. Dexoygenated blood is carried to the lungs by pulmonary artery.
  4. Deoxygenated blood is carried to heart by superior and inferior vena cavae. Oxygenated blood is carried to the body organs by systemic or dorsal aorta.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 10.
Observe the cardiac cycle given below and answer the following questions
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 16

  1. Which phases of cardiac cycle are shown in the above diagrammatic representation ?
  2. How much time is taken for entire heart to be in diastole?
  3. How much longer is ventricular systole as compared to atrial systole?

Answer:

  1. There are four main phases of cardiac cycle shown in the above diagram. They are
    (1) AS : Arterial systole. (2) AD : Atrial diastole. (3) VS : Ventricular systole.
    (4) VD : Ventricular diastole which is along with joint diastole.
  2. Diastole of entire heart is called joint diastole, which is for about 0.4 second.
  3. Ventricular systole is almost for the double time than the atrial systole. Atrial systole is for 0.15 second whereas ventricular systole is for 0.3 second.

Long Answer Questions

Question 1.
Describe the respiratory system of human.
Answer:
Respiratory system of human : Human respiratory system consists of nostrils, nasal chambers, pharynx, larynx, trachea, bronchi, bronchioles, lungs, diaphragm and intercostal muscles.
1. Nostrils and nasal chambers:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 17

  1. Oxygen rich air is taken in the body through the nostrils or external nares. They are external opening of the nose. Carbon dioxide and water vapour are also released out of the body through the same passage i.e. the nostrils.
  2. Internal nares open into the pharynx. The space between external and internal nares is knows as nasal chamber which is lined internally by mucous membrane and ciliated epithelium.
  3. Nasal chamber is divided into two parts by a cartilage called mesethmoid. Each part of these halves is further divided into three regions, viz. vestibule, respiratory part and sensory part.
  4. Vestibule is the anteriormost part of nasal chamber. In the vestibule fine hairs are present. They filter out the dust particles and prevent them from going inside.
  5. Respiratory part is the second region which is richly supplied with the capillaries. Air is made warm and moist in this region.
  6. Sensory part is lined by sensory epithelium. It is concerned with the detection of smell.

2. Pharynx:

  1. Pharynx is a short and vertical tube measuring about 12 cm in length. In pharynx the respiratory and food passages cross each other.
  2. The upper part of pharynx is known as naso-pharynx which conducts the air. The lower part is called laryngo-pharynx or oro¬pharynx which conducts food to the oesophagus.
  3. Tonsils that are made up of lymphatic tissue are present in the pharynx. They kill the bacteria by trapping them in the mucus.

3. Larynx:

  1. Larynx produces sound. In males it increases in size at puberty. This is termed as Adam’s apple. It is clearly seen in the neck region.
  2. From pharynx air enters the larynx. The opening through which it enters is called glottis. Glottis is guarded by a flap called epiglottis.
  3. Epiglottis prevents the entry of food particles into the trachea.
  4. TWo folds of elastic tissue called vocal cords are seen along the side of glottis. When they vibrate the sound is produced.

4. Trachea:

  1. The trachea or wind pipe is about 12 cm long and 2.5 cm wide.
  2. It is situated in front of the oesophagus and runs downwards in the thorax through the neck.
  3. The trachea is made up of fibrous muscular tissue wall which is supported by ‘C’-shaped cartilages. These cartilaginous rings are 16 to 20 in number.
  4. Internally the tracheal wall bears ciliated epithelium and mucous glands.
  5. When any foreign particle enters the trachea inadvertently. It is thrown out by coughing action.
  6. Mucous and ciliary action remove the dust particles and push them upwards to the larynx. These particles are then gulped and taken into the oesophagus.

5. Bronchi and bronchioles:

  1. At the distal end, the trachea divides into two bronchi (Singular – bronchus). Bronchi lie below the sternum or breast bone.
  2. Each bronchus has a complete ring of cartilage for support. The two bronchi enter into the lungs on either side.
  3. After entering into the lungs each bronchus divides into secondary and tertiary bronchi. The tertiary bronchi divide and re-divide to form minute bronchioles.
  4. Bronchioles do not have cartilages in their walls. Each bronchiole ends into a balloon like alveolus.
  5. Owing to the presence of alveoli the lungs become spongy and elastic.

6. Lungs:

  1. Lungs are principal respiratory organs located in the thoracic cavity.
  2. They are pinkish, soft, hollow, paired, elastic and distensible organs.
  3. Each lung is enclosed in a pleural sac which consists of two membranes, viz. an outer parietal and inner visceral.
  4. The parietal and visceral membranes enclose pleural cavity which is filled with pleural fluid. The pleural fluid lubricates and prevents friction when pleural membranes slide on each other.
  5. Lungs are highly vascular as they are richly supplied with blood capillaries.
  6. The left lung has two lobes while the right lung has three lobes. Each lobe has many bronchioles and alveolar sacs. The alveolar sacs are spherical and thin walled.
  7. Each alveolar sac contains about 20 alveoli. The alveoli appear as a bunch of grapes. The lobule in the lung thus consists of alveolar ducts, alveolar sacs and alveoli.
  8. Each alveolus has thin and elastic walls. It is about 0.1 mm in diameter. Alveoli are covered by network of capillaries from pulmonary artery and pulmonary vein. A network of pulmonary capillaries supply the alveolus.
  9. The alveolar wall is 0.0001 mm thick and made up of simple, non-ciliated, squamous epithelium. It has collagen and elastin fibres.
  10. Every lung has about 700 million alveoli. They increase the surface area of the lungs for exchange of gases.

Question 2.
Describe the process of respiration in man.
OR
Describe the mechanism of respiration in human beings.
Answer:
Respiration includes breathing, external respiration, internal respiration and cellular respiration.
A. Breathing : During breathing air comes in and goes out of the lungs. The rate of gaseous exchange is speeded up by breathing. Breathing involves two processes, viz. inspiration and expiration
1. Inspiration:

  1. Inspiration is the process in which the air containing oxygen is taken inside the lung.
  2. Inspiration is the active process which is possible due to intercostal muscles, sternum and diaphragm.
  3. During inspiration, intercostal muscles contract, ribs are pulled outward as a result of which the space in the thoracic cavity is increased.
  4. At the same time the lower part of the breast bone is raised and diaphragm flattens by contraction.
  5. The volume of thoracic cavity is thus increased.
  6. Pressure in the lungs decreases as the lungs expand and their volume is increased. Owing to this the atmospheric air enters inside the body through respiratory passage and reaches the lungs.

2. Expiration:

  1. Expiration is the process in which air containing carbon dioxide and water vapour is expelled out of the lungs.
  2. Expiration is a passive process.
  3. During expiration, intercostal muscles relax and ribs are pulled inwards.
  4. The diaphragm relaxes and becomes dome shaped. Intercostal muscles contract simultaneously and due to these events, the volume of the thoracic cavity is reduced.
  5. The pressure on the lungs is thus increased as a result of which they are compressed.
  6. Due to this, air rushes out of the lungs and is expelled out through the nose.
    Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 18

B. External respiration : The process of external respiration takes place in the lungs where oxygen from the lungs diffuses into the blood capillaries present in the lung tissue. Similarly carbon dioxide from the blood capillaries diffuses out and enters in the alveoli in the lungs.

C. Internal respiration : Internal respiration takes place in the cells of the body. Oxygen brought by the blood is given to the cells and the tissues during internal respiration. Similarly carbon dioxide passes into the blood cells from the cells and tissues.

D. Cellular respiration : Cellular respiration takes place in the mitochondria of the cell, where oxygen is utilised to liberate energy in the form of ATP molecules.

Question 3.
Describe the respiratory disorders.
Answer:

  1. Following are some respiratory disorders : Emphysema is caused due to alveolar abnormalities. Chronic bronchitis results into coughing and shortness of breath.
  2. Viral and bacterial respiratory diseases. Acute bronchitis, sinusitis, laryngitis and pneumonia are some of the inflammatory diseases caused either due to virus or due to bacteria.
  3. Allergens like pollen or pet dander can cause asthma. In asthma constriction of bronchioles takes place causing periodic wheezing and difficulty in breathing.
  4. Occupational hazards cause respiratory diseases like silicosis or asbestosis. In these disorders there is inflammation fibrosis leading to lung damage.

Treatments of respiratory diseases:

  1. Bacterial diseases can be completely cured by specific antibiotics.
  2. Viral diseases need to be taken care of by using vaporizers and decongestants.
  3. Asthma needs treatment by inhalers and nebulizers.
  4. For occupational disorders proper mask and other protective gear is a must.
  5. Lethal diseases like pneumonia should be controlled by medication and rest.

Question 4.
How does transport of O2 and CO2 take place in man?
Answer:
1. Transport of O2:

  1. Only 3% of the total oxygen is transported in a dissolved state by the plasma.
  2. The remaining 97% is transported in the form of oxyhaemoglobin in the RBCs.
  3. Hemoglobin present is RBCs combines with oxygen to form oxyhaemoglobin.
    Hb + 4O2 → Hb (4O2)
  4. Oxyhaemoglobin is transported from lungs to the tissues where it readily dissociates to release O2.
  5. Binding of oxygen with haemoglobin in the alveoli and release of oxygen into the tissue cells depends upon the difference in partial pressure of O2 and CO2.

2. Transport of CO2: Carbon dioxide is transported by RBCs and plasma in three different forms.

  1. By plasma in solution form (7%) : About 7% of CO2 is transported in a dissolved form as carbonic acid (which can be broken down into CO2 and H2O).
    CO2 + H2O = H2CO3.
  2. By bicarbonate ions (70%) : Nearly 70% of carbon dioxide is transported in the form of sodium bicarbonate/potassium bicarbonate in the plasma.
  3. RBCs contains an enzyme, carbonic anhydrase. In the presence of this enzyme CO2 combines with water to form carbonic acid.
  4. Carbonic anhydrase also brings about dissociation of carbonic acid immediately tending to large accumulation of HCO3- ions inside the RBCs.
    CO2 + H2O Carbonic anhydrese H2CO2 Carbonic anhydrase H+ + HCO3-
  5. The bicarbonate ions moves out of RBCs and this would bring about imbalance of the charge inside the RBCs.
  6. To maintain the ionic balance, Cl ions diffuse from plasma into the RBCs. This movement of chloride ions is known as chloride shift or Hamburger’s phenomenon.
  7. HCO3- ions from the plasma then joins to Na+/K+ forming NaHCO3/KHCO3 (to maintain PH of blood).
    HCO3- + Na+ → NaHCO3 Sodium bicarbonate
  8. H+ is taken up by haemoglobin to form Reduced Hb (HHb).
  9. At the level of the lungs due to the low partial pressure of the alveolar air, hydrogen ion and bicarbonate ions recombine to form carbonic acid and in presence of carbonic anhydrase it again yields carbon dioxide and water.
    H+ + HCO3- Carbonic anhydrase H2CO3 Carbonic anhydrase CO2 + H2O.

3. By red blood cells (23%):

  1. Carbon dioxide binds with the amino group of the haemoglobin and form a loosely bound compound carbaminohaemoglobin Hb + CO2 = HbCO2
  2. Due to low partial pressure of CO2 at alveolus carbaminohaemoglobin decomposes releasing the carbon dioxide.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 5.
Describe different types of leucocytes.
OR
Describe five types of leucocytes, with the help of diagrams. Add a note on their functions.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 19

  1. Leucocytes or White Blood Corpuscles (WBCs) are colourless, nucleated, amoeboid and phagocytic cells.
  2. Their size ranges between 8 to 15 pm. Total WBC count is 5000 to 9000 WBCs/cu mm of blood. The average life span of a WBC is about 3 to 4 days.
  3. They are formed by leucopoiesis in red bone marrow, spleen, lymph nodes, tonsils, thymus and Payer’s patches, whereas the dead WBCs are destroyed by phagocytosis in blood, liver and lymph nodes.
  4. Leucocytes are mainly divided into two types, viz., granulocytes and agranulocytes.
  5. Granulocytes : Granulocytes are cells with granular cytoplasm and lobed nucleus. Based on their staining properties and shape of nucleus, they are of three types, viz. neutrophils, eosinophils and basophils.

(I) Neutrophils:

  1. In neutrophils, the cytoplasmic granules take up neutral stains.
  2. Their nucleus is three to five lobed.
  3. It may undergo changes in structure hence they are called polymorphonuclear leucocytes or polymorphs.
  4. Neutrophils are about 70% of total WBCs.
  5. They are phagocytic in function and engulf microorganisms.

(II) Eosinophils or acidophils:

  1. Cytoplasmic granules of eosinophils take up acidic dyes such as eosin. They have bilobed nucleus.
  2. Eosinophils are about 3% of total WBCs.
  3. They are non-phagocytic in nature.
  4. Their number increases (i.e. eosinophilia) during allergic conditions.
  5. They have antihistamine property.

(III) Basophils:

  1. The cytoplasmic granules of basophils take up basic stains such as methylene blue.
  2. They have twisted nucleus.
  3. In size, they are smallest and constitute about 0.5% of total WBCs.
  4. They too are non-phagocytic.
  5. Their function is to release heparin which acts as an anticoagulant and histamine that is involved in inflammatory and allergic reaction.

Agranulocytes : There are two types of agranulocytes, viz. monocytes and lymphocytes. Agranulocytes do not show cytoplasmic granules and their nucleus is not lobed. They are of two types, viz. lymphocytes and monocytes.
(I) Lymphocytes:

  1. Agranulocytes with a large round nucleus are called lymphocyte.
  2. They are about 30% of total WBCs.
  3. Agranulocytes are responsible for immune response of the body by producing antibodies.

(II) Monocytes:

  1. Largest of all WBCs having large kidney shaped nucleus are monocytes. They are about 5% of total WBCs.
  2. They are phagocytic in function.
  3. They can differentiate into macrophages for engulfing microorganisms and removing cell debris. Hence they are also called scavengers.
  4. At the site of infections they are seen in more enlarged form.

Question 6.
Give an account of external features of the human heart.
Answer:
(1) The heart is hollow, muscular, conical organ about the size of one’s fist with broad base and narrow apex tilted towards left measuring about 12 cm in length. 9 cm in breadth and weighing about 250 to 300 grams.

(2) The human heart has four chambers, two atria which are superior, small, thin walled receiving chambers and two ventricles which are inferior, large, thick walled, distributing chambers.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 20

(3) Externally there is a transverse groove between the atria and the ventricles which is known as atrioventricular groove or coronary sulcus.

(4) Between the right and left ventricles there is interventricular sulcus (pi. sulci). In these sulci the coronary arteries and coronary veins are present.

(5) Oxygenated blood to the heart is supplied by coronary arteries while coronary veins collect deoxygenated blood from the heart. The coronary veins join to form coronary sinus which opens into the right atrium.

(6) Right atrium is larger in size than the left atrium. Deoxygenated blood from all over the body is brought through superior vena cava and inferior vena cava and poured into right atrium. Oxygenated blood from lungs is brought to heart by two pairs of pulmonary veins which carry it to the left atrium.

(7) Pulmonary trunk is seen arising from the right ventricle, which carries deoxygenated blood to lungs. While systemic aorta arises from the left ventricle and carries oxygenated blood to all parts of the body.

(8) The pulmonary trunk and systemic aorta are connected by ligamentum arteriosum that represents remnant of ductus arteriosus of foetus.

Question 7.
With the help of well labelled diagram describe the internal structure of human heart.
OR
Sketch and label internal view of heart.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 21
The heart shows four chambers with two atria and two ventricles.

I. Atria:
1. Right atrium:

  1. There are two atria which are separated from each other by interatrial septum. They are thin walled receiving chambers on the upper side.
  2. The right atrium receives deoxygenated blood from upper part of body through superior vena cava and from the lower part of the body by inferior vena cava. In the right atrium opens the coronary sinus.
  3. Eustachian valve guards the opening of inferior vena cava while opening of coronary sinus is guarded by Thebesian valve.
  4. On the right side of interatrial septum is seen an oval depression called the fossa ovalis. In the interatrial septum of the foetus there is an oval opening called foramen ovale. Fossa ovalis is remnant of this foramen ovale.
  5. Right atrium opens into the right ventricle.

2. Left atrium:

  1. The oxygenated blood from the lungs is brought into left atrium through four openings of pulmonary veins.
  2. Left atrium opens into the left ventricle.

II. Ventricles:

  1. There are two ventricles which are separated from each other by interventricular septum. They are two thick walled distributing chambers situated on the lower side of the heart.
  2. Left ventricle has thickest wall as it pumps blood to all parts of the body.
  3. The inner surface of the ventricle is thrown into a series of irregular muscular ridges called columnae carnae or trabeculae carnae.
  4. Each atrium opens into the ventricle of its side through atrioventricular aperture. These apertures are guarded by valves made up of connective tissue. The right atrioventricular valve has three flaps hence called tricuspid valve. Left atrioventricular valve has two flaps hence called bicuspid valve or mitral valve.
  5. Bicuspid and tricuspid valves are attached to papillary muscles of ventricles by chordae tendinae. The chordae tendinae prevent the valves from turning back into the atria during the contraction of ventricles.
  6. From the right ventricle arises pulmonary trunk which carries deoxygenated blood to lungs for oxygenation.
  7. From the left ventricle arise systemic aorta which distributes oxygenated blood to all parts of the body.
  8. Pulmonary aorta and systemic aorta has three semilunar valves at the base which prevent backward flow of blood during ventricular diastole.

Question 8.
With the help of suitable diagram, describe the conducting system of human heart.
Answer:
(1) The human heart is myogenic.

(2) Conducting system of the heart consists of sinoatrial node, atrioventricular node, bundles of His and Purkinje’s fibre system.

(3) The pacemaker of the heart is sinoatrial node because here the heartbeat originates. Pacemaker has power of generation of wave of contraction. This is modified cardiac tissue, also called a nodal tissue.

(4) SA node is situated in the wall of right atrium near the opening of superior vena cava. The wave of contraction generated by SA node is conducted by cardiac muscle fibres to both the atria. This results in contraction resulting into atrial systole.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 22

(5) The atrioventricular node (AV node) is located in the wall of right atrium near the opening of coronary sinus. AV node receives the wave of contraction generated by SA node through intermodal pathways.

(6) Bundle of His arises from AV node and divides into right and left bundle branches. These are located in the interventricular septum.

(7) The bundle branches further form Purkinje fibres which penetrate into myocardium of ventricles.

(8) The bundle of His and Purkinje fibres conduct the wave of contraction from AV node to myocardium of ventricles causing ventricular systole.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 9.
Describe the detail cardiac cycle.
OR
Explain the working of heart.
Answer:
(1) The working of heart or cardiac cycle is formed by atrial systole, ventricular systole and joint diastole. It takes place in 0.8 second.

(2) During atrial systole from right atrium, the deoxygenated blood is poured into right ventricle through atrioventricular aperture. Similarly, from left atrium, the oxygenated blood enters the left ventricle through atrioventricular aperture. This entire atrial systole lasts for 0.1 second.

(3) After auricular systole follows the ventricular systole. During ventricular systole, the deoxygenated blood from the right ventricle enters the pulmonary trunk, which carries blood to lungs for oxygenation. At the same time, the oxygenated blood from the left ventricle enters the aorta which is then supplied to all parts of the body. The ventricular systole lasts for 0.3 second.

(4) Joint diastole or complete cardiac diastole is the phase taking place after the systole, when the entire heart undergoes relaxation, for 0.4 second.

(5) During joint diastole, the right atrium receives deoxygenated blood from all parts of the body through superior vena cava, inferior vena cava and coronary sinus. The left atrium receives oxygenated blood from the lungs through two pairs of pulmonary veins.

Question 10.
What is blood clotting? How and when does it occur?
Answer:

  1. Blood clotting is coagulation of blood in order to stop the blood flow and resuting blood loss at the time of injury.
  2. When the blood vessel is intact, blood does not clot due to the presence of active anticoagulants like heparin and antithrombin. But when there is an injury causing rupture of a blood vessel, bleeding starts.
  3. This bleeding is stopped by the process of blood clotting during which liquid blood is converted into semisolid jelly.

The events occurring during blood clotting are as follows:

  1. Release of thromboplastin from thrombocytes and injured tissue.
  2. Formation of enzyme prothrombinase in the blood due to initiation of thromboplastin.
  3. Conversion of inactive prothrombin into active thrombin by prothrombinase in the presence of Ca ions.
  4. Conversion of soluble fibrinogen into insoluble fibrin by thrombin.
  5. Formation of a clot by enmeshing platelets, other blood cells and plasma in the fibrin fibres enmesh.

These reactions occur in 2 to 8 minutes. Therefore, clotting time is said to be 2 to 8 minutes.

Question 11.
What is repolarization and depolarization ?
Answer:
Repolarization is a stage of an action potential in which the cell experiences a reduction of voltage due to the efflux of potassium (K+) ions along its electrochemical gradient. This phase occurs after the cell reaches its highest voltage from depolarization.

Depolarization occurs in the four chambers of the heart : both atria first and then both ventricles. The SA node sends the depolarization wave to the atrioventricular (AV) node which-with about a 100 minutes delay to let the atria finish contracting-then causes contraction in both ventricles, seen in the QRS wave.

Question 12.
What is the correlation between depolarization and repolarization as well as contraction and relaxation of the heart?
Answer:
Depolarization and Repolarization:

  1. When cardiac cells are at rest, they are polarized, meaning no electrical activity takes place.
  2. The cell membrane of the cardiac muscle cell separates different concentrations of ions, such as sodium, potassium, and calcium. This is called the resting potential.
  3. Electrical impulses are generated by specialized cardiac cells automatically.
  4. Once an electrical cell generates an electrical impulse, this electrical impulse causes the ions to cross the cell membrane and causes the action potential, also called depolarization.
  5. The movement of ions across the cell membrane through sodium, potassium and calcium channels, is the drive that causes contraction of the cardiac cells/muscle.
  6. Depolarization with corresponding contraction of myocardial muscle moves as a wave through the heart. Depolarization thus corresponds with contraction of heart.
  7. Repolarization is the return of the ions to their previous resting state, which corresponds with relaxation of the myocardial muscle. Repolarization thus corresponds with relaxation of heart.
  8. Depolarization and repolarization are electrical activities which cause muscular activity.
  9. The electrical changes in the myocardial cell during the depolarization – repolarization cycle is detected on ECG.

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

Question 13.
How are the signals detected and amplified by electrocardiograph?
Answer:

  1. The action potential created by contractions of the heart wall spreads electrical currents from the heart throughout the body.
  2. The spreading electrical currents create different potentials at points in the body, which can be sensed by electrodes placed on the skin.
  3. The electrodes are made of metals and salts and they act as biological transducers.
  4. Ten electrodes are attached to different points on the body while taking ECG.
  5. There Eire three main leads responsible for measuring the electrical potential difference between arms and legs.
  6. Electrical potential difference between electrodes is recorded.
  7. As in all ECG lead measurements, the electrode connected to the right leg is considered the ground node.
  8. These ECG signals are acquired using a biopotential amplifier and then displayed using instrumentation software. This is recorded on ECG machine or electrocardiograph. The recorded ECG is anailysed by an expert.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Multiple choice questions

Question 1.
For the following, which is the aspect of growth? Example – Increase in length, size and number of cells.
(a) Quantitative
(b) Qualitative
(c) Both (a) and (b)
(d) Three dimensional
Answer:
(a) Quantitative

Question 2.
In vascular plants, growth takes place due to ………………..
(a) conducting tissues
(b) embryo
(c) meristems
(d) stem cell
Answer:
(c) meristems

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 3.
What is growth?
(a) Temporary, irreversible increase in an organism.
(b) Permanent, reversible increase in an organism.
(c) Permanent, irreversible decrease in an organism.
(d) Permanent, irreversible increase in an organism.
Answer:
(d) Permanent, irreversible, increase in an organism

Question 4.
Based on location, plants have these three types of meristems.
(a) Basal, Intercalary and Lateral
(b) Apical, Basal and Lateral
(c) Apical, Interfascicular and Lateral
(d) Apical, Intercalary, Lateral
Answer:
(d) Apical, Intercalary, Lateral

Question 5.
Meristem cells are ………………..
(a) thin walled, vacuolated with prominent nucleus
(b) thick walled, non-vacuolated without nucleus
(c) thin walled, non-vacuolated with prominent nucleus
(d) thick walled, vacuolated with prominent nucleus
Answer:
(c) thin walled, non-vacuolated with prominent nucleus

Question 6.
Select correct sequence of phases of growth.
(a) Phase of formation, phase of elongation, phase of maturation
(b) Phase of cell division, phase of enlargement, phase of elongation
(c) Phase of formation, phase of maturation, phase of elongation
(d) Phase of enlargement, phase of cell division, phase of maturation
Answer:
(a) Phase of formation, phase of elongation, phase of maturation

Question 7.
In which phase of growth, growth rate is at accelerated pace?
(a) Lag phase
(b) Log phase
(c) Steady phase
(d) Stationary phase
Answer:
(b) Log phase

Question 8.
Water is essential for growth because it is necessary ………………..
(a) for turgidity
(b) as nutrient
(c) as raw material
(d) for gravity
Answer:
(a) for turgidity

Question 9.
Which equipment is suitable for measuring linear growth of shoot?
(a) Horizontal microscope
(b) Spectroscope
(c) Crescograph
(d) Auxanometer
Answer:
(d) Auxanometer

Question 10.
Which is correct expression of absolute growth rate (AGR)?
(a) AGR = \(\frac { dn }{ dt }\)
(b) AGR = \(\frac { dt }{ dn}\)
(c) AGR = \(\frac { RGR }{ n }\)
(d) AGR = \(\frac { n }{ RGR }\)
Answer:
(a) AGR = \(\frac { dn }{ dt }\)

Question 11.
Arithmetic growth in plants shows ……………….. graph.
(a) Sigmoid
(b) J-shaped
(c) linear
(d) elliptical
Answer:
(c) linear

Question 12.
What is grand period of growth?
(a) The total time required for all phases to occur
(b) The total time required for exponential phase
(c) The total time required for Lag and Log phase together
(d) The toted time required for stationary phase
Answer:
(a) The total time required for all phases to occur

Question 13.
Which tissue is formed by process of de-differentiation?
(a) Intrafascicular cambium
(b) Secondary phloem
(c) Interfascicular cambium
(d) Secondary xylem
Answer:
(c) Interfascicular cambium

Question 14.
The example of environmental plasticity- heterophylly observed is ………………..
(a) Cotton
(b) Coriander
(c) Larkspur
(d) Buttercup
Answer:
(d) Buttercup

Question 15.
Synthesis of IAA takes place from amino acid ………………..
(a) Methionine
(b) Tryptophan
(c) Valine
(d) Aspartic acid
Answer:
(b) Tryptophan

Question 16.
Find the odd one out.
(a) IAA
(b) 2, 4-D
(c) NAA
(d) IBA
Answer:
(a) IAA

Question 17.
The selective herbicide is ………………..
(a) IBA
(b) GA3
(c) 2, 4 D
(d) NAA
Answer:
(c) 2, 4 D

Question 18.
This hormone promotes rooting in artificial method of cutting ………………..
(a) Gibberellin
(b) Auxin
(c) Cytokinin
(d) Dormin
Answer:
(b) Auxin

Question 19.
Chemically the peculiar structure of gibberellins is ……………….. ring.
la) pyrole ring
(b) purine ring
(c) gibbeane ring
(d) pyrimidine
Answer:
(c) gibbeane ring

Question 20.
First natural cytokinin was obtained from ……………….. by Letham.
(a) Maize grains
(b) Coconut milk
(c) Rice seedling
(d) Tomato
Answer:
(a) Maize grains

Question 21.
A low ratio of cytokinin to auxin induces ……………….. in plants.
(a) rooting
(b) shooting
(c) bud formation
(d) flowering
Answer:
(a) rooting

Question 22.
Apical dominance : Auxin : : Fruit ripening : ?
(a) Gibberellin
(b) Cytokinin
(c) Ethylene
(d) Abscissic acid
Answer:
(c) Ethylene

Question 23.
Which hormone is known as stress hormone ?
(a) Auxin
(b) Gibberellin
(c) Ethylene
(d) Abscissic acid
Answer:
(d) Abscissic acid

Question 24.
Abscissic acid is synthesised from ………………..
(a) Methionine
(b) Malic acid
(c) Mevalonic acid
(d) Mucin
Answer:
(c) Mevalonic acid

Question 25.
Photoperiodic response is because of pigment ………………..
(a) Cytochrome
(b) Phytochrome
(c) Anthocyanin
(d) Phycobilin
Answer:
(b) Phytochrome

Question 26.
The favourable temperature for vernalization is ………………..
(a) 1 to 6 °C
(b) 11 to 16 °C
(c) 10 to 16 °C
(d) – 1 to 1 °C
Answer:
(a) 1 to 6 °C

Question 27.
Identify the group of non-mineral elements needed by plants.
(a) PO4, CO3, SO4
(b) C, H, O
(c) N, P K
(d) C, H, N
Answer:
(b) C, H, O

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 28.
The deficiency symptoms of these minerals are visible in young leaves ………………..
(a) Ca, CO
(b) S, P
(c) Ca, S
(d) Zn, Mg
Answer:
(c) Ca, S

Question 29.
The symptom chlorosis is observed as ………………..
(a) yellowing of leaf
(b) premature leaf fall
(c) malformation of leaf
(d) localized death of tissue
Answer:
(a) yellowing of leaf

Question 30.
…………….. is constituent of chlorophyll.
(a) Mn
(b) Mg
(c) Mo
(d) Fe
Answer:
(b) Mg

Question 31.
This is essential for O2 evolution in photosynthesis and for proper solute concentration
(a) Ca
(b) Cu
(c) Cl
(d) Co
Answer:
(c) Cl

Question 32.
For active absorption of mineral uptake, energy is supplied by ………………..
(a) respiration
(b) chemosynthesis
(c) photosynthesis
(d) transpiration
Answer:
(a) respiration

Question 33.
Cyanobacteria fix nitrogen in specialised cells called ………………..
(a) Velamen
(b) Haustoria
(c) Heteocysts
(d) Hormogonia
Answer:
(c) Heteocysts

Question 34.
In root nodule, symbiotic nitrogen fixer organism is ………………..
(a) Rhizopus
(b) Pseudomonas
(c) Rhizobium
(d) Nitrosomonas
Answer:
(c) Rhizobium

Question 35.
In plant body, amides are transported through ………………..
(a) sieve tubes
(b) xylem vessels
(c) phloem parenchyma
(d) plasmodesmata
Answer:
(b) xylem vessels

Question 36.
Building blocks of proteins are ………………..
(a) amides
(b) amino acids
(c) carboxylic acid
(d) nitrates
Answer:
(b) amino acids

Question 37.
Flowering plants Aster, Dahlia and Chrysanthemum are ………………..
(a) SDP
(b) LDP
(c) DNP
(d) SDP or LDP
Answer:
(a) SDP

Question 38.
Experimental material of Garner and Allard for discovery of photoperiodism was ………………..
(a) Cucumber and Tomato
(b) Dahlia and Aster
(c) Soybean and Tobacco
(d) Cabbage and Spinach
Answer:
(c) Soybean and Tobacco

Question 39.
Which of the following is used for the production of long seedless grapes ?
(a) Auxin
(b) Cytokinin
(c) Ethylene
(d) Gibberellin
Answer:
(d) Gibberellin

Question 40.
In vernalization, the cold stimulus is perceived by ………………..
(a) axillary bud
(b) floral bud
(c) leaves
(d) apical bud (shoot apex)
Answer:
(d) apical bud (shoot apex)

Question 41.
Xanthium is ………………..
(a) SDP
(b) LDP
(c) DNP
(d) not a flowering plant
Answer:
(a) SDP

Question 42.
Growth starts slowly during the ………………..
(a) lag phase
(b) exponential phase
(c) maturation phase
(d) stationary phase
Answer:
(a) lag phase

Question 43.
Cytokinins induce the formation of ………………..
(a) shoot apex
(b) intrafascicular cambium
(c) cork cambium
(d) interfascicular cambium
Answer:
(d) interfascicular cambium

Question 44.
Bakane disease in rice is associated with the discovery of ………………..
(a) cytokinins
(b) gibberellins
(c) auxins
(d) ethylene
Answer:
(b) gibberellins

Question 45.
ABA is also known as ………………..
(a) antitoxin
(b) antivirulent
(c) antioxidant
(d) antigibber ellin
Answer:
(d) antigibberellin

Question 46.
Gibberellins were first discovered from ………………..
(a) bacteria
(b) fungi
(c) algae
(d) gymnosperms
Answer:
(b) fungi

Question 47.
Which of the following is trace element?
(a) Mg
(b) Nitrogen
(c) Sulphur
(d) Mn
Answer:
(d) Mn

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 48.
Which of the following is a macronutrient?
(a) Ca
(b) Mn
(c) Zn
(d) Mo
Answer:
(a) Ca

Question 49.
Nitrogen is an important constituent of ………………..
(a) carbohydrates
(b) sugars
(c) proteins
(d) polyphosphates
Answer:
(c) proteins

Question 50.
Deficiency of phosphorus causes ………………..
(a) stunted growth
(b) inward rolling of leaf margin
(c) brittle cell walls
(d) necrotic spots
Answer:
(a) stunted growth

Question 51.
Which of the following is not an essential element for plant?
(a) Sulphur
(b) Boron
(c) Iron
(d) Cadmium
Answer:
(d) Cadmium

Question 52.
…………….. is a constituent of middle lamella.
(a) Mg
(b) K
(c) Ca
(d) P
Answer:
(c) Ca

Match the columns

Question 1.

Column A (Phase of growth) Column B (Condition)
(1) Lag phase (a) Growth rate faster
(2) Log phase (b) Growth rate steady state
(3) Stationary phase (c) Growth rate slow

Answer:

Column A (Phase of growth) Column B (Condition)
(1) Lag phase (c) Growth rate slow
(2) Log phase (a) Growth rate faster
(3) Stationary phase (b) Growth rate steady state

Question 2.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 1
Answer:
(1) -(c) Arithmetic growth
(2) – (a) Rate of growth against time
(3) -(b) Geometric growth curve

Question 3.

Column A Column B
(1) Use of measuring scale (a) Record of primary growth
(2) Crescograph (b) Increase in height of plant
(3) Auxanometer (c) Measure growth in field
(4) Horizontal microscope (d) Measurement of linear growth of shoot

Answer:

Column A Column B
(1) Use of measuring scale (b) Increase in height of plant
(2) Crescograph (a) Record of primary growth
(3) Auxanometer (d) Measurement of linear growth of shoot
(4) Horizontal microscope (c) Measure growth in field

Question 4.

Column A Column B
(1) Auxin (a) Bolting in rosette plants
(2) Cytokinin (b) Stimulate flowering in SDP
(3) Gibberellins (c) Promotion of growth of lateral buds
(4) Abscissic acid (d) Apical dominance

Answer:

Column A Column B
(1) Auxin (d) Apical dominance
(2) Cytokinin (c) Promotion of growth of lateral buds
(3) Gibberellins (a) Bolting in rosette plants
(4) Abscissic acid (b) Stimulate flowering in SDP

Question 5.

Column A Column B (Symptoms observed)
(1) Deficiency of Cu (a) Malformed leaves
(2) Deficiency of Bo (b) Leaves with yellow edges
(3) Deficiency of Zn (c) Brown heart disease
(4) Deficiency of K (d) Die back of shoot

Answer:

Column A Column B (Symptoms observed)
(1) Deficiency of Cu (d) Die back of shoot
(2) Deficiency of Bo (c) Brown heart disease
(3) Deficiency of Zn (a) Malformed leaves
(4) Deficiency of K (b) Leaves with yellow edges

Question 6.

Column A Column B (Organisms)
1. Symbiotic nitrogen fixation a. Nitrobacter
2. Denitrification b. Cyanobacteria
3. Free living nitrogen fixers c. Rhizobium
4. Nitrification d. Paracoccus

Answer:

Column A Column B (Organisms)
1. Symbiotic nitrogen fixation c. Rhizobium
2. Denitrification d. Paracoccus
3. Free living nitrogen fixers b. Cyanobacteria
4. Nitrification a. Nitrobacter

Classify the following to form Column B as per the category given in Column A.

Question 1.
Classify the given plant growth regulators as per their specific control of event in plant life cycle in Column A and complete Column B.
(IAA, GA, Cytokinin, Abscissic acid)

Column A Column B
(1) Shedding of leaves ————-
(2) Induce flowering in LDP ————
(3) Apical dominance ————-
(4) Induce RNA synthesis ————-

Answer:

Column A Column B
(1) Shedding of leaves Abscissic acid
(2) Induce flowering in LDP GA
(3) Apical dominance IAA
(4) Induce RNA synthesis Cytokinin

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 2.
Classify the given specific effects of different types of auxins in Column A and complete Column B with the examples.
(IAA, NAA, 2, 4-D, IBA, 2, 4, 5-T)

Column A Column B
(1) Selective synthetic herbicide ————-
(2) Seedless fruits ————
(3) Flowering in pineapple ————-
(4) Synthetic auxin ————-
(5) Agent orange ————

Answer:

Column A Column B
(1) Selective synthetic herbicide 2, 4-D
(2) Seedless fruits IAA
(3) Flowering in pineapple NAA
(4) Synthetic auxin IBA
(5) Agent orange 2, 4, 5 – T

Question 3.
Classify the given plant hormones as their specific effect observed in plants in Column A and complete Column B.
(ABA, GA, Ethylene, Kinetin)

Column A Column B
(1) Bolting of rosette plants ————-
(2) Epinasty ————
(3) Closure of stomata ————-
(4) Proliferation of callus ————-

Answer:

Column A Column B
(1) Bolting of rosette plants GA
(2) Epinasty Ethylene
(3) Closure of stomata ABA
(4) Proliferation of callus Kinetin

Question 4.
Classify the given disease to their cause given in Column B.
(Ethylene, GA, Deficiency of BO, Deficiency of Cu)

Column A Column B
(1) Brown heart disease ————-
(2) Bakane disease of Rice ————
(3) Die back of shoot ————-
(4) Degreening of Banana ————-

Answer:

Column A Column B
(1) Brown heart disease Deficiency of BO
(2) Bakane disease of Rice GA
(3) Die back of shoot Deficiency of Cu
(4) Degreening of Banana Ethylene

Question 5.
Classify the given organisms related to Nitrogen cycle in Column B.
[Nitrobacter, Rhizobium, Pseudomonas, Nitrosococcus)

Column A Column B
(1) Symbiont in root nodule ————-
(2) Conversion of nitrite to nitrate ————
(3) Denitrification process ————-
(4) Conversion of ammonia to nitrite ————-

Answer:

Column A Column B
(1) Symbiont in root nodule Rhizobium
(2) Conversion of nitrite to nitrate Nitrobacter
(3) Denitrification process Pseudomonas
(4) Conversion of ammonia to nitrite Nitrosococcus

Very short answer questions

Question 1.
Enlist the types of meristems that we observe in plants.
Answer:
In plants, there are apical, intercalary and lateral meristems.

Question 2.
Give characteristic features of meristematic cells.
Answer:
Meristematic cells are thin walled, non- vacuolated with prominent nuclei having granular cytoplasm and are capable of cell division.

Question 3.
What is the role of oxygen in growth?
Answer:
Oxygen is required for respiration of cells and release of energy for the process of growth.

Question 4.
What is the role of water for growth?
Answer:
Water maintains turgidity of the cell and is chief component of protoplasm as well as it is a medium for various biochemical reactions.

Question 5.
Mention the mathematical formula for rate of absolute growth.
Answer:
Absolute Growth Rate = AGR = \(\frac { dn }{ dt }\) where dn is cell number and dt is time interval.

Question 6.
What is exponential phase of growth curve?
Answer:
In exponential phase or log phase, growth rate is faster. It accelerates and reaches its maximum.

Question 7.
Is there any relation between phases of growth and regions of growth curve?
Answer:
Yes, there is relation between the two as initially growth is slow, which accelerates and ultimately it slows down and becomes steady which is observed in growth curve.

Question 8.
Which plant organ does show both arithmetic and geometric growth?
Answer:
Embryo that develops from zygote inside the seed shows initially the growth which is geometric and later on arithmetic.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 9.
Describe the example of plasticity related to internal stimuli.
Answer:
In plants like cotton, coriander and larkspur heterophylly is observed where leaves in juvenile stage and adult stage show different forms.

Question 10.
What is the role of growth hormones in plants?
Answer:
Growth hormones inhibit, promote or modify the growth in plants.

Question 11.
What is the peculiarity of growth hormones ?
Answer:
Growth hormones are needed in very small amount to evoke the response and they act at a site away from their place of production.

Question 12.
Which is the first hormone to be discovered in plants?
Answer:
Auxin IAA i.e. Indole acetic acid is the first hormone to be discovered in plants.

Question 13.
Give the full form of 2, 4 – D.
Answer:
It is synthetic auxin 2, 4-dichlorophenoxy acetic acid.

Question 14.
What is the effect of NAA and 2, 4-D foliar spray?
Answer:
Foliar spray of synthetic hormones induce flowering in plants litchi and pineapple and prevent premature fruit drop in apples, pear and oranges.

Question 15.
How cytokinins control apical dominance ?
Answer:
Cytokinins promote growth of lateral buds by cell division and thus control apical dominance.

Question 16.
What was the discovery of Richmond and Lang?
Answer:
Richmond and Lang discovered that cytokinins delay the process of ageing and senescence, abscission in plant organs.

Question 17.
What is ‘epinasty’?
Answer:
It is an effect of ethylene where it causes drooping of leaves and flowers.

Question 18.
Why is auxin called a growth regulator?
Answer:
Auxin is synthesized at meristematic region of plants and it controls cell enlargement, cell elongation and stimulates growth of stem and root, apical dominance. Hence it is growth promoting hormone.

Question 19.
What is effect of gibberellin application on apple?
Answer:
Gibberellin causes parthenocarpy in apple.

Question 20.
How can we overcome apical dominance?
Answer:
By application of cytokinin we can overcome apical dominance effect.

Question 21.
Which is standard bioassay method for auxins?
Answer:
Avena curvature test/Avena coleoptile test is a standard bioassay method for auxins.

Question 22.
ABA is called as stress hormone why?
Answer:
Answer: ABA induces dormancy in seeds by inhibiting growth. Thus plants can tide over adverse environmental conditions. Hence it is called as stress hormone.

Question 23.
What was the plant material for study of photoperiodism by Garner and Allard?
Answer:
The flowering response of Soybean and Maryland mammoth variety of tobacco was studied by Garner and Allard.

Question 24.
What is photomorphogenesis?
Answer:
Control of morphogenesis by light and phytochrome pigment is called photomorphogenesis.

Question 25.
What is critical concentration of minerals ?
Answer:
The concentration of the essential elements below which plant growth is retarded is termed as critical concentration.

Question 26.
What is a role of Sulphur in plants?
Answer:
Sulphur is constituent of amino acids, proteins, vitamins (mainly thaimine, biotin CoA) and Ferredoxin.

Question 27.
What is a role of nitrogen in plants?
Answer:
Nitrogen is constituent of amino acids, proteins, nucleic acid, vitamins, hormones, coenzymes, ATP and chlorophyll molecule.

Question 28.
Which is the process by which mainly we get nitrogen in human tissues?
Answer:
Industrial nitrogen fixation by Haber – Bosch Nitrate process is responsible for nitrogen found in human tissues.

Question 29.
Give equation of Haber – Bosch process.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 2

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 30.
What is nitrogen assimilation?
Answer:
Nitrogen present in the soil as nitrates, nitrites and ammonia is absorbed by plants and converted into nitrogenous organic compounds is nitrogen assimilation.

Give definitions of the following

Question 1.
Growth
Answer:
Growth can be defined as permanent, irreversible increase in the bulk of an organism with change in form.

Question 2.
Efficiency index
Answer:
The increased growth per unit time is called efficiency index.

Question 3.
Absolute growth rate (AGR)
Answer:
The measurement and comparison of total growth per unit time is called absolute growth rate.

Question 4.
Relative growth rate (RGR)
Answer:
The growth of a particular system per unit time expressed on a common basis or alternately it is the ratio of growth in the given time per initial growth.

Question 5.
Differentiation
Answer:
A permanent change in structure and function of cells that leads to their maturation is called differentiation.

Question 6.
Redifferentiation
Answer:
When cells produced from de-differentiation lose their capacity to divide and mature for specific function it is known as re-differentiation.

Question 7.
Development
Answer:
The progressive changes in shape, form and degree of complexity which includes growth, maturation and morphogenesis is referred as development.

Question 8.
Growth Hormone or Growth regulators
Answer:
The internal factors that influence growth by inhibiting, promoting or modifying it are called growth hormones or regulators.

Question 9.
Apical dominance
Answer:
In higher plants growing apical bud inhibits the growth of lateral buds. This is known as apical dominance.

Question 10.
Critical photoperiod
Answer:
That length of photoperiod above or below which the plant shows flowering is critical photoperiod.

Question 11.
Photomorphogenesis
Answer:
The control of morphogenesis of plants by light and phytochrome is photomorphogenesis.

Question 12.
Symptom or hunger sign
Answer:
Any visible deviation from the normal structure and function of the plant is called symptom or hunger sign.

Question 13.
Active absorption of minerals
Answer:
Uptake of mineral ions against concentration gradient, with expenditure of energy (ATP) is called active absorption.

Question 14.
Nitrogen fixation
Answer:
Free nitrogen of air N2 is converted to nitrogenous salts so that it is made available to plants is called nitrogen fixation.

Question 15.
Nitrification
Answer:
Soil microbes, mainly : chemoautotrophs convert ammonia into j nitrate, the form of nitrogen which can be used by plants, this process is nitrification.

Question 16.
Amidation
Answer:
Ammonia may be absorbed by amino acids to produce amides. This process is called amidation.

Name the following

Question 1.
Instrument developed by Indian physiologist to measure growth.
Answer:
Crescograph developed by Sir J.C. Bose.

Question 2.
Instrument to measure linear growth of shoot.
Answer:
Auxanometer.

Question 3.
Type of growth curve for geometric growth.
Answer:
J-shaped or exponential growth curve.

Question 4.
The process by which cork cambium is formed.
Answer:
Dedifferentiation.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 5.
Ability of plant to form different kinds of structures.
Answer:
Plasticity.

Question 6.
Condition where plant exhibits different types of leaves on same plant.
Answer:
Heterophylly.

Question 7.
Growth promoter hormones.
Answer:
Auxins, gibberellins and cytokinins.

Question 8.
Growth inhibitors of plants.
Answer:
Ethylene and abscissic acid.

Question 9.
First hormone that is discovered in plant and its precursor.
Answer:
Amino acid tryptophan is precursor of Indole acetic acid (IAA).

Question 10.
A synthetic hormone that acts as selective herbicide.
Answer:
2, 4 – D (dichlorophenoxy acetic acid).

Question 11.
A plant hormone discovered from fungus.
Answer:
Gibberellic acid (GA).

Question 12.
A plant hormone also present in fish (Herring) sperm DNA.
Answer:
Kinetin (cytokinin).

Question 13.
A plant hormone also present in urine of person suffering from Pellagra.
Answer:
Auxin.

Question 14.
Most widely used source of ethylene for fruit ripening.
Answer:
Ethephon – 2 chloroethyl phosphoric acid.

Question 15.
A precursor from which GA and ABA are synthesized.
Answer:
Mevalonic acid.

Question 16.
Plant antitranspirant.
Answer:
Abscissic acid.

Question 17.
Chemical stimulant of low temperature effect on flowering of plants.
Answer:
Vernalin.

Question 18.
Methods of synthesis of amino acids.
Answer:
Reductive animation and Transamination.

Question 19.
Common amides present in plants.
Answer:
Asparagine and Glutamine.

Question 20.
Nitrogen fixing prokaryotic organisms.
Answer:
Diazotrophs or Nitrogen fixers.

Question 21.
Special structures of cyanobacteria where N2 fixation occurs.
Answer:
Heterocysts.

Question 22.
A synthetic cytokinin hormone.
Answer:
6 – benzyl adenine.

Give Functions/Significance/Importance of the following

Question 1.
Meristems
Answer:
In plants meristems are situated at specific regions where growth takes place by constant and continuous addition of new cells. Meristems have capacity to divide (Mitotic divisions).

Question 2.
Synthetic auxin 2, 4-D
Answer:
It is selective herbicide which kills dicot weeds and foliar spray of 2, 4-D induces flowering in litchi and pineapple, prevents premature fruit drop in apples, pear, oranges.

Question 3.
Coconut milk
Answer:
Coconut milk contains natural cytokinin substance kinetin which is used as nutritional supplement for callus tissue culture where proliferation is noticed due to promotion of cell division.

Question 4.
Ethylene/Ethephon
Answer:
It is used for fruit ripening and as it causes degreening effect by increasing activity of chlorophyllase enzyme for banana and citrus fruits.

Question 5.
Abscissic acid/ABA
Answer:
It is natural growth inhibiting substance in plants and it acts as plant anti Iranspirant causing closure of stomata. It is stress hormone that induces plant to bear the adverse environmental conditions like drought.

Question 6.
Phytochrome
Answer:
It is a proteinaceous pigment present in leaves which perceives stimulus of light for flowering. As it is interconvertible in two forms, it promotes flowering in SDP and inhibits flowering in LDE.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 7.
Vernalization
Answer:
It is response of flowering to low temperature treatment which helps in cultivation of crops in regions where they do not occur naturally and also crops can be produced earlier.

Question 8.
Nitrogen cycle
Answer:
Nitrogen is essential macronutrient for plant growth, it is constituent of amino acids, proteins, nucleic acids, vitamins, hormones, ATP coenzymes, chlorophyll molecule. It is limiting nutrient for plant productivity and agricultural ecosystem. Through food chain it moves to consumers, i.e. animals and decomposers.

Distinguish between the following

Question 1.
Phase of cell division and Phase of cell enlargement
Answer:

Phase of cell division Phase of cell enlargement
1. In this phase cells of meristems divide by mitotic division. 1. In this phase newly formed cells become vacuolated and turgid, osmotically active.
2. Rate of growth at slow pace. 2. Rate of growth at accelerated pace.
3. This is described as lag phase. 3. This is described as log phase.
4. There is no synthesis of any new material. 4. Synthesis of new wall materials and other materials takes place.

Question 2.
Long day plants and Short day plants
Answer:

Long day plants Short day plants
1. Plants that flower only when they are exposed to light period longer than their critical photoperiod are called long day plants (LDP). 1. Plants that flower only when they are exposed to light period shorter than the critical photoperiod are called short day plants (SDP).
2. The plants usually flower in summer. 2. The plants usually flower in winter or late summer.
3. These plants require short night period for flowering. Hence, they are also known as short night plants. 3. These plants require long night period for flowering. Hence, they are also known as long night plants.
4. Plants such as Pea, Radish, Sugar, Beet, Cabbage, Spinach, Wheat, poppy are LDP 4. Plants such as Dahlia, Aster, Tobacco, Chrysanthemum, Soybean (Glycine max) Cocklebur (Xanthium) are SDP

Question 3.
Passive absorption of Minerals and Active absorption of minerals
Answer:

Passive absorption of minerals Active absorption of minerals
1. The movement of mineral ions into root cells due to diffusion, without expenditure of energy is called passive absorption. 1. The uptake of mineral ions by root cells that requires expenditure of ATP energy is known as active absorption.
2. The movement is according to concentration gradient. 2. The movement of mineral ions is against concentration gradient.
3. It takes place by direct ion exchange, diffusion, indirect ion exchange – mass flow and Donnan equilibrium. 3. It takes place by mineral ion accumulation in root hair and carrier concept.

Given reasons

Question 1.
Water is essential for growth in plants.
Answer:

  1. Meristematic cells divide and form new cells.
  2. Absorption of water is necessary for maintaining turgidity in the newly formed cells.
  3. Turgidity results in enlargement of cells in phase of cell elongation.
  4. Water is essential component of protoplasm of cells.
  5. biochemical reactions. Therefore it is essential for growth in plants.

Question 2.
2, 4-D is used as herbicide.
Answer:

  1. 2, 4-D is a synthetic auxin which kills dicot weeds.
  2. Our most of the food crops are cereals, i.e. monocot plants.
  3. Weeds are unwanted plants which otherwise lower the productivity hence to kill them. Selective herbicide is used.

Question 3.
In morphogenesis of plants cytokinin auxin ratio is important.
Answer:

  1. Auxins and cytokinins are growth promoting substances which stimulate cell division and cell enlargement.
  2. A high cytokinin promotes shooting in plants.
  3. A low ratio of cytokinin to auxin induces root development.
  4. A high ratio of cytokinin to auxin induces growth of buds and shoot development.
  5. Thus cytokinin and auxin ratio and their interactions control morphogenesis in plants.

Question 4.
ABA is described as an antitranspirant.
Answer:

  1. ABA is a growth inhibiting hormone.
  2. ABA is responsible for causing efflux of K+ ions from guard cells of stomata.
  3. As a result of this, osmotic changes occur and guard cells become flaccid resulting in closure of stomata.
  4. Transpiration mainly occurs through open stomata and due to closure the activity is checked. Hence it is described as antitranspirant.

Question 5.
Some deficiency symptoms of mineral are visible in young leaves while some appear in older leaves.
Answer:

  1. When mineral element is present below a certain critical concentration it is said to be deficient.
  2. Symptoms are indicated in the form of certain morphological changes on the mobility of element.
  3. These symptoms depend on the mobility of element inside the plant body.
  4. When the element is relatively immobile like S, Ca then the symptoms appear first in young leaves.
  5. When the elements are actively mobilised inside plant body, they are transported to young tissues then the symptoms are visible in older, i.e. senescent leaves e.g. N, Mg, K.

Question 6.
In Donnan equilibrium of passive absorption of minerals concentration of cations increases inside the cell.
Answer:

  1. Minerals exist in soil in the form of charged particles.
  2. Certain negatively charged (anions) get accumulated on the inner side of cell membrane after their entry inside cell.
  3. These anions cannot diffuse out through semipermeable cell membrane.
  4. Thus additional mobile cautions are needed inside the cell to balance these fixed anions.
  5. Hence, the concentration of cations increases inside the cell.

Question 7.
A sudden drop in active absorption of minerals is noticed if roots are deprived of oxygen supply.
Answer:

  1. Absorption of mineral ions from soil against concentration gradient is known as active absorption.
  2. It requires energy (ATP) for absorption by absorbing root cell.
  3. The source of energy is respiration of cells for supply of ATE
  4. When the roots are deprived of oxygen, their respiration process is affected and thus energy is not supplied in required amount. Hence, a sudden drop in absorption of minerals is noticed.

Question 8.
Nitrogen is a limiting nutrient in the agricultural system.
Answer:

  1. Nitrogen is a major nutrient for plant growth.
  2. Proper carbon/nitrogen ratio in soil is necessary for plant growth.
  3. It is component of proteins in the form of amino acids.
  4. Proteins are synthesised from photosynthetic products sugars.
  5. Nitrogen exists in atmosphere but it is inert, non-reactive.
  6. Plants need nitrogen in a reactive form usually nitrate in soil.
  7. This supply need to be maintained through biological and physical nitrogen fixation.
  8. Otherwise productivity is affected hence it is limiting nutrient in the agricultural ecosystem.

Question 9.
Cucumber and sunflower are regarded as photoneutral plants.
Answer:

  1. In cucumber and sunflower, the flower is not controlled by light period.
  2. Both these plants flower in all light periods.
  3. Cucumber and sunflower, therefore, are regarded as photoneutral plants.

Write short notes on the following

Question 1.
Meristems
Answer:

  1. In vascular plants, growth is indeterminate and occurs at specific regions where meristems are located.
  2. Meristems are of three types based on their location – apical, intercalary and lateral.
  3. Meristems are thin walled cells with prominent nucleus with granular cytoplasm, non-vacuolated.
  4. Mitotic divisions take place in meristematic cells.

Question 2.
Phase of cell formation
Answer:

  1. Formative phase is the first phase of growth.
  2. During this phase, the meristematic cells undergo mitosis to form new cells.
  3. During formative phase, the rate of growth is slow.
  4. The phase of cell formation is also called lag phase.
  5. This phase is also known as phases of cell division.

Question 3.
Development
Answer:

  1. Development is progressive changes taking place in shape, form and degree of complexity in an organism.
  2. In plants, it includes all the changes taking place in sequence from seed germination to senescence or death of plant.
  3. Development is an orderly process.
  4. It includes growth, morphogenesis, maturation and senescence.

Question 4.
Plasticity
Answer:

  1. Plasticity is the capacity of plant being molded or formed.
  2. It is ability of plant to develop different kinds of structures in response to environmental factors or stimuli.
  3. Different kinds of structures can be developed in plants due to internal stimuli in different phases, i.e. juvenile and adult.
  4. Heterophylly is shown in plant in different phases or in different environmental conditions.
  5. In coriander and cotton plants, two different kinds of leaves are observed in young (juvenile) and mature (adult) plant.
  6. In buttercup, two different kinds of leaves are observed in terrestrial (on land) and aquatic (in water) habitat.

Question 5.
Phytohormones/Plant Growth Regulators
Answer:

  1. Phytohormones or plant growth regulators are internal factor that influence growth.
  2. They inhibit, promote or modify the plant growth.
  3. Plant hormones are organic substances produced naturally in plants and required in small amount.
  4. Their place of production and site of the activity are different.
  5. Auxins, gibberellins, cytokinins are growth promoters and ethylene, abscissic acid are growth inhibitors.

Question 6.
Phytochrome
Answer:

  1. These are proteinaceous pigments present
  2. Phytochrome exists in two interconvertible forms. Pr and Pfr.
  3. Phytochromes are located in cell membranes of chlorophyllous cells.
  4. When Pfr absorbs red light it is converted to Pr.
  5. Pfr is accumulated in plants during daytime and inhibits flowering in SDP but initiates flowering in LDP

Question 7.
Venralization
Answer:

  1. Effect of temperature on flowering of plants is known as vernalization.
  2. For inducing early flowering pretreatment of seeds or seedlings is done at 1 to 6 °C for about a months duration.
  3. Shoot apical meristem is believed to be site of vernalization stimulus.
  4. Vernalization stimulus is in a form of chemical named vernalin.
  5. Vernalization is effective in ereals (wheat) and crucifers.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 8.
Apical dominance
Answer:

  1. The presence of apical bud inhibits the growth of lateral buds. This phenomenon in which the apical (terminal) bud is active and lateral buds remain inactive is called apical dominance.
  2. It is believed that apical dominance is controlled by an auxin which is synthesized in the apical bud.
  3. From the apical bud, the auxin migrates to the lateral buds and inhibits their growth.
  4. When apical bud is removed, the lateral buds grow and form branches.
  5. For producing more branches therefore, the apical buds are removed.
  6. Cytokinins reverse apical dominance effect by promoting growth of lateral buds by cell division.

Question 9.
Toxicity of micronutrients or mineral toxicity
Answer:

  1. Micronutrients are required in minute quantities by plants.
  2. Their moderate decrease causes deficiency symptoms while their moderate increase causes toxicity.
  3. The reduction in dry weight of a tissue by 10% by any mineral is known as toxicity.
  4. It is not easy to identify toxicity symptoms.
  5. Most of the time, the excess of an element inhibits the uptake of another element resulting in causing the deficiency symptom of that element.
  6. Manganese inhibits calcium translocation towards apex of stem and exhibits symptoms of chlorosis with grey spots appearing on leaves.
  7. This is because manganese competes with iron and magnesium for uptake.
  8. Therefore what we see as symptoms of manganese toxicity, may be the deficiency symptoms of Fe, Mg and Ca.

Question 10.
Day neutral plants (DNP)
Answer:

  1. Day neutral plants are those in which flowering is not affected by the duration of the light period.
  2. Day neutral plants flower in all photoperiods.
  3. Day neutral plants are also known as photoneutral or intermediate plants.
  4. Plants such as cucumber, shoe-flower, sunflower, tomato, maize, balsam, etc. are day neutral plants.

Short Answer Questions

Question 1.
What is de-differentiation?
Answer:

  1. The differentiated cells which are formed may again gain the capacity to divide as per need.
  2. Permanent cells (mature cells) undergo de-differentiation and become meristematic.
  3. This acquired feature of living permanent cells is known as de-differentiation.
  4. e.g. Formation of interfascicular cambium and cork cambium from parenchyma cells of medullary rays and outer cortical cells respectively.

Question 2.
Discuss about natural Auxin.
Answer:

  1. F.W. Went named the growth promoting substance as Auxin.
  2. Auxin was also isolated from urine of patient of pellagra.
  3. Indole 3 acetic acid was first hormone discovered in plants.
  4. IAA is most common and natural hormone synthesized at growing tips and responsible for cell elongation.
  5. It is synthesized from tryptophan and shows polar transport – Basipetal transport.

Question 3.
Discuss about discovery of Gibberellins.
Answer:

  1. Gibberellins are growth promoting hormones and were isolated form fungus Gibberella Jujikuroi by Kurasawa.
  2. Rice plants when infected with this fungus show stem elongation i.e. Bakane disease.
  3. Yabuta and Sumuki isolated gibberellins in crystalline form, from fungal culture and named it gibberellins.
  4. Gibberellins are synthesized from mevalonic acid at young leaves, seeds, root and stem tips and show non-polar transport.

Question 4.
Discuss about discovery of cytokinin.
Answer:

  1. Cytokinins are growth promoting hormones that stimulate cell division.
  2. Skoog and Miller discovered first cytokinin when they were investigating nutritional requirements of tobacco callus culture.
  3. It was observed that the callus proliferated when there was addition of coconut milk as supplement.
  4. The degraded sample of herring (fish) sperm DNA also showed similar growth of tobacco callus. They named the substance as kinetin.

Question 5.
Discuss about discovery of abscissic acid.
Answer:

  1. Abscissic acid (ABA) is a natural growth inhibiting hormone.
  2. It was observed by Carns and Addicott that shedding of cotton balls occur due to chemical substance abscission I and II.
  3. From the buds of Acer, Wareing isolated substance that causes bud dormancy and named it as dormin.
  4. These two chemical substances were identical and now known as abscissic acid.
  5. ABA is synthesized in leaves, fruits and seeds from mevalonic acid.

Question 6.
What is day neutral plant (DNP)? Give any two examples.
Answer:

  1. The plants which do not require specific duration of light period or dark period flowering are day neutral plants (DNP).
  2. They flower throughout the year, as they do not need specific photoperiod.
  3. The flowering in these plants is independent of photoperiod.
  4. Examples – cucumber, tomato, cotton, sunflower, maize and balsam.

Question 7.
Discuss about discovery of phytochromes.
Answer:

  1. Phytochromes are proteinaceous pigments present in cell membrane of green cells.
  2. Phytochromes receive photoperiodic stimulus and induce flowering in plants in response to light duration.
  3. Hendrick and Borthwick observed that in SD plants flowering is inhibited if dark period is interrupted by flash of red light (660 nm).
  4. If flash of far red light (730 nm) is given then again flowering is observed in SD plants.
  5. Pigment system of plant receives photoperiodic stimulus and these pigments exist in two interconvertible forms Pfr and Pr

Question 8.
What is mineral nutrition of plants?
Answer:

  1. Plants require inorganic materials for the synthesis of food.
  2. These elements are obtained by plants in the form of minerals mainly form soil.
  3. Chemical analysis of plant ash reveals that about 40 different minerals are needed by plants which are taken from surroundings, (air, soil and water)
  4. These minerals are absorbed in dissolved form, i.e. ionic form through root system mainly root hairs.
  5. Some minerals are required in large amounts (major) while some are needed only in traces or small amounts (minor).

Question 9.
What are symptoms of mineral deficiency in plants?
Answer:
Any visible deviation from the normal structure and function of plants due to lack or unavailability of particular element below its critical concentration is deficiency symptom of that mineral element.

The common symptoms observed in plants are as follows:

  1. Stunting : Retarded growth and thus stem appears short and condensed.
  2. Chlorosis : This is loss or lack of chlorophyll that result in yellowing of leaf.
  3. Necrosis : It is localized death of tissue. Mottling : This is appearance of green or non-green patches or spots on leaves. Abscission : This is premature fall of leaves, buds, fruits and flowers.

Question 10.
Enlist the role of following minerals and the symptoms caused due to their deficiency : (a) Calcium (b) Boron and (c) Chlorine.
Answer:
(a) Calcium:
Role : Involved in selective permeability of cell membranes, activator of certain enzymes, required as calcium pectate in middle lamella of cell wall at root and stem apex (for cell division).
Deficiency symptom : stunted growth.

(b) Boron:
Role : Required for uptake and utilization of Calcium (Ca2+), pollen germination, cell differentiation, carbohydrate translocation. Deficiency symptom : Brown heart disease

(c) Chlorine:
Role : Na+ and K+ help to determine solute concentration and anion – cation balance in cells, necessary for oxygen evolution in photosynthesis.
Deficiency symptom : Poor growth of plant.

Question 11.
What is Donnan Equilibrium?
Answer:

  1. Donnan equilibrium is a process of passive absorption of minerals in plants which is without any expenditure of energy.
  2. It is assumed that certain anions after their entry by diffusion into the cell get fixed on inner side of cell membrane.
  3. Additional mobile cations are needed to balance this fixed anions as they cannot diffuse outside.
  4. Concentration of cations thus increases due to accumulation.
  5. This passive absorption of anions or cations from exterior against their concentration gradients so as to neutralize the effect of cations or anion is known as Donnan equilibrium.

Question 12.
Explain physical nitrogen fixation.
Answer:

  1. Conversion of free nitrogen of air into nitrogenous compounds that are made available to plants for uptake is known as : nitrogen fixation.
  2. Physical nitrogen fixation occurs in step- : wise manner and it takes place in atmosphere j and soil.
  3. Under the influence of electric discharge, lightning and thunder, atmospheric nitrogen combines with oxygen to form nitric oxide.
  4. Nitric oxide is then oxidized to nitrogen peroxide in presence of oxygen.
  5. Nitrogen peroxide combines with rainwater to form nitrous and nitric acid which come on ground as acid rains.
  6. On ground, alkali radicals (mainly of Ca, K) react with nitric acid to produce nitrites and nitrates which are absorbable forms for plants.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 13.
Give equations of physical nitrogen fixation.
Answer:
(1) Physical nitrogen fixation occurs in stepwise manner in atmosphere and on land (soil)
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 3

Question 14.
Give the equations of amino acid synthesis.
Answer:
1. Macromolecule proteins are made up of building blocks of amino acids.
2. Amino acids are synthesized by two methods – Reductive animation and transamination.
3. Reductive amination – Ammonia reacts with alpha keto glutaric acid to form glutamic acid.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 4
4. Transamination – Amino group of one amino acid is transferred to other carboxylic acid at keto position.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 5

Question 15.
Explain lag phase, log phase and steady phase of growth.
Answer:
1. In plants, growth curve is always sigmoid, i.e. S-shaped. This is because growth starts slowly during formative phase, becomes rapid during elongation phase and finally slows down to a steady state during the maturation phase.

2. The standard growth curve shows three phases, viz. lag phase, log phase and stationary phase.
(i) Lag phase or initial growth phase : This is the initial phase of growth. During this phase of growth, the rate of growth is slow. It corresponds to formative phase of growth where new cells are formed due to cell division.

(ii) Log phase or exponential phase : This is the second phase of growth. During this phase, the growth is rapid and maximum. It corresponds to the phase of cell elongation.

(iii) Stationary phase or steady phase : The stationary phase is the third and last phase of growth. In this phase, growth slows down and becomes steady. The cells undergo differentiation during stationary phase.

Chart based or Table based questions

Question 1.
Complete the chart of plasticity.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 6
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 7

Question 2.
Complete the flow chart of development.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 8
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 9

Question 3.
Complete the table of growth hormones.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 10
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 11

Question 4.
Complete the table of mineral nutrition of plants.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 12
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 13

Diagram based questions

Question 1.
Draw diagram of photoperiodic response of SDP and LDP.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 14

Question 2.
With the help of diagram show arithmetic growth and geometric growth.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 15

Question 3.
Draw the sigmoid growth curve.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 16

Question 4.
Observe the diagram and answer the questions related to it.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 17

  1. From which cells newly formed cells are added? How?
  2. How is rate of growth in zone of cell elongation?
  3. What is peculiarity of zone of differentiation? How is rate of growth in this region?

Answer:

  1. Meristematlc cells add new cells by mitotic division.
  2. Rate of growth is at accelerated pace.
  3. The rate of growth is at steady state and cells become specialised to perform specific function become mature.

Question 5.
Observe the adjacent graph indicating growth. What is correct labelling of A, B and C respectively?
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 18
Answer:
A. Stationary Phase
B. Exponential Phase
C. Lag Phase

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 6.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 19
(1) Observe the above graph indicating increase in height of plants. Which type of growth indicates this pattern of graph?
(2) Give its mathematical expression.
Answer:
(1) Arithmetic growth

(2) Lt = Lo + rt where
Lt = Length at time t,
Lo = Length at time zero r = growth rate,
t = time of growth

Question 7.
Observe the figure given below of two different leaves ‘A’ and ‘B’ Which leaf shows much higher relative growth rate ?
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 20
Answer:
Relative Growth Rate (RGR) = Growth per unit time as % of intial size.
RGR = \(\frac { Growth per unit time }{ intial size }\) × 100
Leaf A = 10 – 5 = 5, \(\frac { 5 }{ 5 }\) × 100 = 100%
Leaf B = 55 – 50 = 5, \(\frac { 5 }{ 50 }\) × 100 = 10%
Hence fig. A shows more relative growth.

Question 8.
Observe the diagrams A and ‘B’ showing growth in two leaves. Which diagram shows more relative growth?
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 21
Answer:
Relative Growth Rate (RGR)
= Growth per unit time as % of intial size.
RGR = \(\frac { Growth per unit time }{ intial size }\) × 100
Leaf A = 20 – 10 = 10, \(\frac { 10 }{ 10 }\) × 100 = 100%
Leaf B = 60 – 50 = 10, \(\frac { 10 }{ 50 }\) × 100 = 20%
Hence Diagram A shows more relative growth.

Question 9.
Identify the type of growth curves observed in plants.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 22
Answer:
Constant linear growth curve for the arithmetic growth.
Exponential (J shaped) growth curve for the geometric growth.
Sigmoid growth curve related to distinct phases of growth.

Question 10.
Fill in the blanks in the given nitrogen cycle.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 23
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 24

Long answer questions

Question 1.
What is hydroponics? How is it useful in identifying the role of nutrients?
Answer:

  1. Growing plants in aqueous (soilless) medium is known as hydroponics. [Greek word hudor = water and ponos = work]
  2. It is technique of growing plants by supplying all necessary nutrients in the water supply given to plant.
  3. A nutrient medium is prepared by dissolving necessary salts of micronutrients and macronutricnts In desired quantity and roots of plants are suspended in this liquid with appropriate support.
  4. Hydroponics is of great use in studying the deficiency symptoms of different mineral nutrients.
  5. The plants uptake mineral nutrients in the form of dissolved ions with the help of root hairs from the surrounding medium or nutrient solution supplied.
  6. While preparing the required nutrient medium particular nutrient can be totally avoided and then the effect of lack of that nntr’cnt can be studied in variation of plant growth.
  7. Any visible change noticed from normal structure and function of the plant is the symptom or hunger sign considered.
  8. For e.g. Yellowing of leaf is observed due to loss of chlorophyll pigments or Chlorosis is noticed if Magnesium is lacking as it is a structural component of chlorophyll pigment.

Question 2.
Explain the active absorption of minerals.
Answer:

  1. Plants absorb minerals from the soil with their root system.
  2. Minerals are absorbed from the soil in the form of charged particles, positively charged cations and negatively charged anions.
  3. The absorption of minerals against the concentration gradient which requires expenditure of metabolic energy is called active absorption.
  4. The ATP energy derived from respiration in root cells is utilized for active absorption.
  5. Ions get accumulated in the root hair against the concentration gradient.
  6. These ions pass into cortical cells and finally reach xylem of roots.
  7. Along with the water these minerals are carried to other parts of plant.

Question 3.
What is growth? What are its characteristics ?
Answer:
Growth : Growth is a “vital process which brings about an irreversible increase in an organism or its part with respect to its size, weight, form and volume.”

Characteristics of growth:

  1. Growth is a permanent increase in size, weight, shape, volume and dry weight of a plant.
  2. The change occurring due to growth is permanent and irreversible.
  3. Growth is an intrinsic process caused due to internal activities.
  4. Growth occurs by cell division and cell elongation followed by cell maturation which lead to the formation of different types of tissues.
  5. Growth in plants is mostly localized, i.e. restricted to some regions of plants possessing meristematic tissues or meris terns.
  6. Growth has a qualitative aspect where development takes place in an orderly manner and differentiation leads to higher and more complex state.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 4.
Describe the phases of growth.
Answer:
There are three phases of growth, viz, formation phase, elongation phase and maturation phase.
(1) Formative phase (Phase of cell division) : This is the first phase of growth. In this phase, the meristematic cells undergo mitosis to produce new cells. Owing to the formation of new cells, there occurs a slight increase in the size of the organ.

(2) Elongation phase (Phase of cell enlargement) : This is the second phase of growth. In this phase, the new cells that are formed, undergo enlargement as a result of which the size and volume of the cells increase. Enlargement of cells occur mostly in linear direction as a result of which the elongation of the root and stem takes place. The enlargement of cells causes a considerable increase in size and weight of an organ and a plant as a whole.

(3) Maturation phase (Phase of cell maturation and differentiation) : Maturation phase is the third and last phase of growth. In this phase, the elongated cells undergo maturation and differentiation to form various types of plant tissues like parenchyma, sclerenchyma, xylem and phloem.

Question 5.
Explain the various conditions of growth that are essential.
Answer:

  1. For a proper growth of plant various environmental and physiological conditions are necessary.
  2. Carbon/Nitrogen ratio in soil is having effect on growth as both carbon and nitrogen are structural elements in carbohydrates, proteins and other biomolecules.
  3. Water is essential component of protoplasm and required for turgidity of cells during cell enlargement phase. It is a medium in which various biochemical reactions ocfcur.
  4. Nutrients are necessary for proper growth. Macronutrients and micronutrients have their specific role.
  5. Temperature of 25 – 35 °C is optimum for growth.
  6. Light is essential for seed germination and photosynthesis.
  7. Oxygen is necessary for respiration and supply of energy.
  8. Gravitational force decides direction of growth for root system and shoot system.
  9. Growth hormones are organic compounds that are involved in various physiological aspects and control of growth.

Question 6.
What are plant growth regulators? Give the characteristics of plant growth regulators.
Answer:
Plant growth regulators : Plant growth regulators or phytohormones or plant hormones as they are also called are organic compounds synthesised by the plants which promote, inhibit or control the growth and other physiological processes.

Characteristics of plant growth regulators:

  1. Plant growth regulators are organic compounds other than nutrients.
  2. They are synthesised at the apices of root, stem and leaves from where they are transported to other parts of plants where they produce their effects.
  3. They are required in minute quantities.
  4. A single plant growth regulator can control or regulate the various aspects of growth.

Question 7.
Enlist the five main types of growth regulators and state the role of abscissic acid in plants.
Answer:
Five main types of growth regulators:

  1. Auxins
  2. Gibberellins
  3. Cytokinins
  4. Ethylene and
  5. Abscissic acid.

Role of abscissic acid (ABA) in plants:

  1. Abscissic acid influences abscission and dormancy.
  2. ABA accelerates senescence of leaves, flowers and fruits.
  3. It is a stress hormone as it is produced during drought and other unfavourable climatic conditions.
  4. ABA induces dormancy in seeds, buds and tubers.
  5. It acts as growth inhibitor as it retards growth.
  6. ABA plays an important role in closing of stomata to check transpiration.
  7. It inhibits and delays cell division and suppresses cambial activity by inhibiting mitosis in vascular cambium.
  8. ABA inhibits flowering in LDP and stimulates flowering in short day plants (SDP).

Question 8.
Write an account of auxins as growth regulators.
Answer:

  1. Auxins are plant growth regulators produced naturally by plants.
  2. They are weak organic acids capable of promoting cell elongation during the growth of stem and root.
  3. Auxins are synthesized in shoot and root apices besides young leaf primordia.
  4. Auxins may be natural or synthetic.
  5. Naturally occurring auxins are indole-3- acetic acid (IAA) and its derivatives.
  6. NAA, 2, 4-D and 2,4, 5-T are synthetic auxins.
  7. Auxins in higher concentration promote the growth of stem.
  8. Auxins play an important role in initiation and promotion of cell division.
  9. Auxins help in the formation of adventitious roots from cuttings when applied in lower concentration.
  10. Auxins play an important role in apical dominance.
  11. Auxin prevents abscission by preventing the action of hydrolytic enzymes in abscission layer.
  12. Auxins are used to produce parthenocarpic fruits in plants like orange, apple, tomato and grapes.

Question 9.
Give an account of physiological effects and application of auxin with examples.
Answer:

  1. Auxins are growth promoting substances synthesized at meristematic regions of plants.
  2. The primary effect of auxin is cell enlargement and it stimulates growth of stem and root.
  3. Apical dominance – The phenomenon where growing apical bud inhibits the growth of lateral bud is apical dominance which is controlled by auxin synthesized at apical bud.
  4. Owing to activity of inducing multiplication of cells it is used in plant tissue culture to produce callus.
  5. Auxin stimulates formation of lateral and adventitious roots hence used for rooting propagation of cuttings.
  6. 2, 4-D is a synthetic herbicide which kills dicot weeds.
  7. Induced parthenocarpy in fruits-oranges, banana, grapes, lemons is by application of auxin.
  8. Foliar spray of NAA and 2, 4-D induces flowering in litchi and pineapple.
  9. Premature fruit drop of apples, pear and oranges is prevented.
  10. Auxins break seed dormancy and promote germination.
  11. Auxins promote early differentiation of xylem and phloem, cell elongation, increase rate of respiration, prevent formation of abscission layer.

Question 10.
Explain the application of gibberellins.
Answer:

  1. Gibberellins are growth promoting hormone and it is present in root tips, stem tips and seeds.
  2. Gibberellins break dormancy of bud, dormancy of seed.
  3. They promote seed germination in cereals by activating or synthesising enzyme amylase to produce sugar.
  4. Gibberellins induce elongation of the cells in stem hence increase in internode length is noticed.
  5. In rosette plants like cabbage it causes ‘bolting’ that is increase in internode length before flowering.
  6. Gibberellins are more effective in inducing parthenocarpy than auxins in plants like tomato, apple and pear.
  7. It is also used to increase fruit size and length of bunches in grapes.
  8. By its application genetically dwarf plants can be converted to phenotypically tall e.g. Maize.
  9. It overcomes requirement of vernalization, delays senescence and prevents abscission.
  10. Application of gibberellins causes production of male flowers on female plants.

Question 11.
Describe about the physiological effects and applications of cytokinin.
Answer:

  1. Cytokinins are growth promoting hormone that promotes cell division. Kinetin, zeatin are examples of cytokinin.
  2. They promote cell division as well as cell enlargement.
  3. High cytokinin promotes shoot development.
  4. Growth of lateral buds is promoted by cytokinins. Thus it controls apical dominance.
  5. Process of ageing and senescence, abscission of plant organs is delayed by their application.
  6. It promotes formation of interfascicular cambium.
  7. It has a role in breaking seed dormancy and promotes seed germination.
  8. Cytokinins induce RNA synthesis.
  9. Cytokinin and auxin ratio and their interactions control morphogenesis and cell differentiation.

Question 12.
Discuss about physiological effects and applications of ethylene.
Answer:

  1. Ethylene is a gaseous growth inhibitor hormone.
  2. It promotes ripening of fruits like bananas, apples and mangoes. The commercial application of ethephon is done.
  3. It initiates growth of lateral roots.
  4. Dormancy of buds and seeds is broken by its application.
  5. It accelerates formation of abscission layer and thus abscission of leaves, flowers and fruits is observed.
  6. Ethylene is responsible for checking growth of lateral buds thus causes apical dominance and retards flowering.
  7. Process of senescence of plant organs is enhanced.
  8. Epinasty, i.e. drooping of leaves and flowers results due to its application in some plants.
  9. It increases activity of chlorophyllase enzyme causing degreening effect in banana and Citrus fruits.

Question 13.
Discuss about experiment of Hendricks and Borthwick for discovery of phytochromes.
Answer:

  1. Phytochrome pigments receive photoperiodic stimulus and control flowering in plants.
  2. Hendricks and Borthwick observed that in SDP flowering is inhibited if continuous dark period is interrupted even by a short duration or flash of red light of wavelength 660 nm.
  3. If this interruption is again exposed to flash of far red light of wavelength 730 nm, then these plants flower.
  4. From this they concluded that some pigment system in plant receives the photoperiodic stimulus.
  5. These pigment proteins are called phytochromes and it exists in two interconvertible forms – Pr and Pfr.
  6. These pigments are located in cell membranes of green cells.
  7. Pfr is biologically active form and during daytime it gets accumulated in the plant.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Question 14.
Give schematic representation of nitrogen cycle and enlist important steps of this cycle.
Answer:

  1. Atmospheric nitrogen is a source of nitrogen cycle.
  2. The important steps of the cycle are Nitrogen fixation, Nitrification, Ammonification, Nitrogen assimilation by plants. Amino acid synthesis and amidation, Denitrification and sedimentation.
  3. Amino acid are building blocks of proteins. Amides are amino acid with two amino groups.
  4. Schematic representation of Nitrogen
    Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 25

Question 15.
What is nitrogen cycle? Describe it briefly.
Answer:

  1. The cyclic movement of nitrogen between the atmosphere, biosphere and geosphere in different forms is called nitrogen cycle.
  2. Nitrogen cycle is one of the most important biogeochemical cycles.
  3. The nitrogen cycle involves many processes such as cycling of nitrogen through the biosphere, atmosphere and geosphere, nitrogen fixation, nitrogen uptake, formation of biomass, ammonification, nitrification and denitrification, etc.
  4. Bacteria such as Nitrosomonas, Nitrosococcus and Nitrobacter are the nitrifying bacteria which play an important role in nitrification.
  5. Denitrification is carried out by the bacteria Pseudomonas denitrifficans. From this it is obvious that bacteria play a major role in nitrogen cycle.
  6. Nitrogen fixation occurs by physical, industrial and biological methods. Prokaryotic organism play an important role in biological nitrogen fixation.

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 6 Plant Water Relation Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 6 Plant Water Relation

Multiple Choice Questions

Question 1.
What is the reason behind various properties of water?
(a) Physical state
(b) Hydrogen bonding
(c) Colour
(d) Fluidity
Answer:
(b) Hydrogen bonding

Question 2.
Which special type of tissue is present in epiphytic roots?
(a) Velamen
(b) Lenticel
(c) Aerenchyma
(d) Haustoria
Answer:
(a) Velamen

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 3.
In which zone of root we come across root hairs?
(a) meristematic region
(b) zone of elongation
(c) zone of absorption
(d) maturation zone
Answer:
(c) zone of absorption

Question 4.
In double layered cell wall of root hair, outer layer is of ……………..
(a) cellulose
(b) pectin
(c) cutin
(d) suberin
Answer:
(b) pectin

Question 5.
What is rhizosphere?
(a) Root ball formed by growth of roots
(b) Covering of root tip
(c) Microenvironment around root
(d) region of root hairs
Answer:
(c) Microenvironment around root

Question 6.
A root hair is derived from …………….. cell.
(a) epidermal
(b) cortical
(c) endodermal
(d) pericycle
Answer:
(a) epiderinal

Question 7.
When we keep raisins in water they swell up, due to ……………..
(a) exosmosis
(b) plasmolysis
(c) imbibition
(d) diffusion
Answer:
(c) imbibition

Question 8.
This phenomenon is associated with exchange of gases ……………..
(a) osmosis
(b) diffusion
(c) imbibition
(d) plasmolysis
Answer:
(b) diffusion

Question 9.
What is correct expression for diffusion pressure deficit (D.PD.)?
(a) D.PD. = O.R – S.E
(b) D.PD. = T.E – O.E
(c) D.PD. = W.E – T.R
(d) D.PD. = O.R – T.R
Answer:
(d) D.PD. = O.P – T.P

Question 10.
What is true for a turgid cell?
(a) T.R is zero
(b) T.R = S.R
(c) D.P.D. is zero
(d) W.R = S.R
Answer:
(c) D.P.D. is zero

Question 11.
The particles which easily diffuse through cell membrane are ……………..
(a) lipid soluble
(b) water soluble
(c) hydrophilic
(d) lipophobic
Answer:
(a) lipid soluble

Question 12.
Which proteins help in facilitated diffusion process?
(a) cutin
(b) mucin
(c) aquaporins
(d) lipoproteins
Answer:
(c) aquaporins

Question 13.
D.PD. is now known as ……………..
(a) water potential
(b) solute potential
(c) pressure potential
(d) osmotic potential
Answer:
(a) water potential

Question 14.
Water always flows from ……………..
(a) more negative water potential to less negative water potential
(b) high water potential area to low water potential area
(c) low water potential area to high water potential area
(d) negative water potential area to area of zero water potential
Answer:
(b) high water potential area to low water potential area

Question 15.
Plasmolysed cell becoming turgid is process of ……………..
(a) replasmolysis
(b) incipient plasmolysis
(c) deplasmolysis
(d) exosmosis
Answer:
(c) deplasmolysis

Question 16.
In a fully turgid cell ……………..
(a) T.P = O.P
(b) T.P = S.P
(c) O.P = S.P
(d) T.RP = 0 (zero)
Answer:
(a) T.P = O.P

Question 17.
Casparian strip of endodermis has material ……………..
(a) pectin
(b) suberin
(c) cutin
(d) porin
Answer:
(b) suberin

Question 18.
Water from pericycle is forced into xylem due to ……………..
(a) aquaporin
(b) plasmodesmata
(c) root pressure
(d) ion-channels
Answer:
(c) root pressure

Question 19.
Water absorbed from root hair when passes through intercellular spaces and cell wall it is …………….. pathway.
(a) apoplast
(b) symplast
(c) transmembrane
(d) plasmodesmata
Answer:
(a) apoplast

Question 20.
In root system, secondary roots arise from ……………..
(a) cortical cells
(b) endodermis
(c) pericycle
(d) passage cell
Answer:
(c) pericycle

Question 21.
Active absorption of water occurs during ……………..
(a) early morning
(b) daytime
(c) bright sunlight
(d) night time
Answer:
(d) night-time

Question 22.
The value of root pressure is usually about ……………..
(a) +1 to +12 bars
(b) -1 to +1 bars
(c) +1 to +2 bars
(d) +1 to +21 bars
Answer:
(c) +1 to +2 bars

Question 23.
The ascent of sap in plants takes place through ……………..
(a) xylem
(b) phloem
(c) parenchyma
(d) endodermis
Answer:
(a) xylem

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 24.
From the following, which water will show greatest water potential (ψ)
(a) pure water
(b) salt water
(c) sugar water
(d) salt + sugar + water
Answer:
(a) pure water

Question 25.
Due to entry of water in a cell, the pressure potential ……………..
(a) increases
(b) decreases
(c) remains unaffected
(d) becomes zero
Answer:
(a) increases

Question 26.
Which mineral element is not remobilized in plants?
(a) P
(b) K
(c) S
(d) Ca
Answer:
(d) Ca

Question 27.
Phloem sap analysis is done using isotope ……………..
(a) 16O
(b) 14C
(c) 15N
(d) 35S
Answer:
(b) 14C

Question 28.
In plants, food is mainly translocated in form of ……………..
(a) starch
(b) glucose
(c) sucrose
(d) amino acids
Answer:
(c) sucrose

Question 29.
Transport of food through phloem is ……………..
(a) unidirectional
(b) bidirectional
(c) three dimensional
(d) absent
Answer:
(b) bidirectional

Question 30.
Guttation occurs through ……………..
(a) stomata
(b) lenticels
(c) hydathodes
(d) velamen
Answer:
(c) hydathodes

Question 31.
Amount of cuticular transpiration is about ……………..
(a) 0.1-1%
(b) 8 – 10%
(c) 90 – 93%
(d) 2 – 8%
Answer:
(b) 8 – 10%

Question 32.
Epistomatic leaf shows ……………..
(a) stomata on upper side
(b) stomata on lower side
(c) stomata on both surfaces
(d) absence of stomata
Answer:
(a) stomata on upper side

Question 33.
Dumbbell shaped guard cells are found in ……………..
(a) most dicots
(b) grasses
(c) gymnosperms
(d) desert plants
Answer:
(b) grasses

Question 34.
Stomatal transpiration occurs during night time in ……………..
(a) most dicots
(b) grasses
(c) gymnosperms
(d) desert plants
Answer:
(d) desert plants

Question 35.
Reservoir of K+ ions is ……………..
(a) guard cells
(b) epidermal cells
(c) subsidiary cells
(d) mesophyll
Answer:
(c) subsidiary cells

Question 36.
Guard cells are surrounded by ……………..
(a) epidermal hairs
(b) mesophyll cells
(c) accessory cells
(d) lenticels
Answer:
(c) accessory cells

Question 37.
When guard cells close stomata at night which acid prevents uptake of K+ and C ions?
(a) Abscissic acid
(b) Pyruvic acid
(c) Malic acid
(d) Acetic acid
Answer:
(a) Abscissic acid

Question 38.
Which type of injury is noticed in plants if there is excessive transpiration?
(a) Burning
(b) Chlorosis
(c) Necrosis
(d) Wilting
Answer:
(d) Wilting

Question 39.
Maximum transpiration occurs through ……………..
(a) cuticle
(b) lenticels
(c) stomata
(d) bark
Answer:
(c) stomata

Question 40.
What will be the condition of guard cells during night-time?
(a) Show increased turgor pressure
(b) Become flaccid
(c) Increase uptake of K+ and Cl ions
(d) Starch converted to sugar
Answer:
(b) Become flaccid

Question 41.
Cell A has water potential – 10 bars and cell B has – 5 bars, the movement of water will occur from ……………..
(a) A to B
(b) B to A
(c) No movement
(d) Either A to B or B to A
Answer:
(b) B to A

Question 42.
What is the correct symbol and unit of water potential?
(a) ψ and ha
(b) ω and ha
(c) ω and Pa
(d) ψ and Pa
Answer:
(d) ψ and Pa

Question 43.
When root system absorbs water, which is the first physical process concerned with this activity?
(a) Osmosis
(b) Imbibition
(c) Facilitated diffusion
(d) Diffusion
Answer:
(b) Imbibition

Match the columns

Question 1.

Column A (Scientist) Column B (Theory)
(1) Munch (a) Proton transport theory
(2) Bohem (b) Pressure flow theory
(3) J. Pristley (c) Capillary theory
(4) Levitt (d) Root Pressure theory

Answer:

Column A (Scientist) Column B (Theory)
(1) Munch (b) Pressure flow theory
(2) Bohem (c) Capillary theory
(3) J. Pristley (d) Root Pressure theory
(4) Levitt (a) Proton transport theory

Question 2.

Column A (Scientist) Column B (Theory)
(1) Dixon and Joly (a) Starch-sugar                inter conversion
(2) Steward (b) Osmotic absorption theory
(3) Atkins and Pristley (c) Non-osmotic absorption theory
(4) Kramer and Thimann (d) Cohesion tension theory

Answer:

Column A (Scientist) Column B (Theory)
(1) Dixon and Joly (d) Cohesion tension theory
(2) Steward (a) Starch-sugar inter conversion
(3) Atkins and Pristley (b) Osmotic absorption theory
(4) Kramer and Thimann (c) Non-osmotic absorption theory

Question 3.

Column A Column B (New Terminology)
(1) D.P.D. (a) Osmotic potential
(2) O.E (b) Pressure potential
(3) T.P (c) Water potential

Answer:

Column A Column B (New Terminology)
(1) D.P.D. (c) Water potential
(2) O.E (a) Osmotic potential
(3) T.P (b) Pressure potential

Classify the following to form Column B as per the category given in Column A

Question 1.
B, Co, Mn, P, N, S

Column A Column B
Macro elements ——–, ———-, ——–
Micro elements ——–, ———-, ——–

Answer:

Column A Column B
Macro elements P, N, S
Micro elements B, Co, Mn

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 2.
Capillary water, Combined water, Hygroscopic water, Gravitational water.

Column A Column B
(1) Water adsorbed on soil particles —————
(2) Water penetrated deep in soil ————–
(3) Water present as hydrated oxides ————-
(4) Water present between soil particles ————-

Answer:

Column A Column B
(1) Water adsorbed on soil particles Hygroscopic water
(2) Water penetrated deep in soil Gravitational water
(3) Water present as hydrated oxides Combined water
(4) Water present between soil particles Capillary water

Very short answer question

Question 1.
Why water acts as a thermal buffer?
Answer:
The water has high specific heat, high heat of vaporization and high heat of fusion hence it acts as a thermal buffer.

Question 2.
Why water can easily rise in capillaries?
Answer:
Water has high surface tension and high adhesive and cohesive force hence it can easily rise in capillaries.

Question 3.
Enlist the kind of water available in soil environment.
Answer:
Soil environment has gravitational water, hygroscopic water, combined water and capillary water.

Question 4.
Mention examples of imbibition process.
Answer:
Soaking of seeds, swelling up of raisins, kneading of flour and warping of wooden doors in rainy season.

Question 5.
What is the mechanism of imbibition?
Answer:
When imbibition takes place water molecules (imbibate) get tightly adsorbed on imbibant without formation of solution.

Question 6.
Why water enters plant cell by process of diffusion?
Answer:
Water present around cell wall has more diffusion pressure than inner cell sap hence water moves in the cell through freely permeable cell wall by diffusion.

Question 7.
What is isotonic condition in osmotic system?
Answer:
A condition where concentration of solution has neither gain nor loss of water in an osmotic system is isotonic.

Question 8.
What is effect on protoplasm when cell is plasmolysed ?
Answer:
When cell is plasmolysed, protoplast of cell shrinks and recedes from the cell wall thus gap is observed between cell wall and protoplast.

Question 9.
What is apoplast pathway?
Answer:
When water absorbed by root hair passes across the roots through cell wall and intercellular spaces of root cortex it is called apoplast pathway.

Question 10.
Which experiment is a proof for existence of root pressure?
Answer:
When a stem of potted plant is cut above the soil, xylem sap is seen oozing through cut end, that proves presence of root pressure.

Question 11.
Which analysis indicates that minerals are absorbed by plants?
Answer:
Analysis of plant ash contents is indication of absorbed minerals.

Question 12.
Which ions are readily remobilized in plants?
Answer:
Ions of phosphorus, sulphur and nitrogen are remobilized from older plants, parts (leaves) to younger parts.

Question 13.
What is radial and tangential translocation?
Answer:
When lateral translocation of food occurs in root or stem, transport from phloem to pith is radial translocation while that from phloem to cortex is tangential translocation.

Question 14.
What is loading of vein?
Answer:
In turgid cell, due to increased turgor pressure of photosynthetic cell, sugar is forced into the sieve tube of the vein, which is known as loading of vein.

Question 15.
What is unloading of vein?
Answer:
At the sink end, turgor pressure is lowered and hence turgor pressure gradient is developed from sieve-tube which translocates food passively along concentration gradient, this is vein unloading.

Question 16.
What is peculiarity of wall of guard cells?
Answer:
The inner wall of guard cell that faces opening is thick and inelastic while its outer wall or lateral wall is thin and elastic.

Question 17.
Give reaction of starch-sugar interconversion theory.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 1

Question 18.
How do land plants absorb water and mineral salts from the soil?
Answer:
Land plants absorb water and mineral salts from soil with the help of their roots.

Question 19.
In which form the water is lost from leaves to the atmosphere?
Answer:
The water is lost in the form of vapour from leaves to the atmosphere.

Question 20.
How does plants lift the water from soil up to canopy without any pump?
Answer:
Plants have vascular tissue system of xylem, mainly through vessels and tracheids. Water is conducted upwards due to pull created in this continuous channel of water.

Question 21.
What is a peculiarity of epiphytic plants like orchids?
Answer:
The epiphytic plants like orchids have epiphytic roots with special water vapour absorbing layer of velamen tissue that absorbs water vapour from air.

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 22.
What are root hairs?
Answer:
Root hairs are the extensions of epiblema or epidermal cells in the region of absorption.

Question 23.
Why do the wooden doors become very hard to close and open in rainy season?
Answer:
During rainy season wooden doors swell up due to imbibition of water. The humidity or moisture content of air increases and wood which is hydrophilic takes up this moisture.

Question 24
How does the water come out through the surface of porous earthen pot?
Answer:
Earthen pot has tiny pores through which water diffuses out. Due to evaporation of this water from surface water kept inside becomes cool.

Question 25.
Describe external structure of root hair.
Answer:
A root hair is a cytoplasmic extension of epiblema cell which is colourless, unbranched, ephemeral, very delicate tubular structure of about 1 to 10 mm long.

Question 26.
What is gravitational water?
Answer:
Water present in the soil that percolates deep inside due to gravitational force is called gravitational water.

Question 27.
How can we describe osmotic movement of water related to energy?
Answer:
The osmotic movement of water is on the basis of free energy which is used to do work.

Question 28.
Which process is responsible for transport of minerals and to make them available to cells?
Answer:
The minerals absorbed by roots are transported upwards through sap and from veins by process of diffusion cells uptake them.

Question 29.
Give examples of vertical translocation of food in downward direction and upward direction.
Answer:
Food is translocated in downward direction from leaves (source) to root, while during germination of seed, bulbil, corm it is in upward direction.

Question 30.
What is translocation of food?
Answer:
The transport of food from one part of plant to the other part, i.e. source to sink is called translocation of food.

Question 31.
Which plant tissues are involved in transport of water, minerals and food?
Answer:
A complex plant tissue xylem, mainly with its tracheids and vessels is involved in transport of water and minerals while complex tissue phloem with sieve tubes and companion cells is involved in transport of food.

Question 32.
Which is a direct pathway of transport available in plants through root system?
Answer:
Secondary roots that originate from pericycle that is outer layer of vascular cylinder, bypass endodermis with Casparian strip allow direct apoplast pathway to enter xylem and phloem.

Question 33.
For which type of plants root pressure theory of ascent of sap is applicable?
Answer:
Root pressure theory is applicable to plants having height up to 10 to 20 metres.

Question 34.
What is hydrogen bond?
Answer:
Hydrogen bond results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom like O or N.

Question 35.
When you burn an incense stick in one corner of room, its fragrance spreads all over the room in a short time. How does it happen?
Answer:
The process of diffusion is responsible for spreading the fragrance of an incense stick in one corner of room.

Give definitions of the following

Question 1.
Facilitated diffusion
Answer:
The passive absorption of solutes when mediated by a carrier is called facilitated diffusion.

Question 2.
Osmotic pressure / Osmotic potential
Answer:
The pressure exerted due to osmosis is osmotic pressure now termed as osmotic potential.

Question 3.
Deplasmolysis
Answer:
When plasmolysed cell is placed in hypotonic solution, endo-osmosis occurs making cell turgid again then this is called deplasmolysis.

Question 4.
Translocation of food
Answer:
The movement of food from one part of plant to the other part is called translocation of food.

Name the following

Question 1.
Condition of cell wall and cell membrane : based on permeability.
Answer:
Cell wall freely permeable and Cell membrane semipermeable.

Question 2.
Weak solution having low osmotic concentration.
Answer:
Hypotonic

Question 3.
Strong solution having high osmotic concentration.
Answer:
Hypertonic

Question 4.
A suberised layer on endodermis.
Answer:
Casparian strip

Question 5.
One example of each epistomatic, hypostomatic and amphistomatic leaf.
Answer:

  1. Epistomatic leaf – Lotus
  2. Hypostomatic – Nerium.
  3. Amphistomatic – Grass

Question 6.
Pathway for entry of water into xylem from endodermis.
Answer:
Symplast pathway

Question 7.
A material deposited on endodermis which forms barrier.
Answer:
Suberin

Question 8.
Water imbibed or adsorbed on soil particles.
Answer:
Hygroscopic water

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 9.
Water potential was previously known as.
Answer:
D.ED. or Diffusion pressure deficit

Question 10.
Free energy per molecule in chemical system.
Answer:
Chemical potential

Question 11.
The substance synthesized in cell responsible for increasing osmotic concentration as per Munch theory.
Answer:
Glucose

Question 12.
A waxy substance present in layer on outer surface of epidermis.
Answer:
Cutin in layer cuticle

Question 13.
Anatomical structure through which guttation occurs.
Answer:
Hydathode

Question 14.
Type of leaves in hydrophytes based on distribution of stomata.
Answer:
Epistomatic

Question 15.
Type of leaves in xerophytes based on distribution of stomata.
Answer:
Hypostomatic

Question 16.
The form in which food is transported in plants and then stored in plants.
Answer:
Transported in the form of sugar, sucrose and stored as starch.

Question 17.
Chief vascular element concerned with transport of food in plant.
Answer:
Sieve tubes of phloem.

Give functions/significance of the following

Question 1.
Zone of absorption of root.
Answer:
This zone has unicellular root hairs which absorb water available in rhizosphere.

Question 2.
Diffusion.
Answer:

  1. Necessary for absorption of water by root hairs
  2. For absorption of minerals
  3. Conduction of water against gravity
  4. Exchange of gases
  5. Transport and distribution of food.

Question 3.
Turgor pressure.
Answer:

  1. Keeps cells and cell organelles stretched
  2. Provides support
  3. During growth essential for cell enlargement
  4. Maintains shape of cell
  5. Facilitates opening and closing of stomata

Question 4.
Osmosis.
Answer:

  1. Absorption of water into root
  2. Maintains turgidity of cell
  3. Facilitates cell to cell movement of absorbed water
  4. Resistance to drought or frost injury
  5. Helps in drooping movement of leaflet.
    E.g. Touch me not plant

Question 5.
Root pressure.
Answer:
Hydrostatic pressure developed in living cells of root helps in forcing water from pericycle into xylem
Helps in upward conduction of water against gravity

Question 6.
Transpiration.
Answer:

  1. Removal of excess of water
  2. Helps in passive absorption of water and minerals
  3. Helps in ascent of sap – transpiration pull
  4. Maintains turgor of cells
  5. Imparts cooling effect by reducing temperature 90% – 93% is stomatal transpiration and hence when stomata are open gaseous exchange takes place.

Question 7.
Guard cells.
Answer:

  1. They contain few chloroplasts so capable of photosynthesis.
  2. They can change their size hence opening and closing of stomatal aperture.

Distinguish between the following

Question 1.
Diffusion and Osmosis.
Answer:

Diffusion Osmosis
1. Diffusion is the movement of ions, atoms or molecules of solute, liquids or gases. 1. Osmosis Is the special type of diffusion of water or solvent.
2. Diffusion involves the flow of diffusing particles in both the directions. 2. Osmosis involves the unidirectional flow of solvent molecules.
3. Diffusion does not occur through a semi- permeable membrance. 3. Osmosis occurs through a semi-permeable membrance.
4. Diffusion occurs from a place of higher concentration of diffusing particles to a place of lower concentration. 4. Osmosis occurs from a solution of lower concentration to a solution of higher concentration.

Question 2.
Active absorption and Passive absorption.
Answer:

Active absorption Passive absorption
1. Active absorption is a physiological process. 1. Passive absorption is a physical process.
2. Active absorption takes place by the process of osmosis. 2. Passive absorption takes place by suction force.
3. It involves the expenditure of energy on the part of absorbing cells. 3. It does not involve the expenditure of energy on the part of absorbing cells.
4. Active absorption takes place against the concentration gradient. 4. Passive absorption takes place along the concentration gradient.

Question 3.
Cuticular transpiration and Stomatal transpiration.
Answer:

Cuticular transpiration Stomatal transpiration
1. Cuticular traspiration takes place through the cuticle. 1. Stomatal traspiration takes place through the stomata.
2. Cuticular transpiration accounts for 8 to 10% of total loss of water from plants. 2. Stomatal transpiration accounts for 80 to 90% of total loss of water from plants.
3. Cuticular transpiration depends upon the thickness of the cuticle. 3. Stomatal traspiration depends upon the number and size of stomata.

Give reasons or explain the statements

Question 1.
Water is considered as ‘elixir of life’.
Answer:

  1. Water plays an important role in living organisms.
  2. About 90 – 95% water is present in cell which is functional and structural unit of living organisms.
  3. It helps in maintaining turgidity and shape of cells and cell organelles.
  4. Due to its various properties, it is medium of biochemical reactions, transporting medium and thermal buffer also.
  5. Therefore it is absolutely necessary for life i.e. ‘elixir of life’.

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 2.
Water is a best transporting medium.
Answer:

  1. Water is in liquid state at room temperature.
  2. It is best solvent for most of the solutes. Thus called universal solvent.
  3. It is inert inorganic compound with neutral pH i.e. pH 7 when in pure form.
  4. Hence it is best medium for dissolved minerals.

Question 3.
Water is significant molecule that connects physical world with biological processes.
Answer:

  1. Water is an important constituent of cell. About 90 – 95% of protoplasm is water.
  2. Water in liquid state is best solvent in which various minerals and food molecules are dissolved and transported.
  3. Water acts as the thermal buffer has high specific heat.
  4. Water molecules have high adhesive and cohesive forces of attraction.
  5. It can rise in capillaries due to high surface tension and adhesive forces, e.g. Ascent of sap in plants.
  6. Due to all these important factors it is a significant molecule connecting physical world with biological processes.

Question 4.
Absorption of water by roots from soil is with physical processes inhibition, diffusion and osmosis.
Answer:

  1. Water is absorbed from rhizosphere with the help of unicellular root hairs.
  2. Root hairs have plasma membrane and thin, double layer cell wall of pectin and cellulose.
  3. During imbibition water molecules get tightly adsorbed to the wall of hydrophilic colloids.
  4. Cell wall is freely permeable membrane hence through diffusion water passes into the cell.
  5. Osmosis is a special kind of diffusion of solvent through a semipermeable membrane and as plasma membrane is semipermeable, water enters cell by osmotic mechanism.
  6. Hence all these three physical processes occur sequentially when water is absorbed by roots.

Question 5.
Additional apoplastic pathway through secondary roots is beneficial to plants.
Answer:

  1. Secondary roots develop from the pericycle which is inside endodermis.
  2. Protoxylem is situated close to pericycle in root.
  3. Endodermis have suberized layer Casparian strip which forces water to move in the symplast so that it can enter the vascular xylem.
  4. Since secondary roots originate form pericycle, they bypass the Casparian strip.
  5. Therefore, a direct pathway to xylem and phloem is available without moving into symplast.

Write shorts notes

Question 1.
Role of water or Biological importance of water.
Answer:

  1. Water is absolutely necessary for life.
  2. It is a major constituent of protoplasm.
  3. If provides aqueous medium for various metabolic reactions that take place in plant.
  4. It is raw material for photosynthesis.
  5. It helps in maintaining the turgidity of cells.
  6. It is excellent solvent for various organic materials.
  7. It is transporting medium for dissolved minerals.
  8. It is thermal buffer.

Question 2.
Properties of water.
Answer:

  1. Water is a compound and it is in liquid state at room temperature.
  2. It is an inert inorganic compound with neutral pH.
  3. It has high specific heat, high heat of vapourization, high heat of fusion.
  4. It has high surface tension.
  5. Water molecule has good adhesive and cohesive forces of attraction.
  6. The various properties of water are result of weak hydrogen bonding between the water molecules.

Question 3.
Lenticular transpiration.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 2

  1. Small, raised structures, which are composed of loose cells with large intercellular spaces situated on bark are lenticels.
  2. They are observed on bark of old stem, root and on woody pericarp of fruits.
  3. Lenticular transpiration is of very small amount about 0.1% to 1% of total transpiration.
  4. It is very slow process but occurs throughout the day.

Question 4.
Structure of stomatal apparatus.
Answer:

  1. Stomata are minute openings mainly located in the epidermal surfaces of young stem and leaves.
  2. It is composed of two guard cells and accessory cells which form the opening stomatal pore for transpiration.
  3. Guard cells are kidney shaped in dicot plants and dumbbell shaped in grasses.
  4. Guard cells have unevenly thickened wall, inner wall thick, inelastic while outer wall is thin and elastic.
  5. Guard cells are nucleated cells with few chloroplasts and hence can perform photosynthesis.
  6. Accessory cells are specialized cells that surround the guard cells and are reservoirs of K+ ions.

Question 5.
Apoplast pathway.
Answer:

  1. The movement of water across the root cells occur by two pathways viz. apoplast and symplast.
  2. Apoplast movement of water in plants occurs exclusively through cell walls and intercellular spaces in living cells of cortex.
  3. Apoplast pathway is up to the endodermis as the suberized layer of Casparian strip obstructs the movement.
  4. Additional apoplastic route giving direct contact to xylem and phloem is through secondary roots which originate at pericycle and bypass endodermis.
  5. The cellulosic walls of root hair, cortical cells are hydrophilic and permeable to water.

Question 6.
Water potential.
Answer:

  1. Chemical potential of water is called water potential.
  2. The Greek letter (ψ) psi represents water potential.
  3. The unit of measurement is in bars/pascals (Pa) or atmosphere.
  4. Water potential of pure water is always zero.
  5. Water potential of protoplasm is equal to D.PD. but it has a negative value.
  6. Water always moves from less negative potential to more negative water potential.

Short answer questions

Question 1.
Enlist macronutrients and micronutrients required for plant growth.
Answer:

  1. Some minerals which are required in large amounts for plant metabolism are macronutrients. E.g. C, H, O, E N, S, Mg, Ca, K. etc.
  2. Some minerals which are required in small amounts for plant metabolism are micronutrients. E.g. Cu, Co, Mn, B, Zn, Cl, etc.

Question 2.
Explain about the factors that affect water absorption.
Answer:

  1. Water is absorbed by unicellular root hairs from soil.
  2. Presence of capillary water in soil is needed as this water from soil is absorbed by root hairs.
  3. Soil temperature of 20-30°C favours water absorption.
  4. Rate of absorption is decreased by high concentration of solutes in soil.
  5. Soil should be properly aerated, poorly aerated soil shows poor absorption rate.
  6. Increased transpiration accelerates rate of absorption of water.

Question 3.
Explain mechanism of sugar transport through phloem.
Answer:

  1. The part of plant where food is synthesized is known as source and where it is utilized is sink.
  2. Food is translocated through phloem tissue in the form of sucrose along the concentration gradient from source to sink.
  3. Munch’s pressure flow theory or mass flow hypothesis is widely accepted theory for sugar transport.
  4. Glucose is synthesized in photosynthetic cells hence endo-osmosis occurs due to increased osmotic concentration.
  5. With this turgor pressure increases and sugar is forced in sieve tube of vein from photosynthetic cell – Vein loading.
  6. At the sink, sugar is utilized and excess amount converted to starch, hence osmotic concentration is lowered and exosmosis takes place.
  7. Turgor pressure is lowered hence turgor pressure gradient is set and food is translocated passively against concentration gradient – Vein unloading.

Question 4.
Explain the principles of cohesion tension theory and its limitations?
Answer:

  1. Cohesion tension theory is widely accepted theory for translocation of water proposed by Dixon and Joly.
  2. It is based on principles cohesion and adhesion of water molecules along with transpiration pull.
  3. Strong attractive force between water molecules is cohesive force and the strong force of attraction between water molecules and wall of lumen of xylem vessels is called adhesive force.
  4. Owing to these cohesive and adhesive forces a continuous column of water is maintained in xylem elements from root to aerial parts, tips of leaves.
  5. Transpiration pull is developed in the xylem vessel in leaves.
  6. This pull is transmitted downwards and due to suction pressure, water is pulled passively against gravity, i.e. ascent of sap.

Limitations of theory-

  1. For activity of transpiration pull, water column should be maintained constantly and continuously. Owing to changes in temperature during day and night, gas bubbles may be formed in the water channel. This will break the continuity.
  2. According to this theory, tracheids are more efficient than vessels because of their tapering end walls which support water column. Vessels are tubular structure with open ends.

Question 5.
What is the meaning of specific heat, heat of vaporization and heat of fusion?
Answer:

  1. Specific heat : The specific heat is the amount of heat per unit mass required to raise the temperature by one degree. The specific heat of water is 1 calorie/gram degree C.
  2. Heat of vapourization : It is the amount of energy that must be added to a liquid substance to transform a quantity of that substance to gas. It is also known as heat of evaporation.
  3. Heat of fusion : It is amount of energy typically heat, provided to a specific quantity of the substance to change its state from solid to liquid at constant pressure.

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 6.
What are adhesive and cohesive forces?
Answer:

  1. Adhesive and cohesive forces are attractive forces between molecules of same substance.
  2. Cohesive forces exist between molecules of the same type. E.g. between water-water molecule.
  3. Adhesive forces exist between dissimilar molecules. E.g. water molecule and lignin deposited wall of xylem element.

Chart based/Table based questions

Question 1.
Complete the table based on types of solution.
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 3
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 4

Question 2.
Complete the table.

Event Physical process
1. Soaking of Seeds —————-
2. Water entering guard cells —————-
3. Exchange of gases —————
4. Loss of water in liquid form —————
5. Water coming out through earthen pot —————
6. Loss of water in vapour form —————
7. Absorption of solutes passively by carrier —————
8. Spreading of fragrance of incense stick —————

Answer:

Event Physical process
1. Soaking of Seeds Imbibition
2. Water entering guard cells Osmosis
3. Exchange of gases Diffusion
4. Loss of water in liquid form Guttation
5. Water coming out through earthen pot Diffusion
6. Loss of water in vapour form Transpiration
7. Absorption of solutes passively by carrier Facilitated Diffusion
8. Spreading of fragrance of incense stick Diffusion

Diagram based questions

Question 1.
Zones of root
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 5

Question 2.
Structure of root hair
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 6

Question 3.
Diffusion of water
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 7

Question 4.
Pathway for water uptake
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 8

Question 5.
L. S. of sieve tube (Transport of food)
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 9

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

Question 6.
Structure of stomata and Types of guard cells
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 10

Long answer questions

Question 1.
Write a note on macronutrients and micronutrients required for plant growth.
Answer:

  1. Plants absorb mincral nutrients from their surroundings.
  2. For a proper growth of plants about 35 to 40 different elements are required.
  3. Plants absorb these nutrients in ionic or dissolved form from soil with their root system e.g. Phosphorus as PO4. Sulphur as SO42- etc.
  4. Based on their requirement in quantity. they are classified as major nutrients or macronutrients and those needed In small amounts are minor or micronutrients.
  5. Macroclements are required in large amounts, as they play nutritive and structural roles e.g. C, H, O. R Mg. N, K. S and Ca. -Ca pectate cell wall component. Mg component of chlorophyll.
  6. C. H, O are non-mineral major elements obtained from air and water e.g. CO2 is source of carbon, Hydrogen from water.
  7. Microelements arc required In traces as they mainly have catalytic role as co-factors or activators of enzymes.
  8. Microelernents may be needed for certain activity in life cycle of plant e.g. B for pollen germination, Si has protective role during stress conditions and fungal attacks, Al enhances availability of phosphorus.
  9. The Important micronutrients for plant growth are Mn. B. Cu, Zn, Cl.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 5 Origin and Evolution of Life Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 5 Origin and Evolution of Life

Multiple Choice Questions

Question 1.
Which statement is true for the theory of spontaneous generation?
(a) Life came from outer space.
(b) Life can arise from dead matter.
(c) Life can arise from non-living things only
(d) Life arises spontaneously by miracle
Answer:
(c) Life can arise from non-living things only

Question 2.
When man makes the selection during animal husbandry and plant breeding programmes, then it is an example of ……………….
(a) reverse evolution
(b) artificial selection
(c) mutation
(d) natural selection
Answer:
(b) artificial selection

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 3.
Palaeontological evidences for evolution are ……………….
obtained in the form of
(a) development of embryo
(b) homologous organs
(c) fossils
(d) analogous organs
Answer:
(c) fossils

Question 4.
The homologous organs such as bones of forelimbs of whale, bat, cheetah and man are similar in structure, because ……………….
(a) one organism has given rise to another
(b) they share a common ancestor
(c) they perform the same function
(d) they have biochemical similarities
Answer:
(b) they share a common ancestor

Question 5.
Which type of evolution gives rise to analogous organs?
(a) divergent evolution
(b) parallel evolution
(c) genetic drift
(d) convergent evolution
Answer:
(d) convergent evolution

Question 6.
Hardy Weinberg’s equation of genetic equilibrium i.e. (p + q)² = p² + 2pq + q² = 1 is used in ……………….
(a) population genetics
(b) Mendelian genetics
(c) biometrics
(d) molecular genetics
Answer:
(a) population genetics

Question 7.
Which type of rocks show maximum fossils ?
(a) Sedimentary rocks
(b) Igneous rocks
(c) Metamorphic rocks
(d) Any type of rock
Answer:
(a) Sedimentary rocks

Question 8.
Industrial melanism observed in moth, Biston bitularia shows ………………. type of natural selection.
(a) stabilising
(b) directional
(c) disruptive
(d) artificial
Answer:
(b) directional

Question 9.
Variations during mutations of meiotic recombinations are ……………….
(a) random and directionless
(b) random and directional
(c) small and directional
(d) random, small and directional
Answer:
(a) random and directionless

Question 10.
Theory of special creation is based on ………………. beliefs.
(a) scientific
(b) religious
(c) traditional
(d) mythological
Answer:
(b) religious

Question 11.
Which sentence holds true for theory of biogenesis?
(a) Living organisms arise from non-living things.
(b) Theory of biogenesis can explain origin of life.
(c) Continuity of life can be explained by this theory.
(d) This theory was disproved by Louis Pasteur.
Answer:
(c) Continuity of life can be explained by this theory

Question 12.
Who gave the Big-Bang theory that explained the origin of life ?
(a) Georges Lemaitre
(b) Oparin and Haldane
(c) Charles Darwin
(d) J.B.S. Haldane
Answer:
(a) Georges Lemaitre

Question 13.
Primitive atmosphere of the earth was of type ……………….
la) oxidizing
(b) reducing
(c) aerobic
(d) oxido-reducing
Answer:
(b) reducing

Question 14.
What is the meaning of protobiogenesis?
(a) The origin of life on the earth.
(b) The origin of protozoans on the earth.
(c) The origin of protists on the earth.
(d) The origin of protons of the earth.
Answer:
(a) The origin of life on the earth

Question 15.
What are the first form of life on the earth called ?
(a) Pre-cells or Protobionts
(b) Protoproteins
(c) Coacervates
(d) Chromophores
Answer:
(a) Pre-cells or Protobionts

Question 16.
Which of the following is a landmark in the origin of life ?
(a) Formation of oxygen
(b) Formation of carbohydrates
(c) Formation of proteins
(d) Formation of water
Answer:
(c) Formation of proteins

Question 17.
The first chemicals formed on the earth were ………………. etc.
(a) oxygen, CFC, ozone
(b) DNA. RNA and nucleotides
(c) salt, sugar, proteins
(d) water, ammonia, methane
Answer:
(d) water, ammonia, methane

Question 18.
The unique feature of hot dilute soup is that there was no free ………………. in it.
(a) nitrogen
(b) oxygen
(c) carbon
(d) sulphur
Answer:
(b) oxygen

Question 19.
On which plant did Hugo de Vries work during his experimentations?
(a) Hibiscus rosa sinensis
(b) Oenothera lamarkiana
(c) Mirabilis jalapa
(d) Pisum sativum
Answer:
(b) Oenothera lamarkiana

Question 20.
Which one out of the following is a connecting link between fish and amphibian ?
(a) Archaeopteryx
(b) Seymouria
(c) Ichthyostega
(d) Dinosaurus
Answer:
(c) Ichthyostega

Question 21.
Homologous organs always lead to ………………. evolution.
(a) convergent
(b) divergent
(c) parallel
(d) radiating
Answer:
(b) divergent

Question 22.
Find the odd one out
(a) Caecum
(b) Nictitating membrane
(c) Coccyx
(d) Sacrum
Answer:
(d) Sacrum

Question 23.
The most common types of fossils are ……………….
(a) moulds
(b) casts
(c) actual remains
(d) model
Answer:
(c) actual remains

Question 24.
In geological time scale which period showed dominance of reptiles ?
(a) Triassic
(b) Jurassic
(c) Cretaceous
(d) Eocene
Answer:
(b) Jurassic

Question 25.
Which was the period of beginning of modern birds ?
(a) Triassic
(b) Jurassic
(c) Cretaceous
(d) Eocene
Answer:
(c) Cretaceous

Question 26.
Which epoch showed mammals at height of evolution ?
(a) Eocene
(b) Oligocene
(c) Miocene
(d) Pliocene
Answer:
(c) Mlocene

Question 27.
When did Holocene began ? (mya = million years ago)
(a) 2 mya
(b) 1 mya
(c) 0.5 mya
(d) 0.01 mya
Answer:
(d) 0.01 mya

Question 28.
Cichlid fishes in Lake Victoria are representatives of ………………. type of speciation.
(a) allopatric
(b) sympatric
(c) isolation
(d) random
Answer:
(b) sympatric

Question 29.
Which is the offspring of male horse and female donkey?
(a) Mule
(b) Hinny
(c) Marino
(d) Baroque
Answer:
(b) Hinny

Question 30.
Which one out of the following is a living fossil ?
(a) Coelacanth
(b) Lung fish
(c) Shark
(d) Rays
Answer:
(a) Coelacanth

Question 31.
Giant cephalopods like Nautilus were present in ………………. period.
(a) Silurian
(b) Devonian
(c) Ordovician
(d) Permian
Answer:
(c) Ordovician

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 32.
First modern birds were formed during ………………. period.
(a) Triassic
(b) Jurassic
(c) Cretaceous
(d) Eocene
Answer:
(c) Cretaceous

Question 33.
When did man emerge during evolutionary time period ?
(a) Eocene
(b) Oligocene
(c) Miocene
(d) Pliocene
Answer:
(d) Pliocene

Question 34.
Find the odd monkey out:
(a) Baboons
(b) Gibbons
(c) Macaques
(d) Langurs
Answer:
(b) Gibbons

Question 35.
Who was man with ape-brain ?
(a) Pithecanthropus
(b) Dryopithecus
(c) Australopithecus
(d) Neanderthal man
Answer:
(c) Australopithecus

Question 36.
Who was the first true man ?
(a) Pithecanthropus
(b) Cro-magnon
(c) Australopithecus
(d) Neanderthal man
Answer:
(a) Pithecanthropus

Question 37.
Which one of the following is not present in human beings ?
(a) S curves in vertebral column
(b) Orthognathus face
(c) Simian gap
(d) Chin
Answer:
(c) Simian gap

Question 38.
Which is the correct sequence of human evolution ?
(a) Australopithecus → Ramapithecus → Homo sapiens → Homo habilis
(b) Homo erectus → Homo habilis → Homo sapiens
(c) Ramapithecus → Homo habilis → Homo erectus → Homo sapiens
(d) Australopithecus → Ramapithecus → Homo erectus → Homo habilis → Homo sapiens.
Answer:
(c) Ramapithecus – Homo hablUs -‘ Homo erectus Homo sapiens

Question 39.
Fossils of Homo erectus were obtained in ………………. and ……………….
(a) Kenya, Shivalik hills
(b) Java, Peking
(c) Africa, Asia
(d) Neanderthal valley, Taung
Answer:
(b) Java, Peking

Question 40.
Dentition more like that of the modern man was seen for the first time in ……………….
(a) Dryopithecus
(b) Ramapithecus
(c) Australopithecus
(d) Homo habilis
Answer:
(d) Homo habilis

Question 41.
Lemur and Tarsier belongs to ……………….
(a) Prosimi
(b) Hyalobatidae
(c) Pongidae
(d) Homonidae
Answer:
(a) Prosimi

Match the columns

Question 1.

Geological time Animal life
(1) Cambrian (a) Amphibians
(2) Ordovician (b) First terrestrial animals
(3) Silurian (c) Jawless fishes
(4) Devonian (d) Trilobite

Answer:

Geological time Animal life
(1) Cambrian (d) Trilobite
(2) Ordovician (c) Jawless fishes
(3) Silurian (b) First terrestrial animals
(4) Devonian (a) Amphibians

Question 2.

Human stage Cranial capacity in CC
(1) Homo sapiens (a) 650-800
(2) Homo neanderthalensis (b) 900
(3) Homo habilis (c) 1400
(4) Homo erectus (d) 1450

Answer:

Human stage Cranial capacity in CC
(1) Homo sapiens (d) 1450
(2) Homo neanderthalensis (c) 1400
(3) Homo habilis (a) 650-800
(4) Homo erectus (b) 900

Classify the following to form Column B as per the category given in Column A

Question 1.
Nictitating membrane, Seymouria, Lung fish, Flipper of whale and wing of bird, Wing of insect and wing of bird, wisdom
tooth, Eye of octopus, an eye of mammal, vertebrate heart and brain.

Column I Column II
(1) Homologous organs ————-
(2) Analogous organs ————-
(3) Vestigial organs ————-
(4) Connecting links ————-

Answer:

Column I Column II
(1) Homologous organs Flipper of whale and wing of bird, Vertebrate heart and brain
(2) Analogous organs Wing of insect and wing of bird. Eye of octopus, an eye of mammal
(3) Vestigial organs Nictitating membrane, wisdom tooth
(4) Connecting links Seymouria, Lung fish

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 2.
Origin of conifers, Abundance of trilobites, Diversification of fishes, All types of marine algae, Formation of forests, Rise of dinosaurs, Extinction

Geological time period Major events
Cambrian ————-
Devonian ————-
Permian ————-
Triassic ————-

Answer:

Geological time period Major events
Cambrian Abundance of trilobites, All types of marine algae
Devonian Diversification of fishes, Formation of forests
Permian Origin of conifers. Rise of modern insects.
Triassic Rise of dinosaurs, Extinction of seed ferns

Very short answer questions

Question 1.
What is protobiogenesis?
Answer:
The origin of life on the earth is called protobiogenesis.

Question 2.
What are panspermia?
Answer:
Panspermia or cosmozoa are considered to be spores through which life came on the earth from distant planets.

Question 3.
What are protobionts?
Answer:
Protobionts were first form of life which were formed by nucleic acids along with other inorganic and organic molecules. They have some properties of living form.

Question 4.
What are eobionts?
Answer:
Eobionts or protocells are the first primitive living system which are formed by colloidal aggregations of lipids and proteinoids.

Question 5.
When did universe originate? How?
Answer:
Universe originated about 20 billion years ago by huge titanic explosion called big- bang.

Question 6.
What were the different energy sources during primitive times when earth was cooling?
Answer:
The different available energy sources during primitive times were ultra-violet rays, radiations, lightning and volcanic activities.

Question 7.
How was first cell formed on the primitive earth?
Answer:
In the protocell, when RNA or DNA system developed and they started regulating various metabolic activities, then it was called a first cell.

Question 8.
How was first cell formed on the primitive earth?
Answer:
In the protocell, when RNA or DNA system developed and they started regulating various metabolic activities, then it was called a first cell.

Question 9.
How was first cell on the earth in its metabolism ?
Answer:
First cell was anaerobic, heterotrophic and obtained energy by chemoheterotrophic processes.

Question 10.
What are ribozymes?
Answer:
Ribozymes are catalytic RNA which act as biocatalyst.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 11.
Who disproved Lamarck’s theory?
Answer:
August Weismann disproved Lamarck’s theory.

Question 12.
Write theory of germplasm as suggested by Wallace.
Answer:
The theory of germplasm says that variations produced in germ cells or germplasm are inherited to next generations, but the somatic variations present in somatoplasm or somatic cells are not inherited.

Question 13.
Who are the main contributors of modem : synthetic theory of evolution?
Answer:
R. Fischer, J. B. S. Haldane, T. Dobzhansky, Huxley, E. Mayr, Simpson, Stebbins, Fisher, Sewall Wright, Medel, T. H. Morgan, etc. are the main contributors of modern theory of evolution.

Question 14.
What is variation?
Answer:
The variations are the differences that occur in morphology, physiology, nutrition, habit, behavioural patterns, etc.

Question 15.
What is mutation?
Answer:
Sudden, large, inheritable and drastic change occurring in the genetic constitution is called mutation.

Question 16.
What is gene frequency?
Answer:
Gene frequency is the relative frequency of an allele or a gene at a particular locus in a population, as compared to other genes, expressed as a fraction or percentage.

Question 17.
What is the other name for genetic drift?
Answer:
Sewall wright effect is the other name for genetic drift.

Question 18.
Why is genetic drift called founder’s effect?
Answer:
The allelic frequency of new population which undergoes genetic drift becomes different from the original one, thus the original drifted population becomes different and are called ‘founders’. Since founders are formed therefore genetic drift is called founder’s effect.

Question 19.
What are fossils?
Answer:
Fossils are the dead remains of plants and animals from prehistoric times, which are found in different forms such as moulds, casts, actual remains or compression seen in various geological layers.

Question 20.
What is embryology?
Answer:
Embryology is the branch of biology and medicine which deals with study of embryos and their development.

Give definition of the following

Question 1.
Gene flow
Answer:
The transfer of genes between two genetically different populations among themselves is called gene flow.

Question 2.
Genetic drift
Answer:
Any random fluctuation (alteration) in allele frequency, occurring in the natural population by pure chance, is called genetic drift.

Question 3.
Chromosomal aberrations
Answer:
The structural, morphological change in chromosome due to rearrangement, is called chromosomal aberrations.

Question 4.
Mendelian population
Answer:
Small interbreeding group of a population is defined as Mendelian population.

Question 5.
Crossing over
Answer:
Exchange of genetic material between non-sister chromatids of homologous chromosomes in sexually reproducing organisms, during gamete formation is called crossing over.

Question 6.
Saltation
Answer:
Saltation is defined as single step large mutation.

Name the following

Question 1.
The book written by Charles Darwin after returning from voyage.
Answer:
Origin of species by natural selection.

Question 2.
Five main postulates of Darwinism.
Answer:

  1. Overproduction
  2. Struggle for existence
  3. Organic variations
  4. Natural selection
  5. Origin of next species.

Question 3.
Five key factors of evolution as suggested by Stebbins.
Answer:

  1. Gene mutations
  2. Mutations in the chromosome structure and number
  3. Genetic recombinations
  4. Natural selection
  5. Reproductive isolation.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 4.
Name the four types of chromosomal aberrations.
Answer:

  1. Deletion
  2. Translocation
  3. Duplication
  4. Inversion

Question 5.
Name of the insect that displayed industrial melanism and the name of the scientist who studied them.
Answer:
Kettlewell reported that Biston betularia or peppered moth displayed industrial melanism.

Question 6.
Example of homologous organs in plants.
Answer:
Thorns of Bougainvillea and tendrils of Cucurbita.

Question 7.
Example of analogous organs in plants.
Answer:
Sweet potato and potato.

Question 8.
Pre-mating isolating mechanisms.
Answer:

  1. Habitat or ecological isolating mechanism
  2. Seasonal or temporal isolating mechanism
  3. Ethological isolating mechanism
  4. Mechanical isolating mechanism

Question 9.
Post-mating isolating mechanisms.
Answer:

  1. Gamete mortality
  2. Zygote mortality
  3. Hybrid sterility

Question 10.
Name of connecting link between reptiles and birds.
Answer:
Archaeopteryx is the connecting link between reptiles and birds.

Question 11.
Name of connecting link between amphibians and reptiles.
Answer:
Seymouria is the connecting link between amphibians and reptiles.

Question 12.
Name of connecting link between fish and amphibians.
Answer:
Ichthyostegia is a missing link between fish and amphibians.

Question 13.
Name the period that was dominant for Amphibia.
Answer:
Carboniferous is dominant period for Amphibia.

Question 14.
Name the epoch when mammals were at the height of evolution.
Answer:
Miocene epoch.

Question 15.
Name the era when birds began to origin.
Answer:
Mesozoic era.

Question 16.
Name the three subclasses of Class Mammalia.
Answer:

  1. Marsupials
  2. Monotremes
  3. Eutheria

Question 17.
Name the three subfamilies of family Hominoidea.
Answer:

  1. Hyalobatidae
  2. Pongidae
  3. Hominidae

Give the significance of the following

Question 1.
Significance of Natural selection
Answer:

  1. Natural selection is the main driving force behind the evolution.
  2. Natural selection favours those genetic variations which have better fitness value.
  3. Such organisms are at selective advantage and they produce more offspring than the rest. Such organisms have greater survival and reproductive capacity.
  4. In this way natural selection helps in the evolution of new species.
  5. Natural selection favours differential reproduction of gene and brings about changes in the gene frequency.
  6. Natural selection brings about evolutionary changes.
  7. Natural selection also eliminates the genes carrying harmful mutations. This is called mutation balance in which allele frequency of harmful recessives remain constant generation after generation.

Distinguish between the following

Question 1.
Gene flow and Genetic drift.
Answer:

Gene flow Genetic drift
1. Gene flow is the alteration in the gene frequency due to migrations. 1. Genetic drift is alteration in the gene frequency by pure chance.
2. Gene flow occurs due to exchange of genes in the adjacent populations through interbreeding. 2. Genetic drift occurs due to accidental and sudden elimination of a particular gene.
3. Larger populations tend to show more migrations and hence more gene flow. 3. Smaller populations have greater chances of genetic drift.
4. Gene flow occurs due to emigration and immigration. 4. Genetic drift occurs only within the specified population.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 2.
Directional, Stabilizing and Disruptive selection.
Answer:

Directional selection Stabilizing selection Disruptive selection
1. Natural selection operating in a linear direction is called directional selection. 1. Natural selection operating to balance or stabilize the population is called stabilizing selection. 1. Natural selection which disrupts the mean characteristics of a population is called disruptive selection.
2. In directional selection, more individuals acquire characters other than the mean character value. 2. In stabilizing selection, more individuals of a population acquire a mean character value. 2. In disruptive selection, more number of individuals acquire extreme or peripheral character value.
3. Directional selection eliminates one of the extremes of the phenotypic range and favour the other. 3. Stabilizing selection tends to favour the intermediate forms and eliminate both the phenotypic extremes. 3. Disruptive selection favours extreme phenotypes and eliminate intermediate.
4. It streamlines variations. 4. It reduces variations. 4. It increases variations.
5. This kind of selection is the most common. 5. This kind of selection is common. 5. This kind of selection is rare.
6. Directional selection operates for many generations, it results in an evolutionary trend within a population and shifting a peak in one direction.

E.g. Industrial melanism, DDT resistance in mosquito, etc.

6. This selection leads to evolutionary change but tend to maintain phenotypic stability within population.

E.g. All the populations which have adapted to their environment.

6. It ensures the effect on the entire gene pool of a population, considering all mating types or systems.

E.g. African seed cracker finches with different sized beaks

Question 3.
Homologous organs and Analogous organs.
Answer:

Homologous organs Analogous organs
1. Homologous organs are structurally similar to one another. 1. Analogous organs are structurally dissimilar to one another.
2. Homologous organs are functionally dissimilar from one another. 2. Analogous organs are functionally similar to one another.
3. Homologous organs help in tracing the evolutionary relationships. 3. Analogous organs do not help in tracing the evolutionary relationships.
4. Homologous organs lead to divergent evolution. 4. Analogous organs lead to convergent evolution.
5. Animals residing in different habitats but having closer evolutionary relationship show homologous organs.

E.g. Forelimbs of frog, lizard, bird, bat, whale, man, etc.

5. Diverse animals residing in the similar habitat show analogous organs.

E.g. Wing of a bird and wing of an insect.

Question 4.
Allopatric and Sympatric speciation.
Answer:

Allopatric speciation Sympatric speciation
1. Allopatric speciation is the formation of a new species due to separation of a segment of population from the original population. 1. Sympatric speciation is the formation of species within single population.
2. There is geographical barrier cutting across the species range during such speciation. 2. There is no geographical isolation during sympatric speciation.
3. Allopatric speciation does not have physiological barrier. 3. Sympatric speciation is due to physiological or reproductive isolating barrier.
4. Migration of individual are also helpful in allopatric speciation. E.g. African elephant and Indian elephant. 4. Mutations are helpful in sympatric speciation. E.g. Cichlid fishes in Lake Victoria.

Give scientific reasons

Question 1.
Simple organic molecules did not show decomposition in primitive oceans.
Answer:
Simple organic molecules which were formed during chemical evolution, accumulated at the bottom of water bodies. At that time there was no free oxygen and enzymes. Therefore, simple organic molecules did not show decomposition in primitive oceans.

Question 2.
Archaeopteryx is called connecting link between reptiles and birds.
Answer:
(1) Archaeopteryx shows reptilian as well as avian characters.

(2) Reptilian characters are as follows:

  • Jaws with homodont (all similar) teeth. Bones are nonpneumatic i.e. solid.
  • Ribs have a single head. Sternum without keel.
  • Abdominal ribs present which are like the crocodilian ribs.
  • Forearms have three digits ending in distinct claws while the hind limb has four digits ending in clawed digits.

(3) Avian characters shown by it are as follows:

  • Forearms modified into wings.
  • Feathery exoskeleton.
  • Skull bones are completely fused.
  • Cranium is rounded with large orbits and a single condyle.
  • Jaws are modified into beak.
  • Limb bones have first toe in opposable manner. Foot present with clawed digits. Since it showed characters of both the classes, it is considered as the connecting link between the two.

Question 3.
Birds are glorified reptiles.
Answer:
Huxley, the evolutionary biologist gave this statement after studying the characters of birds and reptiles. The fossil bird, Archaeopteryx was discovered which showed characters of both Reptilia and Aves. It showed transformation of reptilian characters into bird characters. Hence, birds are said to be glorified reptiles with feathery exo-skeleton and other glorious characteristics.

Question 4.
Analogous organs do not have significant role in evolution.
Answer:
Analogous organs lead to convergent evolution, i.e. different organisms show same superficial structural similarities due to similar functions or habitat. But anatomically and structurally they are different. These organs do not help to trace the common ancestry. Therefore, they are said to have no significant role in evolution.

Question 5.
Australopithecus is described as a man with ape brain.
Answer:
(1) Australopithecus can be considered as a connecting link between ape and man due to the following ape-like and man like characteristics shown by it.

(2) The ape-like characteristics of Australopithecus:

  • The jaws and teeth were larger than those of modern man.
  • The face was prognathous, i.e. it had a muzzle like slope
  • The chin was absent
  • The eye-brow ridges projected over the eyes
  • Their cranial capacity ranged from 450-600 c.c.

(3) The man-like characteristics of Australopithecus:

  • It walked nearly or completely straight due to erect posture.
  • The vertebral column had a distinct lumbar curve with broad basin-like pelvic girdle.
  • Dentition was man-like with the smoothly rounded parabolic dental arch.
  • A simian gap was absent. Australopithecus is therefore, rightly described as a man with ape brain.

Write short notes on the following

Question 1.
Evidences of Darwinism.
Answer:
(1) Height of neck of Giraffe : Long-necked Giraffe came into existence in the following way. Long-necked Giraffe could pluck and eat more leaves from tall trees and woody climbers. So it was well adapted to the environment. Short-necked one could not get food and thus perished in the struggle. This adaptation was transmitted to their offspring.

(2) Black colour peppered moths : The example of industrial melanism seen in U.K. is an excellent example of natural selection in action. Black coloured moths evolved gradually as new species from the previous white coloured forms.

(3) DDT resistance in mosquitoes : Intensive DDT spraying destroyed all types of mosquitoes. Some mosquitoes developed resistance to DDT and survived in spite of DDT spray. They reproduced more and were thus selected naturally.

Question 2.
Drawbacks and Objections to Darwinism.
Answer:

  1. Darwin took into consideration minute fluctuating variation as principal factors. But these are neither heritable nor are part of evolution.
  2. Darwin did not distinguish somatic and germinal variation and considered all variations are heritable.
    ‘Arrival of the fittest’ was not explained by him.
  3. Darwin was unable to explain the cause, origin and inheritance of variations and of vestigial organs.
  4. He also could not explain extinction of species.
  5. Gradual accumulation of useful variations forms the new species, but their intermediate forms were not recognised.
  6. Darwin could not explain existence of neutral flowers and the sterility of hybrids.

Question 3.
The main features of mutation theory.
Answer:

  1. Mutations are large, sudden and discontinuous variations in a population.
  2. Changes caused due to mutations are inheritable.
  3. The raw material for organic evolution is provided by mutations.
  4. Mutation can be useful or harmful. Useful mutations are at evolutionary advantage as they are selected by nature.
  5. Accumulation of the useful mutations over a period of time leads to the origin and establishment of new species.
  6. Harmful or non-adaptive mutation may persist or get eliminated by nature.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 4.
Directional selection.
Answer:

  1. Natural selection bringing about directional change without disrupting the balance is called directional selection.
  2. In a population when more individuals acquire characters which are other than the mean character value, then it is called directional selection.
  3. Natural selection usually acts to eliminate one of the extremes of the phenotypic range and favour the other. E.g. systematic elimination of homozygous recessives.
  4. Directional selection operates for many generations, it results in an evolutionary trend within a population and shifting a peak in one direction.
    E.g. Industrial melanism, DDT resistant mosquito, etc.

Question 5.
Stabilizing selection/Balancing selection.
Answer:

  1. Stabilizing selection is the type of natural selection which balances the population, hence it is also known as balancing selection.
  2. In such population more individuals acquire a mean character value.
  3. Such selection tends to favour the intermediate forms and eliminate both the phenotypic extremes.
    E.g. More number of infants with intermediate weight survive better as compared to overweight or underweight infants.
  4. Stabilizing selection reduces variations.
  5. It tends to maintain phenotypic stability within population, and does not bring about drastic evolutionary changes.
  6. A population showing stabilizing selection is well-adapted to its environment.

Question 6.
Homologous organs.
Answer:

  1. The structural similarities between the homologous organs indicate that they have a common ancestry.
  2. Different homologous organs indicate divergent evolution or adaptive radiation.
  3. Homologous organs help in tracing the phylogenetic relationships.
  4. Homologous organs are those organs which are structurally similar but functionally dissimilar.

E.g.
(i) Forelimbs of frog, lizard, bird, bat, whale and man are homologous to each other. All the limbs are morphologically similar in construction such as similar limb bones but are dissimilar in function. Frog limbs are meant for hopping, lizard limbs help in crawling, birds and bats fly with the . help of forelimbs while whale uses it for swimming and man for handling the objects.

(ii) Vertebrate heart and brain. & In plants, thorns of Bougainvillea and tendrils of Cucurbita represent homology.

Question 7.
Analogous organs.
Answer:

  1. Analogous organs are similar in function but dissimilar in structural details.
  2. They do not help to trace the relationship in the evolution but help to understand the convergent evolution.
  3. Structural modifications in the organs are due to similar habitat.

E.g.
(i) Wing of an insect and wing of a bird, both are useful in flight so they are functionally similar but are structurally different. Insect wing is formed by exoskeleton expansion while bird wing is the modified forelimb.

(ii) Eye of the Molluscan octopus and of eye of mammals. They differ in their retinal position, structure of lens and origin of different eye parts, but both perform function of vision.

(iii) The flippers of penguins (birds) and dolphins (mammals).

(iv) Sweet potato which is a root modification and potato which is a stem modification, both perform similar function of storing starchy food.

Question 8.
Vestigial organ.
Answer:

  1. Vestigial organs are rudimentary organs which are imperfectly developed and non¬functional, degenerate structures.
  2. These organs in animals become functionless thus their presence in the body is not required.
  3. But they are simply present as they descend down during evolution and continue to exist.
  4. In the process of evolution, they may disappear totally.
  5. They indicate evolutionary line as they were once functional in the ancestors.

Examples of vestigial organs in human beings:

  1. Caecum and vermiform appendix : These are functional in herbivorous animals where they help in cellulose digestion. In humans they are functionless.
  2. Nictitating membrane situated in the eyes of humans. It is a remnant of third eyelid.
  3. Coccyx or tail vertebrae which shows remnant of tail, Wisdom teeth or 3rd molars. These organs indicate that human beings descended from ape like ancestors.

Question 9.
Types of fossils.
Answer:
There are four main types of fossils : actual remains, moulds, casts and compressions.
1. Actual remains : The most common type of fossil is actual remains in which the plants, animals and human bodies are seen embedded in permafrost of arctic or alpine snow. Due to severe cold temperature, the bodies remain preserved in the actual state, E.g., Fossil of Woolly Mammoth in Siberia. Many insects and smaller arthropods remained embedded and thus preserved in amber or hardened resin.

2. Moulds : Hardened encasements formed in the outer parts of organic remains of animals or plants form moulds. The organisms later decays leaving cavities or the impression in permanent form. E.g. Footprints.

3. Casts : Casts are hardened pieces of mineral matter which is deposited in the cavities of moulds.

4. Compressions : A thin carbon film indicates the outline of external features of ancient organism, but other structural details are not seen.

Question 10.
Major changes that occurred in human evolution.
Answer:

  1. Major changes that took place in evolution of man are as follows :
  2. Increase in size and complexity of brain and enhanced intelligence.
  3. Increase in cranial capacity.
  4. Bipedal locomotion.
  5. Opposable thumb.
  6. Erect posture.
  7. Shortening of forelimbs and lengthening of hind limbs.
  8. Development of chin. Orthognathous face.
  9. Broadening of pelvic girdle and development of lumbar curvature.
  10. Social and cultural development such as articulated speech, art, development of tools, etc.

Question 11.
Dryopithecus.
Answer:

  1. Dryopithecus is also called Proconsul. Leakey discovered the fossils of Dryopithecus, on an island in Lake Victoria of Africa. Also the fossil was found in Haritalyanga in Bilaspur district of Himachal Pradesh.
  2. It was a group of apes that lived in Miocene epoch about 20 to 25 million years ago.’
    Several species of Dryopithecus are available, the important among these is African fossil D. africanus.
  3. Dryopithecus has a close similarity to chimpanzee and also walked like a modern chimpanzee.
  4. The structure of its limbs and wrists show that knuckle walking was lesser in it. It used the flat of its hands like a monkey.
  5. It had arms and legs of the same length and had a semi-erect posture.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 12.
Ramapithecus.
Answer:

  1. Ramapithecus was on direct line of evolution of man.
  2. It was called an ape-man like primate.
  3. Its fossils were obtained in the form of teeth and jaw bones in the rocks of Siwalik Hills in India by Lewis and also in Kenya.
  4. It existed during late Miocene and early Pliocene epoch about 14 to 12 million years ago.
  5. It walked erect on its hind limbs.
  6. It had close similarity with chimpanzee.
  7. Some scientists believe that Dryopithecus evolved into Ramapithecus.

Question 13.
Australopithecus.
Answer:

  1. Australopithecus is considered as connecting link between ape and man.
  2. Its fossils were obtained from Toung valley in South Africa, from Ethiopia and Tanzania.
  3. It was in late Pliocene or early Pleistocene epoch about 4 to 1.8 million years ago.
  4. It was about 4 feet tall. It had prognathus face, with larger jaws. Chin was absent. Lumbar curvature was present.
  5. It walked upright.
  6. The cranial capacity was about 450 to 600 CC. Therefore, it was called man with ape brain.

Question 14.
Homo habilis.
Answer:

  1. Homo habilis is described as Handy man. His fossils were obtained from Olduvai Gorge in Tanzania, Africa.
  2. He existed in late Pliocene or early Pleistocene about 2.5 to 1.4 million years ago.
  3. He was lightly built.
  4. Fossil of lower jaw was obtained which showed that his dentition was more like modern man with small molars.
  5. He walked erect. His cranial capacity was 640 to 800 cc.
  6. He did not eat meat and made stone tools.

Question 15.
Homo erectus.
Answer:

  1. Homo erectus was also known as Java man or Peking Man due to his fossils obtained from these areas.
  2. He was also called ape man.
  3. He lived in the middle Pleistocene epoch about 1.5 million years ago.
  4. He was 5 feet in height with prognathous face, massive jaws, huge teeth and bony eye brow ridges.
    Chin was absent.
  5. He walked erect.
  6. The cranial capacity was 900 cc.
  7. He was omnivorous and probably used fire and ate meat.

Question 16.
Neanderthal man.
Answer:

  1. The scientific name of Neanderthal man is Homo neanderthalensis. He is described as advanced prehistoric man.
  2. It was called Neanderthal man because its first fossil was collected from Neanderthal valley in Germany by Fuhlrott (1856).
  3. It was heavily built and short and had outwardly curved thigh bones.
  4. The facial features were as follows : prominent brow ridges, thick skull bones, low and slanting forehead, deep jaw without a chin, etc.
  5. Neanderthal man existed in late Pleistocene epoch about 1,00,000 to 40,000 years ago. It was widely spread in Europe, Asia and North America. It became extinct about 25,000 years ago.
  6. The cranial capacity of Neanderthal man was about 1400 cc, which was roughly equal to that of modern man. He used hide for dressing.
  7. It showed intellectual development in constructing and using flint tools and fire.
  8. The Neanderthal men used to bury their dead bodies along with their tools and perform ceremonies.

Short answer questions

Question 1.
Enlist the steps in the process of chemical evolution.
Answer:

  1. Origin of Earth and Primitive atmosphere.
  2. Formation of ammonia, water and methane.
  3. Formation of simple organic molecules.
  4. Formation of complex organic molecules.
  5. Formation of Nucleic acids.
  6. Formation of Protobionts or Procells.
  7. Formation of first cell.

Question 2.
When did Earth originate? Which transformations took place later?
Answer:

  1. Earth originated about 4.6 billion years ago as a part of the solar system.
  2. When it was formed, it was a rotating cloud of hot gases and cosmic dust. It was then appearing like a nebula.
  3. Later the condensation and cooling started which resulted in stratification.
  4. Heavier elements like nickel and iron settled to the core. Lighter elements like helium, hydrogen, nitrogen, oxygen, carbon, etc. remained on the surface and they formed the primitive atmosphere.
  5. This atmosphere of the earth was of a reducing type, devoid of free oxygen and very hot.

Question 3.
How were simple organic molecules formed on the earth?
Answer:
1. Initially earth’s temperature was very high but as the cooling process started, lighter elements reacted chemically with each other.

2. The early atmosphere was rich in hydrogen, carbon, nitrogen and sulphur. Hydrogen was most active and hence it reacted with other elements to form chemicals on earth like CH4, NH3, H20 and H2S.

3. With decreasing temperature of the earth, steam condensed into water that resulted in heavy rainfall. This constantiy falling rainwater got accumulated on the land to form different water bodies and especially oceans. It also cooled down the earth.

4. The early molecules of hydrocarbons, ammonia, methane and water underwent reactions like condensation, polymerisation, oxidation and reduction due to different energy sources such as ultra-violet rays, radiations, lightning and volcanic activities.

5. These reactions resulted in formation of simple organic molecules like monosaccharides, amino acids, purines, pyrimidines, fatty acids, glycerol, etc.

Question 4.
How were complex organic molecules formed during chemical evolution?
Answer:

  1. The primitive broth in which simple organic molecules were suspended, was neutral and free from oxygen.
  2. In this broth polymerisation took place and simple organic molecules aggregated to form new complex organic molecules like polysaccharides, fats, proteins, nucleosides and nucleotides.
  3. Protoproteins were formed by polymerisation of amino acids. These protoproteins later formed proteins.
  4. Formation of protein molecules is considered as landmark in the origin of life. Later the enzymes were formed which accelerated the rate of other chemical reactions.

Question 5.
How were protobionts formed with the help of nucleic acids during chemical evolution?
Answer:

  1. By the reaction between phosphoric acid, sugar and nitrogenous bases (purines and pyrimidines), nucleotides may have been formed.
  2. These nucleotides joined together to form nucleic acids such as RNA and DNA.
  3. Nucleic acids acquired self-replicating ability which is a fundamental property of living form.
  4. They later formed protobionts. They were the first form of life formed by nucleic acids along with inorganic and organic molecules.
  5. Protobionts were the prebiotic chemical aggregates having some properties of living system. Aggregation of organic molecules due to coacervation formed these protobionts.

Question 6.
Why variations are seen in population?
Answer:
Variations are seen in population due to gene flow, genetic drift, genetic recombinations that occur at the time of gamete formation, crossing over and sudden drastic changes like gene mutations or chromosomal aberrations. All the above factors are constantly operating over every population. Due to these evolutionary processes, variation take place in a population.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 7.
In which conditions the gene frequency of a population will remain constant?
Answer:
In the condition of no migrations of the organisms, no mutations, no sexually reproduction consisting of crossing over, no genetic drift, no recombinations and variation, the gene frequency of a population will remain constant. Such hypothetical conditions will never exist because even one set of sexually reproducing organisms forms an offspring which is slightly different from its parents. This means that there is constant change of gene frequency.

Question 8.
What is carbon dating and how does it work?
Answer:
Carbon dating is the method to find out the age of the fossil or any other organic matter. In carbon dating, the relative proportions of the carbon isotopes, carbon-12 and carbon-14 which are present in the organic matter, is estimated. The ratio between them changes as radioactive carbon-14 decays and is not replaced by exchange with the atmosphere. From these findings the age of that organic matter can be concluded.

Question 9.
What is a connecting link? Give suitable examples of connecting links.
Answer:

  1. A connecting link is an intermediate or transitional state between two systematic groups of organisms.
  2. It bears characters common to both these groups on either side of its position. Thus it represents an evolutionary line.
  3. Connecting links are also called a missing link.
    E.g. Archaeopteryx, the extinct bird is a connecting link between Reptiles and Aves.
  4. Seymouria is a connecting link between Amphibia and Reptilia.
  5. Ichthyostega is a connecting link between Pisces and Amphibia.

Question 10.
What is geological time scale? How is it divided ?
Answer:

  1. Geological time scale is the arrangement of major divisions of geological time into eras, periods and epochs on the time scale.
  2. This division is based on the study of fossilized organisms obtained from the different strata of the earth.
  3. The characteristic significant events that occurred in the organization of organisms helped the geologists to understand the geological time scale.
  4. The major divisions of geological time are called eras.
  5. The eras are divided into periods and the periods into epochs.
  6. By studying fossils in the earth crust, the evolutionary changes in the organisms have been traced out.

Question 11.
What is meant by palaeontological evidences ?
Answer:

  1. Palaeontology means the study of fossils. Palaeontological evidences are the fossilized forms of various organisms which are obtained from different strata of the earth. They represent the dead remains of plants and animals that lived in the past in various geological layers.
  2. The older and more primitive forms of life are excavated from the lower strata of the soil whereas the recent ones are situated on the upper layers of the soil.
  3. Fossils are formed in variety of materials such as sedimentary rocks, amber, volcanic gas, ice, peat bogs, soil, etc.
  4. They provide the true, direct and reliable evidences of evolution.

Question 12.
What are the molecular evidences that show the evolution?
Answer:

  1. Different organisms have basic similarities in their molecules and the cellular constituents.
  2. All living organisms have the same basic structural and functional unit, i.e. cell.
  3. Cell organelles such as endoplasmic reticulum, Golgi bodies, mitochondria, etc. are present in different types of organisms.
  4. Proteins and gene performing different functions have the same basic pattern which shows a common ancestry.
  5. Catabolic activities of liberating energy, synthesis of macromolecules such as proteins, carbohydrates, nucleic acids, etc. are similar in different organisms.
  6. ATP is the common energy currency of all the organisms.
  7. All the above facts are called molecular evidences in favour of evolution.

Question 13.
Arrange the following stages of the human evolution in the order of their increasing cranial capacity, (a) Neanderthal man (b) Cro-Magnon man (c) Homo erectus (d) Homo habilis.
Answer:

  1. Homo habilis (650-800 cc)
  2. Homo erectus (850-1200 cc)
  3. Neanderthal man (1400 cc)
  4. Cro-Magnon man (1450 cc)

Question 14.
Since your earlier school days you have been solving mysteries/puzzles labelled as use your brain power. Did you ever wonder why human brain has such a capacity? Why and how we evolved along these lines? What is the extent of similarity between humans, chimpanzees and monkeys?
Answer:
Human beings have extremely well- developed brain. Especially the cerebral hemispheres are very large constituting 85% of the brain weight. Due to such cerebrum, there are many, neurons. They are responsible for faculties such as speech, memory, emotions, thought process. The mind or psyche is well developed due to over developed cerebral hemispheres.

That’s why human brain has tremendous capacity of Chimpanzees are also comparatively more intelligent than the monkeys. However, the development of speech and language is lacking in them. The cranial capacity of chimpanzee and monkey is 275-500 cc and 45-50 cc respectively, whereas humans have 1450-1500 cc. Larger the brain, more is the intelligence and all other mental faculties which only humans show.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 15.
Even though the cranium of elephant is larger than that of man, humans are considered more intelligent than elephant. Why is it so?
Answer:
Elephant’s brain is large weighing about 5 kg. But elephant’s body weight too is very high. The proportion of body brain weight is highest in human beings. Therefore, humans are considered more intelligent than elephant. Moreover, the cerebral cortex of elephant is not very well developed, instead they have well developed cerebellum which helps in locomotion and balancing their huge bodies. Human brain has very well-developed cerebrum which brings about cognitive behaviour and intelligence.

Chart based/Table based questions

Question 1.
Give the graphical representation of Hardy-Weinberg’s principle in the form of Punnet square.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 1
Genotypes = AA + 2 Aa + aa
Gene frequencies = p² + 2pq + q²

Question 2.
Make a chart showing the types of isolating mechanisms.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 2

Diagram based questions

Question 1.
Give diagrammatic representation to show RNA world.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 3

Question 2.
Sketch and label four types of chromosomal aberrations
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 4

Question 3.
Sketch the graphs to show directional and stabilizing selection.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 5

Long Answer Questions

Question 1.
Write about four old theories which suggested about how did life originate on the earth.
Answer:
1. Theory of special creation : Theory of special creation is the oldest theory which is based on religious beliefs. According to this theory, all the living organisms were created by supernatural power. However, since there are no scientific proofs to this theory, it is not accepted.

2. Cosmozoic theory/Theory of Panspermia : This theory says that life did not originate on the earth but it was exported from the other planets in the form of biological spores or microorganisms which were named as cosmozoa or panspermia. They may have descended to the earth from other planets. Recently, NASA has reported fossils of bacteria-like organisms on a piece of Martian rock recovered from Antarctica. Such facts may throw some light on the cosmozoic theory.

3. Theory of spontaneous generation or Abiogenesis : There was a belief that life originated from non-living material spontaneously. This theory was later disproved by Louis Pasteur.

4. Theory of biogenesis : This theory says that living organisms can originate only from pre-existing living beings. It is same as reproduction. But this theory of biogenesis was unable to explain origin of life on earth. It explains only the continuity of life.

Question 2.
Haldane described ‘Hot dilute soup’ in his theory. Describe how this soup led to formation of some important molecules.
Answer:
(1) The primitive sea containing molecules of organic substances without free oxygen was described as ‘hot dilute soup or primitive broth’ by Haldane. He proposed the theory of chemical evolution.

(2) According to this theory, the chemical evolution took place in the following steps : (a) Origin of earth and its primitive atmosphere, (b) Formation of ammonia, water and methane. These molecules dissolved in rainwater and formed the seas, (c) Then synthesis of simple organic compounds took place, followed by formation of complex organic compounds such as nucleic acids.

(3) The early molecules underwent chemical reactions such as condensation, polymerization, oxidation and reduction.

(4) The biologically important molecules such as monosaccharides, amino acids, purine, pyrimidine, fatty acids and glycerol were formed due to these reactions, utilizing the sources of energy on the primitive earth.

(5) Since oxygen was lacking, there was no degradation. Enzymes were also absent and hence there was formation of complex molecules in the hot dilute soup.

(6) This further led to the formation of pre-cells or protobiont. These aggregates were called coacervates by Oparin or microspheres by Sidney Fox. This further gave rise to first cells on the earth.

Question 3.
Explain the process of formation of eobionts.
Answer:

  1. Protobionts or coacervates were colloidal aggregations of hydrophobic proteins and lipids (lipoid bubbles).
  2. They grew in size by taking up material from surrounding aqueous medium.
  3. During their growth they became thermodynamically unstable and split into smaller units. These were called microspheres.
  4. They were proteinoids formed from colloidal hydrophilic complexes surrounded by water molecules.
  5. These bodies were like primitive cells having outer double-membrane. Across this membrane diffusion and osmosis may have occurred. They were more stable than coacervates.
  6. Coacervates and microspheres were non-living colloidal aggregations of lipids and proteinoids respectively.
  7. But they showed growth and division like living cells.
  8. These colloidal aggregations turned into first primitive living system called eobionts or protocell.

Question 4.
Describe RNA World hypothesis.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 6
(1) RNA world hypothesis is based on discovery of catalytic RNA or ribozymes. It was proposed by Carl Woese, Francis Crick and Leslie Orgel in 1960 whereas Ribozymes were discovered by Sidney Altman and Thomas Cech in 1980.

(2) According to this hypothesis, early life must have been based most probably on RNA.

(3) Factors supporting this hypothesis are:

  • RNA is found abundantly in all living cells.
  • It is structurally related to DNA.
  • Chains of RNA can evolve or undergo mutations, replicate and catalyse reactions.
  • Biomolecules like Acetyl-Co-A have a nucleotide in their molecular structure.
  • Ribosome acts as a protein assembly unit in the cell and is seen in many types of cells.
  • In ribosomes, translation process is catalysed by RNA.

(4) The primitive molecules underwent repeated replication and mutation forming varieties of RNA molecules with varying sizes and catalytic properties.

(5) They later developed their own protein coats and machinery to survive the assembly of primitive cell.

(6) From them DNA was developed which was double stranded stable structure.

(7) It further kept on evolving giving rise to rich biodiversity on earth.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 5.
Explain in brief Darwinism and its five main postulates.
Answer:
Darwinism means theories of natural selection and speciation as put forth by Charles Darwin. The five main postulates of his theories are as follows: Overproduction or prodigality, Struggle for existence, Organic variations, Natural selection, Origin of new species (speciation).
1. Overproduction (Prodigality of nature) : There is a natural tendency to produce more number of progeny in geometric ratio for continuing the species. E.g. Salmon fish produces about 28 lakh eggs in a single season. Single pair of elephants would produce 19,000,000 elephants. But the size of given species in a given area remains relatively constant because of fluctuations that occur seasonally.

2. Struggle for existence Due to over¬production there is struggle for existence between the members of population for limited supply of food or to overcome adverse environmental conditions or for a space or to escape from enemies, etc.

3. Organic variations : There are differences in morphology, physiology, nutrition, habit, behavioural patterns, etc., among the members of same species or members of different species. These variations act as raw material for evolution.

4. Natural selection : Some organisms possess better variations to get adapted and survive under existing environmental conditions, while some do not have. Better adapted organisms are selected by the nature while those with unfavourable variations perish. The principle by which useful variations are preserved by nature, is called ‘Natural Selection’. It is also called ‘survival of fittest’ by H. Spencer.

5. Origin of new species (speciation) : Favourable variations are transmitted from generation to generation, resulting into better adapted generations. Gradually these adaptations with few new modifications become fixed in the life cycle, forming a new species.

Question 6.
Explain modern Synthetic Theory of Evolution in brief.
Answer:
(1) Modern synthetic theory of evolution is the result of modification of Darwinism and theory of mutations by taking into consideration studies of genetics, ecology, anatomy, geography and palaeontology.

(2) Five key factors of modern synthetic theory are gene mutations, mutations in the chromosome structure and number, genetic recombinations, natural selection and reproductive isolation. All these finally contribute in the evolution of new species or process of speciation.

(3) Population or Mendelian population is the small group of ‘interbreeding populations’. For every Mendelian population there is a gene pool which is constituted by total number of genotypes in it. The genotype of an organism in a population is constant, but the gene pool constantly undergoes change due to different factors such as mutations, recombination, gene flow, genetic drift, etc.

(4) Every gene has two alleles. The proportion of a particular allele in the gene pool, to the total number of alleles at a given locus, is called gene frequency. Thus any change in the gene frequency in the gene pool affects population.

(5) The five main factors are broadly divided into three main concepts as follows:
(i) Genetic variations caused due to various aspects of mutation, recombination and migration. Such variations cause change in the gene frequency. Gene mutations or point mutation change the phenotype of the organism, leading to variation. Recombination is caused due to crossing over in which new genetic combinations are produced. Sexual reproduction due to fertilization of gametes also cause recombinations. All these lead to variations, Gene flow is movement of genes into or out of the population, either due to migrations or dispersal of gametes.

Gene flow therefore change the gene frequencies of the population. Genetic drift is a random change which occurs by pure chance. It occurs in small populations but change the gene frequency. Chromosomal aberrations are structural or morphological changes in the chromosomes causing rearrangement of the sequence of genes.

(ii) Natural selection is said to be the main driving force in evolution. It brings about evolutionary changes by selecting favourable gene combinations by differential reproduction of genes. This brings about changes in gene frequency from one generation to next generation.

(iii) Isolation means the separation of the population of a particular species into smaller units which prevents interbreeding between them. This over a long time period leads to speciation or formation of new species.

Question 7.
What are different types of chromosomal aberrations?
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 7
Chromosomal aberrations:
(1) The structural, morphological change, which take place in chromosome due to rearrangement, is called chromosomal aberrations.

(2) The aberrations change the sequence of the genes. This causes variations. Chromosomal aberrations are mainly of following four types:

  1. Deletion : Loss of genes from chromosome.
  2. Duplication : Genes are repeated or doubled in number on chromosome.
  3. Inversion : A particular segment of chromosome is broken and gets reattached to the same chromosome in an inverted position due to 180° twist. There is no loss or gain of gene complement of the chromosome.
  4. Translocation : Transfer or transposition of a part of chromosome or a set of genes to a non-homologous chromosome is called translocation. It is effected naturally by the transposons present in the cell.

Question 8.
What are the different pre-zygotic isolating mechanisms?
Answer:
(1) Pre-zygotic or pre-mating isolating mechanisms do not allow individuals to mate with each other at all.

(2) By various mechanisms the two groups remain isolated. These mechanisms are of following types:
(i) Habitat isolation : Habitat isolation is the phenomenon in which members of a population living in the same region occupy different habitats. Hence the potential mates do not interbreed among themselves.

(ii) Seasonal isolation : In seasonal isolation, members of a population share the same region but attaining sexual maturity at the different times of the year. They thus remain isolated reproductively preventing interbreeding among themselves.

(iii) Ethological isolation : Ethological isolation is seen when members of two populations have different mating behaviours. This prevents interbreeding.

(iv) Mechanical isolation : Mechanical isolation is seen when the members of two populations have differences in the structure of reproductive organs. Due to such differences interbreeding is not possible.

Question 9.
What are the different post-zygotic isolating mechanisms?
Answer:

  1. In post-zygotic or post-mating isolating mechanisms, the two individuals can mate but the result of mating is not favourable.
  2. Thus the populations remain isolated without the actual genetic exchange.

Post-mating isolating mechanisms are divided into the following categories:

  1. Gamete mortality : In gamete mortality, there is death of gametes. Sperm transfer may take place but the egg is not fertilized due to gamete mortality.
  2. Zygote mortality : In zygote mortality, the zygote is formed but it fails to thrive. Though the egg is fertilized the zygote does not survive.
  3. Hybrid sterility : In this isolation, there is the formation of hybrid as the gametes or zygotes do not die but the hybrid formed is sterile. Sterile hybrid cannot contribute genetically to further generations.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

Question 10.
What is Hardy-Weinberg equilibrium? Explain it in brief.
Answer:

  1. Hardy and Weinberg were two scientists who proposed a concept of genetic equilibrium popularly known as Hardy- Weinberg principle or equilibrium.
  2. This principle states that gene, allele or genotype frequencies remain the same from generation to generation unless disturbed by factors like mutation, non-random mating, genetic drift, etc.
  3. For explaining the concept of equilibrium they assumed that there are two alleles located at a single locus (A and a).
  4. Their respective frequencies are p and q.
  5. The frequency of genotype AA is p, for 2Aa is 2pq and for aa is q.
  6. The equilibrium equation is p² + 2pq + q² = 1
  7. It says that if sum total of gene frequencies is 1, then sum total of genotype frequencies is also equal to 1. When the equilibrium is disturbed then only evolution occurs.

Question 11.
Human being is said to be most evolved, intelligent living being. Yet we are not self-sufficient. Think of various aspects for which we depend on other living beings for our survival.
Answer:
Human brain is evolved and super- specialised but yet in many aspects human beings are much dependent on other natural factors. Human body is not with any protective exoskeleton, or organs of offence and defence.

Unless well dressed, he cannot cope up with severe cold temperatures as he lacks natural protective fur. He cannot run fast as the other animals can. Neither he can digest uncooked food. He has overcome all his shortcomings by using his brain power.

He has managed to take fur and feathers from other animals by killing them. He also uses other natural fibres from plants to cover his body. He has also finished fish from the oceans by over-exploitation and polluting the natural habitats of these creatures.

He takes meat from other animals by killing them and for the purpose he domesticates them to satisfy his hunger. Before technical age, animals were used as beasts of burden and as means of transport. Thus human history has shown excessive use of horses, camels, elephants, etc.

Man snatches milk from other animals like cows and buffaloes which is for their young ones. But the entire diary industry and human needs for dairy products have been taken care of by these herbivores.

Apart from all such uses, man also uses other animals for experimentations and pharmaceutical industries. In this way, man has mastered other animal kingdom due to his intelligence.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 4 Molecular Basis of Inheritance Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Multiple Choice Questions

Question 1.
How many of the following characteristics are shown by the R-strain of Streptococcus pneumonia? Avirulent, Smooth, Pathogenic, Capsulated ………………..
(a) One
(b) TWo
(c) Three
(d) Four
Answer:
(a) One

Question 2.
Griffith obtained …………….. from the blood of the dead mice.
(a) dead S-strain bacteria
(b) live R-strain bacteria
(c) dead R-strain bacteria
(d) live S-strain bacteria
Answer:
(d) live S-strain bacteria

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 3.
Oswald T. Avery, Colin M. MacLeod and Maclyn McCarty demonstrated that ………………..
(a) transformation of live S-strain bacteria into R-strain type was because of DNA of bacteria of S-strain.
(b) the transforming substance was either a protein or RNA.
(c) only DNA was able to transform harmless R-strain into virulent S-strain.
(d) when DNA isolated from S-strain bacteria, was digested with DNase, the transformation occurred.
Answer:
(c) only DNA was able to transform harmless R-strain into virulent S-strain

Question 4.
Which of the following was NOT observed in Hershey and Chase experiment?
(a) Viruses grown in the presence of radioactive sulphur, had radioactive protein but not radioactive DNA.
(b) Radioactive ‘P’ remained in suspension.
(c) Only radioactive ‘P’ was found inside the bacterial cells in the pellet.
(d) Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive proteins.
Answer:
(b) Radioactive ‘P’ remained in suspension.

Question 5.
Enzymes like ……………….. and DNA topoisomerase-I, play important role in maintaining super-coiled state in prokaryotic DNA.
(a) DNA ligase
(b) DNA gyrase
(c) RNA polymerase
(d) None of these
Answer:
(b) DNA gyrase

Question 6.
Histone octamer of nucleosome has two molecules, each of ……………….. proteins.
(a) H2A, H2B, H3 and H4
(b) H2A, H2B, H3 and H1
(c) H2A, H2B. H3A and H3B
(d) H1A, H2B, H3A and H4
Answer:
(a) H2A, H2B, H3 and H4

Question 7.
Select the CORRECT statement.
(a) Euchromatin is mainly located near centromere and telomeres.
(b) Heterochromatin replicates at faster rate than euchromatin.
(c) Heterochromatin has 2 to 3 times more DNA than in the euchromatin.
(d) Heterochromatin is lightly stained region of chromonema.
Answer:
(c) Heterochromatin has 2 to 3 times more DNA than in the euchromatin

Question 8.
A DNA molecule in which both strands have 14N is allowed to replicate in an environment containing 15N. What will be the exact number of DNA molecules that contain the 14N after three replications?
(a) One
(b) Two
(c) Four
(d) Eight
Answer:
(b) Two

Question 9.
As the base sequence present on one strand of DNA decides the base sequence of other 8 strand, this strand is considered as ………………..
(a) descending strand
(b) leading strand
(c) lagging strand
(d) complementary strand
Answer:
(d) complementary strand

Question 10.
In prokaryotes ……………….. recognizes the promoter sequence.
(a) alpha factor
(b) rho factor
(c) theta factor
(d) sigma factor
Answer:
(d) sigma factor

Question 11.
If the base sequence in DNA is 5′ AAAA 3′, then the base sequence in m-RNA is ………………..
(a) 5′ UUUU 3′
(b) 3′ UUUU 5′
(c) 5′ AAAA 3′
(d) 3′ TTTT 5′
Answer:
(c) 5′ AAAA 3′

Question 12.
During capping, methylated guanosine tri¬phosphate is added to 5′ end of ………………..
(a) m-RNA
(b) t-RNA
(c) hnRNA
(d) r-RNA
Answer:
(c) hnRNA

Question 13.
If each codon has two nucleotides, then there will be ……………….. codons, which can encode for only …………….. different types of amino acids.
(a) 16, 16
(b) 16, 20
(c) 20, 16
(d) 64, 64
Answer:
(a) 16, 16

Question 14.
What would happen if in a gene encoding a polypeptide of 50 amino acids, 25th codon (UAU) is mutated to UAA?
(a) A polypeptide of 24 amino acids is formed.
(b) Two polypeptides of 24 and 25 amino acids will be formed.
(c) A polypeptide of 49 amino acids is formed.
(d) A polypeptide of 25 amino acids is formed.
Answer:
(a) A polypeptide of 24 amino acids is formed.

Question 15.
A strand of DNA has following base sequence – 3′ AAAAGTGAATAGTGA 5′. On transcription it produces an m-RNA. Which of the following anticodon of t-RNA recognizes the third codon of this m-RNA?
(a) AAA
(b) CUG
(c) AAG
(d) CTG
Answer:
(c) AAG

Question 16.
Polynucleotide chain consisting of only CUA repeats will give polypeptide chain with only one amino acid ………………..
(a) tryptophan
(b) leucine
(c) serine
(d) methionine
Answer:
(b) leucine

Question 17.
Select the INCORRECT statement.
(a) Dr. Khorana prepared polyribo-nucleotides chains with known repeated sequences of two or three nucleotides by using synthetic DNA.
(b) M. Nirenberg and Matthaei synthesized artificial m-RNA which was a homopolymer of uracil ribonucleotides.
(c) Enzyme polynucleotide phosphorylase polymerizes RNA with defined sequences in a template-dependent manner.
(d) Evidence for triplet nature of geneticcode, was given by Crick (1961) using “frame-shift mutation”.
Answer:
(c) Enzyme polynucleotide phosphorylase polymerizes RNA with defined sequences in a template-dependent manner.

Question 18.
…………… is/are based on complementarity principle.
(a) Replication and translation
(b) Replication and transcription
(c) Translation
(d) Only replication
Answer:
(b) Replication and transcription

Question 19.
Cysteine has codons, while isoleucin has ……………….. codons.
(a) two, three
(b) three, two
(c) two, four
(d) four, two
Answer:
(a) two, three

Question 20.
Out of 64 codons, only 61 code for the 20 different amino acids. This is known as ……………….. of genetic code.
(a) non-ambiguity
(b) overlapping nature
(c) ambiguity
(d) degeneracy
Answer:
(d) degeneracy

Question 21.
Mutation that results in Sickle-cell anaemia is a ………………..
(a) deletion
(b) frame-shiftmutation
(c) point mutation
(d) insertion
Answer:
(c) point mutation

Question 22.
Initiator charged t-RNA occupies the ……………….. of ribosome first.
(a) A-site
(b) P-site
(c) E-site
(d) either A-site or P-site
Answer:
(b) P-site

Question 23.
It takes ……………….. for formation of peptide bond.
(a) 10 seconds
(b) 0.1 second
(c) less than 0.1 second
(d) 60 seconds
Answer:
(c) less than 0.1 second

Question 24.
Anticodon and codon bind by ………………..
(a) glycosidic bond
(b) hydrogen bond
(c) phosphodiester bond
(d) none of these
Answer:
(b) hydrogen bond

Question 25.
The UTRs are present at ………………..
(a) 5′-end, before start codon and at 3′-end, after stop codon of m-RNA
(b) 5′-end, before start codon and at 3′-end, after stop codon of t-RNA
(c) only at 3′-end, after stop codon of m -RNA
(d) only at 5′-end, before start codon of m-RNA
Answer:
(a) 5′-end, before start codon and at 3′-end, after stop codon of m-RNA

Question 26.
The action of structural genes is regulated by …………….. site with the help of a …………….. protein.
(a) operator, inducer
(b) operator, repressor
(c) regulator, repressor
(d) regulator, inducer
Answer:
(b) operator, repressor

Question 27.
Repressor protein is produced by the action of ………………..
(a) gene z
(b) gene y
(c) gene i
(d) gene o
Answer:
(c) gene i

Question 28.
Select the correct pair.
(a) Gene z – Transacetylase
(b) Gene y – Beta-galactocidase
(c) Gene a – Beta-galactoside permease
(d) Gene I – Repressor
Answer:
(d) gene I – repressor

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 29.
Structural genomics involves ……………….. of genome.
(a) mapping
(b) sequencing
(c) analysis
(d) all of these
Answer:
(d) all of these

Question 30.
The technique of transferring DNA fragments separated on agarose gel to a synthetic nitrocellulose membrane is known as ………………..
(a) Southern blotting
(b) Autoradiography
(c) Southern hybridization
(d) None of these
Answer:
(a) Southern blotting

Question 31.
Sequence of various steps in DNA fingerprinting is ………………..
i. Southern blotting.
ii. Restriction digestion
iii. Agarose gel electrophoresis
iv. DNA isolation
v. Photography.
vi. Selection of DNA probe
vii. Hybridization
(a) iv, iii, ii, i, v, vi, vii
(b) iv. v, iii, i, vi, vii, ii
(c) iv, ii, iii, i, vi, vii, v
(d) ii, iii, iv, i, vi, vii,
Answer:
(c) iv, ii, iii, i, vi. vii, v

Match the Columns

Question 1.

Column A Column B
(1) Frederick Griffith (a) Test tube assay
(2) Avery, McCarty and MacLeod (b) Streptococcus pneumoniae
(3) Alfred Hershey and Martha Chase (c) E. coli
(4) Meselson and Stahl (d) Bacteriophages

Answer:

Column A Column B
(1) Frederick Griffith (b) Streptococcus pneumoniae
(2) Avery, McCarty and MacLeod (a) Test tube assay
(3) Alfred Hershey and Martha Chase (d) Bacteriophages
(4) Meselson and Stahl (c) E. coli

Classify the following to form Column B as per the category given in Column A

Question 1.
(i) UUU
(ii) CUA
(iii) UAA
(iv) AUG
(v) UAG
(vi) UGA

Column A Column B
1. Initiator codon ————–
2. Stop codons ————-
3. Codon that codes for Phenyl alanin ————–
4. Codon that codes for leucine ————-

Answer:

Column A Column B
1. Initiator codon (iv) AUG
2. Stop codons (iii) UAA, (v) UAG, (vi) UGA
3. Codon that codes for Phenyl alanin (i) UUU
4. Codon that codes for leucine (ii) CUA

Question 2.
(i) Severo Ochoa
(ii) F. Jacob and J. Monod
(iii) Temin and Baltimore
(iv) H. Winkler
(v) T. H. Roderick
(vi) R Kornberg

Column A Column B
(1) Lac Operon ————–
(2) Central dogma in retroviruses ————-
(3) Coined the term Genome ————–
(4) Coined the term Genomics ————-
(5) Enzymatic synthesis of RNA ————-
(6) DNA is associated with histones and non-histones ————–

Answer:

Column A Column B
(1) Lac Operon (ii) E Jacob and J. Monod
(2) Central dogma in retroviruses (iii) Temin and Baltimore
(3) Coined the term Genome (iv) H. Winkler
(4) Coined the term Genomics (v) T. H. Roderick
(5) Enzymatic synthesis of RNA (i) Severo Ochoa
(6) DNA is associated with histones and non-histones (vi) R. Kornberg

Very Short Answer Questions

Question 1.
What are the two types of bacteria used by F. Griffith and which one out of these is avirulent?
Answer:
S-type and R-type strains of Streptococcus penumoniae were used by F. Griffith and out of these R-type is avirulent.

Question 2.
Enlist the characteristics of S-strain pneumoniae.
Answer:
S-strain pneumoniae are virulent, smooth and encapsulated.

Question 3.
What is the bacteriophage?
Answer:
Bacteriophage is a virus that infects bacterium and injects its genetic material in the bacterium.

Question 4.
What is the length of DNA double helix molecule in a typical mammalian cell?
Answer:
The length of DNA double helix molecule in a typical mammalian cell is approximately 2.2 meters.

Question 5.
What is the approximate size of a typical nucleus ?
Answer:
Approximate size of a typical nucleus is 10-6 m.

Question 6.
What is size of E. coli cell?
Answer:
The size of E. coli cell size is 2-3 µm.

Question 7.
What determines the charge on protein molecules?
Answer:
A protein acquires its charge depending upon the abundance of amino acid residues with charged side chains.

Question 8.
What is nucleosome core?
Answer:
Nucleosome core is a histone octamer.

Question 9.
Where is H1 histone present?
Answer:
H1 histone binds the DNA thread where it enters and leaves the nucleosome.

Question 10.
How are solenoid fibres formed?
Answer:
Six nucleosomes get coiled and then form solenoid that looks like coiled telephone wire of 30 nm diameter (300Å).

Question 11.
How is chromatin fibre formed?
Answer:
Supercoiling of solenoid fibre forms a looped structure called chromatin fibre.

Question 12.
What is NHC?
Answer:
NHC stands for Nonhistone Chromosomal proteins.

Question 13.
List as many different enzyme activities required during DNA synthesis as you can.
Answer:
Phosphorylase, Helicase, DNA polymerase, Primase, DNA ligase, Super helix relaxing enzyme, Topoisomerase (gyrase) are different enzymes required during DNA synthesis.

Question 14.
How many replicons are present in prokaryotes and eukaryotes respectively?
Answer:
Prokaryotes have one replicon. Several replicons in tandem are present in eukaryotes.

Question 15.
What is the function of SSBP?
Answer:
During replication of DNA SSBP proteins remain attached to both the separated strands and prevent them from coiling back.

Question 16.
During which phases of cell cycle, transcription occurs in the nucleus?
Answer:
Transcription occurs in the nucleus during G1 and G2 phases of cell cycle.

Question 17.
Which strand of transcription unit gets transcribed ?
Answer:
DNA strand having 3’ → 5’ polarity acts as template strand and it gets transcribed.

Question 18.
What is a cryptogram?
Answer:
Cryptogram is a genetic code consisting of triplet codons on m-RNA that code for a specific amino acids.

Question 19.
What is meant by the polarity of genetic code?
Answer:
Genetic code is always read in 5′ → 3’ direction. This is called polarity of genetic code.

Question 20.
Which mutation can result in changes in the reading frame?
Answer:
Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion.

Question 21.
Which mutation can result in insertion or deletion of amino acids, but reading frame remains unaltered?
Answer:
Insertion or deletion of three or multiples of three bases results in insertion or deletion of amino acids and reading frame remains unaltered from that point onwards.

Question 22.
Which molecule serves as an intermediate molecules between DNA and protein during proteins synthesis?
Answer:
RNA serves as an intermediate molecule between DNA and protein during proteins synthesis.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 23.
What is the function of a groove present between two subunits of ribosome in eukaryotes ?
Answer:
The groove present between two subunits of ribosomes in eukaryotes protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.

Question 24.
Enlist different steps of protein synthesis.
Answer:
Steps in protein synthesis are:

  1. Transcription
  2. Activation of amino acids and formation of charged t-RNAs,
  3. Synthesis of polypeptide chain:
  4. initiation
  5. elongation and
  6. termination of polypeptide chain.

Question 25.
What is translocation?
Answer:
During elongation of polypeptide chain, the ribosome moves along the m-RNA in stepwise manner from start codon to stop codon (5′ → 3′), 1 codon ahead each time t, his movement is called translocation and due to this t-RNA carrying a dipeptide at A-site of the ribosome moves to the p-site.

Question 26.
What is meant by inducible enzymes?
Answer:
Bacteria like E.coli adapt to their chemical environment by synthesizing certain enzymes depending upon the substrate present. Such adaptive enzyme is called inducible enzyme.

Question 27.
What is meant by induction and inducer?
Answer:
A set of genes are switched on when a new substrate is to be metabolized. This phenomenon is called induction and small molecule responsible for this is known as inducer.

Question 28.
What is the role of a repressor gene?
Answer:
The role of a repressor gene is to produce repressor protein. Repressor binds with operator gene and this prevents transcription of structural genes in the operon.

Question 29.
Which molecule does act as inducer molecule in lac operon?
Answer:
Allolactose acts as inducer molecule in lac operon.

Question 30.
In which condition, lac operon is switched off?
Answer:
If E.coli bacteria do not have lactose in the surrounding medium as a source of energy, lac operon is switched off.

Question 31.
What lac operon consists of?
Answer:
Lac operon consists of a regulator promoter, operator and three structural genes z, y and a.

Question 32.
Which gene acts as a regulatory gene in lac operon?
Answer:
Repressor protein is produced by the action of gene i (inhibitor). This gene acts as a regulator gene.

Question 33.
When was Human Genome Project started ? When was it completed ?
Answer:
The Human Genome Project was started in 1990 and was completed in 2003.

Question 34.
What is functional genomics?
Answer:
Functional genomics is a branch of genomics that involves the study of functions of all gene sequences and their expressions in organisms.

Question 35.
What is the advantage of sequencing of genomes of non-human organisms?
Answer:
Sequencing of genomes of non-human model organisms allows researchers to study gene functions in these organisms. Since human beings possess many genes which are like those of flies, roundworms and mice, comparative studies will lead to greater understanding of human evolution.

Question 36.
What are VNTRs?
Answer:
Variable Number of Tandem Repeats (VNTRs) are unusual sequences of 20-100 base pairs, which are repeated several times and are arranged tandency.

Question 37.
Do different organisms have the same DNA?
Answer:
Different organisms differ in their DNA sequence.

Question 38.
What is the amino acid sequence encoded by base sequence UCA, UUU, UCC, GGG, AGU of an m-RNA segment?
Answer:
The amino acid sequence: Ser-Phe – Ser – Gly- Ser

Give definitions of the following

Question 1.
Replicon
Answer:
The unit of DNA in which replication occurs is known as replicon.

Question 2.
Transcription
Answer:
Transcription is defined as the process of copying of genetic information from template strand of DNA into a complementary single stranded RNA transcript.

Question 3.
Gene
Answer:
Gene is defined as the DNA sequence coding for m-RNA/ t-RNA or r-RNA.

Question 4.
Cistron
Answer:
Cistron is defined as a segment of DNA coding for a polypeptide.

Question 5.
Monocistronic gene
Answer:
Gene is called monocistronic, when there is a single structural gene in one transcription unit.

Question 6.
Polycistronic gene
Answer:
Gene is called polycistronic, when there is a set of various structural genes in one transcription unit.

Question 7.
Interrupted or Split genes
Answer:
Interrupted or split genes are the structural genes in eukaryotes which have both exons and introns.

Question 8.
Exons
Answer:
Exons are the coding sequences or express sequences in DNA/hnRNA/ m-RNA.

Question 9.
Introns
Answer:
Introns are the non-coding sequences in DNA or hnRNA.

Question 10.
Anticodon
Answer:
Anticodon is a triplet of nucleotides present on the anticodon loop of t-RNA, which is complementary to codon on m-RNA.

Question 11.
Mutation
Answer:
Mutation is a sudden heritable change in the DNA sequence that results in the change of genotype.

Question 12.
Translation
Answer:
Translation is the process in which sequence of codons of m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.

Question 13.
Genomics
Genomics is the study of genomes through analysis, sequencing and mapping of genes along with the study of their functions.

Question 14.
Repressors
Answer:
Repressors are proteins which are able to bind the operator region of operon and prevent the RNA polymerase from transcribing the operon.

Name the following

Question 1.
Enzyme that cleaves DNA.
Answer:
DNase

Question 2.
Enzyme that cleaves proteins.
Answer:
Protease

Question 3.
Enzyme involved in activation of nucleotides.
Answer:
Phosphorylase

Question 4.
Enzyme involved in unwinding of DNA.
Answer:
Helicase

Question 5.
Enzyme involved in synthesis of DNA.
Answer:
DNA polymerase

Question 6.
Enzyme involved in joining of Okazaki fragments.
Answer:
DNA ligase

Question 7.
Enzyme involved in synthesis of RNA primer.
Answer:
Primase

Question 8.
Enzyme involved in removal of RNA primer.
Answer:
DNA polymerase

Question 9.
Enzyme involved in replacement of gaps in prokaryotes.
Answer:
DNA polymerase – I

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 10.
Enzyme involved in replacement of gaps in eukaryotes.
Answer:
DNA polymerase α

Question 11.
Enzyme involved in formation of double helix in daughter DNA molecules.
Answer:
Topoisomerase

Question 12.
Enzyme involved in releasing strain created by unwinding of DNA.
Answer:
Super helix relaxing enzyme

Question 13.
Enzyme involved in synthesis of hnRNA, m-RNA.
Answer:
RNA polymerase – II

Question 14.
Enzyme involved in synthesis of t-RNA, snRNA.
Answer:
RNA polymerase – III

Question 15.
Enzyme involved in synthesis of r-RNA
Answer:
RNA polymerase – I

Question 16.
Enzyme involved in polymerizing RNA in template independent manner.
Answer:
Polynucleotide phosphorylase

Question 17.
Enzyme involved in peptide bond synthesis.
Answer:
Ribozyme

Question 18.
Enzyme involved in Cutting DNA at specific sites.
Answer:
Restriction endonuclease

Question 19.
Name the initiator codon of protein synthesis.
Answer:
AUG is the initiator codon of protein synthesis.

Question 20.
Name three binding sites of ribosome.
Answer:
Three binding sites for t-RNA on ribosomes are P-site (peptidy t-RNA-site), A-site (aminoacyl – t-RNA-site) and E-site (exit site).

Question 21.
Name the different structural genes in sequence of lac operon.
Answer:
There are 3 structural genes in the sequence lac-Z, lac-Y and lac-A.

Question 22.
Name the organisms whose genomes have been sequenced?
Answer:
The genomes of several organisms such as bacteria e.g. E.coli, Caenorhabditis elegans (a free living non-pathogenic nematode), Saccharomyces cerevisiae (yeast), Drosophila (fruit fly), plants (rice and Arabidopsis), Mus musculus (mouse), etc. have been sequenced.

Give Significance/Functions of the following

Question 1.
DNA.
Answer:

  1. DNA regulates and controls all the cellular activities.
  2. It replicates and gets distributed equally to the daughter cells when the cell divides.
  3. It is a carrier of genetic information.
  4. Heterocatalytic function : DNA directs the synthesis of chemical molecules other than itself. E.g. Synthesis of RNA (transcription), synthesis of protein (Translation), etc.
  5. Autocatalytic function : DNA directs the synthesis of DNA itself. E.g. Replication.
  6. DNA is a master molecule of a cell that initiates, guides, regulates and controls the process of protein synthesis.

Question 2.
Proteins.
Answer:
Proteins serve as structural components, enzymes and hormones.

Distinguish between the following.

Question 1.
Euchromatin and Heterochromatin.
Answer:

Euchromatin Heterochromatin.
1. Euchromatin is loosely packed region of the chromatain. 1. Heterochromatin is densely packed region of the chromatin.
2. Euchromatin stains lightly. 2. Heterochromatin stains darkly.
3. Euchromatin is transcriptionally active region of the chromatin. 3. Heterochromatin is transcriptionally inactive region of the chromatin.

Question 2.
DNA in prokaryotic and eukaryotic cells.
Answer:

DNA in prokaryotes DNA in eukaryotic
1. It is present in the cytoplasm. 1. It is present in the nucleus.
2. It is not associated with histones. 2. It is associated with histones.
3. It is circular. 3. It is linear.
4. Genes do not contain introns. 4. Genes contain introns along with exons.
5. Genes are polycostronic. 5. Genes are monocistronic.

Question 3.
DNA and RNA
Answer:

DNA RNA
1. DNA is deoxyribonucleic acid. 1. RNA is ribonucleic acid.
2. DNA is double stranded, helical molecule. 2. RNA is single stranded molecule.
3. In DNA, there is deoxyribose sugar. 3. In RNA, there is ribose sugar.
4. The pyrimidine nitrogen bases are cytosine and thymine. 4. The pyrimidine nitrogen bases are cytosine and uracil.
5. DNA is the genetic material in all types of organisms. 5. RNA is genetic material in few viruses only.
6. In eukaryotic cells, DNA is present in nucleus. 6. In eukaryotic cells, RNA is present in nucleus as well as cytoplasm.
7. The number of purine : pyrimidine ratio is always 1 : 1 in DNA molecule. 7. The number of purine : pyrimidine ratio may not be 1 : 1 in RNA molecule.
8. DNA sends the codon for the synthesis of proteins, but otherwise it does not participate in the protein synthesis. 8. RNA takes part in the protein synthesis through transcription and translation.

Question 4.
m-RNA, t-RNA and r-RNA.
Answer:

m-RNA t-RNA r-RNA.
1. m-RNA is a simple molecule which shows linear structure without any folds. 1. t-RNA is a single stranded molecule. There is regular pattern of folding shown by this molecule. 1. r-RNA is a single stranded RNA that is variously folded upon itself. In the folded regions it shows complementary base pairing.
2. It performs the function of transcription during protein synthesis. 2. It performs the function of transferring the amino acids during translation. 2. This RNA remains associated with the ribosomes permanently. It gives the binding site for m-RNA during the process of protein synthesis. It also orients the m-RNA molecule so as to read the message on codons properly.
3. Of the total cellular RNA, m-RNA forms 3-5%. 3. Of the total cellular RNA, t-RNA forms about 10-20%. 3. Of the total cellular RNA, r-RNA forms 80%
4. It has molecular weight of about 5,00,000. 4. t-RNA is the smallest RNA having only 73-93 nucleotides and has molecular weight of about 23,000 to 30,000 daltons. 4. Molecular weight is about 40,000 to 100,000 daltons.

Give Reasons

Question 1.
Nuclein was called as nucleic acid.
Answer:
Nuclein had acidic properties and it was isolated from nucleus. Hence, it was called as nucleic acid.

Question 2.
Initially proteins (and not DNA) were considered as genetic material.
Answer:

  1. Proteins are large, complex molecules and store information required to govern cell metabolism. Hence it was assumed that variations found in species were caused by proteins.
  2. On the other hand, DNA was considered as a small, simple molecule whose composition does not vary much among species.
  3. Variations in the DNA molecules are different than the variation in shape, electrical charge and function shown by proteins.
  4. Hence, initially proteins (and not DNA) were considered as genetic material.

Question 3.
On injecting a mixture of heat-killed S-bacteria and live R bacteria, the mice died.
Answer:

  1. Griffith obtained live S-strain bacteria from the blood of the dead mice.
  2. In a mixture of live R-bacteria and heat killed S-bacteria, live R-strain bacteria picked up something (transforming principle) from the heat-killed S bacterium and got changed into S-type.
  3. Transforming principle allowed R-type bacteria to synthesize capsule and thus they became virulent.
  4. Hence, on injecting a mixture of heat-killed S bacteria and live R bacteria, the mice died.

Question 4.
Viruses obtained by infecting bacteria having radioactive phosphorus contained radioactive DNA (labelled DNA), but not radioactive proteins.
Answer:
Viruses obtained by infecting bacteria having radioactive phosphorus, contained radioactive DNA (labelled DNA). but not radioactive proteins because DNA contains phosphorus (labelled DNA) but proteins do not.

Question 5.
Viruses obtained by infecting bacteria having radioactive sulphur contained radioactive protein but not radioactive DNA.
Answer:
Viruses obtained by infecting bacteria having radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur and proteins contain sulphur.

Question 6.
In bacteria, m-RNA does not require any processing.
Answer:
In bacteria, m-RNA does not require any processing because it has no introns and it is synthesized in cytoplasm.

Question 7.
Eukaryotic DNA is condensed and supercoiled.
Answer:

  1. In a typical mammalian cell, length of DNA double helix is approximately 2.2 metres.
  2. The size of typical nucleus is approximately 10-6 m
  3. Such a long DNA molecule has to be fitted in small nuclear space.
  4. Therefore, DNA is highly condensed, coiled and supercoiled so that it can be accommodated in the nucleus.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 8.
During translation, complementarity principle is not applicable.
Answer:
During translation, complementarity principle is not applicable as, genetic information is transferred from a polymer of nucleotides to a polymer of amino acids.

Question 9.
Protein synthesis is the most important and essential activity in the living cells.
Answer:

  1. Proteins play a significant role in the metabolism of living cells.
  2. The actual phenotypic expression of living cells is dependent on the biochemical reactions.
  3. Each biochemical reaction needs a specific enzyme for its initiation and completion. All the enzymes are proteins.
  4. In a cell there are many structural proteins too. Thousands of structural and catalytic proteins are constantly required within the cell at all times.
  5. Many functional proteins like hormones are also important for metabolism. Thus, for the synthesis of all such proteins, protein synthesis has become the most important and essential activity of the living cell.

Question 10.
Only 20 amino acids are considered as standard.
Answer:

  1. It was believed that there are total 20 amino acids in the living world. But 21st amino acid called selenocysteine was discovered later.
  2. This amino acid is coded by UGA which is usually a termination codon.
  3. In both prokaryotic and eukaryotic cells polypeptide chains contain 100-300 amino acids and they are formed by specific arrangement of 21 amino acids.
  4. But formation of selenocysteine requires the availability of element selenium in the cells.
  5. Therefore, only 20 amino acids are considered as standard.

Write Short Notes on the following

Question 1.
Friedrich Miescher’s nuclein
Answer:

  1. Nuclein is an acidic substance, having high phosphorus content and it was isolated by Friedrich Miescher in 1869, from the nuclei of pus cells.
  2. As Nuclein had acidic properties and it was isolated from nucleus, it was called as nucleic acid.
  3. E Miescher started working with white blood cells (the major component of pus). He used a salt solution to wash the pus off the bandages. He lysed the cells by adding a weak alkaline solution and isolated nucleic acid from nuclei that precipitated out of the solution.
  4. By the early 1900s, it was known that Miescher’s nuclein was a mixture of proteins and nucleic acids (DNA and RNA).

Question 2.
Packaging in Prokaryotes
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 1

  1. Size of cell in E. coli size is 2-3µ.
  2. The nucleoid is small, circular, highly folded, naked DNA (1100 µm long in perimeter and contains about 4.6 million base pairs).
  3. When the negatively charged DNA becomes circular, the size reduces to 350 µm in diameter.
  4. Folding/looping (40-50 domains (loops) further reduce it to 30 µm in diameter.
  5. RNA connectors assist in loop formation.
  6. Further coiling and super coiling of each domain reduces the size to 2 µm in diameter.
  7. This coiling (packaging) is assisted by positively charged HU (Histone like DNA binding proteins) proteins and enzymes like DNA gyrase and DNA topoisomerase-I, which maintain supercoiled state.

Question 3.
Experimental confirmation of semiconservative replication of DNA.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 2
(1) Matthew Meselson and Franklin Stahl (1958) used equilibrium – density – gradient – centrifugation technique to experimentally prove semiconservative DNA replication.

(2) They cultured bacteria E.coli in the medium containing 14N (light nitrogen). They obtained equilibrium density gradient band by using 6M CsCl2. The position of this band is recorded.

(3) E. coli cells were then transferred to 15N medium (heavy isotopic nitrogen) and allowed to replicate for several generations. At equilibrium point density gradient band was obtained, by using 6M CsCl2. The position of this band is recorded.

(4) The heavy DNA (15N) molecule can be distinguished from normal DNA by centrifugation in a 6M Cesium chloride (CsCl2) density gradient. At the equilibrium point 15N DNA will form a band. In this both the strands of DNA are labelled with 15N.

(5) Such E. coli cells were then transferred to another medium containing 14N i.e. normal (light) nitrogen. After first generation, the density gradient band for 14N 15N was obtained and its position was recorded. After second generation, two density gradient bands were obtained – one at 14N 15N position and other at 14N position.

(6) The position of bands after two generations clearly proved that DNA replication is semiconservative.

Question 4.
Central Dogma of molecular biology.
Answer:
(1) Central dogma of molecular biology was postulated by EH.C. Crick in 1958.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 3

(2) DNA gets transcribe to form m-RNA, m-RNA acts as a messenger and gets translated to form a polypeptide chain (protein) having specific amino acid sequence.

(3) This unidirectional flow of information from DNA to RNA and from RNA to proteins is referred as central dogma of molecular biology.

(4) Temin (1970) and Baltimore (1970) : Central dogma in retroviruses.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 4

Question 5.
Processing of hnRNA.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 5

  1. Eukaryotes have split genes.
  2. Primary transcript or hnRNA is non-functional and contains both exons and introns.
  3. Processing of hnRNA results in functional m-RNA.
  4. The fully processed hnRNA is called m-RNA.
  5. m-RNA comes out of the nucleus through nuclear pore for getting translated.

hnRNA undergoes capping, tailing and splicing.

  1. Capping : Methylated guanosine tri¬phosphate is added to 5’ end of hnRNA.
  2. Tailing : Polyadenylation take place at 3′ end.
  3. Splicing : It is removal of introns.
  4. DNA ligase joins exons in a definite sequence (order).

Question 6.
Cracking of genetic code.
Answer:

  1. M. Nirenberg and Matthaei synthesized artificial poly-U m-RNA.
  2. Using this synthetic poly-U m-RNA and cell free system of protein synthesis, a small polypeptide consisting of only amino acid phenylalanine was obtained.
  3. It suggested that UUU codes for phenylalanine.
  4. Dr. Har Gobind Khorana devised a technique for synthesis of artificial m-RNA with repeated sequences of known nucleotides.
  5. He synthesized artificial RNA consisting of known repeated sequences of two or three nucleotides. E.g. CUC, UCU, CUC, UCU, by using synthetic DNA.
  6. This resulted in formation of polypeptide chain consisting of alternate amino acids leucine and serine.
  7. Synthetic RNA consisting of CUA, CUA, CUA, CUA repeats gave polypeptide chain with only one amino acid – leucine.
  8. Severo Ochoa established that the enzyme (polynucleotide phosphorylase) also helps in polymerizing RNA of defined sequences in a template-independent manner (i.e. enzymatic synthesis of RNA).
  9. Thus all the 64 codons in the dictionary of genetic code were deciphered.

Question 7.
Types of mutations.
Answer:

  1. Chromosomal mutations : Loss (deletion) or gain (insertion/duplication) of a segment of DNA results in alteration in the chromosome.
  2. Point mutations : They involve change in a single base pair of DNA. E.g. Mutation that results in Sickle-cell anaemia.
  3. Deletion or insertion of base pairs of DNA : It causes frame-shift mutations or deletion mutation.
  4. Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion.
  5. Insertion or deletion of three or multiples of three bases (insert or delete) results in insertion or deletion of amino acids and reading frame remains unaltered from that point onwards.

Question 8.
Transfer-RNA.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 6
(1) As t-RNA can read the codon and also can bind with the amino acid, t-RNA is considered as an adapter molecule.

(2) Clover leaf structure (2 dimensional) of t-RNA:

  • t-RNA has four arms – DHU arm (has amino acyl binding loop), middle arm (has anticodon loop), Tif/C arm (has ribosome binding loop) and variable arm.
  • It has G nucleotide at 5’ end.
  • Amino acid acceptor end (3’ end) having unpaired CCA bases (i.e. amino acid binding site).

(3) For every amino acid, there is specific t-RNA.
(4) Initiator t-RNA is specific for methionine.
(5) There are no t-RNAs for stop codons.
(6) In the actual structure, the t-RNA molecule looks like inverted L (3 dimensional : structure)

Question 9.
UTRs.
Answer:

  1. m-RNA has some additional sequences that are not translated. These sequences are referred as untranslated regions (UTR).
  2. The UTRs are present at both 5′-end (before start codon) and at 3′-end (after stop codon).
  3. They are required for efficient translation process.

Short Answer Questions

Question 1.
Why are Okazaki fragments formed on lagging strand only?
Answer:

  1. The lagging template is the template strand with free 5’ end.
  2. The replication always starts at C-3 end of template strand and proceeds towards C-5 end.
  3. Both the strands of the parental DNA are antiparallel and new strands are always formed in 5′ → 3′ direction, i.e. DNA polymerase synthesizes new strand in only one direction i.e. 5′ → 3′ direction.
  4. Hence, the lagging templates becomes available for replication only discontinuously in small patches.
  5. The new lagging strand develops discontinuously away from the replicating fork in the form of small Okazaki fragments.
  6. Hence, Okazaki fragments formed on lagging strand only.

Question 2.
Why t-RNA is called as adapter molecule?
Answer:
t-RNA can read the codon on m-RNA. It also can bind with the amino acid at 3’ end and transport it to m-RNA-ribosome complex during translation. It can bind with specific codon which is complementary to its anticodon. So t-RNA is considered as an adapter molecule.

Question 3.
Why DNA replication is called semiconservative replication?
Answer:

  1. In each of the two daughter DNA molecules thus formed, one strand is parental and the other one is newly synthesized.
  2. 50% is contributed by mother DNA.
  3. Hence, DNA replication is described as semiconservative replication.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 4.
What are the functions of three types of RNAs in bacteria? Which enzyme is involved in transcription of DNA to form RNAs in bacteria? What is its function?
Answer:

  1. In bacteria, m-RNA provides the encoded message for protein synthesis; t-RNA brings specific amino acid to the site of translation; r-RNA plays role in providing binding site to m-RNA and t-RNA.
  2. There is single DNA dependent-RNA polymerase that catalyses transcription of all 3 types of RNA in bacteria.
  3. RNA polymerase binds to promoter and initiates transcription (initiation) and synthesizes RNA.

Question 5.
Explain why it was suggested that codon is a sequence of three consecutive nucleotides on m-RNA.
Answer:

  1. If each codon has only one nucleotide, then there will be 41 = 4 codons, which can encode for only four different types of amino acids.
  2. If each codon has two nucleotides, then there will be 4² = 16 codons, which can encode for only 16 different types of amino acids.
  3. If each codon has three nucleotides, then there will be 4³ = 64 codons, which are sufficient to specify 20 different types of amino acids.

Question 6.
Explain Genetic code is Non-overlapping.
Answer:

  1. Each single base is a part of only one codon.
  2. Adjacent codons do not overlap.
  3. If it had been overlapping type, with 6 bases, there would be 4 amino acid molecules in a chain.
  4. Experimental evidence favours non-overlapping nature of genetic code.

Question 7.
How degeneracy of the code is explained by Wobble hypothesis?
Answer:

  1. In 1966, Crick proposed Wobble hypothesis.
  2. According to this hypothesis, in codon- anticodon base pairing, the third base may not be complementary.
  3. The third base of codon is called wobble base and this position is called wobble position.
  4. This results in economy of t-RNA as anticodon of a t-RNA may bind with codon even when only first two bases are complementary.
  5. For example, GUU, GUC, GUA and GUG codons code for amino acid Valine.
  6. Degeneracy of genetic code means many codons can code for same amino acid.
  7. Thus, the degeneracy of genetic code gets explained by Wobble hypothesis.

Question 8.
a. What is meant by universal genetic code? Give example.
b. Why genetic code is called Non- ambiguous?
Answer:
a. Universal genetic code means that the specific codon codes for same amino acid in all living organisms, e.g. codon AUG always specifies amino acid methionine.

b. Genetic code is called non-ambiguous because, each codon specifies a particular amino acid.

Question 9.
Give examples of termination codons. Why are they known as termination codons?
Answer:

  1. UAA, UAG and UGA are known as termination codons.
  2. They do not code for any amino acid.
  3. They terminate or stop the process of elongation of polypeptide chain.
  4. Hence, they are known as termination codons.

Question 10.
How many amino acids are required for protein synthesis? From where are they obtained?
Answer:

  1. About 20 different types of amino acids are required for protein synthesis.
  2. They are available in the cytoplasm.

Question 11.
How DNA regulates protein synthesis?
Answer:

  1. DNA regulates protein synthesis by coding for the specific sequence of amino acids in a protein.
  2. This control is possible through transcription of m-RNA.
  3. Genetic code is specific for particular amino acid.

Question 12.
What is the role of ribosomes in protein synthesis?
Answer:

  1. Ribosomes serve as site for protein synthesis.
  2. A ribosome has one binding site for m-RNA and 3 binding sites for t-RNA. They are P-site (peptidylt-RNA-site), A-site (aminoacyl t-RNA-site) and E-site (exit site).
  3. In Eukaryotes, a groove which is present between two subunits of ribosomes, protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.

Question 13.
Give examples of coordinated regulation or expression, of several sets of genes.
Answer:
Examples of coordinated regulation or expression of several sets of gene are:

  1. If E.colt bacteria do not have lactose in the surrounding medium as a source of energy, then structural genes in Lac operon do not get transcribed and enzyme like β -galactosidase is not synthesized.
  2. The development and differentiation of embryo into an adult organism.

Question 14.
Explain with example what is meant by positive control of gene regulation.
Answer:

  1. A set of genes will be switched on when a substrate is to be metabolized.
  2. This phenomenon is called induction and small molecule responsible for this, is known as inducer.
  3. It is positive control of gene regulation.
  4. For example, lactose acts as an inducer in Lac operon.

Question 15.
If operator gene is deleted due to mutation, how will E.coli metabolise lactose?
Answer:
If operator gene is deleted due to mutation, lac operon cannot be regulated. It will get transcribed continuously and enzymes required for lactose metabolism will get synthesized continuously.

Question 16.
Explain in brief the process of initiation during protein synthesis.
Answer:
Initiation of synthesis of polypeptide chain takes place as follows:

  1. Small subunit of ribosome binds to the m-RNA at 5’ end.
  2. Start codon is positioned properly at P-site.
  3. Initiator t-RNA, (carrying methionine in eukaryotes or formyl methionine in prokaryotes) binds with initiation codon (AUG) of m-RNA by it’s anti-codon (UAC). Codon-anti-codon pairing involves formation of hydrogen bonds.
  4. The large subunit and smaller subunit of ribosome bind in the presence of Mg++.
  5. Thus, initiator charged t-RNA occupies the P-site and A-site is vacant for next charged t-RNA.

Question 17.
What are the application of genomics?
Answer:
Applications of genomics are as follows:

  1. Structural and functional genomics is used in the improvement of crop plant, human health and livestock.
  2. The knowledge and understanding acquired by genomics research can be applied in medicine, biotechnology and social sciences.
  3. It helps in the treatment of genetic disorders through gene therapy.
  4. It helps in the development of transgenic crops having desirable characters.
  5. Genetic markers have applications in forensic analysis.
  6. Genomics can lead to introduction of new gene in microbes to produce enzymes, therapeutic proteins and biofuels.

Question 18.
Why is HGP important?
Answer:

  1. HGP is associated with rapid development of Bioinformatics.
  2. Knowledge gained about the functions of genes and proteins has a major impact in the fields like Medicine. Biotechnology and the Life sciences.
  3. It has helped in identifying the genes that are associated with genetic characteristics.
  4. The genetic basis of many hereditary diseases can be understood.
  5. It has increased the understanding of gene structure and function in other species. As human beings have many of the genes same as those of flies, roundworms and mice, such studies will enhance understanding of human evolution.

Chart based / Table based Questions

Question 1.
Complete the following table:

Organism Diploid chromosome number
Mouse ————-
Fruitfly ————
Roundworm ————-
Yeast ————–

Answer:

Organism Diploid chromosome number
Mouse 40
Fruitfly 8
Roundworm 12
Yeast 32

Diagram Based Questions

Question 1.
a. Identify A and B in the following diagram.
b. Name the scientist who conducted this experiment.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 7
Answer:
a. A : Smooth strain (III-S)
B : Rough Strain and Heat-killed Smooth Strain

b. F. Griffith conducted the experiment shown in the diagram.

Question 2.
a. Identify A and B in the given diagram.
b. What is the conclusion of the given experiment?
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 8
Answer:
a. A : Rough, nonvirulent R-strain B : Heat-killed virulent S-strain
b. When DNA of heat-killed S-strain bacteria is treated with DNase, mouse remains alive as transformation does not take place. This proves that DNA is the genetic material.

Question 3.
Identify A, B, C and name the scientists who carried out the experiment given in the diagram.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 9
Answer:
Answer: A : Infection B : Blending C : Centrifugation Scientists who carried out experiment are Alfred Hershey and Martha Chase.

Question 4.
(1) Identify A, B and C.
(2) What are their sizes?
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 10
Answer:
(1) A : Circular, unfolded chromosome
b : Folded chromosome (40 to 50 loops)
c : Supercoiled, folded chromosome

(2) (A) 350 µ (B) 30 µ (C) 2 µ

Question 5.
Draw a labelled diagram of Nucleosome with H1 histone.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 11

Question 6.
Identify A and B in the given diagram.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 12
Answer:
A : Nucleosome
B : Linker DNA

Question 7.
(a) Identify A in the given diagram.
(b) What is the dimension denoted in ‘B’
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 13
Answer:
A : Solenoid B : 300 Å

Question 8.
Sketch and label process of formation of beads on strings/fibres from chain of nucleosomes.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 14

Question 9.
a. Identify A. Which enzyme joined them?
b. Identify B. What is its function ?
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 15
Answer:
a. A is Okazaki fragment. These fragments are joined by DNA ligase enzyme
b. B is RNA primer. It provides 3’ OH to which nucleotide gets attached by ester bond.

Question 10.
Draw a labelled diagram showing semi-conservative replication of DNA.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 16

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 11.
Sketch and label, Meselson and Stahls experiment.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 17

Question 12.
Identify A, B and C in the following diagram.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 18
Answer:
A : Transcription
B : Translation and
C : Reverse Transcription

Question 13.
a. Draw a labelled diagram of transcription unit.
b. What is the sequence of m-RNA and coding strand if sequence of template strand of DNA is
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 19
Answer:
a. Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 20
b. Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 21

Question 14.
a. The following diagram shows processing of ………….
b. What is capping?
c. Identify A, B and C.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 22
Answer:
a. hnRNA
b. Capping is addition of methylated guanosine tri-phosphate at 5′ end of hnRNA.
c. A : Exon and B : Intron C : m-RNA

Question 15.
Draw a labelled diagram of t-RNA carrying Glutamic acid.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 23

Question 16.
Observe the diagrams (a), (b) and (c)

  1. Which step of protein synthesis is shown in the following diagrams?
  2. During initiation, which subunit of ribosome binds with m-RNA?
  3. What are the three binding sites for t-RNA on ribosomes?
  4. On which site of ribosome second and subsequent t-RNA arrives?
  5. Which link is binding amino acids in diagram (b)?
  6. Which chain is being released from ribosome in diagram (c) ?
    Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 24

Answer:

  1. Translation
  2. 30S or 40S
  3. P site, A site and E site
  4. A site
  5. Peptide link
  6. Polypeptide chain

Question 17.
Draw a labelled diagram – Working of lac operon.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 25

Question 18.
What is the process shown in the following diagram? Mention all the steps given in the diagram in a proper sequence.
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 26
Answer:
The process shown in the above diagram is DNA fingerprinting. The sequence of steps in the process are:

  1. Isolation of DNA
  2. Restriction digestion
  3. Gel electrophoresis
  4. Southern blotting
  5. Selection of DNA probe
  6. Hybridization
  7. Photography,

Long Answer Questions

Question 1.
Describe Griffith’s transformation experiment.
OR
In the light of Griffith’s experiment, explain the action of two strains of streptococcus pneumoniae and give his conclusion.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 27
(1) In 1928, Frederick Griffith, carried out experiments with bacterium Streptococcus pneumoniae (which causes pneumonia in humans and other mammals).

(2) Griffith used two strains of Streptococcus pneumonia:

  • S-type (Virulent, smooth, pathogenic and encapsulated).
  • R-type (Non-virulent, rough, non- pathogenic and non-capsulated).

(3) Experiments carried out by E Griffith:

  • Mice were injected with R-strain bacteria and they survived (no pneumonia).
  • Mice injected with S-strain bacteria developed pneumonia and died.
  • When heat-killed S-strain bacteria were injected in mice, the mice survived.
  • On injecting a mixture of heat-killed S-bacteria and live R bacteria, the mice died.

(4) Griffith obtained live S-strain bacteria from the blood of the dead mice.

(5) He concluded that the live R-strain bacteria must have picked up something (which he called transforming principle) from the heat killed S bacterium, and got changed into S-type. Transforming principle allowed R-type to synthesize capsule and it became virulent.

(6) Thus, F. Griffith first demonstrated genetic transformations.

Question 2.
Describe Avery, McCarty and MacLeod’s experiments.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 28

  • In 1944, Oswald T. Avery, Colin M. MacLeod and Maclyn McCarty proved that the DNA is a genetic material (transforming principle).
  • They purified DNA, RNA, proteins (enzymes) and other materials from heat killed S-strain cells and mixed them with cells of R-strain bacteria separately to confirm which one could transform living R cells into S cells.
  • Only DNA was able to transform harmless R-strain into virulent S-strain.
  • They also demonstrated that proteases, RNases did not affect transformation. Thus it was proved that the transforming substance was neither a protein nor-RNA.
  • When DNA was digested with DNase, there was no transformation.
  • These experiments proved that the transformation of Live R-strain bacteria into S-strain type was because of DNA of bacteria of S-strain.
  • Thus, they proved that the DNA is transforming principle.

Question 3.
Explain how Hershey – Chase experimentally proved that DNA is the genetic material.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 29
(1) Hershey and Chase worked with bacteriophages (viruses that infect bacteria and which are composed of DNA and protein coat).

(2) They cultured E. coli bacteria in medium containing radioactive phosphorus 32p. By infecting these bacteria with bacteriophages, Hershey and Chase could develop bacteriophages having DNA labelled with 32p. as DNA contains phosphorus and proteins do not.

(3) They also cultured E. coli bacteria in medium containing radioactive sulphur 35s. By infecting these bacteria with bacteriophages, they developed
bacteriophages whose protein coat was labelled with 35s, as proteins contain sulphur and DNA does not.
[Note : Viruses cannot be cultivated in medium.]

(4) Experiment involved three steps.

  • Infection : Both the types of radioactive phages were allowed to infect E.coli bacteria grown on the medium containing normal ‘P‘ and ‘S’.
  • Blending : Then bacterial cultures were agitated in blender to break contact between bacteria and parts of viruses that did not enter bacterial cells.
  • Centrifugation : It was done to separate bacterial cells as a pellet. Parts of viruses which did not enter bacteria remained in the suspension.

(5) Observation:

  • Radioactive ‘S’ remained in suspension.
  • Only radioactive ‘P’ was found inside the bacterial cell in the pellet.

(6) Thus it was proved that DNA is the genetic material which enters bacterial cell and not protein.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Question 4.
Explain the formation of beads on string, solenoid fibre, chromatin fibre and chromosome.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 30

  • Beads on string (11 nm in diameter) : Under an electron microscope, nucleosomes in thread like chromatin look like ’beads-on- string’.
  • Solenoid fibre (30 nm in diameter) : Six such nucleosomes get coiled and then form solenoid that looks like coiled telephone wire.
  • Chromatin fibre (200 nm in diameter) : Further supercoiling tends to form a looped structure called chromatin fibre.
  • Chromosome (1400 nm in diameter) : Chromatin fibre further coils and condenses at metaphase stage to form the chromosomes. Each chromatid is 700 nm in diameter.
  • Non-Histone Chromosomal Proteins (NHC) are the additional sets of proteins that contribute to the packaging of chromatin at a higher level.

Question 5.
Explain the components of a transcription unit with the help of a diagram?
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 31
Transcription unit (Each transcribed segment of DNA) consists of the promoter, the structural gene and the terminator.
(1) The promoter:

  • The promoter is located towards 5′ end of structural gene, i.e. upstream.
  • It is a DNA sequence that provides binding site for enzyme RNA polymerase.
  • In prokaryotes, sigma factor sub unit of the enzyme recognizes the promoter.

(2) Structural genes:

  • Template strand (Antisense strand) : DNA strand having 3’→ 5’ polarity acts as template strand as DNA dependent- RNA polymerase catalyses polymerization in 5’ → 3’ direction.
  • Sense strand : The other strand of DNA having 5’ → 3’ polarity is complementary to template strand. It is called as sense strand. The sequence of bases in this strand, is same as in RNA (where Thymine is replaced by Uracil). It is the actual coding strand.

(3) The terminator:

  • The terminator is located at 3’ end of coding strand, i.e. downstream.
  • It defines the end of the transcription process.

Question 6.
What have we learnt from the Human Genome Project?
Answer:
We have learnt following salient features of human genome from the Human Genome Project.

  1. The human genome contains 3164.7 million nucleotide bases.
  2. The average gene consists of 3000 bases.
  3. Largest known human gene is dystrophin at 2.4 million bases.
  4. Total number of genes is estimated to be 3000.
  5. 99.9% nucleotide bases are exactly the same in all people.
  6. The function of about 50% of the discovered genes are unknown.
  7. Less than 2% of the genome codes for proteins.
  8. Repeated sequences make up a very large portion of the human genome. They can shed light on chromosome structure, dynamics and evolution.
  9. Chromosome 1 has most genes (2968) and the Y has the fewest (231).
  10. Single nucleotide polymorphism have been identified at about 1.4 million locations. It is useful in finding chromosomal locations for disease-associated sequences and tracing human history.

Question 7.
Describe the steps involved in DNA fingerprinting ?
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance 32
Steps involved in DNA fingerprinting are as follows:
1. Isolation of DNA : The DNA can be isolated even from the small amount of tissue like blood, hair roots, skin, etc.

2. Restriction digestion:

  • The isolated DNA is treated with restriction enzymes which cut the DNA at specific sites to form small fragments of variable lengths.
  • Variations in the lengths of restriction fragments are known as Restriction Fragment Length Polymorphism (RFLP).

3. Gel electrophoresis:

  • The DNA samples are loaded on agarose gel and electrophoresis is carried out.
  • Negatively charged DNA fragments move to the positive pole.
  • Separation of fragments depends on their length and it results in formation of bands.
  • dsDNA is then denatured into ssDNA by alkali treatment.

4. Southern blotting : The separated DNA fragments are transferred to a nylon membrane or a nitrocellulose membrane.

Maharashtra Board Class 12 Biology Important Questions Chapter 4 Molecular Basis of Inheritance

5. Selection of DNA probe:

  • DNA Probe is a known sequence of single- stranded DNA.
  • It is obtained from organisms or prepared by cDNA preparation method.
  • The DNA probe is labelled with radioactive isotopes.

6. Hybridization:

  • In this process, probe is added to the nitrocellulose membrane containing DNA fragments.
  • The single-stranded DNA probe pairs with the complementary base sequence of the DNA strand.
  • As a result DNA-DNA hybrids are formed on the nitrocellulose membrane. Unbound single-stranded DNA probe fragments are washed off.

7. Photography : The nitrocellulose membrane is then kept in contact with X-ray film. DNA bands, due to radioactive probe, give photographic image on X-ray film. This is autoradiography.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 3 Inheritance and Variation Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 3 Inheritance and Variation

Multiple Choice Questions

Question 1.
Which one of the following characters is recessive in the case of the pea plants?
(a) Axial flower
(b) Green pod
(c) Green seed
(d) Inflated pod
Answer:
(c) Green seed

Question 2.
Which of the following trait is dominant in Pisum sativum?
(a) White flowers
(b) Green seeds
(c) Yellow pods
(d) Inflated pods
Answer:
(d) Inflated pods

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 3.
When phenotypic and genotypic ratio is the same, then it is an example of ……………….
(a) incomplete dominance
(b) cytoplasmic inheritance
(c) quantitative inheritance
(d) incomplete dominance and co-dominance
Answer:
(a) incomplete dominance

Question 4.
A pea plant with yellow and round seeds is crossed with another pea plant with green and wrinkled seeds produce 51 yellow round seeds and 49 yellow wrinkled seeds, the genotype of plant with yellow round seeds must be ……………….
(a) YYRr
(b) YyRr
(c) YyRR
(d) YYRR
Answer:
(a) YYRr

Question 5.
When a single gene produces two effects and one of it is lethal, then the ratio is ……………….
(a) 2 : 1
(b) 1 : 1
(c) 1 : 2 : 1
(d) 1 : 1 : 1 : 1
Answer:
(a) 2 : 1

Question 6.
When two genes control a single character and have cumulative effect, the ratio is ……………….
(a) 1 : 1 : 1 : 1
(b) 1 : 4 : 6 : 4 : 1
(c) 1 : 2 : 1
(d) 1 : 6 : 15 : 20 : 15 : 6 : 1
Answer:
(d) 1 : 6 : 15 : 20 : 15 : 6 : 1

Question 7.
Genes located on the same locus but show more than two different phenotypes are called ……………….
(a) polygenes
(b) multiple alleles
(c) co-dominants
(d) pleiotropic genes
Answer:
(b) multiple alleles

Question 8.
Genotype refers to the genetic composition of ……………….
(a) an organism
(b) an organ
(c) chromosomes
(d) germ cells
Answer:
(a) an organism

Question 9.
Individuals having identical alleles of a gene are known as ……………….
(a) homozygous
(b) heterozygous
(c) hybrids
(d) dominants
Answer:
(a) homozygous

Question 10.
If a heterozygous tall plant is crossed with a homozygous dwarf plant, the proportion of dwarf progeny will be ……………….
(a) 100 per cent
(b) 75 per cent
(c) 50 per cent
(d) 25 per cent
Answer:
(c) 50 percent

Question 11.
Inheritance of AB blood group is due to ……………….
(a) incomplete dominance
(b) polyploidy
(c) polygeny
(d) co-dominance
Answer:
(d) co-dominance

Question 12.
The recombination of characters in a dihybrid cross is related to ……………….
(a) law of dominance
(b) incomplete dominance
(c) co-dominance
(d) independent assortment
Answer:
(d) independent assortment

Question 13.
Which one of the following is a true pleiotropic gene?
(a) HbA
(b) Hbs
(c) HbD
(d) HbP
Answer:
(b) Hbs

Question 14.
For demonstrating the law of independent assortment, one should carry out ……………….
(a) back cross
(b) test cross
(c) dihybrid cross
(d) monohybrid cross
Answer:
(c) dihybrid cross

Question 15.
Which one of the following is an example of multiple alleles?
(a) Height in pea plant
(b) Hair colour in cattle
(c) Petal colour in four o’clock plant
(d) Wing-size in Drosophila
Answer:
(d) Wing-size in Drosophila

Question 16.
For the formation of 50 seeds, how many minimum meiotic divisions are necessary?
(a) 25
(b) 50
(c) 75
(d) 63
Answer:
(d) 63

Question 17.
A cross used to verify the unknown genotype of F1 hybrid is a ………………. cross.
(a) test
(b) back
(c) dihybrid
(d) monohybrid
Answer:
(a) test

Question 18.
Appearance of new combinations in F2 generation in a dihybrid cross proves the law of ……………….
(a) dominance
(b) segregation
(c) independent assortment
(d) purity of gametes
Answer:
(c) independent assortment

Question 19.
Genotype of human blood group ‘O’ will be ……………….
(a) IAIA
(b) IAIB
(c) ii
(d) IAi
Answer:
(c) ii

Question 20.
The genotype of human blood group B is ……………….
(a) IAIA
(b) IBi
(c) IAIB
(d) ii
Answer:
(b) IBi

Question 21.
……………… chromosome appears ‘V’-shaped during anaphase.
(a) Metacentric
(b) Acrocentric
(c) Telocentric
(d) Sub-Metacentric
Answer:
(a) Metacentric

Question 22.
The sister chromatids are held together by ……………….
(a) centrioles
(b) chromonemata
(c) chromomere
(d) centromere
Answer:
(d) centromere

Question 23.
Which of the following is not X-linked disorder ?
(a) Haemophilia
(b) Night-blindness
(c) Hypertrichosis
(d) Myopia
Answer:
(c) Hypertrichosis

Question 24.
Which of the following is also called bleeder’s disease ?
(a) Anaemia
(b) Thrombocytopenia
(c) Polycythemia
(d) Haemophilia
Answer:
(d) Haemophilia

Question 25.
The person with Turner’s syndrome has ……………….
(a) 45 autosomes and X sex chromosome
(b) 44 autosomes and XYY sex chromosome
(c) 45 autosomes and Y chromosome
(d) 44 autosomes and X chromosome
Answer:
(d) 44 autosomes and X chromosome

Question 26.
Which of the following is sex chromosomal disorder ?
(a) Colour blindness
(b) Turner’s syndrome
(c) Thalassemia
(d) Down’s syndrome
Answer:
(b) Turner’s syndrome

Question 27.
The word chroma means ……………….
(a) a part of nucleus
(b) a part of chromosome
(c) colour
(d) filamentous body
Answer:
(c) colour

Question 28.
Presence of whole sets of chromosomes is called ……………….
(a) aneuploidy
(b) euploidy
(c) ploidy
(d) chromatography
Answer:
(b) euploidy

Question 29.
The synonymous term for centromere is ……………….
(a) primary constriction
(b) secondary constriction
(c) telomere
(d) satellite
Answer:
(a) primary constriction

Question 30.
Small swellings on the surface of the chromosome are called ……………….
(a) centromeres
(b) chromonemata
(c) chromomeres
(d) telomeres
Answer:
(c) chromomeres

Question 31.
On what basis are the chromosomes usually classified?
(a) On the basis of their function
(b) On the basis of their length
(c) On the basis of the position of the centromere
(d) On the basis of their number
Answer:
(c) On the basis of the position of the centromere

Question 32.
Find the mismatched pair :
(a) Metacentric – V-shaped
(b) Sub-Metacentric – L-shaped
(c) Acrocentric – J-shaped
(d) Telocentric – S-shaped
Answer:
(d) Telocentric – S-shaped

Question 33.
Out of the following combinations which individual will have maximum genetically active DNA?
(a) 44 + XX
(b) 44 + XY
(c) 44 + XYY
(d) Down’s syndrome
Answer:
(a) 44 +XX

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 34.
Crossing over occurs at the time of ……………….
(a) diplotene
(b) pachytene
(c) leptotene
(d) zygotene
Answer:
(b) pachytene

Question 35.
A mature woman has ………………. linkage groups.
(a) 44
(b) 22
(c) 46
(d) 23
Answer:
(d) 23

Question 36.
The pairing of homologous chromosomes is called ……………….
(a) crossing over
(b) terminalisation
(c) synapsis
(d) bivalent
Answer:
(c) synapsis

Question 37.
If only one ‘X’ chromosome is found in a female person, which of the following symptoms will she show?
(a) epicanthal skin fold
(b) webbing of neck
(c) small testis and absence of spermatogenesis
(d) presence of simian crease on the palm
Answer:
(b) webbing of neck

Question 38.
If centromere is situated in the middle of the chromosome, it is called ……………….
(a) metacentric
(b) acrocentric
(c) submetacentric
(d) telocentric
Answer:
(a) metacentric

Question 39.
In which of the following disorders the number of chromosomes present is (extra) 47?
(a) Turner’s syndrome
(b) Cushing’s syndrome
(c) Acquired immuno-deficiency syndrome
(d) Down’s syndrome
Answer:
(d) Down’s syndrome

Question 40.
Myopia is an example of ……………….
(a) complete sex linkage
(b) incomplete sex linkage
(c) recombination
(d) crossing over
Answer:
(a) complete sex linkage

Question 41.
Down’s syndrome is represented by ……………….
(a) n + 1
(b) 2n + 1
(c) 3n + 1
(d) n – 1
Answer:
(b) 2n + 1

Classify the following to form Column B as per the category given in Column A

Question 1.
Types of traits:
[Sickle-cell anaemia, Flower colour of Mirabelis jalapa, Coat colour of cattle, Human blood groups, Widow’s peak, Height in human beings.]

Column A Column B
(1) Co-dominance —————–
(2) Incomplete dominance —————–
(3) Multiple allelism —————-
(4) Pleiotropy —————–
(5) Polygenes —————-
(6) Autosomal dominance ——————

Answer:

Column A Column B
(1) Co-dominance Coat colour of cattle
(2) Incomplete dominance Flower colour of Mirabelis jalapa
(3) Multiple allelism Human blood groups
(4) Pleiotropy Sickle-cell anaemia
(5) Polygenes Height in human beings
(6) Autosomal dominance Widow’s peak

Question 2.
Types of sex-linked genes:
[Haemophilia, Ichthyosis, Nephritis, Myopia, Hypertrichosis, Retinitis pigmentosa]

Column A Column B
(1) Completely X-linked genes —————–
(2) Completely Y-linked genes —————–
(3) Incompletely sex-linked genes —————-

Answer:

Column A Column B
(1) Completely X-linked genes Haemophilia, Myopia
(2) Completely Y-linked genes Ichthyosis,Hypertrichosis
(3) Incompletely sex-linked genes Nephritis, Retinitis pigmentosa

Question 3.
Genetic Disorders
[Turner’s syndrome, Sickle-cell anaemia, Thalassemia, Edward’s syndrome, Klinefelter’s syndrome, Down’s syndrome]

Column A Column B
(A) Autosomal disorder —————–
(B) Sex chromosomal disorder —————–
(C) Mendelian disorder —————-

Answer:

Column A Column B
(A) Autosomal disorder Edward’s syndrome, Down’s syndrome
(B) Sex chromosomal disorder Turner’s syndrome, Klinefelter’s syndrome
(C) Mendelian disorder Sickle-cell anemia, Thalassemia

Very Short Answer Questions

Question 1.
What is hybrid?
Answer:
Hybrid is a heterozygous individual produced from a cross involving two parents differing in one or more contrasting characters.

Question 2.
What are homologues?
Answer:
Homologues are homologous chromosomes which are morphologically similar to each other.

Question 3.
Which law of Mendelian genetics is universally applicable?
Answer:
The law of segregation of Mendelian genetics is universally applicable.

Question 4.
Which law of Mendelian genetics is not universally applicable?
Answer:
The law of independent assortment of Mendelian genetics is not universally applicable.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 5.
Give the alternative term for checker board.
Answer:
Punnett’s square is the alternative term for the checker board.

Question 6.
Give the genotypic dihybrid ratio.
Answer:
1 : 1 : 2 : 2 : 4 : 2 : 2 : 1 : 1 is the genotypic dihybrid ratio.

Question 7.
What is a lethal gene?
Answer:
The gene which causes the death of the bearer is called lethal gene.

Question 8.
A pea plant pure for yellow seed colour is crossed with a pea plant pure for green seed colour. In F1 generation, all pea plants were with yellow seed. Which law of Mendel is applicable?
Answer:
Mendel’s law of dominance is applicable in this example.

Question 9.
Identify which one of the following is a test cross.

  1. Tt × Tt
  2. TT × tt
  3. Tt × tt

Answer:
3. Tt × tt is a test cross.

Question 10.
What colouration do roans possess? Why?
Answer:
Roans possess the mixture of red and white colour side by side due to codominant alleles for red and white traits.

Question 11.
What are polygenes?
Answer:
When a character is controlled by two or more than two pairs of genes, the genes are called polygenes.

Question 12.
In which region of chromosomes does crossing over take place?
Answer:
Crossing over takes place in the homologous region of the chromosomes.

Question 13.
What are the four sequential steps of crossing over?
Answer:
There are four sequential steps such as synapsis, tetrad formation, crossing over and terminalisation.

Question 14.
Give one example of complete linkage.
Answer:
X chromosome of Drosophila males show complete linkage.

Question 15.
What is the number of linkage groups found in honey bee?
Answer:
The number of linkage group corresponds to the haploid number of chromosomes. Honey bee’s haploid chromosomes number is 16 and thus it has 16 linkage groups.

Question 16.
Name the term for genes located on non-homologous region of Y chromosomes.
Answer:
The genes located on non-homologous region of Y chromosomes are known as holandric genes or Y-linked genes.

Question 17.
What are linkage groups?
Answer:
The genes present on the same chromosome and inherited together are called linkage group.

Question 18.
How are RBCs changed due to sickle-cell anaemia ?
Answer:
RBCs undergo change in their shape and look like a sickle, resulting in reduced capacity to carry haemoglobin.

Question 19.
Which part of a chromosome is called nucleolar organizer?
Answer:
The secondary constriction present on the chromatid arms of a chromosome is called nucleolar organizer.

Question 20.
Why is Y chromosome genetically less active?
Answer:
Since Y-chromosome possesses small amount of euchromatin that contains DNA or genes, therefore it is genetically less active.

Question 21.
Why hypertrichosis is called holandric gene?
Answer:
Hypertrichosis is Y linked gene which can be seen only in males, therefore it is called holandric gene.

Question 22.
What is the genetic difference between total colour blindness and red-green colour blindness ?
Answer:
Total colour blindness is due to incomplete sex-linked genes while red-green colour blindness is due to complete sex linkage.

Question 23.
What happens if the gene for production of factor VIII and IX becomes recessive?
Answer:
The person having recessive gene for haemophilia is deficient in clotting factors (VIII or IX) in blood, such person’s blood does not clot and he thus becomes a patient of haemophilia.

Question 24.
What is the cause of Thalassemia?
Answer:
Thalassemia is caused due to deletion or mutation of gene which codes for alpha (α) and beta (β) globin chains, causing abnormal synthesis of haemoglobin. Thus it is a quantitative abnormality of polypeptide globin chain synthesis.

Question 25.
What is monosomy? Give one example of the same.
Answer:
Monosomy is lack of one chromosome from the usual chromosomal complement. Turner’s syndrome is the example of monosomy.

Give Definitions

Question 1.
Factor
Answer:
The unit of heredity which is responsible for the inheritance and expression of a character and which is responsible for the genetic character is called a factor.

Question 2.
Gene
Answer:
The specific segment of DNA or sequence of nucleotides which is responsible for the inheritance and expression of that character is called a gene.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 3.
Alleles or Allelomorphs
Answer:
The two or more alternative forms of a given gene which are present on the identical loci on the homologous chromosomes are called alleles of each other.

Question 4.
Phenotype
Answer:
The external appearance of an individual for any trait is called phenotype for that trait.

Question 5.
Genotype
Answer:
Genetic constitution of an organism with respect to a particular trait is called genotype.

Question 6.
Homologous Chromosomes?
Answer:
The morphologically, genetically and structurally essentially identical chromosomes present in a diploid cell are called homologous chromosomes.

Question 7.
Back cross
Answer:
The cross of Fx progeny with any of the parents, irrespective of being dominant or recessive, is called back cross.

Question 8.
Linkage
Answer:
Linkage is defined as the tendency of the genes to be inherited together because they are present in the same chromosome.

Question 9.
Non-disjunction
Answer:
Non-disjunction is the phenomenon in which chromosomes fail to separate at the time of cell division, resulting in abnormal chromosomal combinations.

Question 10.
Syndrome
Answer:
The appearance of different types of symptoms at the same time in an individual is called a syndrome.

Question 11.
Aneuploidy
Answer:
Addition or deletion of one or two chromosomes in a diploid chromosome set is called aneuploidy.

Distinguish Between

Question 1.
Homozygous and Heterozygous.
Answer:

Homozygous Heterozygous
1. Individuals with similar gene pairs are called homozygous. 1. Individuals with different gene pairs are called heterozygous.
2. Homozygous individuals form only one type of gametes. 2. Heterozygous individuals form more than one type of gametes.
3. Individuals with similar gene pairs TT, tt, RR and rr are homozygous. 3. Individuals with dissimilar gene pairs Tt and Rr are heterozygous.
4. Homozygous are also called pure breed. 4. Heterozygous are referred to as hybrids.

Question 2.
Monohybrid cross and Dihybrid cross.
Answer:

Monohybrid cross Dihybrid cross
1. Crosses involving a single pair of alleles are called monohybrid crosses. 1. Crosses involving two pairs of alleles are called dihybrid crosses.
2. Monohybrid crosses yield a phenotypic ratio of 3 : 1 in the F2 generation. 2. Dihybrid crosses yield a 9 : 3 : 3 : 1 ratio in F2 generation.
3. Monohybrid crosses yield 1 : 2 : 1 genotypic ratio in F2 generation. 3. Dihybrid crosses yield 1 : 1 : 2 : 2 : 4 : 2 : 2 : 1 : 1 genotypic ratio in F2 generation.
4. Application of the law of independent assortment is not applicable in monohybrid crosses. 4. Application of the law of independent assortment is applicable in dihybrid crosses.

Question 3.
Dominant characters and Recessive characters.
Answer:

Dominant characters Recessive characters
1. The characters that are expressed in the F1 generation are called dominant characters. 1. The characters that are not expressed in the F1 generation are called recessive characters. They are prevented from expressing themselves, due to presence of dominant allele.
2. Dominant character is expressed either in homozygous or heterozygous combination. 2. Recessive characters are expressed only when they are in homozygous combination.
3. Dominant characters cannot be masked by recessive characters.
E.g. Round seed and yellow seed are dominant characters in pea plant.
3. Recessive characters are masked by dominant characters.

E.g. Wrinkled seed and green seed are recessive characters in pea plant.

Question 4.
Phenotype and Genotype.
Answer:

Phenotype Genotype
1. Phenotype refers to the outward appearance of an individual such as shape, colour, sex, etc. 1. Genotype refers to the genetic composition of an individual.
2. Phenotype can be observed directly in an individual. 2. Genotype cannot be seen, but can be found out by modern techniques like DNA fingerprinting.
3. Individuals resembling each other may or may not have the same genotype. 3. Individuals possessing the same genotype usually have the same phenotypic expression.
4. The phenotypic ratio obtained in the F2 generation of a monohybrid cross is 3 : 1. 4. The genotypic ratio obtained in the F2 generation of a monohybrid cross is 1 : 2 : 1.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 5.
Incomplete dominance and Co-dominance.
Answer:

Incomplete dominance Co-dominance
1. Incomplete dominance is seen when the phenotypes of the two parents blend together to create a new phenotype for their offspring. 1. Co-dominance is seen when the two parent phenotypes are expressed together in the offspring.
2. Both the genes of an allelomorphic pair express themselves partially in F1 hybrids. 2. Both the genes of an allelomorphic pair express themselves equally in F1 hybrids.
3. In incomplete dominance, a mixture of the alleles in the genotype is seen in the phenotype. 3. In co-dominance, both alleles in the genotype are seen in the phenotype.
4. The phenotypic effect of one allele is more prominent than the other. 4. The phenotypic effect of both the alleles is equally prominent.
5. Blending or intermixing of two alleles can be observed. A white flower and a red flower alleles mix and produce pink flowers.

Example : Pink flowers in Mirabilis jalapa.

5. No intermixing or blending effect of two alleles is observed. The colours don’t mix but are seen in patches.

Example : Roan colour in cattle.

Question 6.
Turner’s syndrome and Klinefelter’s syndrome.
Answer:

Turner’s syndrome Klinefelter’s syndrome
1. Individual with Turner’s syndrome has total 45 chromosomes in each of her cell. 1. Individual with Klinefelter’s syndrome has total 47 chromosomes in each of his cell.
2. Turner’s syndrome is XO female, caused due to monosomy of X-chromosome. 2. Klinefelter’s syndrome is XXY male, caused due to trisomy of X chromosome.
3. The external phenotype is of female. 3. The external phenotype is of male.
4. The stature is short. 4. The stature is tall and thin.
5. Secondary sexual characteristics are not developed in Turner’s syndrome. 5. Secondary sexual characteristics are poorly developed in Klinefelter’s syndrome.

Give Reasons or Explain the Statements

Question 1.
Law of segregation is universally applicable.
Answer:

  1. According to the law of segregation, the members of the allelic pair remain together without mixing with each other.
  2. They segregate or separate when the gametes are formed.
  3. Thus the gametes that are formed receive only one of the two factors.
  4. Now it is known that the organisms are diploid and the gametes produced by them are haploid.
  5. The law of segregation therefore is universally applicable.

Question 2.
Mendel selected garden pea for his breeding experiments.
Answer:
Mendel selected garden pea for his breeding experiments, because:

  1. The pea plants were true breeding varieties.
  2. The pea plants being annual, it was possible to cross and study many generations within a short period.
  3. The pea plants had a number of distinguishable, contrasting characters such as tall habit and dwarf habit, round seed and wrinkled seed.
  4. The pea plants were easy to handle for breeding experiments.

Question 3.
When Mendel crossed a tall pea plant with a dwarf pea plant the offspring obtained from this cross were all tall.
Answer:

  1. The tall habit of the pea plant is dominant over the dwarf habit of the pea plant.
  2. Hence, when Mendel crossed a tall pea plant with a dwarf pea plant, the offspring obtained from this cross were all tall.

Question 4.
A cross between a homozygous tall plant with a homozygous dwarf plant results in two types of tall plants in the F2 generation.
Answer:

  1. A cross between a homozygous tall (TT) and a homozygous dwarf (tt) gives rise to a heterozygous tall (Tt) plant in the F1 generation.
  2. When the F1 plant is selfed (Tt × Tt), it gives rise to three tall plants of which two- are heterozygous (Tt) tall and one is homozygous (TT) tall.
  3. Hence a cross between a homozygous tall plant with a homozygous dwarf plant results in two types of tall plants in the F2 generation.

Question 5.
Possibility of female becoming a haemophilic is extremely rare.
Answer:

  1. Haemophilia is caused due to X-linked recessive gene. Females have double X chromosomes.
  2. Even if she has haemophilic gene on one of her X-chromosome, the dominant gene on other X-chromosome, suppresses its expression. Female therefore, does not become haemophilic.
  3. If she inherits haemophilic gene on both of her X-chromosomes, this combination becomes lethal. Such embryo is aborted. If born, she dies soon. This makes the possibility of female becoming a haemophilic extremely rare.

Question 6.
Human female is referred to as carrier of colour blindness.
Answer:
Human female is referred to as carrier of colour blindness because of the following reasons:

  1. Females possess double X-chromosomes in her gametes.
  2. If one X-chromosome is carrying recessive gene for colour blindness, her other dominant X hides the expression of colour blindness and hence she does not become a patient.
  3. But such female can carry the defective gene to her progeny. Thus she is called carrier of colour-blindness.
  4. A female having one recessive gene on X-chromosome is a carrier female, while a female possessing both recessive genes on both the X-chromosomes will be colour blind which is very rare.

Write Short Notes

Question 1.
Linkage.
Answer:

  1. Linkage is the tendency of genes to be inherited together because they are present in the same chromosome.
  2. All the genes on a chromosome are linked to one another. In the linkage group some of the genes are included.
  3. The number of linkage groups of a particular species corresponds to its haploid number of chromosomes present in the organism.
  4. In human beings, there are 23 linkage groups which correspond to the pairs of chromosomes found in each cell.
  5. Linkage groups can be separated only at the time of crossing over during meiosis. The linkage group can form a new combination of genes after crossing over.
  6. Linkages Eire of two types, viz, complete linkage and incomplete linkage.
  7. Morgan discovered linkage in animals while Bateson and Punnett discovered it in plants.

Question 2.
Multiple alleles.
Answer:

  1. Multiple alleles are more than two alternative alleles of a gene in a population situated on the same locus on a chromosome or its homologue.
  2. Multiple alleles arise by mutations of the wild type of gene. Series of multiple alleles are formed due to several mutations that take place in the wild type of allele. This series show alternative expression.
  3. Different alleles in a series show dominant-recessive relation or may show co-dominance or incomplete dominance among themselves. Among all the wild type is the most dominant one over all other mutant alleles.
  4. In Drosophila, a large number of multiple alleles are known. E.g. The size of wings from normal wings to vestigial wings is due to one allele (vg) in homozygous condition. The normal wing is dominant and wild type while vestigial wing is recessive type.
  5. Human blood groups A, B, AB and O Eire also due to series of multiple alleles.

Question 3.
Autosomal inheritance.
Answer:

  1. Transmission of body characters occurs due to autosomes. They are not concerned with sex determination or sex linkage.
  2. All the body characters from parents are passed on to their offspring through autosomes. This is called autosomal inheritance.
  3. Some autosomal characters are due to dominant genes while some other are due to recessive genes. E.g. Widow’s peak and Huntington’s disease is also autosomal dominant character, etc.
  4. Phenyl ketonuria (PKU), Cystic fibrosis and Sickle-cell anaemia are autosomal recessive traits.

Question 4.
Widow’s peak.
Answer:

  1. Widow’s peak is a prominent ‘V’ shaped hairline on forehead.
  2. It is due to autosomal dominant gene.
  3. Widow’s peak occurs in homozygous dominant (WW) and also heterozygous (Ww) individuals.
  4. Individuals with homozygous recessive (ww) genotype do not have widow’s peak but have a straight hair line.
  5. Both males and females have equal chance of inheritance.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 5.
Environmental sex determination
Answer:

  1. Environmental sex determination is shown by lower organisms such as Bonellia viridis.
  2. In this animal the environmental factors decide the sex of an offspring.
  3. There is extreme sexual dimorphism in this worm. Female is about 10 cm long while male is tiny and parasitic in the reproductive parts of mature female.
  4. If larva is reared in vicinity of mature female then it becomes a male. By settling on the proboscis of mature female, larva becomes parasitic, enters the female’s mouth and then takes permanent shelter in the female uterus. Such males then produce gametes and fertilize the eggs.
  5. If larvae are drifted away from mature female or if they settle on the sea bottom, they develop into females. Thus determination of sex is due to environmental factors.

Question 6.
Y-linked or Holandric genes.
Answer:

  1. Holandric means entirely of male sex. Y-linked genes are called holandric genes because they are located on non-homologous region of Y chromosome.
  2. The Y-linked genes are inherited directly from male to male.
  3. These genes are never seen in females due to lack of Y chromosome in them.
  4. Hyper Mchosis and ichthyosis are examples of holandric genes.
  5. Hypertrichosis means excessive development of hair on pinna of ear. This character is transmitted directly from father to son.
  6. Ichthyosis person with rough skin.

Question 7.
X-body.
Answer:

  1. German biologist, Henking in 1891, was studying spermatogenesis of the squash bug (Anasa tristis).
  2. He noted that 50% of sperms receive the unpaired chromosomes while other 50% sperms do not receive it.
  3. Henking gave a name to this structure as the X-body. He was unable to explain its role in sex determination.
  4. Further investigations by other scientists led to conclusion that the ‘X-body’ of Henking was a chromosome and gave the name as X-Chromosome to X-body.

Question 8.
Thalassemia
Answer:
(1) Thalassemia is an autosomal-eeessive disorder. The synthesis of alpha ciiains are controlled by two genes, (HBA1 and HBA2) on chromosome 16. Beta chain synthesis is controlled by gene HBB located on chromosome 11. Two alpha chains and two beta chains together form four polypeptide chains that make heterotetrameric haemoglobin molecule. But when there is defective gene on either of chromosome 16 or 11, there is quantitative abnormality of polypeptide globin chain synthesis. This results into thalassemia.

(2) Depending upon which chain is affected, thalassemia is classified as, alpha (α) thalassemia and beta (β) thalassemia.

(3) The clinical symptoms of thalassemia are as follows:

  • Pale yellow skin.
  • Anaemia due to inability to synthesize haemoglobin.
  • Slow growth and development.
  • Variation in the shape and size of RBCs.

(4) Patients need regular blood transfusions to cope with the disorder.

Short Answer Questions

Question 1.
Write the statements of three laws of inheritance given by Mendel.
Answer:
(1) Statement of Law of Dominance : When two homozygous individuals with one or more sets of contrasting characters are crossed, the alleles that appear in F1 are dominant and those which do not appear in F1 are recessive.

(2) Statement of Law of Segregation or Law of purity of gametes : When F1 hybrid forms gametes, the alleles segregate from each other and enter in different gametes. The gametes formed are pure because they carry only one either dominant allele or recessive allele each. Due to this the law is also called “Law of purity of gametes”.

(3) Statement of Law of Independent Assortment : When hybrid possessing two (or more) pairs of contrasting alleles forms gametes, these alleles in each pair segregate independently of the other pair.

Question 2.
Why are farmers and gardeners advised to buy new F1 hybrid seeds every year?
Answer:

  1. Farmers use hybrid seeds for agriculture or horticulture. Hybrid seeds are produced by crossing two unrelated parent plants.
  2. Hybrid seed varieties give improved yields and crop vigour to the farmer.
  3. Hybrids are made by crossing two highly inbred ‘parent’ plants. First generation hybrids, however, do not breed true to type, meaning that the seed they set may not grow into crops that are identical to the ‘parent’ plants.
  4. This can result in variations in yield and quality therefore many farmers prefer to buy new hybrid seed each year to ensure consistency in their final product.

Question 3.
What are the main generalizations given after Mendel’s experiments on the pea plant?
Answer:
After the Mendel’s laws of inheritance and his experiments, following generalizations were made:

  1. Single trait is shown due to single gene. Every single gene has two contrasting alleles.
  2. Two alleles are always in interaction in which one is completely dominant while other is completely recessive.
  3. Factors which were later called genes for different traits are always present on different chromosomes. These traits can assort independently of each other.

Question 4.
Mention the types of deviations from Mendel’s finding.
OR
Describe Neo-Mendelism in short.
Answer:
As the science of genetics progressed, many changes were seen from Mendel’s generalizations. These are called as Neo- Mendelism.
The deviations from Mendel’s findings can be categorised under following heads:

  1. Intragenic interactions : These interactions : are seen between the alleles of same gene. e.g. incomplete dominance and co-dominance. They are also seen in multiple allele series of a gene.
  2. Intergenic interactions : Intergenic interactions are between the alleles of different genes present on the same or different chromosomes, e.g. pleiotropy, polygenes, epistasis, supplementary and complementary genes, etc.

Question 5.
Why Drosophila is most suitable organism for genetics experiments?
Answer:
Drosophila is most suitable organism because of the following reasons:

  1. Drosophila cam easily be cultured under laboratory conditions.
  2. Life span of Drosophila is short for about two weeks.
  3. Drosophila has high rate of reproduction and hence newer organisms can be obtained rapidly.

Question 6.
What are the causes of Down’s syndrome?
Answer:

  1. Down’s syndrome is caused due to aneuploidy.
  2. Aneuploidy is due to non-disjunction of chromosome at the time of gamete formation during meiosis. Due to non-disjunction, chromosomes fail to separate.
  3. In addition to a homologous pair of 21st chromosome there is an extra 21st, therefore it is called trisomy (2n + 1) of 21st chromosome.

Question 7.
What are the characteristic symptoms of Down’s syndrome?
Answer:
Symptoms of Down’s syndrome:

  1. Typical facial features.
  2. An epicanthal skin fold, over the inner corner of eyes causing downward slanting eyes.
  3. Typical flat face, rounded flat nose, mouth always open with protruding tongue.
  4. Mental retardation.
  5. Poor skeletal development.
  6. Short stature, relatively small skull and arched palate.
  7. Flat hand with simian crease that runs across the palm.

Question 8.
Write a brief account of Turner’s syndrome.
Answer:

  1. Turner’s syndrome is a genetic disorder caused due to monosomy of X chromosome.
  2. It was first described by H. H. Turner.
  3. Turner’s syndrome is caused due to non-disjunction of sex chromosomes which takes place during gamete formation.
  4. Chromosomal complement of Turner’s syndrome is 44 + XO, having a total of 45 chromosomes.

Symptoms of Turner’s syndrome are as follows:

  1. Female phenotype.
  2. Short stature with webbing of neck.
  3. Low posterior hair line.
  4. Secondary sexual characters fail to develop.
  5. Mental retardation.

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 9.
Write a brief account of Klinefelter’s syndrome.
Answer:

  1. Klinefelter’s syndrome is a genetic disorder caused due to trisomy of X chromosome.
  2. It was first described by Harry Klinefelter.
  3. Klinefelter’s syndrome is caused due to non-disjunction of sex chromosomes which takes place during gamete formation.
  4. Chromosomal complement of Klinefelter’s syndrome is 44+XXY, having a total of 47 chromosomes.

Symptoms of Klinefelter’s syndrome are as follows:

  1. The Klinefelter’s syndrome individuals are tall, thin and eunuchoid.
  2. They are sterile with poorly developed sexual characteristics.
  3. Testes are underdeveloped and small. Spermatogenesis does not take place.
  4. They have subnormal intelligence and show partial mental retardation.

Diagram Based Questions

Question 1.
Give the graphical representation of test cross and back cross.
Answer:
(1) Test cross
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 1
The F2 generation of test cross consists of 50% heterozygous tall plants and 50% homozygous dwarf plants.

(2) Back cross
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 2
F1 crossed back with its dominant parent
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 3

Question 2.
Give a cross for incomplete dominance using a suitable example.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 4
Result:
Genotypic ratio – 1RR : 2 Rr : 11rr
Phenotypic ratio – 1Red : 2 Pink : 1 White

Question 3.
Give a cross of co-dominance using a suitable example.
Answer:
Coat colour in cattle
Red female RR × White male WW
P1 generation : RR × WW
Gametes : R and W
F1 generation all RW Roan coloured
P2 generation RW × RW

R W
R RR RW
W RW WW

Genotypic ratio : 1 RR : 2 RW : 1 WW
Phenotypic ratio : 1 Red : 2 Roan : 1 White

Question 4.
Draw two crosses to show inheritance of colour blindness, (i) A cross between normal female and colour blind male, (ii) A cross of carrier woman with normal man.
Answer:
(i) A cross between normal female and colour-blind male.
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 5

(ii) A cross of carrier female with normal male.
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 6

Question 5.
Draw the following crosses to show inheritance of haemophilia : Normal female with haemophilic male. Show their progeny. If one of their carrier daughters marries a normal male what would be possible genotypes of this generation.
Answer:
(1) Haemophilic male crossed with normal female:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 7

(2) Carrier female crossed with normal male :
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 8

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 6.
Sketch and label structure of X and Y chromosomes.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 9

Question 7.
Give the graphical representation of pleiotropy to show inheritance of Sickle-cell anaemia.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 10

Long Answer Questions

Question 1.
There are 16 possible individuals in F2 generation. Try to find out the phenotypes as well as the genotypic and phenotypic ratios.
Answer:
In the above dihybrid cross there are 4 phenotypes such as yellow round, yellow wrinkled, green round, green wrinkled.
There are 9 different genotypes as follows:
1 : RRYY / 2 : RRYy / 1 : RRyy / 2 : RrYY / 4 : RrYy / 2 : Rryy / 1 : rrYY /2 : rrYy / 1 : rryy

Question 2.
Explain with suitable diagram how test cross is used to find out genotype of dominant plant.
Answer:
Test cross is used to find out the exact genotype by crossing the F1 individual with homozygous recessive one.
E.g. To find out the genotype of unknown violet flower obtained in F1 generation, one can conduct two crosses as follows:
I. Unknown flower considering as RR (homozygous dominant)
RR × rr Homozygous dominant

R R
R Rr Rr
R Rr Rr

II. In such cross, all the flowers will be violet. II. Unknown flower considering as Rr (heterozygous)
Rr × rr Homozygous recessive with heterozygous

R R
R Rr Rr
R Rr Rr

In such cross half the flowers will be violet and half will be white.

Question 3.
Describe briefly Morgan’s Experiments showing linkage and crossing over. (Diagram is not needed)
Answer:
(1) Morgan used Drosophila melanogaster for his experiments.

(2) He carried out several dihybrid cross experiments to study sex-linked genes of Drosophila.

(3) Crosses between yellow-bodied, white-eyed female and brown-bodied, red-eyed males were done in P1 generation. Brown-bodied and red-eyed forms were wild.

(4) Morgan intercrossed their F1 progeny and noted that two genes did not segregate independently of each other and F2 ratio deviated very significantly from Mendelian 9 : 3 : 3 : 1 ratio.

(5) When genes are grouped on the same chromosome, some genes are strongly linked. They show very few recombinations (1.3%).

(6) When genes are loosely linked, i.e. located away from each other on chromosome, they show more (higher) recombinations (37.2%).

(7) For example, the genes for yellow body and white eye were strongly linked and showed only 1.3 per cent recombination (in cross-I).

(8) White-bodied and miniature wings showed 37.2 per cent recombination (in cross-II). Cross I shows crossing over between genes y and w.

(9) Cross II shows crossing over between genes white (w) and miniature wing (m). Here dominant wild type alleles are represented with (+) sign.

(10) Parental combinations occur more due to linkage and new combinations less due to crossing over.

Question 4.
Describe the mechanism of sex determination in human beings with a suitable cross.
Answer:
1. Sex determination in human beings:
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 11
(1) In human beings, the sex is determined with the help of X and Y chromosomes. This chromosomal mechanism of sex determination is of XX-XY type.

(2) In male, the nucleus of each cell contains 46 chromosomes or 23 pairs of chromosomes. Of these 22 pairs are autosomes and one pair of sex chromosomes. Males are thus heteromorphic as they have two different types of sex chromosomes.

(3) Autosomes or somatic chromosomes are responsible for determination of other characters of the body, but not the sex.

(4) In female cells, there are 22 pairs of autosomes and one pair of X chromosomes. Females are thus homomorphic as they have similar sex chromosomes.

(5) Thus the genotypes of female and male are as follows:
Female : 46 chromosomes = 44 autosomes + XX sex chromosomes
Male : 46 chromosomes = 44 autosomes + XY sex chromosomes

(6) During gamete formation, the diploid germ cells in the testes and ovaries undergo meiosis to produce haploid gametes (sperms and eggs). The homologous chromosomes separate and enter into different gametes during this process.

(7) The human male produces two different types of sperms, one containing 22 autosomes and one X chromosome and the other containing 22 autosomes and one Y chromosome. Human males are therefore called heterogametic, i.e. they produce different types of gametes.

(8) The human female produces only one type of eggs containing 22 autosomes and one X chromosome and therefore she is homogametic.

(9) During fertilization, if X containing sperm fertilizes the egg having X chromosome, then a female child with XX chromosomes is conceived.

(10) If Y containing sperm fertilizes the egg having X chromosome then a male child with XY chromosomes is conceived.

(11) The sex of the child thus depends upon the type of sperm fertilizing the egg. The heterogametic parent determines the sex of the child and thus the father is responsible for the determination of the sex of the child and not the mother.
Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 12

Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation

Question 5.
Explain the mechanism of sex determination in case of birds.
Answer:

  1. Sex determination in birds is by ZW-ZZ mechanism.
  2. In birds, males are homogametic while females are heterogametic.
  3. Males produce all similar types of sperms, containing 8 autosomes and ‘Z’ sex chromosome.
  4. Females produce two different types of eggs, one containing 8 autosomes and Z chromosome and the other containing 8 autosomes and W chromosome.
  5. When Z bearing egg is fertilized by a sperm a male offspring is produced. If W bearing egg is fertilized then female offspring is produced.
    Maharashtra Board Class 12 Biology Important Questions Chapter 3 Inheritance and Variation 13

[Note : In domestic fowl chromosome number is 18, with 16 autosomes and two sex chromosomes.]