Laws of Motion Class 9 Science Chapter 1 Questions And Answers Maharashtra Board

Class 9 Science Chapter 1

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 1 Laws of Motion Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 1 Laws of Motion Question Answer Maharashtra Board

Class 9 Science Chapter 1 Laws of Motion Question Answer Maharashtra Board

1. Match the first column with appropriate entries in the second and third columns and remake the table.
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 1

2. Clarify the differences

A. Distance and displacement
Answer:

Distance Displacement
(i) Distance is the length of the actual path travelled by an object.
(ii) It is a scalar quantity.
(iii) It is either equal to or greater than displacement.
(iv) Distance travelled is always positive.
(i) Displacement is the minimum distance between the starting and finishing points.
(ii) It is a vector quantity.
(iii) It is either equal to or less than distance.
(iv) Displacement may be positive or negative or zero.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

B. Uniform and non-uniform motion.
Answer:

Uniform motion Non-uniform motion
(i) If an object covers equal distances in equal intervals of time it is said to be in uniform motion.
(ii) Distance – time graph for uniform motion is a straight line.
(iii) In uniform motion, acceleration is zero.
(i) If an object moves unequal distances in equal intervals of time, its motion is said to be non­uniform.
(ii) Distance – time graph for non-uniform motion is not a straight line.
(iii) In non-uniform motion acceleration is non-zero.

3. Complete the following table.
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 2

4. Complete the sentences and explain them.

a. The minimum distance between the start and finish points of the motion of an object is called the ……….. of the object.
b. Deceleration is ………………………. acceleration
c. When an object is in uniform circular motion, its ………………………. changes at every point.
d. During collision ………………………. remains constant.
e. The working of a rocket depends on Newton’s ………………………. law of motion.

5. Give scientific reasons.

a. When an object falls freely to the ground, its acceleration is uniform.
Answer:

  • When the body falls freely to the ground, there are equal changes in velocity of the body in equal intervals of time.
  • Thus the acceleration of the body is constant, and it possesses uniform acceleration.

b. Even though the magnitudes of action force and reaction force are equal and their directions are opposite, their effects do not get cancelled.
Answer:

  • Action and reaction forces act on different bodies.
  • They don’t act on the same body, hence they cannot cancel each other’s effect.
  • Hence, even though the magnitudes of action force and reaction force are equal, they do not cancel each other.

c. It is easier to stop a tennis ball as compared to a cricket ball, when both are traveling with the same velocity.
Answer:

  • Momentum of an object depends on its mass as well as its velocity.
  • Cricket ball is heavier than a tennis ball. Although they are thrown with the same velocity, cricket ball has more momentum than a tennis ball.
  • The force required to stop a cricket ball is more than a tennis ball.
  • Hence it is easier to stop a tennis ball than a cricket ball moving with same velocity.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

d. The velocity of an object at rest is considered to be uniform.
Answer:

  • When a body is at rest there is no change in velocity.
  • A body with constant velocity is said to be in uniform motion.
  • Hence, the state of rest is an example of uniform motion.

6. Take 5 examples from your surroundings and give an explanation based on Newton’s laws of motion.

7. Solve the following examples.

a) An object moves 18 m in the first 3 s, 22 m in the next 3 s and 14 m in the last 3 s. What is its average speed? (Ans: 6 m/s)
Answer:
Given:
Total distance (d) = 18 + 22 + 14 = 54 m
Total time taken (t) = 3 + 3 + 3 = 9 sec
To find:
Average speed = ?
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 3

The object moves with an average speed of 6 m/s.

b) An object of mass 16 kg is moving with an acceleration of 3 m/s2. Calculate the applied force. If the same force is applied on an object of mass 24 kg, how much will be the acceleration? (Ans: 48 N, 2 m/s2)
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 4

The force acting on the 1 body is 48 N and the acceleration of the 2” body is 2 m/s2

c) A bullet having a mass of 10 g and moving with a speed of 1.5 m/s, penetrates a thick wooden plank of mass 90 g. The plank was initially at rest. The bullet gets embedded in the plank and both move together. Determine their velocity. (Ans: 0.15 m/s)
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 5
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 6
The plank embedded with the bullet moves with a velocity of 0.15 m/s.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

d) A person swims 100 m in the first 40 s, 80 m in the next 40 s and 45 m in the last 20 s. What is the average speed? (Ans: 2.25 m/s2)

Answer:
Given:
Total distance (d) = 100 + 80 + 45 = 225 m
Total time taken (t) = 40 + 40 + 20 = 100 sec
To find:
Average speed =?
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 7
The person swims with an average speed of 2.25 m/s.

Class 9 Science Chapter 1 Laws of Motion Intext Questions and Answers

(i) Who will take less time to reach the school and why?
Answer:
Prashant will take less time as the path followed by him is the shortest.
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 8

(a) Every morning, Swaralee walks round the edge of a circular field having a radius of 100 m. As shown in figure (a), if she starts from the point A and takes one round, how much distance has she walked and what is her displacement?
Answer:
Radius (r) = 100 m
Distance covered = Circumference of the circle
= 2 nr
= 2 x 3.14 x 100
= 628 m
Displacement = 0 m (Shortest distance between initial and final position is zero)

(b) If a car, starting from point P, goes to point Q (see figure 1.9) and then returns to point P, how much distance has it travelled and what is its displacement?
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 9
Answer:
Distance covered = PQ + QP
= 360 + 360
= 720 m
Displacement = 0 m (The shortest distance between initial and final position is zero)

Class 9 Science Chapter 1 Laws of Motion Additional Important Questions and Answers

(A) Choose and write the correct option:

Laws Of Motion Class 9 Questions And Answers Maharashtra Board Question 1.
The displacement that occurs in unit time is called …………….. .
(a) displacement
(b) distance
(c) velocity
(d) acceleration
Answer:
(c) velocity

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Laws Of Motion Class 9 Maharashtra Board Exercise Answers  Question 2
The unit of velocity in the SI system is …………….. .
(a) cm/s
(b) rn/s2
(c) um/s2
(d) rn/s
Answer:
(d) m/s

Laws Of Motion Class 9 Maharashtra Board  Question 3.
v2 = u2 + 2as is the relation between and …………….. .
(a) speed and velocity
(b) distance and acceleration
(c) displacement and velocity
(d) speed and distance
Answer:
(c) displacement and velocity

Class 9 Science Notes Chapter 1 Laws Of Motion Question 4.
…………….. is the relation between displacement and time.
(a) v = u + at
(b) v2 = u2 + 2as
(c) s = ut + 1/2 at2
(d) v = u + 2as
Answer:
(c) s = ut + 1/2 at2

Class 9 Science Chapter 1 Laws Of Motion Question Answer Question 5.
The force necessary to cause an acceleration of 1 m/s2 in an object of mass 1 kg is called …………….. .
(a) 1 dyne
(b) 1 m/s
(c) 1 Newton
(d) 1 cm/s
Answer:
(c) 1 Newton.

9th Science Chapter 1 Laws Of Motion Question 6.
Even if the displacement of an object is zero, the actual distance traversed by it …………….. .
(a) may not be zero.
(b) will be zero
(c) will be constant
(d) will be infinity
Answer:
(a) may not be zero

Question 7.
If the velocity changes by equal amounts in equal time intervals, the object is said to be in …………….. .
(a) uniform acceleration
(b) uniform velocity
(c) non-uniform acceleration
(d) non-uniform motion
Answer:
(a) uniform acceleration

Question 8.
If an object is moving with a uniform velocity …………….. .
(a) its speed remains the same, but direction of motion changes
(b) its speed changes but direction of motion is same
(c) its speed and direction both change
(d) its speed and direction both remain the same
Answer:
(d) its speed and direction both remain the same

Question 9.
is an example of positive acceleration.
(a) A stone is thrown vertically upwards
(b) A stone falls freely towards the earth
(c) Brakes are applied by the truck driver
(d) The train arriving at the station
Answer:
(b) a stone falls freely towards the earth

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 10.
An object continues to remain at rest or in a state of uniform motion along a straight line unless an …………….. acts on it.
(a) internal imbalanced force
(b) external unbalanced force
(c) internal balanced force
(d) external balanced force
Answer:
(b) external unbalanced force

Question 11.
The …………….. is proportional to the applied force and it occurs in the direction of the force.
(a) change of momentum
(b) rate of change of velocity
(c) change of velocity
(d) rate of change of momentum
Answer:
(d) rate of change of momentum

Question 12.
…………….. is a relative concept.
(a) Motion
(b) Direction
(c) Power
(d) Acceleration
Answer:
(a) Motion

Question 13.
A body is said to be in motion if it changes its …………….. with respect to its surroundings.
(a) position
(b) direction
(c) speed
(d) force
Answer:
(a) position

Question 14.
A body is said to be at …………….. if it does not change its position with respect to its surroundings.
(a) Motion
(b) Rest
(c) Gravity
(d) Force
Answer:
(b) Rest

Question 15.
…………….. is the length of the actual path travelled by an object in motion while going from one point to another.
(a) Distance
(b) Displacement
(c) Speed
(d) Velocity
Answer:
(a) Distance

Question 16.
The distance covered by a body in unit time is called its …………….. .
(a) velocity
(b) speed
(c) displacement
(d) rest
Answer:
(b) speed

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 17.
S.I. unit of speed is …………….. and in C.G.S unit it is …………….. .
(a) m/s and cm/s
(b) km/s and cm/s
(c) m/s and mm/s
(d) m/s and nm/s
Answer:
(a) m/s , cm/s

Question 18.
The distance travelled in a particular direction by an object in unit time is called its …………….. .
(a) velocity
(b) speed
(c) displacement
(d) rest
Answer:
(a) velocity

Question 19.
Units of speed and velocity are the
(a) Same
(b) Different
(c) Greater than each other
(d) Unequal
Answer:
(a) same

Question 20.
…………….. . is related to distance, while …………….. is related to displacement.
(a) Gravity and magnetism
(b) Speed and force
(c) Speed and velocity
(d) Motion and rest
Answer:
(c) Speed, velocity

Question 21.
If an object covers equal distances in equal time intervals, it is said to be moving with …………….. speed.
(a) Uniform
(b) Non uniform
(c) Changing
(d) Random
Answer:
(a) uniform

Question 22.
If an object covers unequal distances in equal time Intervals, it is said to be moving with speed.
(a) Uniform
(b) Non uniform
(c) Changing
(d) Random
Answer:
(b) non uniform

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 23.
The rate of change of velocity is called
(a) Speed
(b) Acceleration
(c) Velocity
(d) Rest
Answer:
(b) acceleration

Question 24.
Speed of light in dry air is …………….. m/s.
(a) 3 x 107
(b) 3 x 108
(c) 3 x 109
(d) 3 x 103
Answer:
(b) 3 x 108

Question 25.
When velocity of a body increases, its acceleration is …………….. .
(a) Negative
(b) Zero
(c) Positive
(d) Equal
Answer:
(c) positive

Question 26.
When velocity of a body decreases, its acceleration is …………….. .
(a) Negative
(b) Zero
(c) Positive
(d) Equal
Answer:
(a) negative

Question 27.
Negative acceleration is also called or
(a) Deceleration or retardation
(b) Deceleration or acceleration
(c) acceleration or retardation
(d) Zero
Answer:
(a) deceleration or retardation

Question 28.
In case of motion, object travels equal …………….. in equal intervals of time.
(a) Uniform, distance
(b) Non-Uniform, distance
(c) Uniform, displacement
(d) Uniform, displacement
Answer:
(a) uniform, distances

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 29.
Motion of an object was studied by …………….. .
(a) Sir Albert Einstein
(b) Sir Thomas Edison
(c) Sir Isaac Newton
(d) Sir Ravindranath Tagore
Answer:
(c) Sir Issac Newton

Question 30.
When an object moves in a circular path with uniform speed, its motion is …………….. motion.
(a) Non uniform circular
(b) Random circular
(c) Uniform circular
(d) Uniform linear
Answer:
(c) uniform circular

Question 31.
When a coin moves along a circular path, the direction of its motion at every point is …………….. .
(a) Circular
(b) Concave
(c) Tangential
(d) Convex
Answer:
(c) tangential

Question 32.
For all uniformly accelerated motions, the velocity-time graph is a …………….. .
(a) Curved line
(b) Straight line
(c) Negative line
(d) Positive line
Answer:
(b) straight line

Question 33.
In the distance-time graph, the slope of the straight line indicates …………….. .
(a) Acceleration
(b) Velocity
(c) Speed
(d) Rest
Answer:
(b) velocity

Question 34.
The first equation of motion gives relation between …………….. and time.
(a) Acceleration
(b) Velocity
(c) Speed
(d) Rest
Answer:
(b) velocity

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 35.
Newton’s first law explains the phenomenon of
(a) Rest
(b) Inertia
(c) Speed
(d) Velocity
Answer:
(b) inertia

Question 36.
…………….. cause a change in the state of an object at rest or in uniform motion.
(a) Balanced forces
(b) Zero forces
(c) Unbalanced forces
(d) None of them
Answer:
(c) Unbalanced forces

Question 37.
To describe an object’s momentum, we must specify its …………….. and …………….. .
(a) Mass and displacement
(b) Mass and direction
(c) Mass and velocity
(d) Mass and acceleration
Answer:
(c) mass and velocity

Question 38.
…………….. is the product of mass and velocity of an object.
(a) Speed
(b) Acceleration
(c) Momentum
(d) Force
Answer:
(c) Momentum

Question 39.
The rate of change of momentum is proportional to the applied …………….. .
(a) Balanced force
(b) Unbalanced force
(c) Mass
(d) Velocity
Answer:
(b) unbalanced force

Question 40.
S.I. unit of momentum is
(a) kg cm/s
(b) kg m/s
(c) gm/s
(d) m/s
Answer:
(b) kg m/s

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 41.
…………….. is always conserved in a collision.
(a) Force
(b) Power
(c) Speed
(d) Total momentum
Answer:
(d) Total momentum

Question 42.
When a bullet is fired from the gun, the gun moves in backward direction. This motion is called as …………….. .
(a) Momentum
(b) Velocity
(c) Acceleration
(d) Recoil
Answer:
(d) Recoil

Question 43.
In CGS system, the unit of force is …………….. .
(а) Newton
(b) Watt
(c) Horse power
(d) Dyne
Answer:
(d) Dyne.

 (B) 1. Find the odd man out:

Question 1.
Displacement, Force, Momentum, Mass
Answer:
Mass

Question 2.
Speed, Power, Energy, Acceleration
Answer:
Acceleration

Question 3.
Newton’s 1st law, Newton’s 2nd law, Newton’s 3rd law, Kepler’s laws of motion
Answer:
Newton’s 3rd law

(B) 2. Find out the correlation

Question 1.
Speed zero: Body at rest :: Negative acceleration : Retardation
Answer:
Retardation

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 2.
Displacement : Vector quantity :: Distance : Scalar quantity
Answer:
Vector quantity

Question 3.
When body comes to rest at the end of the motion : Final velocity is zero :: When body is at rest at the starting of motion : Initial velocity is zero
Answer:
Initial velocity is zero

Question 4.
Uniform circular motion: Displacement is zero :: Uniform velocity : Acceleration is zero
Answer:
Displacement is zero

Question 5.
Inertia : Newton’s 1st law :: Rate of change of momentum : Newton’s 2nd law
Answer:
Newton’s 2nd law

Question 6.
Balanced force : Body at rest :: Force equation : Mass x acceleration
Answer:
Body at rest

(B) 3. Distinguish between:

Question 1.
Positive acceleration and Negative acceleration
Answer:

Positive acceleration Negative acceleration
(i) When the velocity of a body increases, acceleration is said to be positive acceleration. (i) When the velocity of a body decreases, acceleration is said to be negative acceleration.

Question 2.
Scalar quantity and Vector quantity
Answer:

Scalar quantity Vector quantity
(i) Scalar quantities are physical quantities having magnitude only. (i) Vector quantities are physical quantities having both magnitude and direction.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 3.
Balanced force and Unbalanced force
Answer:

Balanced force Unbalanced force
(i) Balanced force keeps the body at rest. (i) Balanced force keeps the body at rest.

(B) 4. State whether the following statements are true or false:

Question 1.
The velocity of a body is given by the distance covered by it in unit time in a given direction.
Answer:
True

Question 2.
Displacement is a scalar quantity.
Answer:
False

Question 3.
Uniform acceleration means that the body is moving with a uniform velocity.
Answer:
False

Question 4.
The direction of acceleration can be opposite to that of velocity.
Answer:
True

Question 5.
Work is a vector quantity.
Answer:
False

Question 6.
Displacement is always greater than distance.
Answer:
False

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 7.
The distance and displacement are equal only if, motion is along a straight path.
Answer:
True

Question 8.
If an object experiences acceleration, a force is acting on it.
Answer:
True

Question 9.
A train pulling out from a station is in uniform motion.
Answer:
False

Question 10.
If a bus in motion is suddenly stopped, the passengers fall backwards.
Answer:
False

Question 11.
If a single force is acting on an object, it will always accelerate.
Answer:
True

Question 12.
In circular motion, direction of motion is tangential.
Answer:
True

Question 13.
The inertia of a body is measured in terms of its mass.
Answer:
True

(B) 5. Name the following:

Question 1.
The scientist who summarized motion in a set of equations of motion.
Answer:
Isaac Newton

Question 2.
Motion of an object along a circular path with uniform speed.
Answer:
Uniform circular motion

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 3.
What is the backward motion of the gun called?
Answer:
Recoil

Question 4.
The motion in which the object covers equal distance in equal intervals of time.
Answer:
Uniform motion

Question 5.
S. I. unit of acceleration.
Answer:
m/s2

Question 6.
CGS unit of momentum
Answer:
g cm/s

(B) 6. Answer the following in one sentence:

Question 1.
When is acceleration said to be positive?
Answer:
When the velocity of a body increases, acceleration is said to be positive acceleration.

Question 2.
What is negative acceleration?
Answer:
When the velocity of a body decreases, acceleration is said to be negative acceleration.

Question 3.
What is the direction of velocity of an object performing uniform circular motion?
Answer:
The direction of velocity is along the tangential direction to its position.

Question 4.
Give the mathematical expression used to determine velocity of an object moving with uniform circular motion.
Answer:
is the expression used to determine velocity of a body moving with uniform circular motion.

Question 5.
What kind of force keeps the body at rest?
Answer:
Balanced force keeps the body at rest.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 6.
Which law of motion gives the measure of force?
Answer:
Newton’s second law of motion gives the measure of force.

Question 7.
What are vectors and scalars?
Answer:
Scalars are physical quantities having magnitude only whereas, vectors are physical quantities having both magnitude and direction.

Question 8.
Which of the quantities distance, speed, velocity, time and displacement are scalars and which are vectors?
Answer:
Distance, speed and time are scalars displacement and velocity are vectors.

Give formula:

Question 1.
Force =
Answer:
Mass x Acceleration = ma

Question 2.
Final velocity (v) =
Answer:
Initial Velocity + (Acceleration x Time) = u + at

Question 3.
Displacement (s) =
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 10

Question 4.
Final velocity2 (v2) =
Answer:
Initial Velocity2 + 2 x Acceleration x Displacement = u2 + 2as

Question 5.
velocity of an object moving with uniform circular motion =
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 11

Give scientific reasons:

Question 5.
Motion is relative.
Answer:

  • The motion of an object depends on the observer, hence a body may appear moving for one person and at the same time at rest for another one.
  • Hence, motion is relative.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 6.
Newton’s first law of motion is called as law of inertia.
OR
Heavier objects offer more inertia.
Answer:

  • Inertia is related to the mass of the object.
  • As mass is the quantity of matter in a body, we need to exert more force to push a heavier body.
  • Hence heavier objects offer more inertia.
  • As the same property is described by Newton’s first law of motion, it is called as Law of Inertia.

Question 7.
The launching of a rocket is based on Newton’s third law of motion.
Answer:

  • Newton’s third law of motion states that ‘Every action force has an equal and opposite reaction force which acts simultaneously.’
  • When the fuel in a rocket is ignited, it bums as a result of chemical reaction.
  • The exhaust gases escape with a great force in the backward direction.
  • It exerts an equal and opposite reaction force on the rocket, due to which the rocket moves in the forward direction.
  • Thus, the principle of launching of rocket is based on Newton’s third law of motion.

Solve the following numerical:

Question 1.
An athlete is running on a circular track. He runs a distance of 400 m in 25 s before returning to his original position. What is his average speed and velocity?
Answer:
Given:
Total distance travelled = 400 m
Total displacement = 0, as he returns to his original position.
Total time = 25 seconds.
To find:
Average speed = ?
Average velocity = ?
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 12
The athlete runs at an average speed of 16 m/s and velocity 0 m/s.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 2.
An aeroplane taxies on the runway for 30 s with an acceleration of 3.2 m/s2 before taking off. How much distance would it have covered on the runway?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 13
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 14
The distance covered on the runway is 1440 m.

Question 3.
A kangaroo can jump 2.5 m vertically. What must be the initial velocity of the kangaroo?
Answer:
Given:
a = 9.8 m/s2, s = 2.5 m, v = 0,
To find:
u = ?
Formula:
v2 = u2 + 2as
Solution:
v2 = u2 + 2as
02 = u2 + 2 x (-9.8) (2.5) : (Negative sign is used as the acceleration is in the direction opposite to that of velocity.)
0= u2 – 49 u2
= 49 u
= 7 m/s

The initial velocity of the kangaroo must be 7 m/s.

Question 4.
A motorboat starts from rest and moves with uniform acceleration, if it attains the velocity of 15 m/s in 5s, calculate the acceleration and the distance travelled in that time.
Answer:
Given:
Initial velocity, u = 0
Final velocity, v = 15 m/s time, t = 5 s.
To find:
Acceleration (a) = ?
Distance (s) = ?
Solution:
From the first equation of motion
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 15
From the second equation of motion, the distance covered will be
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 16
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 17

The acceleration is 3 m/s2 and distance travelled is 37.5 m.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 5.
The mass of a cannon is 500 kg and it recoils with a speed of 0.25 m/s. What is the momentum of the cannon?
Answer:
Given:
mass of the cannon = 50 kg
recoil speed = 0.25 m/s
To find:
Momentum = ?
Formula:
Momentum = m x v
Solution:
Momentum = m x v
= 500 x 0.25
= 125 kg m/s

The momentum of cannon is 125 kg m/s

Answer the following in short:

Question 1.
Explain the three different ways to change the velocity.
Answer:
As velocity is related to speed and direction, it can be changed by :

  • Changing the speed while keeping the direction constant.
  • Changing direction while keeping speed constant.
  • Changing both speed as well as direction of motion.

Question 2.
Explain what is positive, negative and zero acceleration.
Answer:

  • Positive Acceleration: When the velocity of an object increases, the acceleration is positive. In this case, the acceleration is in the direction of velocity.
  • Negative Acceleration: When the velocity of an object decreases with time, it has negative acceleration. Negative acceleration is also called deceleration. Its direction is opposite to the direction of velocity.
  • Zero Acceleration: If the velocity of the object does not change with time, it has zero acceleration.

Question 3.
What inference do we draw from the velocity-time graph for a uniformly accelerated motion?
Answer:

  • From velocity-time graph we can infer whether velocity changes by equal amounts in equal intervals of time or not.
  • Thus, for all uniformly accelerated motion, the velocity – time graph is a straight line and slope of the line gives the acceleration.
  • For non-uniformly accelerated motion, velocity-time graph can have any shape according to variation in velocity with respect to time.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 4.
State the three equations of motion and give the relationship explained by them.
Answer:

  • v = u + at: This is the relation between velocity and time.
  • s = ut + 1/2 at2 : This is the relation between displacement and time
  • v2 = u2 + 2as : This is the relation between displacement and velocity.

Question 5.
What are the implications of Newton’s Third Law of motion?
Answer:

  • Action and reaction are terms that express force.
  • These forces act in pairs. One force cannot exist by itself.
  • Action and reaction forces act simultaneously.
  • Action and reaction forces act on different objects. They do not act on the same object and hence cannot cancel each other’s effect.

Question 6.
Explain recoil and recoil velocity. Derive its expression.
Answer:

  • Let us consider the example of a bullet fired from a gun. When a bullet of mass nij is fired from a gun of mass m2, its velocity becomes v2, and its momentum becomes m^. Before firing the bullet, both the gun and the bullet are at rest. Hence, total initial momentum is zero.
  • According to Newton’s third law of motion, the total final momentum also has to be zero. Thus, the forward-moving bullet causes the gun to move backward after firing.
  • This backward motion of the gun is called its recoil. The velocity of recoil, v2 is such that,
    Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 19

Complete the flow chart:

(1) Types of force and their effects
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 20

(2) Newton’s laws
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 21

Distinguish between:

Question 1.
Speed and velocity.
Answer:

Speed Velocity
(i) Speed is the distance covered by a body in unit time.
(ii) It is a scalar quantity.
(iii) Speed is equal to or greater than velocity.
(iv) Speed = distance/time
(v) It is always positive or zero but never negative.
(i) The displacement that occurs in unit time is called velocity.
(ii) It is a vector quantity.
(iii) Velocity is equal to or less than speed.
(iv) Velocity = displacement/time
(v) It may be positive, Zero or negative.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 2.
Balanced force and unbalanced force.
Answer:

Balanced force Unbalanced force
(i) Two equal forces applied on a body in the opposite direction.
(ii) This force does not change the state of rest or the state of uniform motion
(i) Two unequal forces applied on a body.
(ii) This force can change the state of rest or the state of uniform motion of a body in a straight line.

Give examples:

Question 1.
Scalar quantities
Answer:
Time, Volume, Speed, Mass, Temperature, Distance, Entropy, Energy, and Work

Question 2.
Vector quantities
Answer:
Acceleration, Velocity, Momentum, Force, and Weight

Answer the following questions:

Observe the figure and answer the questions:

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 22

(a) Measure the distance between points A and B in different ways as shown in figure (I).
Answer:
Distances measured may be of different lengths depending on the path taken.

(b) Now measure the distance along the dotted line. Which distance is correct according to you and why?
Answer:
Dotted line shows the shortest way of reaching from A to B.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

(c) Observe the following figures. If you increase the number of sides of the polygon and make it infinite, how many times will you have to change the direction? What will be the shape of the path?
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 23
Answer:
If we increase the number of sides of the polygon and make it infinite, then we will have to change the direction an infinite number of times. The shape of the path thus obtained will be a circle.

Observe the figure and answer the questions

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 24

Question 1.
What will be the effect on the velocity of the motorcycle if its speed increases or decreases, but its direction remains unchanged?
Answer:
If the speed is increased the velocity of the motorcycle will increase and if the speed is decreased the velocity of the motorcycle will decrease

Question 2.
In case of a turning on the road, will the velocity and speed be same?
Answer:
As speed is scalar quantity while velocity is vector quantity so by turning velocity will change while speed remains same

Question 3.
If, on a turning, we change the direction as well as the speed of the motorcycle, what will be the effect on its velocity?
Answer:
Its velocity will change because velocity depends on speed as well as direction and here both speed and direction are changed.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Numerical:

Question 1.
A person travels a distance of 72 km in 4 hours. Calculate average speed in m/s.
Answer:
Given :
Total distance (d)
= 72 km
= 72×1000
= 72000 m

Total time taken (t)
= 4 hours
= 4 x 3600 (v lhr = 3600 sec)
= 14400 s

To find:
Average speed = ?
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 25
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 26
The person travels with average speed of 5 m/s

Question 2.
balls have masses of 50 gm and 100 gm and they are moving along the same line in the same direction with velocities of 3 m/s and 1.5 m/s respectively. They collide with each other and after the collision, the first ball moves with a velocity of 2.5 m/s. Calculate the velocity of the other ball after collision.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 27
Final velocity of second ball after collision is 1.75 m/s.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Write laws and explain write implications:

Question 1.
Newton’s third law of motion
Answer:
‘Every action force has an equal and opposite reaction force which acts simultaneously’.

  • Action and reaction are terms that express force.
  • These forces act in pairs. One force cannot exist by itself.
  • Action and reaction forces act simultaneously.
  • Action and reaction forces act on different objects. They do not act on the same object and hence cannot cancel each other’s effect.

Question 2.
Explain Newton’s second law of motion and derive the formula.
Answer:
Newton’s second law explains about change in momentum. It states that The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the force.

Suppose an object of mass ‘m’ has an initial velocity ‘u. When a force ‘F’ is applied in direction of its velocity for time ‘t’, its velocity becomes ‘y’. Then, the total initial momentum of the body = ‘mu’. Its final momentum after time t = ‘mv’.

So, the rate of change of momentum
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 28

Hence by Newtons second la of motion, 4he rate of change of momentum is proportional to the applied force.
∴ ma ∝ F
∴ F ∝ ma
∴ F ∝ kma (k = Constant of proportionaLity and value is 1).
∴ F = ma

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 3.
State the law of conservation of momentum and derive the formula.
Answer:

Let mass of object A and B be m1 and m1 respectively
Let their initial velocity be u1 and u2 Let their final velocity be v1 and v2

We know,
P = mv
Let their initial momentum be m1u1 and m2 u2
Let their final momentum be m1v1 and m2v2

Total initial momentum = (m1u1 + m2u2)
Total final momentum = (m1v1 + m2v2)

If F2 is the force that acts on object B,
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 29

i.e. The magnitude of total of total final momentum = the magnitude of total initial momentum

Complete the paragraph:

Question 1.
Moving Objects
Answer:
‘Distance’ is the length of the actual path travelled by an object in motion while going from one point to another, whereas displacement is the minimum distance between the starting and finishing points. The displacement that occurs in unit time is called velocity. The units of speed and velocity are the same. In the SI system, the unit is m/s while in the CGS system; it is cm/s. Speed is related to distance while velocity is related to the displacement. If the motion is along a straight line, the values of speed and velocity are the same, otherwise they can be different. The first scientist to measure speed as the distance /time was Galileo. The speed of sound in dry air is 343.2 m/s while the speed of light is about 3 x 108 m/s. The speed of revolution of the earth around the sun is about 29770 m/s.

Question 2.
Types of motion
Answer:
If an object covers unequal distances in equal time intervals, it is said to be moving with non-uniform speed. For example the motion of a vehicle being driven through heavy traffic. If an object covers equal distances in equal time intervals, it is said to be moving with uniform speed. The rate of change of velocity is called acceleration. If the velocity changes by equal amounts in equal time intervals, the object is said to be in uniform acceleration. If the velocity changes by unequal amounts in equal time intervals, the object are said to be non-uniform acceleration. The speed of the tip is constant, but the direction of its displacement and therefore, its velocity is constantly changing. As the tip is moving along a circular path, its motion is called uniform circular motion.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 3.
Newton’s laws and conservation of momentum
Answer:
An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced force acts on it. The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the force. Momentum has magnitude as well as direction. Its direction is the same as that of velocity. In SI system, the unit of momentum is kg m/s, while in CGS system, it is g cm/s. If an unbalanced force applied on an object causes a change in the velocity of the object, then it also causes a change in its momentum.

The force necessary to cause a change in the momentum of an object depends upon the rate of change of momentum. Every action force has an equal and opposite reaction force which acts simultaneously. As the mass of the gun is much higher than the mass of the bullet, the velocity of the gun is much smaller than the velocity of the bullet. The magnitude of the momentum of the bullet and that of the gun are equal and their directions are opposite. Thus, the total momentum is constant. Total momentum is also constant during the launch of a rocket.

Answer the following in detail:

Question 1.
What is speed? State its units and types. Explain instantaneous speed and average speed.
Answer:

The speed of a body is the distance travelled in unit time. The units of speed in CGS system is cm/s and in SI system is m/s.
There are two types of speed :

    • Uniform speed : When a body covers equal distance in equal intervals of time throughout its motion, it is said to have uniform speed.
    • Non-uniform or variable speed: When a body covers unequal distance in equal intervals of time it is said to have non-uniform speed.

The speed of the body at any instant is called instantaneous speed. Average speed is the ratio of total distance covered to total time taken.

Question 2.
What is velocity? State its units and types.
Answer:

The velocity of a body is the distance travelled by a body in a particular direction in unit time. Thus, rate of change of displacement is called velocity.

v = s/t
where: s = displacement; t = time taken; v = velocity
(MKS unit: m/s CGS unit: cm/s)
There are two types of velocities :

    • Uniform velocity: If there is equal displacement taking place in equal intervals of time, it is uniform velocity.
    • Non-uniform velocity or variable velocity: If there is unequal displacement in equal intervals of time, it is non-uniform velocity.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 3.
What is acceleration? State its units and types.
Answer:
(i) Acceleration is a rate of change in velocity. It is a vector quantity, \(a=\frac{v-u}{t}\)
where : v = final velocity; u = initial velocity;
a = acceleration
Units of acceleration in SI system is m/s2 and CGS system is cm/s2.

(ii) Types of acceleration: .
(a) Uniform acceleration : If the change in velocity is equal in equal intervals of time, the acceleration is uniform acceleration.
(b) Non-uniform acceleration : If the change in velocity is unequal in equal intervals of time, the acceleration is a non-uniform acceleration.

(iii) Kinds of acceleration:
Positive acceleration : When the velocity of an object goes on increasing, it is said to have Positive acceleration.
Negative acceleration : When the velocity of an object goes on decreasing, it is said to have negative acceleration or retardation or deceleration.
Zero acceleration : If the velocity of the object does not change with time, it has zero acceleration.

Question 4.
Explain Newton’s second law of motion and derive the formula.
Answer:
Newton’s second law explains about change in momentum. It states that ‘The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the force.’

Suppose an object of mass’m’ has an initial velocity ‘u’. When a force ‘F’ is applied in the direction of its velocity for time’t’, its velocity becomes ‘v’. Then, the total initial momentum of the body = ‘mu’. Its final momentum after time t = ‘mv’.

So, the rate of change of momentum
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 30

Hence by Newton’s second law of motion, the rate of change of momentum is proportional to the applied force.
∴ ma ∝ F
∴ F ∝ ma
∴ F = kma (k = Constant of proportionality and value is 1).
∴ F = ma

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

Question 5.
State the law of conservation of momentum and derive the formula.
Answer:
(i) Let mass of objects A and B be and m2 respectively
Let their initial velocity be m1 and u2
Let their final velocity be V1 and v2

(ii) We know,
P = mv
Let their initial momentum be m1u1 and m2u2
Let their final momentum be m1v1 and m2v2

(iii) Total initial momentum = (m1u1 + m2u2)
Total final momentum = (m1v1 + m2v2)

(iv) If F2 is the force that acts on object B,
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 31
i.e. The magnitude of total final momentum = the magnitude of total initial momentum.

Question 6.
Obtain the equations of motion by graphical method:

(a) Equation for velocity-time relation.
Answer:
Velocity-time graph: shows the change in velocity with time of a uniformly accelerated object. The object starts from the point D in the graph with velocity v. Its velocity keeps increasing and after time t it reaches the point B on the graph.

The initial velocity of the object = u = OD
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 32

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

(b) Equation for displacement-time relation.
Answer:
Suppose that an object is in uniform acceleration ‘a’ and it has covered the distance ‘s’ within time’t’. From the graph the distance covered by the object during time’t’ is given by the area of quadrangle DOEB.
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 33

(c) Equation for displacement-velocity relation.
Answer:
We can determine the distance covered by the object in time t from the area of the quadrangle DOEB. DOEB is a trapezium. So we can use the formula for its area.
Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion 34
This is Newton’s third equation of motion.

Read the paragraph and answer the questions:

1. Paragraph

(a) When a bat strikes a ball, the ball exerts an equal and opposite force on the bat. The force : acting on the ball projects it with high velocity, Due to the large mass of bat compared to ball, reaction force on the bat slows down the bat’s motion.
(b) When a bullet is fired from a gun, the gun exerts a force on the bullet in the forward direction. This is the action force. The bullet also exerts an equal force on the gun in the backward direction. This is the reaction force. Due to the large mass of the gun, it j moves only a little distance backward. This backward movement of the gun is called the recoil of the gun.
(c) In a rocket, burning fuel creates a push on the front of the rocket pushing it forward. This creates an equal and opposite push on the exhaust gas backwards.

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

(i) Which of Newton’s law examples are given here?
Answer:
Newton’s 3rd law is stated by the above example

(ii) When a rifle is fired it is pushed back this movement is called what?
Answer:
Movement of rifle getting pushed back after firing is called recoil.

(iii) What does the ball acquire after it gets velocity?
Answer:
The ball acquires momentum after it gets velocity.

(iv) State newton’s 1st law of motion
Answer:
An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced force acts on it.

(v) Which force is required to produce motion in an object?
Answer:
Unbalanced force is required to produce motion in an object.

2. Paragraph 2
constant velocity. If that velocity is zero, then the object remains at rest. If an external force is applied, the velocity will change because of the force. The second law explains how the velocity of an object changes when it is subjected to an external force. The law defines a force to be equal to changes in momentum (mass times velocity) per change in time. Newton also developed the calculus of mathematics, and the “changes” “expressed in j the second law are most accurately defined in differential forms. (Calculus can also be used to determine the velocity and location variations experienced by an object subjected to an external force.) For an object with a constant mass the second law states that the force F is the product of an objects mass and its acceleration a:
F = m * a
For an external applied force the change in velocity depends on the mass of the object.

A force will cause in velocity; and likewise, a change in velocity will generate a force. The equation works both ways.
The third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal force on object: A. Notice that the forces are exerted on different objects. The third law can be used to explain the generation of lift by a wing and the production of thrust by a jet engine

(i) A chalk kept on the table remains in the position of rest until moved by the teacher. Which law of motion is followed in this situation?
Answer:
Newton’s 1st law is followed in this situation e

(ii) What will happen to momentum if the mass and acceleration both are doubled?
Answer:
If the mass and acceleration both are doubled then the momentum will be increased four times

(iii) What will happen to momentum if the mass and acceleration both are halved?
Answer:
If the mass and acceleration both are doubled then the momentum will be decreased one four times

(iv) A moving stone filled truck collides with a moving car coming from opposite direction. Why is it observed that only the car is pushed backward?
Answer:
A stone filled truck as more mass than a car hence it has more momentum thus it is observed that only the car is pushed backward

Maharashtra Board Class 9 Science Solutions Chapter 1 Laws of Motion

(v) What will happen to the force if the jet engines do not produce enough thrust to push the aeroplane in mid-air?
Answer:
If the jet engines do not produce enough thrust to push the aeroplane in mid-air the force of aeroplane and resistance of air will become balanced thus putting the aeroplane at rest which will result in a crash.

9th Std Science Questions And Answers:

Carbon: An Important Element Class 9 Science Chapter 13 Questions And Answers Maharashtra Board

Class 9 Science Chapter 13

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 13 Carbon: An Important Element Question Answer Maharashtra Board

Class 9 Science Chapter 13 Carbon: An Important Element Question Answer Maharashtra Board

1. Select the proper option and complete the statements
(single, all, double, ionic, carbon, give and take, hydrogen, multiple, share, most, covalent)

a. A carbon atom forms a ….……………. bond with other atoms. In this bond the two atoms ….electrons.
Answer:
covalent, share

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

b. All the carbon bonds in a saturated hydrocarbon ….……………. electrons.
Answer:
share

c. At least one carbon bond in an unsaturated hydrocarbon is ….…………… .
Answer:
multiple

d. ….……………. is the essential element in all the organic compounds.
Answer:
Carbon

e. The element hydrogen is present in ….. organic compound.
Answer:
all

2. Answer the following questions

a. Why are carbon and its compounds used as fuels?

Answer:
(i) The name ‘carbon’ is derived from Latin word ‘carbo’meaning coal. In the earth’s crust, carbon is present to an extent of approximately 0.27% in the form of carbonate, coal and petroleum. One of the non-crystalline and amorphous form of carbon is coal. Coal is a fossil fuel.

(ii) Peat, lignite, bituminous and anthracite are the four types of coal in the increasing order of their carbon content and heat produced respectively. Charcoal and coke are the other amorphous forms of carbon used as fuel.

(iii) Compounds of carbon such as hydrocarbons consist of carbon and hydrogen and they are easily combustible. For example, methane (CH4) which occurs in natural gas is highly inflammable. It bums by reacting with oxygen to give a bluish flame. In this reaction, 213 Kcal/mol of heat is given out. Methane bums completely.

Chemical reaction:
CH4 + 2O2 → CO2 + 2H2O + Heat

(iv) Thus when hydrocarbons are burnt in air, large amount of heat is evolved with formation of carbon dioxide (CO2) and water (H2O). Due to evolution of heat on combustion, carbon and its compounds are used as fuels.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

b. In which compound forms does carbon occur?
Answer:
Carbon in its combined state exists as various compounds such as:

  • Carbon dioxide and in the form of carbonates such as calcium carbonate, marble, calamine (ZnCO3).
  • Fossil fuel – coal, petroleum, natural gas.
  • Carbonaceous nutrients – carbohydrates, proteins, fats.
  • Natural fibres – cotton, wool, silk.
  • Hydrocarbons – compound of carbon and hydrogen.

c. Write the uses of the diamond.
Answer:
Uses of diamonds are:

  • Diamonds are used in glass cutting and rock drilling machines.
  • Diamonds are used in ornaments.
  • Diamond knives are used in the eye surgery.
  • Diamond dust is used for polishing other diamonds.
  • Diamond is used to make windows giving protection from radiation in space and in artificial satellites.

3. Explain the difference:

a. Diamond and graphite.

Answer:

Diamond  Graphite
(i) Diamond is a brilliant, hard and crystalline allotrope of carbon. (i) Graphite is a black, soft, brittle and slippery crystalline allotrope of carbon.
(ii) In diamonds, every carbon atom is bonded to four neighbouring atoms by covalent bonds forming tetragonal three dimensional structure which makes it very hard. (ii) In graphite, every carbon atom is bonded to three other carbon atoms by covalent bonds in such a way that a hexagonal layered structure is formed. A graphite crystal is made of many such layers of carbon atoms. These layers slip over each other on applying pressure.
(iii) Density of diamond is 3.5 g/cm3. (iii) Density of graphite is 1.9 to 2.3 g/cm3.
(iv) Diamond is a bad conductor of electricity as it does not have free electrons. (iv) Inside each layer of graphite, free electrons move continuously within the entire layer. Hence, graphite is a good conductor of electricity.

b. Crystalline and non-crystalline forms of carbon.
Answer:

Crystalline forms of carbon Non-crystalline forms of carbon
(i) A crystalline form has a regular and definite arrangement of atoms.
(ii) They have high melting points and boiling points.
(iii) A crystalline form has a definite geometrical shape, sharp edges and plane surfaces.
(iv) Diamond, graphite and fullerene are different crystalline forms of carbon.
(i) A non-crystalline form does not have a regular and definite arrangement of atoms.
(ii) They have low melting points and boiling points.
(iii) They are amorphous, hence, they do not have definite geometrical shape.
(iv) Coal, charcoal and coke are different non­crystalline/amorphous forms of carbon.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

4. Write scientific reasons

a. Graphite is a conductor of electricity.
Answer:

  • In graphite, each carbon is bonded to three other carbon atoms in such a way that a hexagonal layered structure is formed.
  • Due to this structure, graphite has free electrons available.
  • These free electrons move continuously within the entire layer.
  • Hence, graphite is a good conductor of electricity.

b. Graphite is not used in ornaments.
Answer:

  • Graphite is a black, soft, brittle and dull form of carbon.
  • It is neither malleable nor ductile.
  • These properties of graphite make it unsuitable for making of ornaments.
  • Hence, graphite is not used for making ornaments.

c. Limewater turns milky when CO2 is passed through it.
Answer:

  • Limewater traditionally means a weak solution of the alkali calcium hydroxide Ca(OH)2.
  • When CO2 is passed through limewater, it reacts with calcium hydroxide to form insoluble particulates (precipitate) of Calcium carbonate (CaCO3).
  • Calcium carbonate is weak basic salt and this gives a milky white precipitate.
  • Hence, lime water turns milky when CO2 gas is passed through it.

d. Biogas is an eco-friendly fuel.
Answer:

  • Biogas is formed by the decomposition of animal dung, dry leaves, wet garbage in a biogas plant.
  • This produces methane gas also called biogas.
  • Biogas is a very cheap fuel option which meets the demand for cooking gas.
  • Biogas is eco-friendly as it contains about 55% to 60% of methane and rest is carbon dioxide, hence, on combustion it does not produce harmful gases which cause pollution.
  • Biogas is a fuel which is convenient to use and in addition to this it produces a very good manure as a side product of the process.
  • Hence, biogas is an eco-friendly fuel.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

5. Explain the following.

a. Diamond, graphite and fullerenes are crystalline forms of carbon.
Answer:

  • Carbon exhibits a property of allotropy in which an element exists in more than one form in nature.
  • The chemical properties of these different forms are the same but their physical properties are different.
  • Carbon exists in crystalline as well as non-crystalline (amorphous form).
  • Crystalline form has a regular and definite arrangement of atoms. They have high melting points and boiling points.
  • A crystalline form has a definite geometrical shape, sharp edges and plane surfaces.
  • Carbon has three crystalline allotropes such as diamond, graphite and fullerene.
  • In the structure of diamond, every carbon atom is bonded to four neighbouring atoms by covalent bonds.
  • Therefore, diamond has a tetragonal three dimensional structure which makes it very hard.
  • Brilliant and pure diamond is the hardest natural substance.
  •  In the structure of graphite, every carbon atom in graphite is bonded to three other carbon atoms in such a way that a hexagonal layered structure is formed.
  • A graphite crystal is made of many sheets or layers of carbon atoms.
  • These layers slip over each other on applying pressure. One layer of graphite is called graphene.
  • Fullerene is rarely found in nature. It is found in soot and in interstellar space.
  • The first example of fullerene is Buckminster fullerene (C60).
  • This allotrope of carbon is named fullerene after the architect.
  • Richard Buckminster Fuller because the structure of C60 resembles the structure of the geodesic dome he designed.
  • (xi) Molecules of fullerenes are found in the form of buckyballs and buckytubes.
  • (xii) There are 30 to 900 carbon atoms in one molecule of a fullerene. , C60, C70, C76, C82 and C86 are other examples of fullerene. Their molecules occur in small numbers in soot.

b. Methane is called marsh gas.
Answer:

  • Methane is formed by the decomposition of plant and animal matter in swamps or marshy areas.
  • As methane gas bubbles out from marshy area, it is called as marsh gas.

c. Petrol, diesel, coal are fossil fuels.
Answer:
(i) A fossil fuel is a fuel formed by natural processes, such as anaerobic decomposition of buried dead organisms. Fossil fuels contain high percentage of carbon. The word carbon is derived from the Latin word ‘Carbo’ meaning coal.

(ii) Coal is formed from the remains of trees and other vegetation. Approximately 350 million years ago, these remains were trapped on the bottom of swamps, accumulating layer after layer and creating a dense material called peat. As this peat was buried under more and more ground, the high temperature and pressure transformed it into coal.

(iii) Petrol and diesel are obtained from mineral oil. Mineral oil also called as crude oil or petroleum oil is formed from the remains of plants and animals that lived in the seas millions of years ago. This plant and animal matter has been drawn down and subjected to extremes of temperature and pressure over millions of years ago.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

(iv) Mineral oil is commonly formed in rocks under the sea bed. The word petroleum is derived from Latin word ‘Rock oil’. Petrol and Diesel are obtained from mineral oil by the process called as fractional distillation.
Thus petrol, diesel and coal are fossil fuels.

d. Uses of various allotropes of carbon.

e. Use of CO2 in fire extinguisher.
Answer:

  • CO2 based fire extinguishers do not cause corrosion and are non-conductors of electricity.
  • It is beyond their capacity to extinguish a big fire.
  • Therefore these fire extinguishers are used to extinguish small scale fire of electrical and electronic equipments.

f. Practical uses of CO2.
Answer:
Practical uses of CO2 are:

  • CO2 is used to make aerated drinks.
  • CO2 obtained by chemical reaction or kept under pressure is used in fire extinguishers. Liquified CO2 is used to remove caffeine from coffee.
  • Liquid CO2 is used as solvent in modem eco-friendly dry cleaning.
  • Solid carbon dioxide is used in cold storage and to keep milk and milk products and frozen substances cool during transport. It is also used for getting special effects of a mist in dramas and movies.

6. Write two physical properties each.

a. Diamond
Answer:
Properties of diamond are:

  • Brilliant and pure diamond is the hardest natural substance.
  • The density of diamond is 3.5 g/cm3.
  • The melting point of diamond is 3500 °C.
  • When a diamond is heated at 800 °C in the presence of oxygen, CO2 is given away. In this process no other product besides CO2 is formed.
  • Diamond does not dissolve in any solvent.
  • Acids/bases have no effect on diamond.
  • Diamond is a bad conductor of electricity as it does not have free electrons.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

b. Charcoal
Answer:

  • The charcoal that is made from animals is made from their bones, horns, etc.
  • On the other hand, the charcoal made from plants is formed by combustion of wood in an insufficient supply of air.

c. Fullerene
Answer:
Properties of fullerenes are:

  • Molecules of fullerenes are found in the form of buckyballs and buckytubes.
  • There are 30 to 900 carbon atoms in one molecule of a fullerene.
  • Fullerenes are soluble in organic solvents such as carbon disulphide, chlorobenzene.

7. Complete the following Chemical reactions.

1. ………………..+………………..→ CO + 2H2O + Heat
2. ………………..+………………..→ HCl + Cl + HCl
3. 2 NaOH + CO2 →………………..+………………..
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 1

8. Write answers to the following in detail.

a. What are the different types of coal? What are their uses?
Answer:
Coal is a fossil fuel. It contains carbon, hydrogen and oxygen. It also contains nitrogen, phosphorus and sulphur. It occurs in the solid state. It is of four types.

  • Peat: Formation of peat is the first step in the formation of coal. It contains a high proportion
    of water and less than 60% of carbon. Therefore, not much heat can be obtained from peat.
  • Lignite: Peat was transformed into Lignite due to increased pressure and temperature inside the earth. It contains 60 to 70% of carbon.- Lignite is the second step of the formation of coal.
  • Bituminous coal: Bituminous coal was formed as the third step of formation of coal. It contains 70 to 90% of carbon.
  • Anthracite: Anthracite is known as the pure form of coal. This coal is hard and contains about 95% of carbon.

Uses of coal:

  • Coal is used as fuel in factories and homes.
  • Coal is used to obtain coke, coal gas and coal tar.
  • Coal is used in thermal power plants for generation of electricity.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

b. How will you prove experimentally that graphite is good conductor of electricity?
Answer:
Apparatus required: Lead pendi, electrical wires, battery/cell, small bulb, etc.

Step-I: Remove the lead from a pencil and assemble the apparatus as shown in the diagram.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 28

Step-II:

  • Start the electric current in the circuit, the moment the electric current is passed through the circuit, the bulb glows.
  • This experiment proves that graphite is a good conductor of electricity as graphite has free electrons moving continuously within the entire layer and these free electrons conduct electricity in the lead of the pencil.

c. Explain the properties of carbon.
Answer:
Allotropic nature of Carbon: Some elements occur in nature in more than one form. The chemical properties of these different forms are the same but their physical properties are different. This property of elements is called allotropy. Carbon shows the property of allotropy.

Carbon allotropes are of two types:
(A) Crystalline forms:
Carbon has three crystalline allotropes: Diamond, Graphite and Fullerene. Properties of crystalline forms of carbon are as follows:

  • A crystalline form has a regular and definite arrangement of atoms.
  • They are made up of only carbon atoms.
  • They have high melting points and boiling points.
  • A crystalline form has a definite geometrical shape, sharp edges and plane surfaces.

(B) Amorphous forms or non-crystalline forms: Coal, charcoal, coke are the non-crystalline forms of carbon.
Properties of non-crystalline forms of carbon are as follows:

  • The arrangement of carbon atoms in this form is not regular.
  • Apart from carbon atoms, they also contain hydrogen, oxygen, nitrogen, phosphorus and sulfur atoms.
  • Compared to a crystalline form, they have low melting and boiling points.
  • Most of them are used as fuels.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

d. Classify carbon.
Answer:
Carbon is classified as follows:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 2

9. How will you verify the properties of carbon dioxide?
Answer:
Properties of carbon dioxide can be verified in the following ways:

  • When a burning candle is placed in a gas jar of carbon dioxide, it extinguishes indicating that carbon dioxide is a non-combustible gas and does not support combustion.
  • When carbon dioxide gas is passed through lime water, it turns lime water milky due to the formation of insoluble calcium carbonate.
  • Moist blue litmus turns red in a gas jar of carbon dioxide indicating, it is acidic in nature.
  • Carbon dioxide is fairly soluble in water, it dissolves in water forming carbonic acid.

Class 9 Science Chapter 13 Carbon: An Important Element Intext Questions and Answers

Can you recall?

Question 1.
Whatremains behind on complete combustion of any organic compound?
Answer:
A black colour substance ‘Carbon’ remains behind on complete combustion of any organic compound.

Question 2.
What type of element is carbon? Give some information about it.
Answer:

  • Carbon is a non-metallic element. It is available ‘ abundantly in the nature and occurs in free as well as combined state.
  • Carbon in free state is found as diamond and graphite and in combined state it is present to the extent of approximately 0.27% in the form of carbonate, coal, petroleum.
  • In atmosphere, the proportion of carbon in the form of carbon dioxide is approximately 0.03%.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 3.
What is an element? What are the different types of elements.
Answer:

  • A substance which cannot be decomposed into simple substances by any physical or simple chemical method is called as an Element.
  • An element is composed of atoms of only one kind.
  • The different types of elements are:
    (a) Metals – Examples: Gold, Silver, etc.
    (b) Non-metals – Examples: Carbon, Sulphur, etc.
    (c) Metalloids – Examples: Silicon, Antimony, etc.

Answer the following questions:

Question 1.
Does an electric charge form on atoms when a covalent bond is formed between them? Why is a single bond between two carbon atoms strong and stable?
Answer:

  • No, electric charge is not formed on atoms when a covalent bond is formed between them. This is because covalent bond is formed by sharing of electrons.
  • Therefore, there is no change in number of electrons and protons in these atoms and they remain electrically neutral.
  • Covalent bonds occur when electrons are shared between two atoms. A single covalent bond is formed when only one pair of electrons is shared between atoms.
  • In this sharing, the atomic orbitals directly overlap between the nuclei of two atoms forming the strongest type of covalent bond called as sigma bond.
  • Hence, a single covalent bond between two carbon atoms is strong and stable.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 2.
In which of the solvents – water, kerosene and cooking oil does the coal powder dissolve?
Answer:
Solvents such as water, kerosene and cooking oil do not dissolve coal powder in them.

Question 3.
What inference will you draw about the solubility of carbon?
Answer:
Carbon is insoluble in water, kerosene and cooking oil.

Question 4.
Is the density of CO2 more or less than that of air?
Answer:

  1. Density is defined as mass per unit volume of a substance, expressed as kilograms per cubic meter.
  2. At standard temperature and pressure, the density of air is 1.29 kg/m3. While that of carbon dioxide (CO2) is 1.79 kg/m3. Hence, density of CO2 is more than that of air.

Question 5.
What is a compound? How are compounds formed?
Answer:

  • A compound is a substance formed when two or more elements are chemically bonded with each other in definite proportion by weight.
  • When atoms or two or more different elements chemically react with each other in a definite proportion by weight, a compound is formed. Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element
  • The properties of a compound are altogether different from its constituent elements.
  • Example: Pure water is a compound made up of two elements Hydrogen and Oxygen. Hydrogen (H) atoms and oxygen (O) atoms have chemically reacted with each other in definite proportion to form a compound-water (H2O). The proportion of hydrogen and oxygen in water by volume is 2 : 1 and by weight is 1 :8 respectively.
  • The properties of water are altogether different from the properties of its constituent elements, hydrogen and oxygen.

Class 9 Science Chapter 13 Carbon: An Important Element Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
The organic compound having double or triple bond in them is termed as
(a) unsaturated
(b) inorganic
(c) saturated
(d) complete
Answer:
(a) Unsaturated hydrocarbon

Question 2.
Covalent compounds are generally soluble in solvents.
(a) inorganic
(b) organic
(c) mixed
(d) mineral
Answer:
(b) organic

Question 3.
Methane is also called as
(a) maha gas
(b) mar’s gas
(c) Anthracite
(d) marsh gas
Answer:
(d) Marsh gas

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 4.
is an alio trope of carbon used to make ornaments.
Answer:
Diamond

Question 5.
The number of valence electrons in carbon is
(a) 4
(b) 5
(c) 6
(d) 3
Answer:
(a) 4

Question 6.
In saturated hydrocarbons, two carbon atoms are linked by
(a) double bond
(b) triple bond
(c) multiple bond
(d) single bond
Answer:
(d) single bond

Question 7.
Carbon has 4 valence electrons in its outermost shell, hence, it is
(a) divalent
(b) tetravalent
(c) trivalent
(d) pentavalent
Answer:
(b) tetravalent

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 8.
Methanogenic bacteria act on the organic acids to produce
(a) oxygen gas
(b) nitrogen gas
(c) methane gas
(d) carbon dioxide gas
Answer:
(c) Methane gas

Question 9.
is used to remove caffeine from coffee.
(a) liquefied O2
(b) liquefied N2
(c) liquefied CO2
(d) liquefied CH4
Answer:
(c) Liquified CO2

Question 10.
Exhaled air contains about CO2.
(a) 4%
(b) 3%
(c) 5%
(d) 6%
Answer:
(a) 4%

Question 11.
Electronic configuration of carbon is
(a) (2, 2)
(b) (2,4)
(c) (2, 5,4)
(d) (2, 6)
Answer:
(b) 2, 4

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 12.
Molecular formula of ethane is
(a) C3H4
(b) C2H4
(c) C2H6
(d) C2H2
Answer:
(c) C2H6

Question 13.
Molecular mass of carbon dioxide is
(a) 28
(b) 22
(c) 56
(d) 44
Answer:
(d) 44

Question 14.
On heating a mixture of hydrogen and carbon monoxide gases at 300°C in the presence of nickel (catalyst) is formed.
(a) carbon dioxide gas
(b) oxygen gas
(c) methane gas
(d) ethylene gas
Answer:
(c) methane gas

Question 15.
Ethane with molecular formula C2H6 has covalent bonds.
(a) six
(b) seven
(c) eight
(d) nine
Answer:
(b) seven

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 15.
Weight of Kohinoor Diamond was
(a) 186 carats
(b) 27 carats
(c) 252 carats
(d) 23 carats
Answer:
(a) 186 carats

Question 16.
Covalent compounds have
(a) high melting point
(b) low melting point
(c) moderate melting point
(d) very high melting point
Answer:
(b) low melting point

Question 17.
Methane is
(a) C2H6
(b) C3H8
(C) CH2
(d) CH4
Answer:
(d) CH4

Question 18.
Wohler, a German scientist synthesized the compound from an inorganic compound ammonium cyanate.
(a) methane
(b) ethylene
(c) urea
(d) acetic acid
Answer:
(c) urea

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 19.
Molecular mass of methane is
(a) 19
(b) 16
(c) 17
(d) 18
Answer:
(b) 16

Question 20.
coal contains highest percentage of carbon.
(a) Bituminous
(b) Peat
(c) Anthracite
(d) Lignite
Answer:
(c) Anthracite

Question 21.
H2C = CH2 is ……………………… .
(a) Propane
(b) Ethyne
(c) Ethane
(d) Ethene
Answer:
(d) Ethene

Question 22.
CH3 – C = CH is ……………………… .
(a) Propene
(b) Propyne
(c) Ethyne
(d) Ethene
Answer:
(b) Propyne

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 23.
Carbon dioxide gas is not used in
(a) photosynthesis
(b) aerated drinks
(c) glass cutting
(d) fire extinguishers
Answer:
(c) glass cutting

Question 24.
Melting point of diamond is
(a) 3700 °C
(b) 3500 °C
(c) 4000 °C
(d) 2500 °C
Answer:
(b) 3500 °C

Question 25.
Melting point of C02 is
(a) 26 °C
(b) 56 °C
(c)-56.6°C
(d)-98°C
Answer:
(c) -56.6 °C

Question 26.
Melting point o* methane is
(a) -182.5 °C
(b) -161.5 °C
(c) 182.5 °C
(d) 161.5 °C
Answer:
(a) -182.5 °C

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 27.
Biogas contains about methane.
(a) 55% to 60%
(b) 20% to 25%
(c) 90% to 95%
(d) 40% to 45%
Answer:
(a) 55% to 60%

Question 28.
is not a property of carbon dioxide gas.
(a) Supporting combustion
(b) Odourless
(c) Colourless
(d) Turns blue litmus red
Answer:
(a) Supporting combustion

Name the following:

Question 1.
Industries that use methane in the form of natural gas.
Answer:
Fabric mills, paper mills, food processing industry, petrol purification.

Question 2.
Organic compounds prepared from methane.
Answer:
Ethanol, methyl chloride, methylene chloride, acetylene.

Question 3.
Chemicals used in regular fire extinguisher.
Answer:
Sodium bicarbonate and dilute sulphuric acid.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 4.
Components of biogas
Answer:
Methane, carbon dioxide

Question 5.
Organic solvents is which fullerenes are soluble.
Answer:
Carbon disulfide, chlorobenzene.

Question 6.
Use of carbon dioxide is dramas and movies
Answer:
The special effect of mist.

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) Diamond
(2) Fullerenes
(3) Graphite
(a) Hexagonal layered structure
(b) Tetragonal three-dimensional structure
(c) Geodesic dome

Answer:
(1-b),
(2- c),
(3 – a)

Question 2.

Column ‘A’ Column ‘B’
(1) Peat
(2) Lignite
(3) Bituminous
(4) Anthracite
(a) 60-70% of Carbon
(b) 95% of Carbon
(c) less than 60% of Carbon
(d) 70 – 90% of Carbon

Answer:
(1 – c),
(2 – a),
(3 – d),
(4 – b)

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 3.

Column ‘A’ Column ‘B’
(1) Propyne
(2) Propene
(3) Propane
(a) CH3 – CH2 – CH3
(b) CH3 – C = CH
(c) CH3 – CH = CH2

Answer:
(1 – b),
(2 – c),
(3 – a)

Question 4.

Column ‘A’ Column ‘B’
(1) Fullerene
(2) Diamond
(3) Graphite
(a) Lubricants
(b) Insulator
(c) Ornaments

Answer:
(1 – b),
(2 – c),
(3 – a)

Question 5.

Column ‘A Column ‘B’
(1) Water gas
(2) Methane gas
(3) Producer gas
(4) Carbon dioxide gas
(a) CH4
(b) CO + H2
(c) CO2
(d) CO + H2 + CO2 + N2

Answer:
(1 – b),
(2 – a),
(3 – d),
(4 – c)

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

State whether the following statements are true or false. Correct the false statement.

(1) Study of organic compounds is called as organic chemistry.
(2) Hydrocarbons with double bonds are called as saturated hydrocarbons.
(3) Ethene is saturated hydrocarbon.
(4) Covalent compounds are good conductor of electricity.
(5) Methane is a covalent compound.
(6) Covalent compounds are soluble in organic solvents.
(7) Graphite is used in making lubricants and lead pencils.
(8) The density of diamond is 9.8 g/cm3.
(9) Diamond knives are used in eye surgery.
(10) Fullerenes are used as insulators.
(11) Coke is used in production of water gas i.e. (CO2 + H2O).
(12) Structural formula of propyne is CH3 – C = CH.
(13) COz is used to make aerated drinks.
(14) Methane gas is black in colour.
(15) Production of biogas is aerobic process.
Answer:
(1) True
(2) False. Hydrocarbons with double bonds are called as Unsaturated hydrocarbons.
(3) False. Ethene is Unsaturated hydrocarbon.
(4) False. Covalent compounds are bad conductor of electricity.
(5) True
(6) True
(7) True
(8) False: The density of diamond is 3.5 g/cm3.
(9) True
(10) True
(11) False. Coke is used in production of water gas i.e. (CO + H2).
(12) False. Structural formula of propyne is CH3 – C = CH.
(13) True
(14) False. Methane gas is colourless.
(15) False. Production of biogas is anaerobic process.

Find the odd man out:

Question 1.
Propane, Methane, Ethene, Pentane.
Answer:
Ethene. It is an unsaturated hydrocarbon with double bond between two carbon atoms while rest are saturated hydrocarbons with single bond between two carbon atoms.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 2.
CH4, C2H6, C3H8, CaCO3.
Answer:
CaCO3. It is a salt which is an inorganic’ compound, while rest are hydrocarbon compounds, i.e. organic compounds.

Question 3.
C2H2, C3H8, C2H6, CH4.
Answer:
C2H2. It is an unsaturated hydrocarbon with triple bonds while rest are saturated hydrocarbons with single bonds between two carbon atoms.

Question 4.
Diamond, Fullerene, Graphite, Methane.
Answer:
Methane. It is a marsh gas while rest are allotropes of carbon.

Question 5.
Coal, Petroleum, Natural gas, Cotton
Answer:
Cotton. Cotton is a natural fibre while rest all are fossil fuels.

Question 6.
Cotton, Silk, Proteins, Wool
Answer:
Proteins. Proteins are carbonaceous nutrients while rest all are natural fibres.

Question 7.
Carbohydrates, Coal, Proteins, Fats
Answer:
Coal. Coal is a fossil fuel while rest all are carbonaceous nutrients.

Question 8.
Peat, Charcoal, Lignite, Bituminous
Answer:
Charcoal. Charcoal is a non-crystalline form of carbon while rest all are types of coal.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 9.
Lubricants, Electrodes, Ornaments, Arc Lamps
Answer:
Ornaments. Ornaments are made from diamonds while rest all are made from graphite.

Write the correlated terms:

(1) Propene : Double bond :: Propyne : ……………………. .
(2) Ethane : CH3 – CH3 :: Ethene : ……………………. .
(3) Hydrogen : Monovalent :: Carbon : ……………………. .
(4) Methane : Low melting point i.e. -182.5°C Diamond : ……………………. .
(5) Graphite : Hexagonal structure :: Diamond : ……………………. .
(6) Density of Diamond : 3.5 g/cm3 :: Density of Graphite : ……………………. .
(7) Peat: 60% of carbon :: Anthracite : ……………………. .
(8) Melting point of CO2: -56.6°C :: Melting point of CH4: ……………………. .
Answer:
(1) Triple Bond
(2) CH2 = CH2
(3) Tetravalent
(4) High melting point i.e. 3500°C
(5) Tetragonal structure
(6) 1.9 to 2.3 g/cm3
(7) 95% of carbon
(8) -182.5 °C

Give scientific reasons:

Question 1.
Covalent compounds have low melting and boiling points.
Answer:

  • When substances melt or boil, bonds between the molecules are broken due to heat supplied.
  • On covalent compounds, the intermolecular forces of attraction are weak.
  • Hence, intermolecular forces in covalent compounds are broken easily due to which they have low melting and boiling points.

Question 2.
Graphite is a good conductor of electricity whereas diamond is a non-conductor of electricity.
Answer:

  • In diamond, each carbon atom uses all its four electronic to get bonded to another carbon atom, whereas in graphite each carbon uses there out of four electrons during bonding.
  • Hence in diamond no free electrons are left whereas in graphite free electrons are available.
  • Due to the presence of electron which are free to flow graphite is a good conductor of electricity, whereas due to the absence of free electrons diamond is a non-conductor of electricity.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 3.
Graphite is used as a lubricant.
Answer:

  • Lubricants are used to reduce friction and wear and tear of mechanical parts.
  • Graphite is smooth and slippery and hence is useful in reducing friction. Hence, graphite is used as a lubricant.

Question 4.
Methane is used as a domestic fuel.
Answer:

  • Methane is highly inflammable.
  • It burns by reacting with oxygen to give a bluish flame.
  • It burns completely and producers 213 Kcal/ mol of heat.
  • Being the smallest hydrocarbon, the proportion of CO2 released in the combustion of methane is small.
  • Therefore methane is used as a domestic fuel.

Answer in short:

Question 1.
What was the contribution of chemist Wohier in organic chemistry?
Answer:

  • The German chemist Wohier synthesized an organic compound urca from an inorganic compound ammonium cyanate,
  • Ever since then, many organic compounds have been made from inorganic compounds.
  • Carbon was found to he the main element in all these compounds.
  • Hence, organic chemistry is also referred to as chem is try of carbon corn pounds.

Question 2.
With neat diagram explain the structure of diamond.
Ans.

  • In diamond, every carbon atom is bonded to four neighbouring atoms by covalent bonds.
  • Therefore, diamond has a tetragonal three dimensional structure which makes it very hard.
    Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 29

Question 3.
With neat diagram explain the structure of graphite?
Answer:

  • Every carbon atom in graphite is bonded to three other carbon atoms in such a way that a hexagonal layered structure is formed.
  • A graphite crystal is made of many sheets or layers of carbon atoms.
  • These layers slip over each other on applying pressure.
  • One layer of graphite is called graphene.
    Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 30

Question 4.
Give the properties of graphite.
Answer:
Properties of graphite are:

  • Graphite found in nature is black, soft, brittle and slippery.
  • Inside each layer of graphite, free electrons move continuously within the entire layer. That is why graphite is a good conductor of electricity. Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element
  • Due to the layered structure graphite can be used for writing on paper.
  • The density of graphite is 1.9 to 2.3 g/cm3.
  • Graphite does not dissolve in most solvents.

Question 5.
Give the uses of graphite.
Answer:
Uses of graphite are:

  • Graphite is used for making lubricants.
  • Graphite is used for making carbon electrodes.
  • Graphite is used in pencils for writing.
  • Graphite is used in paints and polish.
  • Graphite is used in arc lamps which gives a very bright light.

Question 6.
Give the uses of fullerenes.
Answer:
Uses of Fullerenes are:

  • Fullerenes are used as insulators.
  • Fullerenes are used as a catalyst in water purification.
  • At a certain temperature, fullerene exhibits superconductivity.

Question 7.
What is Coal?
Answer:

  • Coal is a fossil fuel. It contains carbon, hydrogen and oxygen.
  • It also contains nitrogen, phosphorus and sulphur.
  • It occurs in the solid state. It is of four types – peat, lignite, bituminous and anthracite.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 8.
What is Coke?
Answer:
The pure coal that remains when coal gas has been taken away from coal, is called coke.

Question 9.
Give the uses of Coal.
Answer:
Uses of Coal are:

  • Coal is used as fuel in factories and homes.
  • Coal is used to obtain coke, coal gas and coal tar.
  • Coal is used in thermal power plants for generation of electricity.

Question 10.
Give the uses of Coke.
Answer:
Uses of Coke are:

  • Used as domestic fuel.
  • Coke is used as a reducing agent.
  • Coke is used in production of water gas (CO + H2) and producer gas (CO + H2 + CO2 + N2)

Question 11.
Name the types of coal and give the differences in them.
Answer:
The four types of coal are Peat, Lignite, Bituminous and Anthracite.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 3

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 12.
Give the properties of covalent compounds.
Answer:
Properties of covalent compounds:

  • Covalent compounds have low melting points and boiling points.
  • Generally they are insoluble in water and soluble in organic solvents.
  • They are poor conductors of heat and electricity.

Question 13.
What are saturated hydrocarbons? Give examples.
Answer:

  • The hydrocarbons having only single bonds between carbon atoms are called saturated hydrocarbons.
  • For example, ethane (C2H6) which is (CH3 – CH3), propane (C3H8) which is (CH3 – CH2 – CH3).

Question 14.
What are unsaturated hydrocarbons? Give examples.
Answer:

  • Some hydrocarbons have a multiple bond between two carbon atoms.
  • A multiple bond can be a double bond or a triple bond.
  • Hydrocarbons having at least one multiple bond are called unsaturated hydrocarbons.
  • For example, ethene (H2C = CH2), ethyne (HC = CH), propene (CH3 – CH = CH2), propyne (CH3 – C = CH).

Question 15.
Give the following information of carbon dioxide: Molecular formula, Molecular mass, Melting point, Percentage occurrence in air.
Answer:
Molecular formula – CO2, Molecular mass – 44, Melting point -56.6 °C Percentage occurrence in air – 0.03%.

Question 16.
Give the physical and chemical properties of carbon dioxide.
Answer:
Physical properties of carbon dioxide are:

  • It is an odourless gas.
  • It is a colourless gas.

Chemical properties of carbon dioxide are:

  • It is non-combustible and does not support combustion.
  • It turns lime water milky.
  • It is fairly soluble in water and dissolves in water forming carbonic acid.
  • It turns blue litmus red indicating it is acidic in nature.
  • The colour of universal indicator turns orange/ yellow in C02 indicating its pH value between 4 and 6, i.e. acidic in nature.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 17.
Give the uses of Carbon dioxide.
Answer:
Uses of Carbon dioxide are:

  • CO2 is used to make aerated drinks
  • Solid carbon dioxide is used in cold storage and also to keep milk and milk products and frozen substances cool during transport. It is also used for getting special effects of a mist in dramas and movies.
  • CO2 obtained by chemical reaction or kept under pressure is used in fire extinguishers.
  • Liquified CO2 is used to remove caffeine from coffee.
  • Liquid CO2 is used as solvent in modern eco-friendly dry cleaning.
  • Plants use CO2 in air for photosynthesis.

Question 18.
Give the occurrence of methane.
Answer:
The occurrence of methane is as follows:

  • Methane occurs in natural gas to the extent of 87%.
  • Decomposition of organic matter in the absence of air (anaerobic) produces methane.
  • Methane is present in biogas.
  • Methane is found in coal mines.
  • Methane is found at the surface of marshy places which is why it is also called marsh gas.
  • On heating a mixture of hydrogen and carbon monoxide gases at 300 °C in the presence of nickel (catalyst), methane gas is formed.
  • Fractional distillation of natural gas gives methane in pure form.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 19.
Give the physical properties of methane.
Answer:
Physical properties of methane are:

  • Melting point of methane is (-182.5 °C).
  • Boiling point of methane is (-161.5 °C).
  • It is a colourless gas.
  • The density of liquid methane is less than that of water.
  • Methane is sparingly soluble in water. It is highly soluble in organic solvents like gasoline, ether and alcohol.
  • Methane is in gaseous state at room temperature.

Question 20.
Give the uses of methane.
Answer:
Uses of methane are:

  • Methane in the form of natural gas is used in industries such as fabric mills, paper mills, food processing industry, petrol purification.
  • Being the smallest hydrocarbon, the proportion of CO2 released in the combustion of methane is small and, therefore, it is used as a domestic fuel.
  • Methane is used for production of organic compounds such as ethanol, methyl chloride, methylene chloride and acetylene.

Question 21.
How is methane formed? Give structural formula and electron dot model of methane.
Answer:

  • Atomic number of carbon is 6. The electronic configuration of carbon is (2, 4). The valency of carbon is 4.
  • Carbon atom can form four covalent bonds with other carbon atoms or atoms of different elements.
  • When a carbon atom shares one electron each with four hydrogen atoms and forms four C-H bonds, a methane (CH4) molecule is formed.
    Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 4

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 22.
What are organic and inorganic compounds?
Answer:
Compounds obtained directly or indirectly from plants and animals are called organic compounds and compounds obtained from minerals are called inorganic compounds.

Question 23.
What is allotropy?
Answer:

  • Allotropy – Some elements occur in nature in more than one form.
  • The chemical properties of these different forms are the same but their physical properties are different.
  • This property of elements is called allotropy.

Question 24.
What are basic organic compounds? What are they also called as?
Answer:

  • The compounds formed from only carbon and hydrogen are called basic organic compounds.
  • These are also called hydrocarbons.

Write a balanced chemical equation and explain the following chemical reactions.

Question 1.
Preparation of urea from Ammonium cyanate:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 5
Organic compound Urea can be synthesized from an inorganic compound Ammonium cyanate.

Question 2.
Coal when burnt in air
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 6
When coal is burnt in air, the carbon present in coal combines with oxygen present in air to form carbon dioxide gas.

Question 3.
Calcium carbonate reacts with dilute hydrochloric acid:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 7
When Calcium carbonate reacts with dilute hydrochloric add it forms Caldum chloride, water and Carbon dioxide gas is evolved.

Question 4.
Carbon dioxide is passed through an aqueous solution of Sodium hydroxide:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 8
When Carbon dioxide gas is passed through an aqueous solution of Sodium hydroxide it forms Sodium carbonate and Water.

Question 5.
Carbon dioxide is passed through an aqueous solution of Sodium carbonate:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 9
When carbon dioxide is passed through an aqueous solution of Sodium carbonate, it forms Sodium bicarbonate.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 6.
Carbon dioxide is passed through freshly prepared lime water for a short duration and then for longer duration:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 10
When Carbon dioxide is passed through freshly prepared lime water, it forms water and white precipitate of Calcium carbonate because of which lime water turns milky.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 11
When Carbon dioxide is continuously passed through milky lime water, it forms Calcium bicarbonate which is soluble in water and therefore, water once again turns colourless.

Question 7.
Carbon dioxide gas is dissolved in water.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 12
Carboh dioxide gas is fairly soluble in water, it dissolves in water under pressure to form Carbonic acid.

Question 8.
Sodium bicarbonate reacts with dilute Sulphuric acid, (reaction in fire extinguisher):
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 13
When Sodium bicarbonate reacts with dilute Sulphuric acid it forms Sodium sulphate, water and Carbon dioxide gas.

Question 9.
Methane gas is burnt in air:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 14
Methane is highly inflammable, it burns in air with a bluish flame and combines with oxygen present in air to form Carbon dioxide and water. In these reaction 213 kcal/mol of heat is given out.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 10.
Methane and Chlorine gases react with each other.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 15
Methane and chlorine gases react with each other at the temperature of 250 °C to 400 °C in presence of ultra voilet light and form mainly Methyl chloride (Chloromethane and ‘ Hydrogen chloride)
This reaction is called Chlorination of methane.

Question 11.
Production of methane gas in biogas plant.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 16
In biogas plant, microbes act on the bio-degradable complex organic compounds and produce organic acids.

The methanogenic bacteria act on the organic acids to produce Methane gas and Carbon dioxide gas.

Question 12.
Production of water gas:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 17
When super heated steam is passed over red hot coke at high temperature, it forms carbon monoxide gas and hydrogen gas.

This gaseous product of carbon monoxide gas and hydrogen gas together is called as water gas.

With the help of ne abelled diagram explain the following:

Question 1.
Regular fire extinguisher.
Answer:

  • A fire extinguisher contains Sodium bicarbonate powder. There is also dilute sulphuric acid placed in a glass capsule.
  • On pressing the knob, the capsule breaks and sulphuric acid comes in contact with the sodium bicarbonate. These two react chemically to release CO2 which comes out.
  • CO2 based fire extinguishers do not cause corrosion and are non-conductors of electricity. Therefore, they are used when electrical or electronic equipment catches fire. Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element
  • CO2 based fire extinguishers are used to extinguish small scale fire. It is beyond their capacity to extinguish a big fire.
  • In modern fire extinguishers liquid and solid CO2 is filled under pressure. On reducing the pressure, it becomes gaseous and comes out forcefully through the horn-like hose pipe.
  • Chemical reaction:
    Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 18

Question 2.
Biogas plant.
Answer:

  • Animal dung, dry leaves, wet garbage get decomposed by anaerobic microbes in a biogas plant.
  • This produces methane gas, also called biogas.
  • Biogas is a very cheap fuel option which meets the demand for cooking gas. It is also used for the production of electricity.
  • Biogas contains about 55% to 60% methane and the rest is carbon dioxide.
  • Biogas is a fuel which is convenient to use and, in addition to this, a very good manure is also produced as a side product of the process.
  • Biogas production process – Production of biogas is an anaerobic process. It takes place in two stages.
    (a) Production of acids – The microbes act on the biodegradable complex organic compound and produce organic acids.
    (b) Methane gas production – The methanogenic bacteria act on the organic acids to produce methane gas.

Chemical reaction:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 19

Answer the following

Question 1.
What happens when substances like milk, sugar, wool, dry leaves, hair and seeds are heated strongly in hard glass test tube?
Answer:
When milk, sugar, wool, dry leaves, hair and seeds are heated strongly in hard glass test tube, they get charred and a black substance is left behind. This black substance is carbon.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Answer the following questions:

Question 1.
Take some milk in an evaporating dish. Heat the evaporating dish on a bunsen burner. What remains behind at the bottom of an evaporating dish on complete evaporation of the milk?
Answer:

  • On complete evaporation of milk, a black residue is left behind.
  • This residue is of carbon.

Question 2.
Take small samples of sugar, wool, dry leaves, hair, seeds, split pulses and plastic in separate test tubes. Heat each test tube and observe the changes taking place in the substances. What does the black substance in each test tube indicate?
Answer:

  • On heating the above samples, they get charred and a black substance is left behind.
  • The black substance is carbon.

Question 3.
Apparatus : Coal, match box, moist litmus paper, etc.

Procedure : Ignite the coal. Hold the moist blue litmus paper over the gas released on igniting the coal. Note the observation.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 20

(a) With which gas in the air does the coal react on igniting?
Answer:
Coal reacts with oxygen gas present in the air, on igniting.

(b) What is the substance formed?
Answer:
The substance formed is carbon dioxide.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

(c) What change takes place in the litmus paper?
Answer:
The moist blue litmus paper turns red.

(d) Write down the chemical reaction taking place in the above procedure.
Answer:
(i) Carbon combines with oxygen to form Carbon dioxide gas.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 21
(ii) This C02 combines with water present on moist blue litmus paper to form Carbonic acid which turns blue litmus to red.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 22

Question 4.
Apparatus : Pencil, electrical wires, battery/ cell, small bulb, water, kerosene, test tube, lead pencils, etc.

Procedure : Remove the lead from a pencil and arrange the apparatus as shown in the above diagram. Note your observations for the following.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 23

(a) What is the colour of lead in the pencil? .
Answer:
The colour of the lead in the pencil is black as it is made from graphite which is an allotrophic form of carbon.

(b) Try to break lead with your hand.
Answer:
The lead breaks easily as it is made up of graphite which is brittle in nature.

(c) Start the electric current in the circuit and observe. What did you find?
Answer:
When we start the electric current in the circuit, the bulb in the circuit glows, indicating that the lead in the pencil is a good conductor of electricity.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

(d) Take some water in a test tube. Take some kerosene in another test tube. Put lead dust in both the test tube. What did you observe?
Answer:
(i) Lead dust does not dissolve in water nor in kerosene.
(ii) It remains insoluble in both the test tubes.

Question 5.
Apparatus : Test tube, straw, lime water, etc.

Procedure: Take freshly prepared lime water in a test tube. Blow air in it for some time through the straw and observe the lime water. Note the observation.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 24

(a) What did you see?
Answer:
When we blow air through the straw in freshly prepared lime water, it turns milky.

(b) What might be the reason behind the change?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 25
Freshly prepared lime water is obtained by dissolving lime (CaO) in water (H2O).

When we blow air through the straw in lime water, carbon dioxide (CO2) present in the air reacts with lime water to form white precipitate (insoluble substance) of Calcium Carbonate (CaCO3) due to which lime water turns milky.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 26

Question 6.
With the help of a neat labelled diagram explain the laboratory preparation of Carbon dioxide gas.
Answer:
The laboratory preparation of Carbon dioxide gas is as follows:

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Apparatus : Retort stand, round bottom flask, thistle funnel, gas delivery tube, gas jars.
Chemicals : Calcium carbonate (Pieces of Shahabad tiles/marble pieces/limestone), dilute hydrochloric acid.

Procedure:

  • Assemble the apparatus as shown in the figure. While assembling, place CaCO3 in the round bottom flask.
  • Add dilute HCl in the flask through thistle funnel. See to it that the end of the funnel dips in the acid.
  • CO2 is formed as a result of the reaction between CaCO3 and HCl. Collect this gas in four to five gas jars. The chemical equation of the above reaction is as follows. CaCO3 + 2HCl → CaCl2 + H2O + CO2
    Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 27

9th Std Science Questions And Answers:

Observing Space: Telescopes Class 9 Science Chapter 18 Questions And Answers Maharashtra Board

Class 9 Science Chapter 18

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 18 Observing Space: Telescopes Question Answer Maharashtra Board

Class 9 Science Chapter 18 Observing Space: Telescopes Question Answer Maharashtra Board

1. Fill in the blanks with the proper words.

a. The wavelength of visible light is between …………………..….. and …………………..…. .
b. GMRT is used for …………………..….. waves.
c. A certain X-ray telescope is named after scientist …………………..…. .
d. The first scientist to use a telescope for space observations …………………..…. .
e. The biggest optical telescope in India is situated at …………………..…. .

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

2. Form pairs

Column ‘A’ Column ‘B’
(1) X-rays (a) GMRT
(2) Optical telescope (b) ISRO
(3) Indian radio telescope (c) Hubble
(4) Launching artificial-satellite (d) Chandra

Answer:
(1 – d),
(2 – c),
(3 – a),
(4 – b),

3. What are the difficulties in using ground-based optical telescopes? How are they overcome?
Answer:
Difficulties:

  • The intensity of visible light coming from a heavenly body decreases due to absorption as it passes through the Earth’s atmosphere. Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes
  • Due to changes in atmospheric temperature and pressure light rays change their path.
  • Because of sunlight, we cannot use optical telescopes during the day.
  • During the night, city lights and cloudy weather cause difficulties.

Solution:

  • To reduce these problems, optical telescopes are situated on top of mountains, in uninhabited places.
  • To get rid of all the above problems completely, the telescope is placed above the earth’s atmosphere, in space.

4. Which type of telescopes can be made using a concave mirror, convex mirror, plane mirror, and a lens? Draw diagrams of these telescopes.
Answer:

  • A Newtonian telescope can be made using a concave mirror, plane mirror and a lens.
    Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space Telescopes 4
  • A Cassegrain telescope can be made using a convex mirror, concave mirror and a lens.
    Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space Telescopes 5
  • A refracting telescope can be made using two or more lenses.
    Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space Telescopes 6

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

5. Study the figure and answer the following questions:
Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space Telescopes 3

a. What type of telescope is shown in the figure?
b. Label the main parts of the telescope.
c. Which type of mirror does the telescope use?
d. What other type of telescope uses a curved mirror?
e. Explain the working of the above telescope.
Answer:
(a) The Newtonian telescope (Reflecting telescope)
(b)
Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes 1
(c) Concave mirror
(d) The Cassegrain telescope
(e) Working.

  • Light rays coming from space are reflected by the concave mirror.
  • Before these reflected rays converge at the focus, they are deflected again by a small plane mirror.
  • As a result, they get focused at a point lying on the perpendicular to the axis of the telescope’s cylinder.
  • They pass through the eyepiece and we get a magnified image of the source.

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

6. Answer the following questions.

a. Explain the construction of Galileo’s telescope.
Answer:

  • The Galileo’s telescope is made up of two or more convex lenses.
  • The lens facing the object is called the objective lens which is made as large as possible to collect the maximum amount of light coming from a heavenly object.
  • The lens closer to the eye is called the eyepiece which uses the light collected by the objective lens and produces a large image of the source.
  • Such a telescope is also called a refracting telescope.

b. Explain the construction of a radio telescope.
Answer:

  • The radio telescope is made from one or more dishes of a particular parabolic shape.
  • As in optical telescope, the incident radio waves are reflected by these dishes and converge at the focus.
    Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space Telescopes 2
  • A radio receiver is placed at the focal point.
  • The information gathered by this receiver is passed on to a computer which analyses it and constructs an image of the source.

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

c. Why are optical telescopes located in uninhabited places on mountains?
Answer:

  • The visible light coming from a heavenly body has to pass through the earth’s atmosphere to reach the earth’s surface.
  • During this journey, some of the light is absorbed by the atmosphere and the intensity of the light reaching the earth’s surface decreases.
  • A second problem is caused by the changes in atmospheric pressure and temperature.
  • These changes cause turbulence in the atmosphere which in turn cause the light rays to change their path slightly and thereby shake the position of the image.
  • We cannot use optical telescope during the day. because of sunlight also.
  • During the night too, city lights and cloudy weather can cause difficulties in observing the heavenly bodies.
  • Hence, optical telescopes are located in uninhabited places on mountains.

d. Why can an X-ray telescope not be based on the earth?
Answer:

  • X-rays are emitted from distant heavenly bodies.
  • The rays coming towards the earth will be absorbed by the earth’s atmosphere and will go undetected.
  • Hence, X-ray telescopes are placed in an orbit outside the earth’s atmosphere and not on the earth.

Class 9 Science Chapter 18 Observing Space: Telescopes Intext Questions and Answers

Question 1.
What is meant by space observation? Why is it important?
Answer:
The observation of distant planets, galaxies and other astronomical objects is called as space observation.

  • Various space based satellites support our telecommunication networks by providing us television broadcasting.
  • Meteorology services such as weather forecasting help in disaster management.
  • Observation of positions of stars and other heavenly bodies has helped in the study of solar system, solar winds, pulsars, supernova, etc.

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Question 2.
What is the difference between sky and space?
Answer:
Sky: It is a layer of gas that surrounds the earth in which sunlight is scattered making it visible from the surface of the earth.
Space: The vast emptiness beyond the sky is called as space.

Class 9 Science Chapter 18 Observing Space: Telescopes Additional Important Questions and Answers

Choose and write the correct options:

Question 1.
The first telescope was invented by
(a) Einstein
(b) Galileo
(c) Newton
(d) Mendel
Answer:
(b) Galileo

Question 2.
Eight is an wave.
(a) electric
(b) magnetic
(c) electromagnetic
(d) mechanical
Answer:
(c) electromagnetic

Question 3.
Visible radiation telescopes are also called as
(a) Gamma Ray telescopes
(b) Optical telescopes
(c) X-ray telescopes
(d) Radio telescopes
Answer:
(b) Optical telescope

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Question 4.
Reflecting telescopes are mainly of two types : ……………….. and ………………. .
(a) Newtonian and Galileoan
(b) Optical and Cassegrain
(c) Newtonian and Cassegrain
(d) Optical and Refracting
Answer:
(c) Newtonian and cassegrain

Question 5.
Giant Meterwave Radio Telescope (GMRT) has been erected at near Pune.
(a) Asangaon
(b) Talegaon
(c) Narayangaon
(d) Bhategaon
Answer:
(c) Narayangaon

Question 6.
Visible light coming from heavenly bodies have to pass through the to reach the earth surface.
(a) stratosphere
(b) troposphere
(c) atmosphere
(d) biosphere
Answer:
(c) atmosphere

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Question 7.
To collect the maximum amount of light coming from an object, the objective lens should be made as as possible.
(a) large
(b) small
(c) circular
(d) flat
Answer:
(a) large

Question 8.
The images formed by lenses have errors of colours. This is called
(a) chromatic disruption
(b) chromatic aberration
(c) chromatic reflection
(d) chromatic reaction
Answer:
(b) chromatic aberration

Question 9.
Radio telescope is made up of one or more dishes of shape.
(a) spherical
(b) hyperbolic
(c) parabolic
(d) hexagonal
Answer:
(c) parabolic

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) X-rays (a) GMRT
(2) Optical telescope (b) ISRO
(3) Indian radio telescope (c) Hubble
(4) Launching artificial satellite (d) Chandra
(5) Radio waves (e) telecommunication
(6) INSAT and GSAT (f) longer than 20cm

Answer:
(1 – d),
(2 – c),
(3 – a),
(4 – b),
(5 – f),
(6- e)

Question 2.

Column ‘A’ Column ‘B’
(1) Micro waves
(2) Infrared waves
(3) Visible light rays
(4) Ultraviolet rays
(5) X-rays
(a) 800 nm – 0.3mm
(b) 360 pm – 400 nm
(c) 3 pm – 300 pm
(d) 400 nm – 800nm
(e) 0.3 mm -20cm

Answer:
(1 – e),
(2 – a),
(3 – d),
(4 – b),
(5 – c)

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Define the following:

Question 1.
Visible radiation
Answer:
Light is an electromagnetic wave. Every wave has a characteristic wavelength. Our eyes can see only that light which has wavelengths between 400 nm to 800 nm. Such light is called visible radiation.

Question 2.
Reflecting Telescope
Answer:
When light falls on the mirror, it gets reflected. Telescopes that use concave mirrors are called reflecting telescope.

Question 3.
Optical telescopes
Answer:
Telescopes which are made from regular lenses and mirrors and used to see visible radiations coming from space are called optical telescopes.

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Question 4.
Refracting Telescopes
Answer:
Light rays change their direction as they enter a lens from the atmosphere and again when they enter the atmosphere after passing through the lens. This is called refraction. The telescopes that use such lenses are called refracting telescopes.

Question 5.
Radio telescopes
Answer:
Many heavenly bodies emit radio waves in addition to visible radiation. A special type of telescope which is used to receive these rays are known as radio telescope.

Answer the following:

Question 1.
How do refracting telescopes work?
Answer:

  • Most refracting telescopes are made with two or more lenses.
  • The two lenses are called objective lens and an eyepiece lens.
  • The objective lens should be large so that maximum light can be collected from the source.
  • Using the light collected, the eyepiece lens which is small in size, produces a large image of the source.

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Question 2.
What are the difficulties in using refracting telescopes?
Answer:

  • We require a large objective lens which is very difficult to make. Also large lenses are heavy and tend to get distorted.
  • As the objective and eyepiece lens are placed on the opposite sides of the telescope, the length of the telescope also increases with increase in the size of the lenses. This makes it difficult to manage a large telescope.
  • The images formed by lenses have errors of colours. This is called chromatic aberration.

Question 3.
How does a radio telescope work?
Answer:

  • A radio telescope is made from one or more dishes of a particular parabolic shape.
  • The incident radio waves are reflected by these dishes and converge at a focus.
  • A radio receiver is placed at focal point.
  • The information gathered by this receiver is then passed onto a computer which analyses and constructs the image of the source.

Question 4.
What are the two types of reflecting telescopes?
Answer:

  • The reflecting telescopes are mainly of two types: Newtonian and Cassegrain.
  • The newtonian telescope uses a plane mirror and a concave mirror.
  • The Cassegrain telescope uses a convex and a concave mirror.

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Question 5.
What is the purpose of launching X-Ray telescope Chandra?
Answer:
Chandra was launched to study X-rays coming from heavenly objects. Special mirrors which . can reflect X-rays were used in this telescope.

Answer the following:

Question 1.
State some of the satellites and their uses.
Answer:

  • The INSAT and GSAT series of satellites support our telecommunication network, television broadcasting and meteorological services.
  • It is because of them that telephone, television and internet services are available everywhere in the country.
  • The EDUSAT satellite is used exclusively for education.
  • The IRS satellite series is used for the monitoring and management of natural resources as well as disaster management. Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes
  • Astrosat: It has ultraviolet and X-ray telescopes and detectors. It is a unique system having different kinds of telescopes on a single satellite.

Question 2.
What is the specialty of the artificial satellite Astrosat?
Answer:

  • This satellite was launched by Indian Space Research Organization (ISRO).
  • It has ultraviolet and X-ray telescopes and detectors. Most of the parts used in this satellite are made in India.
  • It is an unique system having different kinds of telescopes on a single satellite. Indian scientists are studying various aspects of the Universe using the data obtained with these telescopes.

Answer in detail:

Question 1.
Explain any two space observation telescope.
Answer:
Hubble Telescope:

  • In 1990, the National Aeronautics and Space Administration launched into space an optical telescope called the Hubble telescope.
  • It has a mirror of diameter 94 inches and is orbiting the earth at a height of 589 km from it.
  • This telescope is still working and has helped to make important discoveries.

Chandra Telescope:

  • In 1999, the National Aeronautics and Space Administration launched an X-ray telescope named Chandra, in space, to study X-rays coming from heavenly objects.
  • Special mirrors which can reflect X-rays were used in this telescope.
  • Chandra has given us very useful information about stars and galaxies.
  • The telescope is named after the famous Indian scientist Subramanian Chandrashekhar.

Question 2.
Write short note : GMRT
Answer:

  • Giant Meterwave Radio Telescope (GMRT) is a large radio telescope erected at Narayangaon near Pune.
  • It uses radio waves having wavelengths of about a metre, coming from heavenly bodies to study those bodies. This telescope is actually a collection of 30 dishes, each having a diameter of 45 m.
  • It is called a giant telescope as the arrangement of the 30 dishes over an area which measures up to 25 km across. It works as a single dish having a diameter of 25 km. Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes
  • GMRT gives the same data that we would have got from a telescope having a single dish of 25 km diameter.
  • Scientists study the solar system, solar winds, pulsars, supernova, interstellar hydrogen clouds, etc. with the help of the GMRT.

9th Std Science Questions And Answers:

Environmental Management Class 9 Science Chapter 9 Questions And Answers Maharashtra Board

Class 9 Science Chapter 9

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 9 Environmental Management Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 9 Environmental Management Question Answer Maharashtra Board

Class 9 Science Chapter 9 Environmental Management Question Answer Maharashtra Board

1. Match the items in column ‘A’ with the proper ones in coloum ‘B’ and explain their impact on the environment.

Column ‘A’ Column ‘B’
1. Harmful waste – a. Glass, rubber, carry bags, etc.
2. Domestic waste – b. Chemicals, pigments, ash, etc.
3. Biomedical waste – c. Radioactive material
4. Industrial waste – d. Left over food, vegetables, peelings of fruits.
5. Urban waste – e. Bandages, cotton, needles, etc.
Answer:
(1 – c),
(2 – d),
(3 – e),
(4 – b),
(5 – a)

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

2. Complete the statements using the given options and justify those statements. (Geographic favourability, climate, weather, observatory)

a. Of the abiotic factors that affect biodiversity by far the most important is ……………… .
Answer:
Climate.
Climate influences our basic needs like food, clothing and shelter as well as our occupations. Various factors of climate like temperature, atmospheric pressure, sunlight, rainfall, humidity etc. will decide which kind of plants and animals can survive there.

b. A description of the climatic conditions of short duration in a particular area is ……………… .
Answer:
Weather.
Atmospheric conditions at a specific time at a particular place are referred to as weather. Weather is related to a specific location and specific time whereas climate is related to a longer duration and larger area.

c. Irrespective of the progress of human beings, we have to think about ……………… .
Answer:
Geographical favourability.
Geographical favourability includes location, availability of water, availability of natural resources and climate of a region. This has a great impact on the progress of human beings. Countries that have more favourable factors will progress more.

d. Establishments where various climatic factors are recorded are called ……………… .
Answer:
Observatories
Most countries in the world have established meteorology departments for recording climatic factors. These departments have observatories which are equipped with modern instruments and technology.

3. Answer the following questions.

a. How is first aid provided to victims of disasters who are injured?
Answer:
(i) Bleeding: If the victim is injured and bleeding through the wound, the wound should be covered with an antiseptic pad and pressure applied on it for 5 minutes with either thumb or palm.

Fracture and impact on vertebrae: If any bone is fractured, it is essential that the fractured part be immobilized. It can be done with the help of any available wooden rods / batons / rulers. If there is an impact on the back or vertebral column; the patient should be kept immobile on a firm stretcher.

Bums: If victims have bum injuries, it is beneficial to hold the injured part under clean and cold flowing water for at least 10 minutes. How is Solid Waste classified? OR What are the sources of Solid Waste?

Domestic waste: Waste food, paper, plastic paper, plastic bags, vegetable waste, fruit skins, glass and sheet metal articles, etc. Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Industrial waste: Chemicals, pigments, sludge, ash, metals, etc.

Hazardous waste: Chemicals generated in various industries, radioactive materials, explosives, infectious materials, etc. Farm/Garden waste: Leaves, flowers, branches of trees, crop residues like straw, animal urine and dung, pesticides, remains of various chemicals and fertilizers, etc.

Electronic waste: Non-functional TV sets, cell phones, music systems, computers and their parts, etc.

Biomedical waste: Bandages, dressings, gloves, needles, saline bottles, medicines, medicine bottles, test tubes, body parts, blood, etc. from clinics, hospitals, blood banks and laboratories. Urban waste: Waste generated through household industries and large commercial and industrial establishments, carry bags, glass, metal pieces and rods, threads, rubber, paper, cans from shops, vegetable and meat markets, construction waste, etc.

(viii) Radioactive waste: Radioactive materials like Strontium-10, Cerium-141, Barium-140 and heavy water, etc. generated from atomic energy plants, uranium mines, atomic research centres, nuclear weapons testing sites, etc.
(ix) Mining waste: Remains of heavy metals like lead, arsenic, cadmium, etc. from mines.

b. State the scientific and eco-friendly methods of waste management.
Answer:
Following are the scientific and eco-friendly methods of waste management:

  • Waste separation: In this method, dry and wet wastes are separated, stored and later collected for proper use.
  • Composting: Decomposition of degradable kitchen waste in small pits in the backyard, garden or terrace is called composting. Good quality manure can be produced by composting leftover food, peels of fruits, vegetables etc.
  • Vermicomposting: Converting solid waste into manure or compost with the help of earthworms is called vermicomposting.
  • Vermicompost manure is highly nutritious and can be used for agricultural purposes and garden plants.
  • Secured landfill: Solid waste is disposed off in secured landfills.
  • The site for secured landfill is selected minimum 2 km away from water bodies and human habitation.
  • Care is taken to see that the site does not fall in sensitive zone.
  • The landfills are layered with clay and plastic and then the garbage is spread and left to decompose naturally.
  • Pyrolysis: In this method, the waste is heated to a high temperature to obtain gas and electricity.
  • Semi-combustible waste is burnt in pyrolysis. It is suitable for municipal solid waste management.
  • Incineration: Biomedical waste is burnt in incinerators to kill the pathogens. Disinfection and sterilization is also done while treating the biomedical waste.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

c. Explain with suitable examples, the relationship between weather forecasting and disaster management.
Answer:

  1. Depending upon the factors such as storms, clouds, rainfall, etc., weather forecasts are made. It is useful in aviation, shipping, fishing, industries as well as during natural calamities like dust storms, sand storms, heavy rainfall, tsunami etc where proper predictions are made with well equipped satellites and highclass technology.
  2. Observatories at several locations are doing excellent work in the analysis of the information received from these satellites.
  3. This data can be used in prevention of losses and danger, improving tolerance, providing relief from disaster, minimising the intensity and extent of harm as well as preparation to face the disaster.

d. Why is e-waste harmful? Express your opinion about this.
Answer:

  • e-waste is electronic waste which includes non-functional TV Sets, cell phones, music systems, computers and their parts, etc.
  • E-waste contains heavy metals like lead, beryllium, mercury and cadmium.
  • These metals accumulate in the soil for long periods and thus affect the biodiversity of the soil.
  • Many a times e-waste like battery etc. contain acids which make the soil acidic.
  • e-waste can cause ground water pollution, which also affects living organisms when they drink this polluted water.
  • Thus, e-waste affects human health and soil microorganisms directly or indirectly.

e. How will you register individual your participation in solid waste management?
Answer:
We can register our individual participation in solid waste management in the following ways:

  • Following the 3R mantra: Reduce (reducing the waste), reuse (reuse of waste) and recycle (recycling of waste).
  • Throwing plastic wrappers of chocolates, ice-creams, biscuits, etc. into dust bins. Avoid littering.
  • Avoiding the use of plastic bags and instead using cloth bags or bags prepared from old sarees, bed-sheets, curtains, etc.
  • Using both sides of a paper for writing. Reusing greeting cards and gift papers.
  • Avoiding use of tissue paper and preferring to use one’s own handkerchief.
  • Using rechargeable batteries instead of lead batteries.
  • Implementing various programmes of solid waste management and educating, encouraging the family and society in this regard. Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management
  • Avoiding ‘use and throw’ type of articles like pens, canned cold drinks and tetra-packs etc.

4. Write notes.
Meteorology, Climatic factors, Monsoon model, Industrial waste, Plastic waste, Principles of first aid.

5. Give examples of the importance of climate in the living world with explanations, in your own words.
Answer:

  1. Daily weather as also long term climatic conditions influence human lifestyle directly or indirectly.
  2. Land, water bodies, plants and animals collectively form the natural environment on earth. This environment is responsible for the development of organisms.
  3. The climate of a particular region helps to determine the diet, clothing, housing, occupations and lifestyle of the people of that region. For example, the characteristic lifestyle of Kashmiri and Rajasthani people.
  4. Salinity of marine water, formation of oceanic currents, water cycle, etc. are all related to various weather and climatic factors.
  5. Various climatic factors bring about the weathering of rocks in the earth’s crust.
  6. Climate plays a very important role in the formation and enrichment of soil.
  7. Microbes in the soil play an important role in formation of organic materials. This process depends upon various climatic factors.

6. Explain with suitable examples, the care to be taken when using the methods of transporting patients.
Answer:

  • Cradle Method: This method is used for children and under-weight victims.
  • Carrying piggy back: This method is used for carrying patients who are unconscious.
  • Human crutch method: If one of the legs is injured, the victim should be supported with minimum load on the other leg. Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management
  • Pulling or lifting method: This method is used for carrying an unconscious patient, through a short distance.
  • Carrying on four-hand chair: This method is used when support is needed for the part below the waist.
  • Carrying on two-hand chair: This method is useful for those patients who cannot use their hands but can hold their body upright.
  • Stretcher: In an emergency, if a conventional stretcher is not available, then a temporary stretcher can be made using bamboo, blanket, etc.

7. Explain the differences.

a. Weather and climate
Answer:

Weather Climate
(i) Atmospheric conditions at a specific time at a particular place are referred to as weather. (i) The climate of a particular region is the average of daily readings of various weather-related parameters recorded for several years.
(ii) Weather can change continuously. (ii) Climate remains constant in a region for a long duration.
(iii) Weather is related to a specific location and specific time. (iii) Climate is related to a longer duration and larger area.
(iv) Changes in the weather may occur for short periods of time. (iv) Changes in the climate take place slowly over a much long duration.

b. Degradable and non-degradable waste
Answer:

Degradable Waste Non-degradable Waste
(i) This type of waste is easily degraded by microbes. (i) This type of waste is not easily degraded by microbes because it takes a very long period of time and the use of various techniques.
(ii) It includes kitchen waste (spoiled food, fruits, vegetables), ash, soil, dung, parts of the plants etc. (ii) It includes plastic, metal and other similar materials.
(iii) It is also called wet solid waste or wet garbage. (iii) It is also called dry solid waste or dry garbage.
(iv) If it is carefully decomposed, we can obtain compost and fuel of good quality from it. (iv) It can be recycled.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Class 9 Science Chapter 9 Environmental Management Intext Questions and Answers

Can you recall?

Environmental Management Class 9 Question 1.
Which natural calamities have you experienced? How did they affect the conditions in your surroundings?
Answer:
In July 2005, heavy rains caused flooding in many parts of Maharashtra including large areas of Mumbai.

They effects were as follows.

  • Many people lost life.
  • There was water clogging in most parts causing severe damage to private property and public property.
  • Hospitals were submerged causing severe inconvenience to patients.
  • The city incurred huge financial loss.

Environmental Management Class 9 Exercise  Question 2.
How will you make a plan to be safe from calamities or to minimize the damage?
Answer:

  1. As a responsible citizen we can contribute by educating the people about all the precautions to be taken when such a calamity occurs.
  2. We can also contribute by minimizing the occurrence of natural calamity like flood by taking measure to reduce land pollution, proper disposing waste material and recycling them.
  3. We can plan to be safe by keeping the following things handy-torch, first-aid kit, ample food supply and water storage, emergency numbers list and extra pair of clothes.

Thus by taking appropriate measures, we can be safe from calamities

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Class 9 Science Chapter 9 Environmental Management  Question 3.
How does the atmosphere affect our daily life?
Answer:
The atmosphere affect our daily life in following ways

  • If protects us from harmfull radiations of sun.
  • Atmosphere helps to sustain life on earth by providing oxygen for human beings and animal to breathe and carbon dioxide to plants.
  • It helps in keeping us warm due to green house effect.

Environmental Management Class 9 Answers Question 4.
Forecasts about which weather related factors are given during the news bulletins on Doordarshan and Akashvani?
Answer:
Forecasts about different weather related factors such as cloud cover, rain, snowfall, wind speed and temperature are given during the news bulletins on Doordarshan and Akashvani.

9 Environmental Management Exercises  Question 5.
What is meant by pollution?
Answer:
Contamination of natural environment that can harmfully affect the ecosystem is called as pollution.

9th Class Science Chapter 9 Environmental Management Exercise Question 6.
What is meant by solid waste?
Answer:
The waste materials generated through the various daily human activities are called solid waste.

Maharashtra Board Class 9 Science Solutions Question 7.
What are the different things included in solid waste?
Answer:
Domestic waste, industrial waste, hazardous waste, electronic waste, biomedical waste, urban waste, radioactive waste and mining waste are the different things included in solid waste.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 8.
Why is it necessary to recycle non-degradable waste?
Answer:
(i) non-degradable waste cannot be easily degraded because it takes a very long period of time and the use of various techniques.
(ii) Therefore, non-degradable waste should be recycled so that it does not accumulate and cause hazards to the environment.

Question 9.
Which materials are included in solid dry waste?
Answer:
Solid dry waste includes paper, plastic, metals, glass, cardboard, thermocol etc.

Answer the following

Question 1.
In which different ways do our surroundings get polluted?
Answer:

  • Air pollution is caused due to emissions from industries, vehicles, burning of fossil fuels, construction, mining and agriculture.
  • Water pollution is caused by domestic sewage and industrial waste water released into the water bodies.
  • Soil pollution is caused due to industrial wastes, domestic waste, chemical fertilizers, biomedical waste and pesticides.
  • Noise pollution is caused by machines, vehicular traffic, loudspeakers and household appliances.

Question 2.
Which factors are affected favourably or unfavourably by climate? What must we do to minimize the effect?
Answer:

  • Climate plays a very important role in our day to day life.
  • It influences our basic needs like food, clothing and shelter as well as our occupations.
  • Climate is especially important for an agrarian country like India. Also climatic factors like direction and speed of the winds, temperature, atmospheric pressure etc. are also considered during construction work.
  • The science of meteorology helps in predicting climatic conditions by satellites in different ways such as prediction of rainfall, air pollution, dust storms, hot and cold waves tsunamic etc.
  • so that all the citizens are well-equipped beforehand and can take preventive measures.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 3.
Into which two categories can the waste materials in the lists above be classified?
Answer:
The waste materials can be classified as biodegradable waste and non-biodegradable waste.

Answer the following:

Question 1.
Make a list of various waste materials and articles in your area and prepare a chart as follows:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management 6

Question 1.
Nowadays, an electronic device – the cell phone – is very popular. From a mobile shop near your house, find out how they dispose off old and broken down cell phones.
Answer:
Old and broken down cell phones are sold to scrap dealers, who sell to a bigger dealer where reusable parts are taken out and useless parts are sent for recycling.

Question 2.
What are the different types of casualties that are seen to occur in different types of disasters?
Answer:

  • Death.
  • Injuries.
  • Loss of limbs or body parts.
  • Burns.
  • Diseases.
  • Fractures.
  • Bleeding.
  • People becoming unconscious.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 3.
Which waste management processes are used in your village/town / city?
Answer:
i. The Municipal Corporation or Municipality collects the various kinds of wastes like dry waste, solid waste, biomedical waste in different coloured containers and transports them to areas where they are treated and disposed off.

ii. Industrial waste is mostly recycled and biomedical waste is treated by the scientific methods mentioned below.

a. Solid waste is disposed off in secured land fills. The site for secured landfill is selected 2 km away from water bodies and away from human habitation. Care is taken to see that the site does not fall in sensitive zone. The landfills are layered with clay and plastic and then the garbage is spread and left.
b. Pyrolysis is done for semi combustible material. Semi combustible materials are heated to high temperature by gas arid electricity.
c. The municipality can also set up biogas plants where the solid waste is converted to biogas by anaerobic fermentation. The biogas can be used to generate power and also a good manure which can be used for agriculture.
(d) Biomedical waste is burnt in incinerators to kill the pathogens. Disinfection and sterilization is also done while treating biomedical waste by the Municipal Corporation.

These are some of the waste management processes used in village/town/city.

Answer the following questions:

Question 1.
Observe the garbage collected in the dustbin of your classroom and make a list of the various materials in it.
Discuss with your teacher, how these materials can be properly disposed off. Can we do the same with the garbage generated in our house? Think about it.
Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management 1
Answer:
(i) Garbage collected in classroom dustbin:

  • Waste paper and paper bits.
  • Pencil shavings
  • Wrappers of chocolates, biscuits etc.
  • Left over food from tiffins and fruit peels.
  • Empty ball pen refills.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

(ii) Out of these left over food from tiffins and fruit peels and pencil shaving can be used to make compost manure in school garden. The remaining waste can be sold to scrap dealers and sent for recycling.
(iii) Yes, we can do the same with the garbage generated in the house.

Question 2.
What is the main difference between what we see in the two pictures alongside (A and B).
Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management 2
Answer:

  • We see that in picture A there is lot of garbage spread around making the place very dirty, whereas in picture B, there is no garbage and the place is absolutely neat and clean.

Question 3.
What should we do to permanently maintain the condition seen in picture B?
Answer:
To permanently maintain the condition seen in picture B, we should follow the 3R mantra (Reduce, Reuse, Recycle). Also, we must avoid littering, throwing plastic bags, wrappers of chocolates, ice-creams, biscuits etc.

Class 9 Science Chapter 9 Environmental Management Additional Important Questions and Answers

Choose and write the correct option

Question 1.
is a long term predominant condition of the atmosphere.
(a) Climate
(b) Weather
(c) Pressure
(d) Biosphere
Answer:
(a) Climate

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 2.
Various climatic factors like are considered during construction of runways, seaports, huge bridges and skyscrapers, etc.
(a) Direction and speed of wind
(b) Temperature
(c) Atmospheric pressure
(d) All of these
Answer:
(d) All of these

Question 3.
The was established by the United Nations Organization on 23rd March 1950.
(a) World Health Organization
(b) National Oceanographic and Atmospheric Administration
(c) World Meteorological Organization
(d) Indian Institute of Tropical Meteorology
Answer:
(c) World Meteorological Organization

Question 4.
is/are related to various weather and climatic factors.
(a) Salinity of marine water
(b) Formation of ocean currents
(c) Water cycle
(d) All of these
Answer:
(d) All of these

Question 5.
is the founder of IMD.
(a) Dr. Vasantrao Govarikar
(b) H.F. Blanford
(c) Sir Gilbert Walker
(d) Dr. Radhakrishnan Nair
Answer:
(b) H. F. Blanford

Question 6.
The monsoon model based upon 16 worldwide parameters was developed by the initiative of
(a) Dr. Vasantrao Govarikar
(b) H.F. Blanford
(c) Sir Gilbert Walker
(d) Virghese Kurien
Answer:
(a) Dr. Vasantrao Govarikar

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 7.
In , forecasts are made taking into account the estimates of current weather related events and ongoing physical activity.
(a) Statistical model
(b) Holistic model
(c) Mathematical model
(d) All of these
Answer:
(c) Mathematical model

Question 8.
In , predictions are based upon those parameters used in other models which have the greatest effect on the monsoon.
(a) Holistic model
(b) Mathematical model
(c) Statistical model
(d) Scientific model
Answer:
(a) Holistic model

Question 9.
Radioactive materials, explosives and infectious materials are classified as waste.
(a) Industrial waste
(b) Biomedical waste
(c) Urban waste
(d) Hazardous waste
Answer:
(d) Hazardous waste

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question10.
are radioactive waste.
(a) Remains of heavy metals like arsenic, cadmium etc.
(b) Strontium-10, Cerium-141, Barium -140
(c) Waste from blood banks and laboratories
(d) All of these
Answer:
(b) Strontium-10, Cerium-141, Barium-140

Question 11.
Waste food, paper, plastic, vegetable and fruit waste etc. are classified as waste.
(a) Industrial waste
(b) Farm waste
(c) Domestic waste
(d) Urban waste
Answer:
(c) Domestic waste

Question 12.
is the 3 R mantra.
(a) Refuse, Research, Recycle
(b) Reduce, Reuse, Recycle
(c) Reduce, Reuse, Refuse
(d) Rethink, Recycle, Reuse
Answer:
(b) Reduce, Reuse, Recycle

Question 13.
is the largest producer of electricity from solid waste.
(a) India
(b) America
(c) Japan
(d) China
Answer:
(b) America

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 14.
Chemicals, pigments, sludge, ash, metals, etc. are classified as waste.
(a) Domestic waste
(b) Industrial waste
(c) Urban waste
(d) Hazardous waste
Answer:
(b) Industrial waste

Question 15.
Period of natural degradation for banana peels is
(a) 1 month
(b) 1-2 weeks
(c) 3-4 weeks
(d) 2 months
Answer:
(c) 3-4 weeks

Question 16.
Period of natural degradation for cloth bags is
(a) 2-3 weeks
(b) 1 month
(c) 5 months
(d) 1 year
Answer:
(b) 1 month

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 17.
Period of natural degradation for wood is
(a) 1 month
(b) 5 months
(c) 10-15 years
(c) 40-50 years
Answer:
(c) 10-15 years

Question 18.
Period of natural degradation for certain plastic bags is
(a) 50-100 years
(b) infinite duration
(c) 10 lakh years
(d) 1000 years

(c) 10 lakh years

Question 19.
Period of natural degradation for thermocol or Styrofoam cup is
(a) 10 lakh years
(b) infinite duration
(c) 200-250 years
(d) 1 year
Answer:
(b) infinite duration

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 20.
If any bone is fractured, it is essential that the fractured part be
(a) mobilized
(b) immobilized
(c) pulled
(d) massaged
Answer:
(b) immobilized

Question 21.
For transporting children and under-weight victims, method is used.
(a) carrying piggy back
(b) human crutch method
(c) cradle method
(d) stretcher
Answer:
(c) cradle method

Question 22.
method is useful to carry patients who are unconscious.
(a) Cradle Method
(b) Carrying piggy back
(c) Human crutch
(d) Carrying on four-hand chair
Answer:
(b) Carrying piggy back

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 23.
method is used for carrying an unconscious patient through a short distance.
(a) Carrying piggy back
(b) Cradle method
(c) Carrying on two-hand chair
(d) Pulling or lifting method
Answer:
(d) Pulling or lifting method

Question 24.
method is useful to carry patients who cannot use their hands but can hold their body upright.
(a) Carrying on four-hand chair
(b) Carrying on two-hand chair
(c) Carrying piggy back
(d) Cradle method
Answer:
(b) Carrying on two-hand chair

v 25.
method is used to carry patients when support is needed for the part below the waist.
(a) Carrying on four-hand chair
(b) Carrying on two-hand chair
(c) Stretcher
(d) Pulling or lifting method
Answer:
(a) Carrying on four-hand chair

Question 26.
For injuries like sprains, twisting and contusion, should be applied on the injured part.
(a) turmeric powder
(b) antiseptic pad
(c) ice-pack
(d) pressure
Answer:
(c) ice-pack

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Find the odd man out:

Question 1.
Waste food, paper, plastic, bandages.
Answer:
Bandages: It is a biomedical waste whereas the others are domestic waste.

Question 2.
Pesticides, fertilizers, crop residue, sludge.
Answer:
Sludge: It is an industrial waste whereas the others are farm waste.

Question 3.
Strontium -10, Cerium – 141, Barium – 140, Cadmium.
Answer:
Cadmium: It is a mining waste whereas the others are radioactive waste.

Question 4.
Banana peels, cloth bag, food waste, plastic bag.
Answer:
Plastic bag: It is a non-biodegradable waste whereas the others are degradable wastes.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 5.
Fruits, ash, metals, vegetables.
Answer:
Metal: It is a non-biodegradable waste whereas the others are degradable wastes.

Complete the analogy:

(1) Specific duration and specific time: Weather : : Longer duration and longer time :
(2) Mumbai: 5000 tons solid waste :: Pune:
(3) Kitchen waste, parts of plants : Wet solid waste :: Plastic, metals :
(4) Largest producer of electricity from solid waste : America : : Production of useful materials from banana peelings :
(5) Cloth bags : 1 month : : Rags :
(6) Tin cans : 50-100 years : : Aluminium cans :
(7) Wood : 10-15 years :: Styrofoam :
Answer:
(1) Climate
(2) 1700 tons solid waste
(3) Dry solid waste
(4) Japan
(5) 5 months
(6) 200-250 years
(7) Infinite duration

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) Mining waste (a) Leaves, flowers, crop residue.
(2) Electronic waste (b) Remains of heavy metals like lead, arsenic, cadmium.
(3) Farm waste (c) Strontium-10,
Cerium-141,
Barium-140
(4) Radioactive waste (d) Cell phones, TV sets, Computers

Answer:
(1 – b),
(2 – d),
(3 – a),
(4 – c)

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 2.

Column ‘A’ Column ‘B’
(1) Banana peels (a) 10 lakh years.
(2) Plastic bags (b) 200-250 years.
(3) Leather shoes (c) 3-4 weeks
(4) Aluminium cans (d) 40-50 years

Answer:
(1 – c),
(2 – a),
(3 – d),
(4 -b)

Question 3.

Column ‘A’ Column ‘B’
(1) Thermocol
(2) Tin cans
(3) Woollen socks
(4) Wood
(a) 1 year.
(b) Infinite duration.
(c) 10-15 years
(d) 50-100 years

Answer:
(1 – b),
(2 – d),
(3 – a),
(4 – c)

State whether the following statements are true or false. Correct the false statements:

Question 1.
Climate is a long term predominant condition of the atmosphere.
Answer:
True.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 2.
Climate changes continuously.
Answer:
False. Climate does not change continuously. It remains constant in a region for a long duration.

Question 3.
If present climatic conditions are analysed with reference to the past climatic conditions, we can predict climatic changes of the future.
Answer:
True.

Question 4.
Forecasting is difficult for places where climatic changes are slow and of a limited nature.
Answer:
False. Forecasting is easy for places where climatic changes are slow and of a limited nature.

Question 5.
Climate plays a very important role in the formation and enrichment of soil.
Answer:
True.

Question 6.
Prediction maps are prepared once in 24 hours.
Answer:
False. Prediction maps are prepared twice in every 24 hours.

Question 7.
The first prediction of monsoon in India was made by Dr. Vasantrao Govariakar.
Answer:
False. The first prediction of monsoon in India was made my H.F. Blanford.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 8.
H.F. Blanford used the rainfall in Kerala as the parameter for prediction of monsoon in India.
Answer:
False. H.F. Blanford used the snowfall in Himalayas as a parameter for prediction of monsoon in India.

Question 9.
In Holistic model, predictions are made taking into account estimates of current weather-related events and ongoing physical interactions between them.
Answer:
False. In Holistic model, predictions are based upon those parameters used in other models which have the greatest effect on monsoon.

Question 10.
Any meteorological model depends upon the inter-relationships between parameters used in that model and the results expected from it.
Answer:
True.

Question 11.
Remains of heavy metals like lead, arsenic, cadmium, etc. from mines are industrial waste.
Answer:
False. Remains of heavy metals like lead, arsenic, cadmium etc. from mines are mining waste.

Question 12.
Use of ‘use and throw’ type of articles like pens, canned cold drinks, tetra packs should be encouraged in waste management.
Answer:
False. Use of ‘use and throw’ type of articles like pens, canned cold drinks, tetra packs should be strictly avoided in waste management.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 13.
We should use tissue paper instead of one’s own handkerchief for effective waste management.
Answer:
False. We should use one’s own handkerchief instead of tissue paper for effective waste management.

Question 14.
Certain plastic bags take 1 month to degrade.
Answer:
False. Certain plastic bags take 10 lakh years to degrade.

Question 15.
Banana peels can degrade in 3-4 weeks.
Answer:
True.

Question 16.
If the victim has burn injuries, it is beneficial to cover the burnt part with blanket.
Answer:
False. If the victim has burn injuries, it is beneficial to hold the injured part under clean and cold flowing water for at least 10 minutes.

Question 17.
CPR helps to bring the circulation to normal.
Answer:
True.

Question 18.
If breathing has stopped, the head should be held in backward sloping position.
Answer:
False. If breathing has stopped, the victim should be given artificial ventilation by mouth to mouth resuscitation.

Question 19.
Japan is the largest producer of electricity from solid waste.
Answer:
False. America is the largest producer of electricity from solid waste.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 20.
China has developed the projects of production of threads, paper and other useful materials from banana peels.
Answer:
False. Japan has developed the projects of production of threads, paper and other useful materials from banana peels.

Define the following:

Question 1.
Weather
Answer:
Atmospheric conditions at a specific time at a particular place are referred to as weather.

Question 2.
Climate
Answer:
The climate of a particular region is the average of daily readings of various weather-related parameters recorded for several years.

Question 3.
Meteorology
Answer:
The science that studies the inter-relationships between the various components of air, natural cycles, geological movements of earth and climate is called meteorology.

Question 4.
Solid waste
Answer:
Waste materials generated through daily human activities are called solid waste.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 5.
Urban waste
Answer:
Waste generated through household industries and large commercial and industrial establishments is called Urban waste.

Answer the following:

Question 1.
What is the necessity of solid waste management?
Answer:
Necessity of solid waste management:

  • For preventing environmental pollution and to keep the surroundings clean.
  • For energy as well as fertilizer production and through that to generate work and employment opportunities.
  • To reduce the strain on natural resources through treatment of solid waste.
  • To improve the health and quality of life and to maintain environmental balance.

Question 2.
What are the harmful effects of solid waste?
Answer:
Harmful effects of solid waste:

  • Effect on biodiversity.
  • Releases bad odour.
  • Produces toxic gases.
  • Leads to degradation of natural beauty.
  • Leads to pollution of air, water and soil.
  • Spreads diseases.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 3.
What first-aid should be given for injuries like sprains, twisting and contusion?
Answer:
For injuries like sprains, twisting and contusion, the ‘RICE’ remedy should be applied:

  • Rest: Allow the victim to sit in a relaxed position.
  • Ice: Apply an ice-pack to the injured part.
  • Compression: After the ice-pack treatment, the injured part should be massaged gently.
  • Elevate: The injured part should be kept in a raised/elevated position.

Question 4.
Why do meteorological models need to be changed continually?
Answer:

  1. Any meteorological model depends upon the inter-relationship between parameters used in that model and the results expected from it.
  2. However, as these inter-relationships with reference to the ocean and atmosphere are never constant, meteorological models need to be changed continually.

Question 5.
What is urban waste? What does it include?
Answer:

  1. Waste generated through household industries and large commercial and industrial establishments is called urban waste.
  2. It includes carry bags, glass, metal pieces and rods, threads, rubber, paper, cans from shops, waste from vegetable and meat markets, construction waste etc.

Question 6.
What does biomedical waste include?
Answer:
biomedical waste includes bandages, dressings, gloves, needles, saline bottle, medicines, medicine bottles, test tubes, body parts, blood etc. from clinics, hospitals, blood banks and laboratories.

Write short notes on:

Question 1.
Meteorology
Answer:

  • The science that studies the inter-relationship between the various components of air, natural cycles, geological movements of the earth and climate is called meteorology.
  • Meteorology includes the study of storms, clouds, rainfall, thunder, lightning etc.
  • Depending upon the study of such factors, weather forecasts are made.
  • They are useful to common people, farmers, fisheries, aviation services, water transport and various other organizations. Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 2.
Climatic factors
Answer:

  • The climate of a particular region is the average of daily readings of various weather-related parameters recorded for several years.
  • Climatic factors include direction and speed of wind, temperature, atmospheric pressure, clouds, rainfall, humidity, visibility etc.
  • These factors influence our basic needs like food, clothing, shelter as well as our occupations.
  • Various climatic factors bring about the weathering of rocks in the earth’s crust.
  • Microbes in the soil play an important role in formation of organic materials. This process also depends upon various climatic factors.

Question 3.
Monsoon model
Answer:

  1. The tradition of forecasting the monsoon season in India is older than 100 years.
  2. After the famine of 1877, H.F. Blanford, the founder of IMD had made such a prediction for the first time taking the snowfall in the Himalayas as a parameter for this prediction.
  3. In the decade of the 1930’s, the then director of IMD, Sir Gilbert Walker had underlined the relationship between various worldwide climatic factors and the Indian monsoon and based on available observations and previous recordings related to this relationship, he put forth a hypothesis regarding the nature of the monsoon.
  4. With the initiative of Dr. Vasantrao Govarikar in the decade of the 1990’s, a monsoon model based upon 16 worldwide climatic parameters was developed. This model was in use from 1990 to 2002.
  5. Presently, new models are being developed at IITM. Work is in progress at two levels, namely designing new models and developing new technology.
  6. The main focus is on the development of the radar system and satellite technology.

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 4.
Plastic waste
Answer:

  • Plastic waste is the accumulation of plastic products in the environment that adversely affects environment, humans and animals.
  • Plastic waste is excessively generated as plastic is inexpensive and durable.
  • Plastic is slow to degrade. It takes around 10 lakh years for certain plastic bags to degrade. .
  • Plastic waste affects land and water.
  • It also affects the health of animals, cattle unknowingly ingest these plastic bags leading to stomach cancer in them.
  • Plastic also releases toxic chemicals which are carcinogenic to humAnswer:
  • To avoid plastic waste, cloth bags should be used instead of plastic bags. Plastic articles should be recycled.

Question 5.
Industrial waste
Answer:

  • Industrial waste is the waste produced by industrial processes or activities.
  • There is a huge variety of industries producing different types of materials and articles. All of these use raw materials and give out a lot of waste.
  • There are hundreds of mines which extract copper, silver, gold, iron, coal etc. Huge quantities of waste are produced while processing them.
  • Cement industries give out solid, liquid and gaseous wastes.
  • While refining crude oil, a lot of poisonous gaseous and liquid wastes are produced.
  • Construction units produce huge quantities of waste stones, pebbles, broken bricks, wood waste etc. Mostly they are dumped in landfills.
  • It also includes chemicals, pigments, sludge, ash, metal, etc. given out from mining, textile, construction, chemical industries.

Question 6.
Principles of first aid
Answer:
Life and Resuscitation – ‘ABC’ is the Basic Principle of first aid which is provided to the victims of disaster.

  1. Airway: If the victim has difficulty in breathing, the head should be held in a backward sloping position or the chin should be raised so that the respiratory passage remains open.
  2. Breathing: If breathing has stopped, the victim should be given artificial ventilation by mouth to mouth resuscitation.
  3. Circulation: If the victim is unconscious, then after giving mouth to mouth respiration twice,

Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management

Question 7.
Statistical Model
Answer:

  • In this model, current climatic observations in a region are compared with earlier parameters such as oceanic temperature, atmospheric” pressure and the nature of the monsoon rainfall for several years.
  • This data is comparatively analysed by statistical methods and predictions are made about the monsoon in the present conditions.

Question 9.
Holistic Model
Answer:

  • In this model, predictions are based upon those parameters used in other models which have the greatest effect on the monsoon.
  • Nowadays, predictions declared by IMD are the collective outcome of various model. This is called a holistic model.

Complete the following concept chart:

Question 1.
Weather-related climatic factors
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management 3

Question 2.
Harmful effects of solid waste
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management 4

Question 3.
Scientific and eco-friendly waste Management
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management 5

Answer in detail:

Question 1.
What are the principles of solid waste management?
Answer:
Principles of solid waste management:

  • Reuse: After use, materials should be reused for some other proper purposes.
  • Refuse: Refusal to use articles made from non- degradable articles like plastic and thermocol.
  • Recycle: Production of useful articles by recycling solid wastes. For example, paper and glass can be recycled.
  • Rethink: Rethinking our habits, activities and their consequences in connection with the use of various articles of daily use. Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management
  • Reduce: Restricting the use of resources to avoid their wastage.
  • Old materials should be reused. One thing should be shared by many, use and throw type of objects should be avoided.
  • Research: Conducting research related to reuse of materials that are temporarily out of use.
  • Regulation and Public awareness: Following the laws and rules related to waste management and motivating others to do the same.

Question 2.
What is disaster management? What actions does it include?
Answer:
Disaster management is action implemented through proper planning, organized activity and co-ordination.

It includes the following:

  • Prevention of loss and danger.
  • Improving tolerance.
  • Providing relief from disaster, minimising the intensity and extent of harm.
  • Preparation to face the disaster.
  • Immediate action in the disaster situation.
  • Assessment of damages and intensity of the disaster.
  • Arranging for rescue work and help.
  • Rehabilitation and rebuilding.

Question 3.
Write a short note on : Indian Meteorological Department
Answer:

  • The Indian Meteorological Department was founded by the British in 1875 at Shimla.
  • Its head office is at Pune and its Regional offices are at Mumbai, Kolkata, Chennai, Nagpur and Delhi.
  • Maps are prepared every day which indicate the daily predictions about the weather.
  • Such maps are prepared and published twice in every 24 hours.
  • In this institute, research goes on continuously on various aspects like instruments for climatic readings, predictions made about climate using radar, predictions about climate related to seismology, predictions regarding rainfall by satellites, air pollution etc. Maharashtra Board Class 9 Science Solutions Chapter 9 Environmental Management
  • The Indian Meteorological Department provides information regarding weather and climatic conditions to other departments like aviation, shipping, agriculture, irrigation, marine oil exploration and production etc.
  • Predictions regarding calamities like dust storms, sand storms, heavy rainfall, hot and cold waves, tsunami, etc. are communicated to various departments, all types of mass communication media and all citizens.
  • India has launched several satellites equipped with highclass technology.
  • Observatories at several locations are doing excellent work in the analysis of the information received from these satellites.

9th Std Science Questions And Answers:

Life Processes in Living Organisms Class 9 Science Chapter 15 Questions And Answers Maharashtra Board

Class 9 Science Chapter 15

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 15 Life Processes in Living Organisms Question Answer Maharashtra Board

Class 9 Science Chapter 15 Life Processes in Living Organisms Question Answer Maharashtra Board

1. Match the pairs and explain.

Column ‘A’ Column ‘B’
(1) Growth of pollen tube towards ovules
(2) Growth of shoot system
(3) Growth of root system
(4) Growth towards the water
(a) Gravitropic movement
(b) Chemotropic movement
(c) Phototropic movement
(d) Growth-irrelevant movement
(e) Hydrotropic movement

Answer:
(1 – b),
(2 – c),
(3 – a),
(4 – e).

Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms

2. Complete the paragraph.

The milk was on the stove. Rasika was engrossed in watching television. She smelled something burning. She ran towards the kitchen. The milk was boiling over. She held the vessel with her bare hands but, screaming, she let it go at once. This activity was controlled by ……………….. cells. Special ends of ……………….. in these cells collected the information, from where it was transferred to the ……………….. and then towards the terminal end of the ……………….. The chemicals produced at the terminal end passed through the minute space i.e. ………………. . In this way, ……………….. were conducted in the body and the process of ……………….. was completed by conducting the impulses from ……………….. to ………………

(Nerve, muscle cell, impulse, dendrite, synapse, axon, reflex action, cell body)
Answer:
nerve, dendrite, cell body, axon, synapse, impulses, reflex action, nerve, muscle cell.

3. Write notes on Root pressure, Transpiration, Nerve cell, Human brain, Reflex action

4. Name the hormones of the following endocrine glands and the function of each.
Pituitary, Thyroid, Adrenal, Thymus, Testis, Ovary.

Life Processes In Living Organisms Class 9 Question 1.
Pituitary:
Answer:

Hormone Function
(i) Growth hormone Stimulates growth of bones.
(ii) Adrenocorticotropic hormone Stimulates adrenal gland.
(iii) Thyroid-stimulating hormone Stimulates thyroid gland.
(iv) Prolactin Stimulates milk production.
(v) Follicle-stimulating hormone Controls growth of gonads.
(vi) Luteinizing hormone Controls menstrual cycle and ovulation.
(vii) Oxytocin Contracts uterus during parturition.
(viii) Antidiuretic hormone Regulates water-level in the body.

Class 9 Science Chapter 15 Life Processes In Living Organisms Question 2.
Thyroid
Answer:

Hormone Function
(i) Thyroxine Controls growth of body and metabolic activities.
(ii) Calcitonin Controls calcium metabolism and calcium level in blood.

Life Processes In Living Organisms Class 9 Exercise Question 3.
Adrenal
Answer:

Hormone Function
(i) Adrenalin and nor-adrenalin Controls behaviour during crisis and emotional situation.
(ii) Corticosteroid Maintains balance of Na+ and K+ and stimulates metabolism.

9th Class Science Chapter 15 Life Processes In Living Organisms Exercise Question 4.
Thymus
Answer:

Hormone Function
Thymosin Control the cells which give rise to immunity.

Question 5.
Testis
Answer:

Hormone Function
Testosterone Stimulates growth of secondary sexual characteristics like beard, mustache, hoarse voice etc. in men.

Std 9 Science Chapter 15 Life Processes In Living Organisms Question 6.
Ovary
Answer:

Hormone Function
(i) Oestrogen Stimulates growth of the endometrium. Stimulates growth of secondary sexual characteristics in women.
(ii) Progesterone Prepares the endometrium for conception and maintains pregnancy.

5. Draw and label the diagrams.
Human endocrine glands, Human brain, Nephron, Nerve cell, Human excretory system.

15. Life Processes In Living Organisms Question 1.
Human endocrine glands
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 1

15 Life Processes In Living Organisms Question 2.
Human brain
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 2

Life Processes In Living Organisms 9th Class Notes Question 3.
Nephron
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 3

Life Processes In Living Organisms Class 9 Questions And Answers Question 4.
Nerve cell
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 4

Life Processes In Living Organisms Class 9 Solutions Question 5.
Human excretory system
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 5

6. Answer the following.

a. Explain chemical co-ordination in humans and give the names and functions of some hormones.
Answer:
Chemical co-ordination in humans :

  • In humans chemical co-ordination is brought about with the help of certain chemical substances called hormones.
  • Hormones are secreted by endocrine glands. These glands are also called ductless glands.
  • These glands do not have any arrangement of their own to either store or carry their secretions. So, the hormones are directly released into the blood circulation.
  • Though the endocrine glands are present at specific locations in our body, their secretions reach all parts of the body via the blood.
  • Endocrine glands along with the nervous systems are responsible for control and co-ordination in our body.
    These two systems help each other to control and integrate the various activities of the body.
  • A marked difference between these two systems is that nerve impulses are fast but short-lived whereas the actions of hormones are very slow but long-lasting.
  • It is very important that hormones are secreted only in the required quantity and there is a special mechanism which controls the quantity and timing of hormone secretion.
  • For example, whenever there is an increase in blood glucose level, certain cells in the pancreas get stimulated and as a response, they release a greater quantity of insulin, thus bringing down the sugar level to normal.
Hormone Function
(1) Growth hormone Stimulates growth of bones.
(2) Luteinizing hormone Controls menstrual cycle and ovulation.
(3) Thyroxine Controls growth of body and metabolic activities.
(4) Insulin Stimulates liver to convert excess blood glucose to glycogen.
(5) Testosterone Stimulates growth of secondary sexual characters like beard, mustache, hoarse voice, etc. in men.

Names and functions of some hormones:

b. Explain the difference between the excretory system of humans and plants.
Answer:

Excretory system of humans Excretory system of plants
(i) In humans, the excretory system carries out the function of the removal of waste from the body. (i) In plants, there is no special system or organ for excretion.
(ii) Waste substances are generally eliminated out of the body (ii) Most of the waste substances are stored in the leaves, flowers, fruits and bark of the stem.
(iii) The excretory products are urea, uric acid, ammonia, etc. (iii) The excretory products are gum, resin, latex of rubber, etc.

c. Explain co-ordination in plants with the help of suitable examples.
Answer:

  • Plants do not have systems like the nervous system or muscular system.
  • In plants, movements are mainly in the form of responses given to the stimuli.
  • Plants show two types of movements – growth relevant movements and growth irrelevant movements.
  • Movement or growth of any part of the plant in response to an external stimulus is called tropism or tropic movement.
  • Examples of growth-relevant movements are :
  • The shoot system of any plant responds to the light stimulus i.e., it grows towards the source of light. This is called phototropic movement.
  • The root system of plants responds to stimuli like gravitation and water. These responses are called gravitropic movement and hydrotropic movement respectively.
  • Movement shown by plants in response to specific chemicals is called chemotropic movement.
  • For example, the growth of the pollen tube towards the ovule.
  • Some specific movements of the plants do not lead to the plant growth.
  • Such movements are called growth-irrelevant movements.
  • As a response to changes in the surroundings, plant hormones bring about various movements in plants.
  • Examples of growth irrelevant movements are closing of leaflets of Touch-me-not (Mimosa) plant on touch.
  • In the plant called Venus fly trap, when an insect visits the flower like trap, the trap closes up and the trapped insect is digested by the plant.
  • The lotus flower opens during day-time while that of the tuberose opens at night.
  • In plants like Touch-me-not (Mimosa), movement also occurs at places other than where it has been touched.
  • Hence, we can infer that the information about the touch must have been released within the plant from one place to another.
  • Plants use electrochemical impulses for transfer of information from one place to another.
  • Plant cells change their shape by increasing or decreasing the water content and thereby, bring about the movements of plants.

7. Explain in your own words with suitable examples.

a. What is meant by co-ordination?
Answer:

  • Several different organ systems function in multicellular organisms.
  • Their life goes on smoothly if there is co-ordination between different organ systems or organs and the stimuli in the surrounding.
  • Depending upon this, we can say that systematic regulation of different processes can be called control and bringing about the different processes in the proper sequence is called co-ordination.
  • If any activity in the body is to be completed successfully, proper co-ordination between different systems, and organs participating at different steps of that activity is necessary.
  • If due to lack of co-ordination, there is confusion at any step, the activity may not get completed. There should be no randomness at any step.
  • There should be proper co-ordination between internal activities of the body resulting from various factors like body temperature, water- level, enzyme-level, etc. or stimuli arising in the surrounding environment.
  • Proper co-ordination between various systems of an organism helps to maintain a state of equilibrium called homeostasis which is necessary for the optimal efficiency of the body.

b. How does excretion occur in human beings?
Answer:

  • Removal of wastes from the body is very important and it is the excretory system that carries out this function.
  • The human excretory system consists of a pair of kidneys, a pair of ureters, the urinary bladder and the urethra. Urine is formed by the kidneys by separating the waste and unwanted excess substances from the blood.
  • The functional unit of the kidney that performs the basic function of filtration is called a nephron. The urea produced in the liver comes into the blood.
  • When the urea-containing blood comes into the glomerulus, it is filtered through capillaries and urea and other similar substances are separated from it.
  • Water molecules and small molecules of some other substances can cross the semipermeable membrane of Bowman’s capsule.
  • The solution accumulated in the cavity of Bowman’s capsule passes into the tubular part of the nephron. Here, molecules of water and some other useful substances are reabsorbed into the blood.
  • Urine is formed from the remaining solution which is full of waste materials. The urine is carried by the ureters and stored in the urinary bladder. Afterwards, urine is given out through the urethra.

c. How is excretion in plants useful to human beings?
Answer:

  • Plants give out oxygen during photosynthesis by diffusion which is useful to human beings for respiration.
  • Some waste materials of plants for example, gum, resin, latex of rubber, etc. are useful to humans.

d. Describe the transportation system in plants.
Answer:

  • Plants have two types of conducting tissues i.e., xylem which conducts water and phloem which conducts food.
  • During transpiration, water is released into the atmosphere.
  • As a result, water level is the epidermal layer of the leaf decreases.
  • Water is brought up to the leaves through the xylem so as to compensate for the lost water.
  • Transpiration helps in absorption of water and minerals and distribution to all parts of the plant whereas root pressure preforms the important role of pushing the water up at night time.
  • The food produced by leaves is transported to each cell of the plant through phloem.
  • When the food material like sucrose is transported towards a part of the plant through phloem, using ATP, the water concentration in that part decreases.
  • As a result, water enters the cell by diffusion. The pressure on the cell wall increases due to increase in the cellular contents.
  • Due to the increased pressure, food is pushed into the neighboring cells where the pressure is low.
  • This process helps the phloem to transport the materials as per the need of the plant.

Class 9 Science Chapter 15 Life Processes in Living Organisms Intext Questions and Answers

Class 9 Science Chapter 15 Life Processes In Living Organisms Exercise  Question 1.
How do the digestive system and respiratory system work?
Answer:
(a) Digestive system:

  • The digestive system is responsible for digestion of food. i.e. breaking down complex insoluble food into simpler soluble form.
  • The digestive system is made up of alimentary canal and digestive glands.
  • The alimentary canal is a long and muscular tube of varying diameter which extends from the mouth to the anus.
  • The digestive glands associated with alimentary canal are salivary glands, gastric glands, liver and pancreas.
  • The process of digestion includes various stages wherein the food passes through the different organs of the alimentary canal and there it is digested part by part with the help of various enzymes secreted from the digestive glands.
  • After the food is digested, the important nutrients are absorbed into the blood and the undigested food and residue of digested food is removed out of the body through the anus.

(b) Respiratory system:

  • Respiration is the process of release of energy from the assimilated food.
  • The respiratory system consists of nose, pharynx, larynx, trachea, bronchi and lungs (alveoli).
  • Oxygen enters our body through breathing.
  • This oxygen is taken to the cells and with the help of oxygen, food is oxidized into the cells to release energy in the form of ATR
  • During this process, carbon dioxide and water vapour are given out as by-products which are given out during breathing.

Thus, the digestive and respiratory systems work in co-ordination with each other to produce energy from food.

Life Processes In Living Organisms Exercise Question 2.
Previously you have performed the activity of observing a branch covered in a plastic bag. What did you observe in that activity?
Answer:

  • When a branch is covered in a plastic bag, we observe that small droplets of water accumulate on the inner side of the plastic bag.
  • This shows that leaves release water in the form of vapour through the stomata during transpiration. These vapours get condensed to form water droplets and accumulate inside the bag.

Life Process In Living Organism Exercise Question 3.
Which are the sensory organs of an organism? What is their function?
Answer:

  • Sensory organs of an organism are eyes, ears, tongue, skin and nose.
  • The eyes help the organism to perceive the sensation of light to see the objects around it.
  • The ears help the organism to perceive the sensation of sound and also perceive the sensation of pressure.
  • The tongue helps to perceive different tastes and also in chewing the food.
  • The skin helps to perceive the sensation of touch, pain, pressure etc.
  • Some insects also possess antennae for sensing the changes in the environment.

15 Life Processes In Living Organisms Class 9 Question 4.
Where are the gustatory and olfactory nerves to be found?
Answer:

  • Gustatory nerves are concerned with sense of taste whereas olfactory nerves are concerned with the sense of smell.
  • Gustatory nerves are found in the tongue from where they carry impulses to the gustatory cortex in the brain.
  • Whereas the olfactory nerves are found in the nasal passage from where they carry impulses to the olfactory lobes in the brain.

Answer the following questions:

15.Life Processes In Living Organisms Question 1.
Why do we eat fruits and vegetables? Do the plants also need minerals like we do?
Answer:

  • We eat fruits and vegetables to get nutrients like carbohydrates, vitamins, minerals and fibers. These nutrients are important for the proper functioning of the body .
  • Plants too need minerals for their growth and development.

Life Processes In Living Organisms Class 9 Notes Question 2.
From where do plants get inorganic substances other than carbon dioxide and oxygen?
Answer:
Plants get inorganic substances like nitrogen, phosphorus, magnesium, manganese, sodium etc. from the soil. Soil is the nearest and richest source of these elements.

9th Std Science Chapter 15 Life Processes In Living Organisms Question 3.
At least a small quantity of garbage or waste is produced every day in each house. What will happen if you keep this garbage for many days in your house?
Answer:

  • If we keep garbage for many days in our house, it will start decaying due to the growth of decomposers like fungi and bacteria.
  • This will result in foul odour and will also lead to spread of diseases.

Life Processes In Living Organisms Exercise Class 9 Question 4.
Sometimes, while eating we bite our own finger or tongue by mistake.
Answer:

  • Sometimes, while eating we over-fill our mouth causing the bite to misalign, thus we bite our own finger or tongue by mistake.
  • If we are dehydrated, the muscles of the mouth and tongue do not function properly causing us to bite our own tongue.
  • If we are distracted with other activities while eating, it can cause us to bite our tongue or finger.
  • Thus, due to lack of co-ordination sometimes while eating, we bite our own finger or tongue by mistake.

9th Class Science 15 Chapter Life Processes In Living Organisms Question 5.
Sometimes, we choke while eating in a hurry.
Answer:

  • There is an elastic flap at the starting of the wind-pipe called epiglottis which opens whenever we breathe in air and closes to prevent the entry of food into the respiratory system.
  • While eating in hurry, due to lack of co-ordination, the epiglottis opens to take in air but food too enters in causing us to choke.

Life Processes In Living Organisms Class 9 Extra Questions Question 6.
An injury to the medulla oblongata can lead to death. Why?
Answer:

  • The medulla oblongata controls involuntary activities like the beating of the heart, blood circulation, breathing etc.
  • An injury to the medulla oblongata will stop these activities, leading to death.

Life Processes In Living Organisms Class 9 Maharashtra Board  Question 7.
You may have seen how a drunken person struggles to maintain his body balance. An excess of alcohol in the body causes one to lose control over it. Why does this happen? Find out the reason with help of the internet.
Answer:

  • The cerebellum co-ordinates voluntary movements and maintains balance of the body.
  • An excess of alcohol causes dysfunctioning of the cerebellum.
  • It results in primary loss of co-ordinated movements and loss of balance.

9th Life Processes In Living Organisms 5 Question 8.
Why does this happen?

(a) Leaves of plants fall off in a particular season.
Answer:

  • Plants lose water through their leaves by transpiration. So some plants shed their leaves in dry season to reduce transpiration and conserve water.
  • In plants, the waste substances are stored in the vacuoles of leaf-cells which are then shed along with the leaves in a specific season.

(b) Fruits, flowers fall off after a certain period of time.
Answer:

  • In plants, waste substances are stored in the vacuoles of fruits and flowers. So, they fall off after some time to excrete these wastes.
  • Also, when fruits and flowers mature, the supply of nutrients to mature fruits and flowers is blocked. This causes abscission in that part of the plant leading to their detachment from the plant body and falling off.

(c) Substances like resin, gum, etc. are given out of the plant body.
Answer:

  • Resin, gum etc. are the waste products of plants. They are given out after some time as a process of excretion in plants.
  • Many plants, particularly, woody plants produce resin and gum in response to injury. These resin and gum act as bandage protecting the plants from invading insects and pathogens.

15 Life Processes In Living Organisms Exercise Question 9.
Which waste products are produced in our body through metabolic activities?
Answer:
The waste products formed in our body through metabolic activities are urea, uric acid, ammonia, carbon dioxide etc.

Life Processes In Living Organisms 9th Class Exercise Question 10.
As compared to the monsoons and winter a very small quantity of urine is produced in the summer season. Why is it so?
Answer:
In summer, we sweat more due to excess heat. So, the body tries to conserve water by reducing urine formation. This is to prevent dehydration.

Question 11.
In adults, the process of urination is under their control but not in infants. Why is it so?
Answer:

  • In adults, the muscles of the bladder are under the voluntary control. So the process of urination is under control.
  • In infants, the bladder muscles are not mature until about two years of age. So, the process of urination is not under control.

Answer the following questions:

Question 1.
Observe the following figures carefully.
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 9
Answer:
Observations:

  • Figure ‘A’ shows chemotropic movement. Pollen tube grows towards the ovule in response to specific chemicals released by the ovule.
  • Figure ‘B’ shows phototropic movement. Shoot grows towards the Sun in response to stimulus of light.
  • Figure ‘C’ shows thigmotropic movement. Tendrils of climbers are sensitive to touth. When they come in contact with an object, they entwine around the object and cling to it.
  • All the above movements are related to growth. Hence, they are called growth relevant movements.

Question 2.
Observe the pictures carefully and think about them.
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 10
Answer:

  • Figure ‘A’ shows touch-me-not (Mimosa) plant. In this plant, the leaflets close in response to stimulus of touch.
  • Figure ‘B’ shows Venus fly trap. In this plant, there is a trap that appears and smells like a flower and deceives the insect. When an insect visits that flower-like trap, the trap closes up and the trapped insect is digested by the plant,
  • Figure ‘C’ shows lotus plant. The lotus flower opens during day-time.?
  • Figure ‘D’ shows Balsam plant. In this plant, the ripened fruit dehisces (bursts open) at the right time to disperse the seeds.

Question 3.
As you watch the match being played on your school ground, you will see the control and co-ordination among the movements of the players. Make a list of all such different actions.
Answer:
In an inter-house basketball match, control and co-ordination is seen in dodging the ball, jumping, running, passing it to the team mates and dribbling the ball.

Question 4.
Observe the figure carefully and as per the numbers in that figure, answer the following questions.
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 11

(a) What is happening at 1 and 2?
Answer:
At 1 stimulus of heat is received by thermo-receptors in the skin. At 2, the hand is withdrawn instantly.

(b) Which nerve carried the impulse to the point marked 3? In which direction is it conducting the impulse?
Answer:
Sensory nerve carried the impulse to the point marked 3. The impulse is conducted from the skin towards the spinal cord.

(c) Which is the nerve shown by 4?
Answer:
The nerve shown by 4 is association neuron. It is present in the gray matter of the spinal cord. It receives the sensory impulse, interprets it and generates impulse.

(d) Which is the organ marked as 5?
Answer:
Spinal cord.

(e) At 6, which nerve is conducting the response impulse?
Answer:
Motor nerve.

(f) At 7, where has the impulse reached? What is its effect?
Answer:
At 7, the impulse has reached the muscles of the hand. As a result, the muscles contract and the hand is withdrawn instantly in response to the stimulus of heat.

Question 5.
Sketch and label:
Answer:

  • Stimulus
  • Hand is withdrawn
  • Sensory neuron
  • Association neuron
  • Spinal cord
  • Motor neuron
  • Muscles

(6) Observe the figure and list down functions for ‘A’ and ‘B’
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 8
Answer:

Functions of Left Brain Functions of right brain
(i) Analytical thinking Holistic thinking
(ii) Logical thinking Intuition
(iii) Language Creativity
(iv) Science and Mathematics Art and music

Answer the following questions:

Question 1.
Take a small plant like balsam or tuberose with its roots intact. Wash and clean its roots. As shown in the fig, keep it in the water containing a stain like safranin or eosin. Observe the stem and the veins of the leaves after 2-3 hours.
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 12
Answer:

  • The stem and veins of the leaves appear pinkish in colour.
  • Due to the root pressure generated, eosin solution is continuously pushed upward.
  • As a result water reaches xylem and the water column is pushed forward.

Question 2.
Take a transverse section of the stem of a plant and observe the stained xylem under a compound microscope.
Answer:
The xylem elements appear pinkish as the eosin solution is conducted upwards through the xylem. From this we infer that the xylem is responsible for the conduction of water and minerals in the plant.

Question 3.
Observe your mother while she cuts the elephant’s foot (Amorphophallus) or arum leaves. Your hands may also begin to itch if you try to cut these leaves. Why does this happen? Try to find out. Ask your mother what she does to prevent the itching.
Answer:

  • In some plants like elephant foot (Amorphophallus) or arum, waste materials are in the form of crystals of calcium oxalate. They are called raphides.
  • As they are needle-shaped, they prickle and cause irritation of the skin leading to itching.
  • This can be prevented by applying tartaric acid (tamarind) or citric acid (lemon) on the hands while cutting these vegetables as the acids dissolve the crystals of calcium oxalate.

Class 9 Science Chapter 15 Life Processes in Living Organisms Additional Important Questions and Answers

Choose and write the correct options.

Question 1.
The transport of food produced in leaves to each cell in the plant body is called …………………………… .
(a) Transpiration
(b) Root pressure
(c) Translocation
(d) Ascent of sap
Answer:
(c) translocation

Question 2.
Kidneys filter blood about ……………………………. times every day.
(a) 100
(b) 200
(c) 300
(d) 400
Answer:
(d) 400

Question 3.
About ……………………………. of blood is sent through the dialysis machine at one time.
(a) 500 ml
(b) 5 l
(c) 200 ml
(d) 400 ml
Answer:
(a) 500 ml

Question 4.
The movement of plant towards the source of light is called ……………………………. movement.
(a) chemotropic
(b) hydrotropic
(c) phototropic
(d) gravitropic
Answer:
(c) phototropic

Question 5.
The movement of root system towards stimulus of gravity is called ……………………………. movement.
(a) phototropic
(b) gravitropic
(c) hydrotropic
(d) chemotropic
Answer:
(b) gravitropic

Question 6.
The movement of root system towards stimulus of water is called ……………………………. movement.
(a) Phototropic
(b) Gravitropic
(c) Hydrotropic
(d) Chemiotropic
Answer:
(c) hydrotropic

Question 7.
Hormone ……………………………. helps in stem elongation.
(a) Auxin
(b) Gibberellin
(c) Cytokinin
(d) Abscisic acid
Answer:
(b) Gibberellin

Question 8.
Hormone ……………………………. helps in cell division.
(a) Auxin
(b) Gibberellin
(c) Cytokinin
(d) Abscisic acid
Answer:
(c) Cytokinin

Question 9.
The hormone ……………………………. is effective in prevention and retardation of growth, leaf wilting, etc.
(a) Auxin
(b) Cytokinin
(c) Gibberellin
(d) Abscisic acid
Answer:
(d) Abscisic acid

Question 10.
……………………………. neurons conduct impulses from the sensory organs to the brain and the spinal cord,
(a) Sensory
(b) Motor
(c) Association
(d) All of these
Answer:
(a) Sensory

Question 11.
The brain of an adult human weighs about …………………………… .
(a) 500 – 600 grams
(b) 2 – 3 kilograms
(c) 1300 -1400 grams
(d) 1500 -1600 grams
Answer:
(c) 1300 – 1400 grams

Question 12.
……………………………. is the largest part of the brain.
(a) Cerebrum
(b) Cerebellum
(c) Medulla oblongata
(d) Spinal cord
Answer:
(a) Cerebrum

Question 13.
There are two triangular swollen structure called ……………………………. on the upper side of the medulla oblongata.
(a) Ventricles
(b) Meninges
(c) Pyramids
(d) Pons
Answer:
(c) Pyramids

Question 14.
Control of voluntary movements is done by …………………………… .
(a) Cerebrum
(b) Cerebellum
(c) Medulla oblongala
(d) Spinal cord
Answer:
(a) Cerebrum

Question 15.
Co-ordination of voluntary movements is done by …………………………… .
(a) cerebrum
(b) cerebellum
(c) medulla oblongata
(d) spinal cord
Answer:
(b) cerebellum

Question 16.
Control of involuntary activities like the beating of the heart, blood circulation, breathing etc. are controlled by …………………………… .
(a) Spinal cord
(b) Cerebrum
(c) Cerebellum
(d) Medulla oblongata
Answer:
(d) Medulla oblongata

Question 17.
The function of cerebrum is to …………………………… .
(a) control of voluntary movements
(b) co-ordination of voluntary movements
(c) control of involuntary activities
(d) maintaining the body balance
Answer:
(a) control of voluntary movements

Question 18.
The function of cerebellum is …………………………… .
(a) to control of voluntary movements
(b) to maintain the body balance
(c) to control of involuntary activities
(d) to co-ordinate of voluntary movements
Answer:
(b) to maintain the body balance

Question 19.
Hormone ……………………………. stimulates the liver to convert excess blood glucose into glycogen.
(a) Glucagon
(b) insulin
(c) Somatostatin
(d) Pancreatic polypeptide
Answer:
(a) Glucagon

Question 20.
Hormone ……………………………. controls behaviour during crisis and emotional situation.
(a) Thyroxine
(b) Parathormone
(c) Adrenalin and Nor adrenalin
(d) Somatostatin
Answer:
(c) Adrenalin and Nor adrenalin

Question 21.
……………………………. controls menstrual cycle and ovulation.
(a) Prolactin
(b) Oxytocin
(c) Luteinizing hormone
(d) Progesterone
Answer:
(c) Luteinizing hormone

Question 22.
……………………………. stimulates growth of secondary sexual characteristics in men.
(a) Thymosin
(b) Testosterone
(c) Oestrogen
(d) Thyroxine
Answer:
(b) Testosterone

Question 23.
……………………………. occurs through stomata.
(a) Translocation
(b) Perspiration
(c) Transpiration
(d) Conduction
Answer:
(c) Transpiration

Question 24.
……………………………. performs the important role of pushing the water up during the night time.
(a) Translocation
(b) Root pressure
(c) Transpiration pull
(d) None of these
Answer:
(b) Root pressure

Question 25.
In plants, gaseous substances are given out by …………………………….
(a) Root pressure
(b) Osmosis
(c) Diffusion
(d) Translocation
Answer:
(c) diffusion

Question 26.
Each kidney has approximately ……………………………. nephrons.
(a) 1 lakh
(b) 10,000
(c) 10 lakh
(d) 50 lakh
Answer:
(c) 10 lakh

Question 27.
Insulin is secreted by the ……………………………. of the pancreas.
(a) Alpha cells
(b) Beta cells
(c) Delta cells
(d) F cells
Answer:
(b) Beta cells

Select the odd man out:

Question 1.
Skin, Brain, Kidneys, Lungs.
Answer:
Brain – It is a controlling organ whereas the others are excretory orgAnswer:

Question 2.
Gum, Resin, Urea, Latex
Answer:
Urea – It is an excretory product of animals whereas the others are excretory products of plants.

Question 3.
Gibberellin, Auxin, Thyroxine, Cytokinin.
Answer:
Thyroxine – It is a hormone in animals whereas the others are hormones in plants.

Question 4.
Dendrite, Cell body, Axon, Synapse.
Answer:
Synapse – It is a minute space between two adjacent neurons whereas the others are parts of the neuron.

Question 5.
Thyroxine, Calcitonin, Insulin, Auxin.
Answer:
Auxin – It is a plant hormone whereas the others are hormones in human beings.

Complete the Analogy:

(1) Water : Xylem :: Food : …………………………….
(2) Movement of root system : Gravitropic :: Movement of shoot system : …………………………….
(3) Brain: Cranial Cavity :: Spinal Cord : …………………………….
(4) Cavities of the brain : Ventricles :: Cavities of the spinal cord : …………………………….
(5) Involuntary activities : Medulla oblongata : : Balance of the body : …………………………….
(6) Stem elongation: Gibberellin :: Cell division : …………………………….
(7) Control of voluntary movements : Cerebrum :: Co-ordination of voluntary movements : …………………………….
(8) Converts glucose to glycogen : Insulin :: Converts glycogen to glucose : …………………………….
(9) Secondary sexual characters in males : Testosterone :: Secondary sexual characters in females : …………………………….
Answer:
(1) Phloem
(2) Phototropic
(3) Vertebral column
(4) Central canal
(5) Cerebellum
(6) Cytokinin
(7) Cerebellum
(8) Glucagon
(9) Oestrogen

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) Gibberellins
(2) Cytokinins
(3) Abscisic acid
(4) Auxin
(a) Enlargement of cells
(b) Helps in stem elogation
(c) Cell division
(d) Wilting of leaf

Answer:
(1 – b),
(2 – c),
(3 – d),
(4 – a).

Question 1.

Column ‘A’ Column B’
(1) Cerebrum
(2) Cerebelleum
(3) Medulla oblongata
(4) Spinal cord
(a) Control of involuntary activities
(b) Decision – making, memory and intellectual activities.
(c) Centre for co­ordination of reflex actions
(d) Maintaining body’s balance

Answer:
(1 – b),
(2 – d),
(3 – a),
(4 – c).

State whether the following statements are true or false. Correct the false statements.

Question 1.
Plants need less energy as compared to animals.
Answer:
True

Question 2.
Xylem conducts food whereas phloem conducts water.
Answer:
False. Xylem conducts water whereas phloem conducts food.

Question 3.
The loss of water from the plants is known as translocation.
Answer:
False. The loss of water from the plants is known as transpiration.

Question 4.
Calcium oxalate crystals present in some plants are called resins.
Answer:
False. Calcium oxalate crystals present in some plants are called raphides.

Question 5.
Root pressure helps to push the water up in plants during night time.
Answer:
True

Question 6.
Translocation is carried out by phloem in downward direction.
Answer:
False. Translocation is carried out by phloem in both upward and downward direction.

Question 7.
In unicellular organisms, waste materials are directly eliminated across the cell surface.
Answer:
True

Question 8.
Urea is produced in the liver.
Answer:
True

Question 9.
Systematic regulation of different processes is called co-ordination.
Answer:
False. Systematic regulation of different processes is called control.

Question 10.
A hormone auxin produced in the apical part of the shoot helps in enlargement of cells.
Answer:
True

Question 11.
Gibberellin helps in prevention and retardation of growth, leaf wilting etc.
Answer:
False. Abscisic acid helps in prevention and retardation of growth, leaf wilting etc.

Question 12.
The movement shown by venus fly trap while trapping an insect is a growth relevant movement.
Answer:
False. The movement shown by venus fly trap while trapping an insect is a growth irrelevant movement.

Question 13.
The lotus flower opens during night time while that of tuberose (polyanthus) opens during day time.
Answer:
False. The lotus flower opens during day time while that of tuberose (polyanthus) opens during night time.

Question 14.
Sensory neurons conduct impulses from the brain and the spinal cord to effector organs like muscles or glands.
Answer:
False. Sensory neurons conduct impulses from the sensory organs to the brain and the spinal cord.

Question 15.
There are 31 pairs of cranial nerves.
Answer:
False. There are 12 pairs of cranial nerves.

Question 16.
Activities like heart beat, blood circulation, breathing, sneezing, etc. are controlled by cerebellum.
Answer:
False. Activities like heart beat, blood circulation, breathing, sneezing, etc. are controlled by cerebellum.

Question 17.
Spinal nerves are associated with head, thorax and abdomen.
Answer:
False. Spinal nerves are associated with arms, legs, skin and some other parts of the body.

Question 18.
Nerve impulses are slow and long lasting whereas action of hormones is fast and short lived.
Answer:
False. Nerve impulses are fast and short lived whereas action of hormones is slow and long lasting.

Question 19.
Reflex actions are controlled by cerebellum.
Answer:
False. Reflex actions are controlled by spinal cord.

Question 20.
Thyroid stimulating hormone is secreted by the pituitary gland.
Answer:
True

Question 21.
The hormone thymosin is secreted by the thyroid gland.
Answer:
False. The hormone thymosin is secreted by the thymus gland.

Question 22.
Parathormone controls metabolism of calcium and phosphorns.
Answer:
True

Question 23.
The right kidney is slightly lower than the left kidney.
Answer:
True

Name the following:

Question 1.
The structural and functional unit of the nervous system.
Answer:
Newtron.

Question 2.
The network of capillaries in the Bowman’s capsule.
Answer:
Glomerulus.

Question 3.
The minute space between two adjacent nervous.
Answer:
Synapse.

Question 4.
The special cells which bring about control and co-ordination.
Answer:
Neurons.

Question 5.
The cavities present in various parts of the brain.
Answer:
Ventricles.

Question 6.
The long tubular cavity of the spinal cord.
Answer:
Central Canal.

Question 7.
The thread like fibrous structure at the end of the spinal cord.
Answer:
Filum terminale.

Give scientific reasons:

Question 1.
It is necessary to remove harmful and waste substances from the body.
Answer:

  • Many harmful and waste substances like urea, uric acid, ammonia, etc. are produced in living organisms.
  • If these substances accumulate in the body or are retained in the body for long, it can lead to serious harm or even death.
  • Hence, it is necessary to remove such harmful and waste substances from the body.

Question 2.
In plants like Touch-me-not (Mimosa), movement also occurs at the places other than where it has been touched.
Answer:

  • In plants like Touch-me-not (Mimosa), the information about the touch is relayed within the plant from one place to another.
  • Plants use electro-chemical impulses for transfer of information from one place to another.
  • Plant cells change their shape by increasing or decreasing their water content and thereby, bring about the movements of plants.
  • Hence, in plants like Touch-me-not (Mimosa) movement also occurs at the places other than where it has been touched.

Question 3.
Endocrine glands are also called ductless glands.
Answer:

  • Hormones are secreted by endocrine glands.
  • These glands do not have any arrangement of their own to store or carry their secretions.
  • Hence, as soon as hormones are produced, they are directly released into the blood circulation.
  • Thus, though these endocrine glands are present at specific locations in our body, their secretions reach all parts of the body via blood.
  • Therefore, endocrine glands are also called ductless glands.

Define the following:

Question 1.
Transportation
Answer:
The process by which a substance either synthesized or absorbed in one part of the body reaches another is called as transportation.

Question 2.
Control
Answer:
Systematic regulation of different processes is called as control.

Question 3.
Coordination
Answer:
Bringing about the different processes in the proper sequence is called as co-ordination.

Question 4.
Tropism or Tropic movement
Answer:
Movement or growth of any part of the plant in response to an external stimulus is called as tropism or tropic movement.

Question 5.
Excretion
Answer:
Removal of waste or harmful substances from the body is called excretion.

Question 6.
Dialysis
Answer:
The process of separating the nitrogenous waste from the blood with the help of a man?made machine is called dialysis.

Question 7.
Reflex action
Answer:
An immediate and involuntary response given to a stimulus from the environment is called a reflex action.

Question 8.
Homeostasis
Answer:
Proper co-ordination between various systems of an organism helps to maintain a state of equilibrium called homeostasis.

Question 9.
Tropism
Answer:
Movement or growth of any part of the plant in response to an external stimulus is called tropism.

Distinguish between:

Question 1.
Nervous control and Chemical control
Answer:

Nervous Control Chemical Control
(i) Nervous control is brought about by special types of cells called neurons. (i) Chemical control is brought about by chemical substances called hormones.
(ii) It involves the nervous system i.e., brain, spinal cord and nerves. (ii) It involves the endocrine glands.
(iii) Nerve impulses are fast and short-lived. (iii) Actions of hormones are very slow and long-lasting.

Question 2.
Growth relevant movements and growth irrelevant movements
Answer:

Growth relevant movements Growth irrelevant movements
(i) Growth relevant movements result in growth of the plants. (i) Growth irrelevant movements do not result in growth of the plants.
(ii) They are also called as tropic movements. (ii) They are also called as nastic movements.
(iii) They are in response to stimulus of light, water, gravity, chemicals, etc. (iii) They are generally in response to stimulus of touch.
(iv) Examples : Bending of stem towards light, movement of roots towards gravity and water, etc. (iv) Example : Closing of leaflets of Mimosa plant, closing of Venus fly trap, etc.

Question 3.
Cerebrum and Cerebellum
Answer:

Cerebrum Cerebellum
(i) It is the largest part of the brain. (i) It is the smaller part of the brain.
(ii) It occupies two thirds of the brain. (ii) It is situated below the cerebrum at the back of the cranial cavity.
(iii) Its surface has deep, irregular ridges and grooves which are called convolutions. (iii) Its surface shows shallow grooves instead of convolutions.
(iv) It controls voluntary movements, concentration, planning, decision-making, memory, intelligence and intellectual activities. (iv) It co-ordinates voluntary movements and maintains the body’s balance.

Write short notes:

Question 1.
Root Pressure
Answer:

  • Root cells are in contact with water and minerals in the soil.
  • Water and minerals enter the cells of the root surface due to difference in concentration. As a result, these cells become turgid.
  • This is called root pressure.
  • Under the effect of this pressure, water and minerals reach the xylem of the roots and to reduce this difference in concentration, they are continuously pushed forward.
  • As a result of this continuous movement, a water column is formed, which is continuously pushed ahead.
  • This pressure, is sufficient to lift the water up in shrubs, small plants and small trees.

Question 2.
Transpiration
Answer:

  • Plants give out water in the form of water vapour through the stomata on their leaves. This is known as transpiration.
  • Two cells called guard cells are present around the stomata. They control the opening and closing of stomata.
  • Water is released into the atmosphere by leaves through the process of evaporation.
  • As a result, water in the epidermal layer of the leaf decreases.
  • Water is brought up to the leaves through the xylem to compensate for the lost water.
  • Transpiration helps in absorption of water and minerals and distribution to all parts of the plant.

Question 3.
Nerve cell
Answer:

  • Nerve cells also called as neurons are special types of cells which conduct impulses from one place to another in the body.
  • Neurons are the structural and functional units of the nervous system.
  • Nerve cells, the largest cell in the human body, may measure upto a few metres in length.
  • Nerve cells have the ability to generate and conduct electrochemical impulse.
  • The cells that support the nerve cells and help in their functioning are called neuroglia. Nerve cells and neuroglial cells together form nerves.

Question 4.
Reflex action
Answer:

  • An immediate and involuntary response given to a stimulus from the environment is called a reflex action.
  • Sometimes we react to an incident without any thinking on our part or control over the reaction.
  • This is a response given to a certain stimulus from the surroundings.
  • In such situations, proper control and co-ordination is achieved even without the intervention of the brain.
  • For example, when our hand touches a hot object, the receptor in the skin detects the stimulus.
  • The sensory neuron sends the impulse to the relay neuron in the spinal cord.
  • The impulses are then sent to motor neurons which cause the muscles of the hand to contract and move away from the hot object.
  • The spinal cord is the center of co-ordination for reflex actions.

Question 5.
Spinal cord
Answer:

  • The spinal cord is the part of the central nervous system and it is held within the vertebral column.
  • It is slightly thick but gradually tapers towards the end.
  • There is a thread like fibrous structure at its end. It is called the filum terminale.
  • The long tubular cavity of the spinal cord is called the central canal. It is filled with cerebrospinal fluid.
  • The spinal cord conducts impulses from the skin towards the brain and from the brain to the muscles and glands.
  • It functions as a centre of co-ordination of reflex actions.

Question 6.
Dialysis
Answer:

  • The efficiency of kidneys can be adversely affected by injury, infection or decreased blood supply.
  • If this happens, excess of toxic substances accumulates in the body and it can lead to death.
  • If kidneys fail, nitrogenous wastes are separated from the blood with the help of a man-made machine.
  • The process of separating the nitrogenous waste from the blood with the help of this machine is called dialysis.
  • About 500 ml of blood is sent at one time through this machine. Purified blood is reinfused into the body of the patient.

Draw neat and labelled diagrams:

Question 1.
Vertical Section of Kidney
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 6

Question 2.
Reflex action
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 15 Life Processes in Living Organisms 7

Name the hormones of the following endocrine glands and the function of each:

Question 1.
Parathyroid
Answer:

Hormone Function
Parathormone Controls metabolism of calcium and phosphorus

Question 2.
Pancreas
Answer:

Hormone Function
(i) Glucagon Stimulates liver to convert glycogen into glucose.
(ii) Insulin Stimulates liver to convert excess blood glucose into glycogen.
(iii) Somatostatin Controls levels of insulin and glucagon.
(iv) Pancreatic Polypeptide Controls movement of intestine and thereby glucose absorption. Controls secretion of pancreatic juice.

Complete the following table of plant hormones with their functions:
Answer:

Plant hormone Functions
(i) Auxin Produced in the apical part of the shoot
Helps in enlargement of cells
(ii) Gibberellins Help in stem elongation
(iii) Cytokinins Help in cell division
(iv) Abscisic acid Effective in prevention and retardation of growth, leaf wilting, etc.

State the functions of different parts of the brain.
Answer:

Parts of the Brain Functions
Cerebrum Control of voluntary movements, concentration, planning, decision­making, memory, intelligence, and intellectual activities.
Cerebellum (i) Co-ordination of voluntary movements.
(ii)  Maintaining the body’s balance.
Medulla oblongata Control of involuntary activities like the beating of the heart, blood circulation, breathing, sneezing, coughing, salivation, etc.
Spinal cord (i) Conduction of impulses from the skin towards the brain.
(ii) Conduction of impulses from brain to muscles and glands.
(iii) Functions as centre of co­ordination of reflex actions.

Answer the following:

Question 1.
How do plants bring about movements in them?
Answer:

  • Plants use electro-chemical impulses for transfer of information from one place to another.
  • Plant cells change their shape by increasing or decreasing their water content and thereby bring about the movement of plants.

Answer the following:

Question 1.
What are growth-irrelevant movements? Give examples.
Answer:

  • Some specific movements of the plants do not lead to the plant growth. Such movements are called growth-irrelevant movements.
  • In the plant called venus fly trap, there is a trap that appears and smells like flowers and deceives insects. When an insect visits the flower-like trap, the trap closes up and the trapped insect is digested by the plant.
  • Fibrils present on the leaves of the insectivorous plant Drosera, bend inward as soon as an insect lands on the leaves and surround the insects from all sides.
  • In Balsam, the ripened fruit dehisces (bursts-open) at the right time to disperse the seeds.

Question 2.
Which are the different types of nerve cells or neurons?
OR
How are nerve cells classified according to their functions?
Answer:
According to their functions, nerve cells are classified into three types:

  • Sensory neurons: They conduct impulses from the sensory organs to the brain and the spinal cord.
  • Motor neurons : They conduct impulses from the brain or spinal cord to effector organs like muscles or glands.
  • Association neurons : Association neurons perform the function of integration in the nervous system.

Question 3.
What are the different parts of the human nervous system?
Answer:
The human nervous system can be divided into the following three parts.

  1. Central nervous system (CNS) : It consists of the brain and the spinal cord. It controls and regulates all activities of the body.
  2. Peripheral nervous system (PNS) : The peripheral nervous system consists of the nerves originating from the central nervous system which connect the central nervous system with all parts of the body. They are of two types cranial nerves and spinal nerves.
  3. Autonomic nervous system (ANS) : The autonomous nervous system consists of the nerves of involuntary organs like the heart, lungs, stomach etc. It is not under the control of our will.

Question 4.
Describe the structure of the central nervous system.
Answer:

  • The central nervous system consists of the brain and spinal card.
  • The organization of the brain is extremely delicate and highly evolved.
  • The brain is the main controlling part of the nervous system and is safely located in the cranial cavity. The spinal cord is protected by the vertebral column.
  • In the space between the delicate central nervous system and its bony covering are the protective layers called the meninges.
  • Cavities present in various parts of the brain are called ventricles whereas the long tubular cavity of the spinal cord is called the central canal.
  • The ventricles, central canal and spaces between the meninges are filled with cerebro?spinal fluid.
  • This fluid supplies nutrients to the central nervous system and protects it from shock.

Question 5.
Which are the two types of peripheral nerves?
Answer:
Peripheral nerves are of two types:
(a) Cranial nerves:

  • Nerves originating from the brain are called cranial nerves.
  • They are associated with various parts in the head.
  • There are 12 pairs of cranial nerves.

(b) Spinal nerves:

  • Nerves originating from the spinal cord are called spinal nerves.
  • They are associated with arms, legs, skin and some other parts of the body.
  • There are 31 pairs of spinal nerves.

Question 6.
How do plants get rid of their excretory products?
Answer:

  • Excretion is a simpler process in plants. There is no special organ or system for excretion in plants.
  • Gases are given out by diffusion.
  • Most of the waste substances of plants are stored in vacuoles of leaf-cells and in flowers, fruits and the bark of the stem. After some time these parts fall off.
  • Some other waste materials are stored in old and worn out xylem in the form of resin and gum.
  • Some waste materials are also given out through roots in the surrounding soil.

Answer in detail:

Question 1.
Explain how food and other substances are transported in plants?
OR
Explain translocation in plants.
Answer:

  • The food produced in leaves is transported to each cell in the plant body.
  • Excess food, except amino acids, is stored in roots, fruits and seeds. This process is called translocation of materials.
  • It is carried out in both the upward and downward directions by the phloem.
  • Translocation is not a simple process, it requires energy. This energy is obtained from ATP.
  • Whenever food material like sucrose is transported towards a part of a plant via the phloem with the help of ATP, the water concentration decreases in the part. As a result, water enters the cell by the process of diffusion.
  • The pressure on the cell wall increases due to increase in cellular contents.
  • Due to the increase pressure, food is pushed into the neighbouring cells where the pressure is low.
  • This process helps the phloem to transport the materials as per the need of the plant. During flowering season, the sugar stored in roots or stem is transported towards the floral buds to make them open and blossom.

Question 2.
What is tropic movement? Describe the different types of tropic movements.
OR
What are growth relevant movements? Describe the different types of growth relevant movements.
Answer:
Movement or growth of any part of the plant in response to an external stimulus is called tropic movement or tropism.

The different types of tropic movements are:

  • Phototropic movement : The shoot system of any plant responds to the light stimulus, i.e., it grows towards the source of light. The movement shown by plants towards the source of light is called phototropic movement.
  • Gravitropic movement : The root system of plants responds to stimulus of gravity. This is called gravitropic movement.
  • Hydrotropic movement : The root system of plants responds to the stimulus of water. This is called hydrotropic movement.
  • Chemotropic movement : Movement shown by plants in response to specific chemicals is called chemotropic movement. For example, the growth of the pollen tube towards the ovule.

All the above-mentioned movements of plants are related with growth, hence are called growth relevant movement.

Question 3.
Write short note on : Human Brain
Answer:

  • The brain is the main controlling part of the nervous system and is safely located in the cranial cavity.
  • The brain is covered by protective layers called the Meninges.
  • Cavities present in various parts of the brain are called ventricles.
  • The ventricles are filled with cerebrospinal fluid. This fluid supplies nutrients to the brain and also protects it from shock.
  • The brain of an adult human weighs about 1300 -1400 grams and consists of approximately 100 million neurons.
  • The left side of our brain controls the right side of our body and right side of our brain controls left side of the body.
  • In addition, the left side of the brain controls our speech and conversation, writing, logical thinking etc., whereas the right side controls artistic abilities.

Human brain is made up of following parts:

(a) Cerebrum:

  • This is the largest part of the brain and consists of two cerebral hemispheres. These hemispheres are joined with each other with the help of tough fibres and nerve tracts.
  • The cerebrum occupies two-thirds of the brain. Hence, it is also called as large brain.
  • Its surface has deep, irregular ridges and grooves which are called convolutions, convolution increases the surface area of the cerebrum and therefore a large number of nerve cells can be accommodated.

(b) Cerebellum:

  • This is the smaller part of the brain situated below the cerebrum at the back of the cranial cavity.
  • Its surface shows shallow grooves instead of deep convolutions.

(c) Medulla Oblongata:

  • This is the hind-most part of the brain.
  • There are two triangular swollen structures called pyramids on the upper side of medulla oblongata.
  • The medulla oblongata continues downwards the spinal cord.

(6) How are nerve impulses conducted in the body?
Answer:

  • Special types of cells which conduct impulses from one place to another in the body are called neurons. Nerve cells have the ability to generate and conduct electro-chemical impulses.
  • The cells that support the nerve cells and help in their functioning are called neuroglia.
  • All the information about our surroundings is collected by the ends or dendrites of the neuron.
  • The chemical process begins at those ends and electric impulses are generated which are conducted from the dendrites to the cell body, from the cell body to the axon and from the axon to its terminal.
  • These impulses are then to be transferred from this nerve cell to the next.
  • Now the impulse that reaches the terminal of an axon, stimulates the nerve cell to secrete certain chemicals.
  • These chemicals pass through a minute space, called the synapse, between two adjacent neurons and generate the impulses in the dendrites of the next neuron.
  • In this way, impulses are conducted in the body and these impulses are finally conveyed by nerve cells to muscles cells or glands.

9th Std Science Questions And Answers:

Current Electricity Class 9 Science Chapter 3 Questions And Answers Maharashtra Board

Class 9 Science Chapter 3

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 3 Current Electricity Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 3 Current Electricity Question Answer Maharashtra Board

Class 9 Science Chapter 3 Current Electricity Question Answer Maharashtra Board

1. The accompanying figure shows some electrical appliances connected in a circuit in a house. Answer the following questions.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 26
A. By which method are the appliances connected?
Answer:
Appliances are connected in parallel.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

B. What must be the potential difference across individual appliances?
Answer:
The potential difference across all appliances is same in parallel connection.

C. Will the current passing through each appliance be the same? Justify your answer.
Answer:
No, as every appliance has a different load (resistance), the current flowing through each appliance will be different.

D. Why are the domestic appliances connected in this way?
Answer:
The domestic appliances are connected in parallel as the potential difference remains same.

E. If the T.V. stops working, will the other appliances also stop working? Explain your answer.
Answer:
No, the other devices will not stop working as the current flowing through them is along different paths.

2. The following figure shows the symbols for components used in the accompanying electrical circuit.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 1
Which law can you prove with the help of the above circuit?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 2
(b) This circuit can be used to prove Ohm’s law.
(c) V = 1R is the expression of Ohm’s law

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

3. Umesh has two bulbs having resistances of 15 W and 30 W. He wants to connect them in a circuit, but if he connects them one at a time the filament gets burnt. Answer the following.

A. Which method should he use to connect the bulbs?
B. What are the characteristics of this way of connecting the bulbs depending on the answer of A above?
C. What will be the effective resistance in the above circuit?

4. The following table shows current in Amperes and potential difference in Volts.

a. Find the average resistance.
b. What will be the nature of the graph between the current and potential difference? (Do not draw a graph.)
c. Which law will the graph prove? Explain the law.

5. Match the pairs

‘A’ Group – ‘B’ Group
1. Free electrons – a. V/ R
2. Current – b. Increases the resistance in the circuit
3. Resistivity – c. Weakly attached
4. Resistances in series – d. VA/LI

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

6. The resistance of a conductor of length x is r. If its area of crosssection is a, what is its resistivity? What is its unit?

7. Resistances R1, R2, R3 and R4 are connected as shown in the figure. S1 and S2 are two keys. Discuss the current flowing in the circuit in the following cases.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 3

a. Both S1 and S2 are closed.
b. Both S1 and S2 are open.
c. S1 is closed but S2 is open.
Answer:
(a) When both S1 and S2 are dosed, the effective resistance of the circuit decreases and hence, current will increase.
(b) When both S1 and S2 are open, the effective resistance of the rircuit increases and hence, current will decrease.
(c) When S2 is closed and S2 is open, the effective resistance of the tircuit decreases and hence current will increase. [Current will be more than case (b) but less than in case (a)]

8. Three resistances x1, x2 and x3 are connected in a circuit in different ways. x is the effective resistance. The properties observed for these different ways of connecting x1, x2 and x3 are given below. Write the way in which they are connected in each case. (I-current, V-potential difference, x-effective resistance)

a. Current I flows through x1, x2 and x3
b. x is larger than x1, x2 and x3
c. x is smaller than x1, x2 and x3
d. The potential difference across x1, x2and x3 is the same
e. x = x1 + x2 + x3
\(\text { f. } x=\frac{1}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}}\)

9. Solve the following problems.

A. The resistance of a 1m long nichrome wire is 6Ω. If we reduce the length of the wire to 70 cm. what will its resistance be? (Answer : 4.2Ω)
Answer:
The resistance of 70cm wire will be 4.2 Ω

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

B. When two resistors are connected in series, their effective resistance is 80Ω. When they are connected in parallel, their effective resistance is 20Ω. What are the values of the two resistances? (Answer : 40Ω, 40Ω)
Answer:
The values of the two resistances R1 and R2 are 40Ω and 40Ω.

C. If a charge of 420 C flows through a conducting wire in 5 minutes what is the value of the current? (Answer : 1.4 A)
Answer:
Given: Electric charge (Q) = 420 C
Time (t) = 5 min = 5 x 60
= 300 sec.
To find: Electric current (1) = ?
Formula:
\(I=\frac{Q}{t}\)
Solution:
\(I=\frac{Q}{t}\)
The current in the circuit is 1.4 A.

Class 9 Science Chapter 3 Current Electricity Intext Questions and Answers

Can you recall?

Question 1.
You must have seen a waterfall. Which way does the water flow?
Answer:
Water flows from a certain height of a mountain towards the ground.

Question 2.
Material: Copper and aluminium wires, glass rod, rubber.
Make connection as shown in figure 3.8. First connect a copper wire between points A and B and measure the current in the circuit. Then in place of the copper wire, connect the aluminium wire, glass rod, rubber, etc one at a time and measure the current each time. Compare the values of the current in different cases.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 4
Also take different metal strips (Iron, Copper, Zinc, and Aluminium) and connect it in slot AB. Now observe the difference in the resistance using Ohm meter.
Answer:
When copper and aluminium wires are connected to the circuit, current flows through it, as both are good conductors of electricity. When glass rod or rubber was connected to the circuit, current does not flow through it, as both are bad conductors of electricity.

Copper displays lowest resistance while the resistance increases with aluminium, zinc and iron respectively.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 3.
Set up the experiment as shown in figure. Then remove the clamp from the rubber tube.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 27

(a) What happens when the clamp is removed?
Answer:
When the clamp is removed, water flows from higher level to lower level.

(b) Does the water stop flowing? Why?
Answer:
Yes, the water stops flowing. This happens when the level of water becomes equal in both the bottles, i.e., there is no difference in the water levels.

(c) What will you do to keep the water flowing for a longer duration?
Answer:
The difference in the water level has to be maintained till that time. The difference must never be zero.

Question 4.
Point out the mistakes in the figure below:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 28
Answer:
A: Wire is broken at the negative terminal. Bulb will not glow as the circuit is incomplete.
B: Wire is disconnected at the negative terminal. Bulb will not glow as the circuit is incomplete.
C: The circuit is complete. Therefore, bulb will glow.
D: Rubber is a bad conductor of electricity. Hence, it will not allow current to flow and the bulb will not glow.

Question 5.
Why are the bulbs in Figures B, C and D not lighting up?
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 29
Answer:

  • In B, the blue wire is broken. Hence circuit is incomplete and current does not flow. Therefore, bulb will not light up.
  • In C, the red wire is broken. Hence circuit is incomplete and current does not flow. Therefore, bulb will not light up.
  • In D, both wires are connected to the same terminal. Hence, there is no potential difference and current does not flow. Therefore, bulb will not light up.

Class 9 Science Chapter 3 Current Electricity Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
1mA = …………… A.
(a) 103
(b) 10-3
(c) 106
(d) 10-6
Answer:
(a) 103

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 2.
To increase the effective resistance in a circuit the resistors are connected in ………….. .
(a) Series
(b) Parallel
(c) Both ways
(d) None of these
Answer:
(a) series

Question 3.
1 kilowatt hr = …………… joules.
(a) 4.6 x 106
(b) 3.6 x 106
(c) 30.6 x 106
(d) 3.6 x 1O5
Answer:
(b) 3.6 x 106

Question 4.
The voltage difference in India between the live and neutral wires is about ………….. .
(a) 110 V
(b) 220 V
(c) 440 V
(d) 60 V
Answer:
(b) 220 V

Question 5.
Resistivity is the specific property of a ………….. .
(a) Area of cross-section
(b) Temperature
(c) Length
(d) Material
Answer:
(d) material

Question 6.
If a P.D. of 12 V is applied across a 3Ω resistor then the current passing through it is ………….. .
(a) 36 A
(b) 4 A
(c) 0.25 A
(d) 15 A
Answer:
(b) 4 A.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 7.
In order to measure the electric current flowing through a circuit, we connect …………… with the circuit.
(a) a voltmeter in parallel
(b) a voltmeter in series
(c) an ammeter in parallel
(d) an ammeter in series
Answer:
(d) an ammeter in series

Question 8.
P and Q are two wires of same length and different cross-sectional areas and made of same material. Name the property which is same for both the wires.
(a) Resistivity
(b) Resistance
(c) Current
(d) Both (a) and (b)
Answer:
(a) Resistivity

Question 9.
The following is true for identical bulbs connected in parallel.
(a) All bulbs glow with unequal brightness.
(b) If one bulb is non-functional, all will stop working.
(c) All bulbs glow with equal brightness.
(d) Bulbs function for longer time.
Answer:
(c) All bulbs glow with equal brightness

Question 10.
The …………… wire is either yellow or green in colour.
(a) Live
(b) Neutral
(c) Earth
(d) Fuse
Answer:
(c) earth

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 11.
A current flows through a circuit due to the difference in …………… between two points in the conductor.
(a) Gravity
(b) Potential
(c) Resistance
(d) Fuse
Answer:
(b) potential

Question 12.
…………… is the amount of charge flowing through a particular cross sectional area in unit time.
(a) Electric current
(b) Ampere
(c) Volt
(d) Force
Answer:
(a) Electric current

Question 13.
The flow of …………… constitutes the electric current in a wire.
(a) Protons
(b) Neutrons
(c) Electrons
(d) Gravitons
Answer:
(c) electrons

Question 14.
The conventional direction of flow of current is from …………… terminal to …………… terminal.
(a) Negative to positive
(b) Neutral to positive
(c) Positive to negative
(d) Positive to neutral
Answer:
(c) positive, negative

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 15.
Current stops flowing when potential difference between two ends of a wire becomes ………….. .
(a) Zero
(b) Positive
(c) Negative
(d) Higher
Answer:
(a) zero

Question 16.
Resistances are connected in …………… so as to pass the same current through them.
(a) Series
(b) Parallel
(c) Reversed
(d) Disconnect
Answer:
(a) series

Question 17.
To decrease the effective resistance in a circuit, the resistances are connected in ………….. .
(a) Series
(b) Parallel
(c) Reversed
(d) Disconnect
Answer:
(b) parallel

Question 18.
1μV = …………… V
(a) 102
(b) 10-6
(c) 106
(d) 103
Answer:
(b) 10-6

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 19.
Good conductors contain a large number of ………….. .
(a) Protons
(b) Neutrons
(c) Electrons
(d) Gravitons
Answer:
(c) free electrons

Question 20.
Electrons flow from …………… terminal to …………… terminal in a conductor when a potential difference is applied.
(a) Negative to positive
(b) Neutral to positive
(c) Positive to negative
(d) Positive to neutral
Answer:
(a) negative, positive

Find the odd one out:

Question 1.
Voltmeter, Ammeter, Galvanometer, Thermometer
Answer:
Thermometer

Question 2.
Rubber, Silver, Copper, Gold
Answer:
Rubber

Question 3.
Wood, Glass, Steel, Rubber
Answer:
Steel

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 4.
Graphite, Diamond, Fullerenes, Coal
Answer:
Fullerenes

Distinguish between:

Question 1.
Voltmeter and Ammeter
Answer:

Voltmeter Ammeter
(i) It is an instrument used to measure the potential difference between two terminals of a cell. (i) It is an instrument to measure the electric current flowing through a circuit.
(ii) It is connected in parallel with the cell. (ii) It is connected in series with the cell.
(iii) It has a very high resistance. (iii) It has a very low resistance.
(iv) Voltmeter has range of volts. (iv) Ammeter has range of amps.

Question 2.
Ohmic conductors and Non-Ohmic conductors
Answer:

Conductors Insulators
(i) Substances which have very low electrical resistances are called conductors. (i) Substances which have extremely high electrical resistances are called Insulators.
(ii) They contain a large number of free electrons. (ii) They contain practically no free electrons.
(iii) Conductors are mostly metals. (iii) Insulators are mostly non metals.
(iv) Conductor example iron, copper. (iv) Insulator example rubber, plastic.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 3.
Conductors and Insulators
Answer:

Resistance Resistivity
(i) The hindrance to the flow of electrons is called resistance. (i) Resistivity is the specific property of the material of a conductor.
(ii) The S.I. unit of resistance is ohm (Q). (ii) The S.I. unit of resistivity is ohm-metre (Q – m).
(iii) It depends on temperature, area of cross-section, length of conductor and material of the conductor. (iii) It depends on material of the conductor.
(iv) Resistance can be changed as it depends of external factor as well. (iv) Resistivity cannot be changed as it depends of internal factors.

Question 4.
Resistance in Series and Resistance in Parallel
Answer:

Resistance in Series Resistance in Parallel
(i) Effective resistance of the resistors is equal to the sum of their individual resistances. (i) Inverse of the effective resistance is equal to the sum of the inverse of individual resistances.
(ii) The same current flows through each resistor. (ii) The total current flowing through the circuit is the sum of the currents flowing through individual resistors.
(iii) The effective resistance is larger than each of the individual resistances. (iii) The effective resistance of resistors connected in parallel is less than the least resistance of individual resistors.
(iv) This arrangement is used to increase the resistance in a circuit. (iv) This arrangement is used to decrease the resistance in a circuit.

Question 5.
Answer:

Electric current Potential difference
(i) The flow of electric charge per unit time is called electric current. (i) The difference in potential between the positive and negative terminal of a cell is the potential difference of that cell.
(ii) The S.I. unit of electric current is ampere. (ii) The S.I. unit of potential difference volt.
(iii) Ammeter is used to measure electric current. (iii) Voltmeter is used to measure electric current.
(iv) Current is represented by: \(\mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}}\) (iv) Potential difference is represented by: \(\mathrm{V}=\frac{\mathrm{W}}{\mathrm{Q}}\)

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Make pair:

Question 1.
Copper : Conductor :: Rubber : ……………….
Answer:
Insulator

Question 2.
Aluminium : ………………. :: Indium oxide : Super Insulator
Answer:
Super conductor

Question 3.
Parallel Connection : \(\frac{1}{\mathrm{R}_{p}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\) :: Series Connection : ……………….
Answer:
Rs = R1 + R2

(4) Electric Current : ………………. :: Electric charge : Coulomb
Answer:
Ampere

(5) Electric resistance : Ohm :: Potential difference : ……………….
Answer:
Volt

State whether the following statements are true or false. Correct the false statements:

(1) The SI unit of charge is volt.
(2) Voltmeter is always connected in series with the device.
(3) The conventional direction of flow of current is from positive terminal to negative terminal.
(4) Silver and copper are good conductors.
(5) Resistivity of pure metals is more than alloys.
(6) Resistance in series arrangement is used to decrease resistance of circuit.
(7) A conducting wire offers less resistance to flow of electrons.
(8) Charges are measured in ampere.
(9) The unit of potential difference is ampere.
(10) Resistance of a conductor is inversely proportional to the length of the conductor.
(11) Ammeter is connected in parallel to the cell to measure current.
(12) Fuse is made of wire having high melting point.
Answer:
(1) False. The SI unit of charge is coulomb.
(2) False. Voltmeter is ahvays connected in parallel with the device.
(3) True
(4) True
(5) False. Resistivity of pure metals is less than alloys.
(6) False. Resistance in series arrangement is used to increase resistance of circuit.
(7) True
(8) False. Charges are measured in coulomb.
(9) False. The unit of potential difference is volt.
(10) False. Resistance of a conductor is directly proportional to the length of the conductor.
(11) False. Ammeter is connected in series to the cell to measure current.
(12) False. Fuse is made of wire having low melting point.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Answer the following in one sentence:

Question 1.
Which is the unit used to measure large voltages?
Answer:
Kilovolts and Megavolts are the units used to measure large voltages.

Question 2.
What is the SI unit of potential difference?
Answer:
The SI unit of potential difference is volt (V).

Question 3.
What is lightning?
Answer:
Lightning is the electric discharge travelling from clouds at high potential to earth’s surface which is at zero potential.

Question 4.
What is the unit of resistivity.
Answer:
The unit of resistivity is ohm metre (Qm).

Question 5.
Which substances are called conductors of electricity?
Answer:
Those substances which have very low electrical resistance are called conductors of electricity.

Question 6.
What is Earth wire?
Answer:
Earth wire is generally yellow or green colour, it is connected to a metal plate buried deep underground near the house and is for safety purpose.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Write formula:
(1) E1ectriccurrent \(=\frac{Q}{t}\)
(2) Electric charge = It
(3) Potential difference = IR
(4) Electric resistance \(=\frac{V}{I}\)
(5) Current \(=\frac{V}{R}\)
(6) Resistivity \(=\frac{RA}{L}\)

Give scientific reasons:

Question 1.
Free electrons are required for conduction of electricity.
Answer:

  • Every atom of a metallic conductor has one or more outermost electrons which are very weakly bound to the nucleus.
  • These are called free electrons. These electrons can easily move from one part of a conductor to its other parts. The negative charge of the electrons also gets transferred as a result of this motion.
  • The free electrons in a conductor are the carriers of negative charge. Hence, free electrons are required for conduction of electricity.

Question 2.
Wood and glass are good insulators.
Answer:

  • Those substances which have infinitely high electrical resistance are called insulators.
  • Wood and glass have high resistance and negligible free electrons for conduction of electricity.
  • Hence, wood and glass are good insulators.

Question 3.
Connecting wires in a circuit are made of copper and aluminium.
Answer:

  • Copper and aluminum are good conductors of electricity.
  • They have low electrical resistance and large number of free electrons.
  • As they are malleable and ductile, they can be drawn into thin wires. Hence, connecting wires in a circuit are made of copper or aluminum.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 4.
A thick wire has a low resistance.
Answer:

  • The resistance (R) of a wire is inversely proportional to the cross-sectional area (A) of a wire. i.e., R i
  • Thus, greater is the cross-sectional area of a conductor (wire), lower is its resistance. Hence, a thick wire has a low resistance.

Question 5.
A series combination of resistances is used to increase the resistance of a circuit.
Answer:

  • When resistances are connected in series, the effective resistance of the resistors is equal to the sum of their individual resistances. Rs = R1 + R2 ………….. Rn
  • The effective resistance is larger than each of the individual resistances. Hence, This arrangement is used to increase the resistance in a circuit.

Question 6.
A parallel combination of resistances decreases the effective resistance of the circuit.
Answer:

  • In a parallel combination, the inverse of the effective resistance is equal to the sum of the inverses of individual resistances. \(\frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}} \ldots \ldots \cdot \frac{1}{\mathrm{R}_{\mathrm{n}}}\)
  • The effective resistance of resistors connected in parallel is less than the individual resistors.
  • Due to this, any addition of an individual resistance in parallel combination will decrease the overall resistance of the circuit. Hence, a parallel combination of resistance decreases the effective resistance of the circuit.

Question 7.
Lightning occurs from sky to earth.
Answer:

  • Lightning is the electric discharge travelling from clouds at high potential to the earth’s surface, which is at zero potential.
  • The earth is always at lower potential as compared to the clouds.
  • Hence, lightning occurs from sky to earth.

Question 8.
In streetlights, bulbs are connected in parallel.
Answer:

  1. Even if any one of the several bulbs connected in parallel becomes non-functional because of some damage to its filament, the circuit does not break as the current flows through the other paths, and the rest of the bulbs light up.
  2. When several bulbs are connected in parallel, they emit the same amount of light as when they are connected individually in the circuit, while bulbs connected in series emit less light than when connected individually. Hence, streetlights are connected in parallel.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Numerical:
Numericals based on the formula: (1) Q= It (2) W= VQ

Question 1.
A current of 0.4 A flows through a conductor for 5 minutes. How much charge would have passed through the conductor?
Answer:
Given: Current (I) = 0.4 A
Time (t) = 5 min = 5 x 60 = 300 s
To find: Charge (Q) =?
Formula: Q = 1 x
Solution: Q = 0.4 x 300
Q= 120 C.
Charge passing through the conductor is 120

Question 2.
Find the amount of work done if 3 C of charge is moved through a potential difference of 9 V.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 5
The work done is 27 joule.

Question 3.
The resistance of the filament of a bulb is 1000Ω. It is drawing a current from a source of 230 V. How much current is flowing through it?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 6
The current flowing through the filament of bulb is 0.23 A.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 4.
The length of a conducting wire is 50 cm and its radius is 0.5 mm. If its resistance is 30Ω, what is the resistivity of its material?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 7
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 8
The resistivity of the wire is 4.71 x 10-5 Qm.

Question 5.
A current of 0.24 A flows through a conductor when a potential difference of 24 V is applied between its two ends. What is its resistance?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 9
The resistance of a conductor is 100Ω.

Question 6.
If three resistors 15Ω, 3Ω and 4Ω each are connected in series, what is the effective resistance in the circuit?
Answer:
Given:
R1 =15Ω
R2 = 3Ω
R3 = 4Ω
Effective resistance in series (Rs) = ?
Rs = R1 + R2 + R3
Rs = 15 + 3 + 4
Rs = 22Ω
The effective resistance in the circuit is 22Ω.

Question 7.
Three resistances 15Ω, 20Ω and 10Ω are connected in parallel. Find the effective resistance of the circuit.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 10
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 11
The effective resistance of the circuit is 4.615 Ω. It is less than the least of the three i.e., 10Ω.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Write a note on the following:

Question 1.
Electric current
Answer:
An electric current is the flow of electrons through a conductor. Quantitatively, current
(I) is defined as the charge passing through a conductor in unit time.
\(T=\frac{Q}{t}\)

Question 2.
1 ampere
Answer:
One ampere current is said to flow in a conductor if one coulomb charge flows through it every second.
\(1 \mathrm{~A}=\frac{1 \mathrm{C}}{1 \mathrm{~s}}\)

Question 3.
1 volt
Answer:
The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to another.
\(1 \mathrm{~V}=\frac{1 \mathrm{~J}}{1 \mathrm{C}}\)

Question 4.
Potential Difference
Answer:
The amount of work done to carry a unit positive charge from point A to point B is called the electric potential difference between the two points.
\(V=\frac{W}{Q}\)

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 5.
Conductor
Answer:
Those substances which have very low resistance are called conductors. Current can flow easily through such materials.

Question 6.
Insulators
Answer:
Those substances which have extremely high resistance and through which current cannot flow are called insulators.

Question 7.
1 ohm
Answer:
If one Ampere current flows through a conductor when one Volt potential difference is applied between its ends, then the resistance of the conductor is one Ohm.
\(\frac{1 \text { Volt }}{1 \text { Ampere }}=1 \mathrm{Ohm}\)

Question 8.
Potential
Answer:
The level of electric charge present is known as potential.

Question 9.
Ohm’s Law
Answer:
If the physical state of a conductor remains constant, the current (I) flowing through it is directly proportional to the potential difference (V) between its two ends.
V = IR

Question 10.
Superconductors
Answer:
The resistance of some conductors becomes nearly zero if their temperature is decreased up to a certain value close to 0 K. Such conductors are called superconductors.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 11.
Non-ohmic conductors
Answer:
Conductors which do not obey Ohm’s law are called non-ohmic conductors.

Complete the flow charts:

(1) Protection from Electricity
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 12

(2) Resistance
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 13

(3) Resistivity
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 14

Write properties/characteristics/advantages of the following:

Superconductors
Answer:
The resistance of these conductors becomes nearly zero if their temperature is decreased up to a certain value close to 0 K. Aluminium is an example of Super Conductor. Superconductors can be used in space missions to increase/ boost the signal strength. They are also used i in the data fibres to increase the speed of data transfer.

Give explanations of the given statements:

Question 1.
Safety precautions are to be taken while using electricity.
Answer:

  • Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket.
  • Before cleaning an electrical appliance it should be switched off and its plug removed from the socket.
  • One’s hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles. Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity
  • As rubber is an insulator, it prevents the current from flowing’ through our body, thereby protecting it.
  • If a person gets an electric shock, you should not touch that person. You should switch off the main switch or remove the plug from the socket if possible.
  • If not, then you should use a wooden pole to push the person away from the electric wire.

Question 2.
In a domestic circuit colour code is followed while setting up electrical wiring.
Answer:

  • The electricity in our homes is brought through the main conducting cable either from the electric pole or from underground cables.
  • Usually, there are three wires in the cable.
    (a) Live wire which brings in the current. It has a red or brown insulation.
    (b) Neutral wire through which the current returns. It is blue or black.
    (c) Earth wire is of yellow or green colour. This is connected to a metal plate buried deep underground near the house and is for safety purposes.
  • In India, the voltage difference between the live and neutral wires is about 220 V.
  • Live and neutral wires are connected to the electric meter through a fuse.
  • They are connected through a main switch, to all the conducting wires inside the home so as to provide electricity to every room.
  • In each separate circuit, various electrical appliances are connected between the live and neutral wires.
  • The different appliances are connected in parallel and the potential difference across every appliance is the same.

Question 3.
Fuse used in electrical circuit can save electrical objects from damage.
Answer:

  • Fuse wire is used to protect domestic appliances.
  • It is made of a mixture of substances and has a specific melting point.
  • It is connected in series to the electric appliances. If for some reason, the current in the circuit increases excessively, the fuse wire gets heated and melts. Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity
  • The circuit gets broken and the flow of current stops, thus protecting the appliance.
  • This wire is fitted in a groove in a body of porcelain-like non-conducting material. For domestic use, fuse wires with upper limits of 1 A, 2 A, 3 A, 4 A, 5 A and 10 A are used.

Question 4.
Bulbs arranged in parallel glow brighter than bulbs arranged in series.
Answer:

  • The amount of light given out by bulbs in parallel combination will be more than that in series combination.
  • In parallel combination the resistance of the overall circuit decreases whereas in series it increases, so the current flowing through the bulbs in parallel circuit is more.
  • Due to this, intensity of light given out by bulbs in parallel combination is more than the bulbs in series combination.

Complete the following table:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 15

Solve the numerical:

Question 1.
The length of a conducting wire is 50 cm and its radius is 0.5 mm. If its resistance is 30Ω, what is the resistivity of its material?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 16
The resistivity of the wire is 4.71 x 10-5 Ωm.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 2.
Determine the current that will flow when a potential difference of 33 V is applied between two ends of an appliance having a resistance of 110 Ω. If the same current is to flow through an appliance having a resistance of 500 Ω, how much potential difference should be applied across its two ends?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 17
The current is 0.3 A and potential difference to be applied is 150 V.

Question 3.
Determine the resistance of a copper wire having a length of 1 km and diameter of 0.5 mm.
Answer:
Given: Resistivity of copper (p)
= 1.7 x 10-8 Ω m
Converting all measures into metres.
Length of wire (L) = 1 km
= 1000 m = 103 m
Diameter of wire (d) = 0.5 mm
= 0.5 x 10-3m
To find: Resistance of wire (R) = ?
Formula:
\(R=\rho \frac{L}{A}\)
Solution:
If d is the diameter of the wire then, its area of cross-section
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 18
The resistance of a copper wire is 85Ω and area of cross section is 0.2 x 10-6 m2

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 5.
Two resistors having resistance of 16 and 14 are connected in series. If a potential difference of 18 V Is applied across them, calculate the current flowing through the circuit and the potential difference across each individual resistor.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 19
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 20
The current in the circuit is 0.6 A and potential across 16 Ω retor is 9.6 volt and 14 Ω resistor is 8.4 voIt.

Question 6.
If the resistors 5 Ω, 10 Ω and 30 Ω are connected in parallel to battery of 12 V, find the effective resistance in the circuit. Calculate the total current and current in each resistor.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 21
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 22
(a) The total current is 4 A and current in each resistor is 2.4 A, 1.2 A and 0.4 A respectively.
(b) The effective resistance in the circuit is 3Ω.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Complete the diagram and answer the questions:

Question 1.
The following figure shows the symbols for components used in the accompanying electrical circuit.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 23
(a) Place them at proper places and complete the circuit.
(b) Which law can you prove with the help of the above circuit?
(c) State expression for Ohm’s law
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 24
(b) This circuit can be used to prove Ohm’s law.
(c) V = IR is the expression of Ohm’s law

Question 3.
Explain with the help of a diagram, what are free electrons and how they move through the conductor?
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 25
Answer:

  • Every atom of a metallic conductor has one or more outermost electrons which are very weakly bound to nucleus.
  • These are called free electrons.
  • These electrons can easily move from one part of a conductor to its other parts.

Complete the paragraph:

Question 1.
If resistors are connected in series,
Answer:
The same current flows through each resistor. The effective resistance of the resistors is equal to the sum of their individual resistances. The potential difference between the two extremes of the arrangement is equal to the sum of the potential differences across individual resistors. The effective resistance is larger than each of the individual resistances. This arrangement is used to increase the resistance in a circuit. This type of connection is used in electrical heating equipment like geysers, iron, and hair dryers.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 2.
If a number of resistors are connected in parallel,
Answer:
The inverse of the effective resistance is equal to the sum of the inverses of individual resistances. The current flowing through an individual resistor is proportional to the r inverse of its resistance and the total current flowing through the circuit is the sum of the currents flowing through individual resistors. The potential difference across all r resistors is the same. The effective resistance of resistors connected in parallel is less than r the least resistance of individual resistors.

This arrangement is used to reduce the resistance in a circuit. Even if any one of the several bulbs connected in parallel becomes non-functional because of some damage to its filament, the circuit does not break as the current flows, through the other paths, and the rest of the bulbs light up. When several bulbs are connected in parallel, they emit the same amount of light as when they are connected individually in the circuit, while bulbs connected in series emit less light than when connected individually.

Read the paragraph and answer the questions.

Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket. Before cleaning an electrical appliance it should be switched off and its plug removed from the socket. One’s hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles. As rubber is an insulator, it prevents | the current from flowing through our body, thereby protecting it. If a person gets an electric shock, you should not touch that person. You should switch off the main switch and if the switch is too far or you do not know where it is located, then you should remove the plug from the socket if possible. If not, then you should use a wooden pole to push the person away from the electric wire.

(i) Why should the electrical sockets be fitted at a certain height?
Answer:
Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside.

(ii) Why plug wires should not be pulled out while removing any electrical device?
Answer:
Plug wires should not be pulled out while removing any electrical device as it may cause the wire to break causing short circuit which can lead to fire or death.

(iii) Why should a person wear footwear with rubber soles while handling electrical appliances. .
Answer:
As rubber is an insulator, it prevents the current from flowing through our body, thereby protecting it. Hence a person should wear footwear with rubber soles while handling electrical appliances.

(iv) Saee is touching an electrical button socket with wet hands what will you advise her and why?
Answer:
We will advise her to dry her hands before touching any electrical sockets or devices as water on the hands can cause an electrical short circuit producing shock to the person touching it.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

(v) Sneha is getting an electrical shock what will you do the save her life?
Answer:
We should switch off the main switch and if the switch is too far or we do not know where it is located, then we should remove the plug from the socket if possible. If not, then we should use a wooden pole to push the person away from the electric wire.

(vi) Give a title to the above passage.
Answer:
Precautions to be taken while using electricity

Answer the questions in details:

Question 1.
Find the expression (i.e., derive the expression) for the resistors connected in series.
Answer:
Expression for the resistance connected in series:
(i) Let R1, R2 and R3 be three resistances connected in series between C and D.
(ii) Let Rs be the effective resistance in circuit and V1, V2 and V3 be the potential difference across R1, R2 and R3 respectively.
(iii) Let the potential difference across CD be V.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 30
(iv) In series combination.
v = v1 + v2 + v3 ……………………(i)
By using Ohm’s law
V = IRs
∴ V1= IR1, V2 = IR2 and V3 = 1R3
Substituting these values in equation (j) we get
IRs = IR1 + IR2 + IR3
∴ Rs = R1 + R2 + R3
For ‘n’ number of resistors conneded in series we get
Rs = R1 + R2 + R3 + …………………. + Rn

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 2.
Find the expression (i.e., derive the expression) for the resistors connected in parallel.
Answer:
Expression for the resistance connected in parallel.
(i) Let R1, R2 and R3 be the three resistances connected in parallel combination between points C and D and let R be their effective resistance.
(ii) Let I1, I2 and I3 be the currents flowing through resistances R1, R2 and R3 respectively. Let I be the current flowing through the circuit and V be the potential difference of the cell.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 31
(iii) For parallel combination of resistances,
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 32
(iv) Substituting the values of (I, I1, I2 and I3) in equation (i) we get
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 33

Question 3.
Find the expression for resistivity of a material.
Answer:
(i) At a given temperature, the resistance (R) of a conductor depends on its length (L), area of cross-section (A) and the material it is made of. If the resistance of a conductor is R, then
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 34
(ii) p is the constant of proportionality and is called the resistivity of the material.
(iii) The unit of resistivity in SI units is Ohm metre (Ω m).
(iv) Resistivity is a specific property of a material and different materials have different resistivity. ’

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Make the concept diagram and explain:

Question 1.
Make the concept diagram of an electrical circuit and explain the working of a fuse.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 35
Answer:

  • Fuse wire is used to protect domestic appliances.
  • It is made of a mixture of substances and has a specific melting point.
  • It is connected in series to the electric appliances. If for some reason, the current in the circuit increases excessively, the fuse wire gets heated and melts. The circuit gets broken and the flow of current stops, thus protecting the appliance.
  • This wire is fitted in a groove in a body of porcelain-like non-conducting material. For domestic use, fuse wires with upper limits of 1A, 2A, 3A, 4A, 5A, and lO Aareused.

(2) Show motion of electrons in an circuit and explain precautions while using an electrical device.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 37

  • Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket.
  • Before cleaning an electrical appliance it should be switched off and its plug removed from the socket.
  • One’s hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles. Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity
  • As rubber is an insulator, it prevents the current from flowing through our body, thereby protecting it.
  • If a person gets an electric shock, you should not touch that person. You should switch off the main switch or remove the plug from the socket if possible.
  • 1f not, then you should use a wooden pole to push the person away from the electric wire.

Q.4.4.Complete the incomplete figure and give an explanation:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 38
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 39
If the physical state of a conductor remains constant, the current (I) flowing through it is directly proportional to the potential difference (V) between its two ends.

I α V
I = kV (k = constant of proportionality)
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 40

This is known as Ohm’s law.
We can obtain the SI unit of resistance from the above formula, Potential difference and current are measured in Volts and Amperes respectively. The unit o resistance is called Ohm. It is indicated by the symbol Ω.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 41
The resistance of one Ohm : If one Ampere current flows through a conductor when one Volt potential difference is applied between its ends, then the resistance of the conductor is one Ohm.

9th Std Science Questions And Answers:

Introduction to Biotechnology Class 9 Science Chapter 17 Questions And Answers Maharashtra Board

Class 9 Science Chapter 17

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 17 Introduction to Biotechnology Question Answer Maharashtra Board

Class 9 Science Chapter 17 Introduction to Biotechnology Question Answer Maharashtra Board

1. Each of the following statements is wrong. Rewrite them correctly by changing either one or two words.

a. Simple squamous epithelium is present in the respiratory tract.
b. Glandular epithelium is present in kidneys.
c. Chlorenchyma helps the plant to float in water.
d. Striated muscles are also called involuntary muscles.
e. Chloroplast is present in permanent tissue.
Answer:
a. False. Ciliated columnar epithelium is present in respiratory tract.
b. False. Cuboidal epithelium is present in kidneys.
c. False. Aerenchyma helps the plant to float in water.
d. False. Striated muscles are also called voluntary muscles.
e. False. Chlorenchyma is present in permanent tissue.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

2. Identify the odd word and explain why it is odd.

a. Xylem, phloem, permanent tissue, meristematic tissue.
Answer:
Meristematic tissue. It has the ability to divide whereas the others have lost the ability to divide.

b. Epithelium, Muscle fibre, nerve fibre, the epidermis.
Answer:
Epidermis. It is a plant tissue whereas the others are animal tissues.

c. Cartilage, bone, tendon, cardiac muscle.
Answer:
Cardiac muscle. It is a muscular tissue whereas the others are connective tissues.

3. Write the names of the following tissues.

a. Tissue lining inner surface of mouth.
Answer:
Squamous epithelium.

b. Tissue joining muscles and bones.
Answer:
Tendon

c. Tissue responsible for increasing height of plants.
Answer:
Apical meristem.

d. Tissue responsible for increasing girth of stem.
Answer:
Lateral meristem.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

4. Write the differences.
Simple tissue and complex tissues in plants.
Answer:

Simple Tissue in plants Complex Tissues in plants
(i) They are made up of only one type of cells. (i) They are made up of more than one type of cells.
(ii) They are found in all parts of the plant. (ii) They are found in the vascular regions of the plant.
(iii) They perform different functions like storage (iii) They mainly perform the function of
of food, support, giving strength etc. conduction of water and food.
(iv) Examples – Parenchyma, Collenchyma, Sclerenchyma. (iv) Example – Xylem and phloem.

5. Write short notes.

a. Meristematic tissue.
Answer:

  • Meristematic tissue is present in specific parts of a plant where growth takes place.
  • Cells of meristematic tissue contain thick cytoplasm, a conspicuous nucleus and a thin cell wall and are compactly packed together.
  • Vacuoles are usually absent in these cells and they are highly active.
  • The main function of meristematic tissue is to bring about plant growth.
  • According to the location, meristematic tissues are of three types: Apical meristem, intercalary meristem and lateral meristem.

b. Xylem
Answer:

  • Xylem is a complex permanent tissue in plants.
  • It consists of thick-walled dead cells.
  • The type of cells in xylem are trachieds, vessels, xylem fibres (dead cells) and xylem parenchyma (living cells).
  • Its structure is like interconnected tubes conduct water and minerals only in upward direction.

c. Striated muscles.
Answer:

  • The cells of striated muscles are long, cylindrical, multinucleate and have no branches.
  • These are alternate dark and light bands on these muscles.
  • As they are attached to bones, they are also called skeletal muscles.
  • They move as per our will, hence they are called voluntary muscles.
  • Striated muscles bring about movements of arms and legs, running, speaking etc.

d. Agro-complementary business.
Answer:
(i) The business that are complementary to agriculture and generate supplementary income for the farmers are called agro complementary business.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

(ii) These include:

  • Animal Husbandry: It is practiced for milk production and for using the cattle as help in farming operations, e.g. cows and buffaloes are raised for milk whereas bulls and male buffaloes for pulling heavy loads.
  • Poultry farming: It is the rearing of egg and meat yielding chickens. Chickens raised for eggs are called layers while those raised for meat are called broilers.
  • Sericulture: It is the rearing of silkworms (moths) for silk production. The silk fibres obtained are processed, reeled and then woven into fabric.

e. Genetic engineering
Answer:

  • Genetic engineering is the deliberate modification of the characteristics of an organism by manipulating its genetic material.
  • An organism that is generated through genetic engineering is called a genetically modified organism (GMO).
  • Genetic engineering is applied in many fields like research, agriculture, industrial biotechnology and medicine.
  • In agriculture, genetic engineering is used in the production of cash crops, improvement in varieties of cash crops, increase in ability of plants to withstand environmental stresses.
  • In medicine, genetic engineering is used for vaccine production, early diagnosis of congenital disease, organ transplant, cancer research, production of artificial skin, cartilage etc. in laboratories.

f. Sericulture
Answer:

  • Sericulture is the rearing of silkworms (moths) for production of silk.
  • Bombyx mori is the most commonly used variety of silkworm for this purpose.
  • The life cycle of silkworm consist of four stages namely egg, larva, pupa and adult.
  • Thousands of eggs deposited by female moths are incubated artificially to shorten the incubation period.
  • Larvae hatching out of eggs are released on mulberry plants.
  • Larvae are nourished by feeding on mulberry leaves.
  • After feeding for 3-4 days, larvae move to branches of mulberry plant.
  • The silk thread is formed from the secretion of their salivary glands.
  • Larvae spin this thread around themselves to form a cocoon. The cocoon may be spherical in nature.
  • Ten days before the pupa turns into an adult, all the cocoons are transferred into boiling water.
  • Due to the boiling water, the pupa dies in the cocoon and silk fibres become loose.
  • These fibres are unwound, processed and reeled. Various kinds of fabric is woven from silk threads.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

6. Explain the meaning of biotechnology and its impact on agricultural management with suitable examples.
Answer:
The techniques of bringing about improvements in living organisms by artificial genetic changes and by hybridization for the welfare of human beings, are together called biotechnology. Impact of biotechnology on agricultural management:

  • Genetically modified crops (GM crops) are being produced by introducing changes in DNA of natural crops.
  • Normally such varieties are not found in nature.
  • Thus, new varieties are produced artificially Different useful characters are introduced in such varieties.
  • Some naturally occurring varieties cannot withstand environmental stress like frequently changing temperature.
  • Wet and dry famines, changing climates etc. However, GM crops can grow in any of such adverse conditions.
  • As GM crops are resistant to insect pests, pathogens, chemical weedicides, etc. the use of harmful chemicals like pesticides can be avoided.
  • Due to use of seeds of GM crops, there is improvement in nutritive value and decrease in loss of crops.

Draw neat and labelled diagrams of the following:
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 2
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 3

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

7. Which two main techniques are used in biotechnology? Why?
Answer:

  • The two main techniques used in biotechnology are – Genetic engineering and tissue culture.
  • These techniques are used to bring about improvements in living organisms by artificial genetic changes and by hybridization for the welfare of human beings.
  • Its uses are in the production of cash crops, improvement in varieties of cash crops, early diagnosis of congenital diseases, organ transplant, cancer research, production of artificial skin, cartilage etc. in laboratories.

8. Discuss ‘Agritourism’ in the class and write a project on an agrotourism centre nearby. Present it in the class in groups.
Answer:

  • In agritourism, plantlets of flowering, medicinal, ornamental, vegetable plants and fruit trees are produced on a large scale by tissue culture technique.
  • By growing some of the plants fully, an agritourism centre can be developed.
  • If sufficient land is available, the emerging field of agritoursim would be a good business.
  • An agritourism centre consists of following:
    (a) Mango, chikoo (sapota), guava, coconut, custard apple and some other regional fruit trees.
    (b) Shade giving local or exotic attractive plants.
    (c) Ornamental and flowering plants.
    (d) Butterfly garden.
    (e) Medicinal plant garden.
    (f) Organic vegetables and fruits.
  • People visit places with such attraction in large numbers.
  • Selling plantlets/seedlings, fruits, vegetables at such places can be quite profitable.

9. Define the term tissue and explain the concept of tissue culture.
Answer:
A group of cells having the same origin, same structure and same function is called tissue. Concept of tissue culture:

  • Ex vivo growth of cells or tissues in an aseptic and nutrient-rich medium is called tissue culture.
  • Nowadays, a complete organism can be developed from a single cell or from tissue with the help of the tissue culture technique. Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology
  • A liquid, solid or gel-like medium prepared from agar, which supplies nutrients and energy necessary for tissue culture is used in this technique.
  • Tissue culture can be used to grow plants on a large scale, which bear flowers, fruits of excellent quality, in shorter durations and are disease free.
  • The various processes involved in tissue culture are:

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 7

10. ‘Rearing of sheep is a livestock’. Justify this statement.
Answer:

  • The term livestock refers to animals reared for profit or for use.
  • Sheep provides us with wool, skin, meat and milk.
  • Therefore, rearing of sheep is a livestock.

Class 9 Science Chapter 17 Introduction to Biotechnology Intext Questions and Answers

Question 1.
Which components bring about important processes in the living organisms?
Answer:
Cells, tissues, organs, organ systems bring out important processes in living organisms.

Question 2.
Which is the smallest structural and functional unit of the body of living organisms?
Answer:
Cell is the smallest structural and functional unit of the body of living organisms.

Question 3.
Which type of muscle is the diaphragm of the respiratory system?
Answer:
Diaphragm is a skeletal muscle. It is an involuntary muscle that regulates breathing, although some voluntary control can be achieved.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 4.
What is the main difference between the growth of animals and plants?
Answer:

  • Growth in animals is uniform whereas the growth in plants occurs in specific parts of the body.
  • Growth in animals occurs for a certain period of life whereas the growth in plants occur throughout their life.

Question 5.
Why does the growth of a plant occur only at specific parts of the plant body?
Answer:
Growth of a plant occurs only at specific parts of the body due to the presence of meristematic tissues which contain dividing cells.

Question 6.
Suppose you want to grow a garden like the one shown in the picture, around your home or school. What would you do to achieve that? By which methods will you cultivate the seedlings?
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 4
Answer:

  • To grow a garden around the school or home, one must cultivate seedlings of different plants and those seedlings must be watered regularly and nurtured properly.
  • Seedlings can be cultivated by the following methods:
    (a) Seed sowing: Seeds are sown, watered and allowed to germinate.
    (b) Transplantation: Seeds are sown somewhere else and then the seedlings are transplanted in the garden.

Question 7.
You must have seen flowers of same variety but of 2 or 3 different colours borne by same plant. How is this possible?
Answer:

  1. The different coloured flowers borne by the same plant are due to the pigments like anthocyanins, carotenoids etc. according to the genetic makeup of the plant.
  2. Also, by using the latest techniques of biotechnology and tissue culture, it is possible to manipulate the genes for flower colour and get the desired flower colour.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 8.
What keeps the various organs and organ systems separate from each other? Why?
Answer:
The epithelial tissue keeps the various organs and organ systems separate from each other by forming a layer on the organs.

Question 9.
Why are epithelial tissues said to be simple tissues?
Answer:
Epithelial tissues are said to be simple tissues as they are made up of only one type of cells.

Question 10.
Why do slim persons feel more cold in winter than those who are obese?
Answer:

  • Below the skin, there is a tissue called adipose tissue.
  • The cells of these tissue are filled with fat droplets.
  • Due to storage of fat, it acts as an insulator and helps to retain heat in the body.
  • Obese people have more fat deposited in the adipose tissue as compared to slim persons. Therefore, slim persons feel more cold in winter than those who are obese.

Question 11.
Why can bones not be folded?
Answer:

  • Bone cells called osteocytes are embedded in solid ground substance made up of calcium phosphate.
  • This makes the bone hard, rigid and non- flexible. Therefore, bones cannot be folded.

Question 12.
Which other industries can be started as an extension of the plant nursery business?
Answer:
Businesses like Agritourism, Ecotourism, forest resorts and organic fruit gardens, yoga and meditation centres can be started as an extension of the plant nursery business.

Question 13.
To which places do people choose to go on vacation in order to relax when they are tired of crowds and stressful life?
Answer:
People choose to go to hill stations, beaches, forest resorts and places where there is lot of greenery to relax when they are tired of crowds and stressful life.

Question 14.
What is the inter-relationship between the two questions (7 and 8) above?
Answer:
The inter-relationship between the above two questions is that businesses like agritourism, ecotourism, forest resorts etc. can fetch a huge profit as people look for these kind of places to relax, to be away from the hustle and bustle of city life and feel close to nature.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 15.
Why are the cocoons transferred to boiling water before the pupa develops into an adult?
Answer:

  • Once the pupa develops into an adult, it will secrete a fluid to dissolve the silk and emerge out.
  • Hence the cocoons are transferred to boiling water before the pupa develops in to an adult to kill the silkworms before transformation is complete.

Question 16.
Why we cannot see our organs like the heart, blood vessels and intestines?
Answer:
We cannot see our organs like the heart, blood vessels and intestines as they are located inside the body.

Question 17.
What is meant by white revolution? Who was its pioneer? What benefits did it bring?
Answer:

  • White revolution was the programme launched by the National Dairy Development Board to increase the milk production in India.
  • Dr. Verghese Kurien was the pioneer of white revolution.
  • It transformed India from a milk-deficient nation to the largest producer of milk. It also helped the dairy farmers in directing their own development and empowering them.

It also helped to reduce the malpractices carried out by milk traders and merchants.

Answer the following questions:

Question 1.
Collect information about GM varieties of crops in your area and make a note of them. Also find out if there are adverse effects of GM crops on human beings and environment.
Answer:

  • Some of the GM crops are:
    (a) Maize: MON 810, MON 863
    (b) Rice: Golden rice
    (c) Brinjal: BT brinjal
    (d) Cotton: BT cotton
  • Effects of GM Crops on human beings: GM crops may cause toxicity and allergic reactions in humAnswer:
  • Effects of GM Crops on environment:
    (a) GM crops may be toxic to pollinators and non-target species like butterflies.
    (b) Many GM crops may be a threat to soil ecosystem as they secrete their toxins into the soil.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
As shown in the figure, place an onion on each gas jar in such a way that its base (roots) will remain dipped in water. Measure and record the length of the roots of both onions on the first, second and third day. On the fourth day, cut off 1 cm of the roots of the onion in flask B Measure the length of the roots of both onions for the next five days and record your observations in the following table.
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 6

Length (cm) Day 1 Day 2 Day 3 Day 4 Day 5
Flask A
Flask B

Answer:
(Students are expected to record their observation in the given table).

(a) Which onion has longer roots? Why?
Answer:
The onion in jar Ahas longer roots as it continues to grow due to the presence of meristematic tissue at root tip which had dividing cells.

(b) Why did the roots of the onion in jar B stop growing?
Answer:
The roots of onion in jar B stopped growing as the meristematic tissue present in the root tips were cut off.

Question 3.
Take a fresh and fleshy leaf of Rhoeo, lily or any other plant. Pull and press it, tearing it obliquely in such a way that its transparent epidermis will be visible at the cut margins. Take the transparent epidermis with the forceps and keep it in dilute safranin solution for 1 minute. Spread it on a slide, cover it with a cover-slip and observe it under a compound microscope.
Answer:

  • The cells of the epidermis are flat and polygonal and there are no intercellular spaces between them.
  • A single continuous layer is formed.
  • The epidermis is covered by a waxy cuticle.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 4.
Establish your own plant nursery near your school or home. Prepare the seedlings of flowering plants, fruit plants, and ornamental plants being grown in your area. Can you start a business in the future with the help of this activity? Think it over.
Answer:
(Students are expected to do this activity on their own.)

Yes, we can start a business of plant nursey in future. If sufficient land is available, the emerging field of agritourism would be a good business.

Question 5.
Observe the skin of the back of your hand with the help of a magnifying lens. Do you see the closely attached squarish and pentagonal shapes?
Answer:
Yes, we can see the squarish and pentagonal shapes. This is the stratified squamous epithelium present on the outer layer of skin.

Question 6.
Observe a permanent slide of blood smear under a compound microscope. What did you see?
Answer:

  • A permanent slide of blood smear under a compound microscope shows different types of cells like RBCs, WBCs and platelets.
  • The different types of WBCs that can be seen are eosin mphocytes.

Question 7.
Visit a modern cowshed nearby and record the following points – The number of cattle, their variety, total milk production, cleanliness in cattle-shed, arrangements for health care of cattle.
Answer:
Students are expected to do this activity on their own.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 8.
Collect more information about animal husbandry.
Answer:

  • The branch of agriculture which deals with the feeding, shelter, health and breeding of domestic animals is called animal husbandry.
  • It is a scientific and systematic management of livestock.
  • Animal husbandry practices include:
    (a) Proper food and clean drinking water.
    (b) Proper shelter
    (c) Proper methods of breeding
    (d) Prevention and cure of disease.
  • Animal husbandary practices serve as an alternative income for the farmers and help to satisfy the need of food for man.

Question 9.
Find out from the internet the average daily milk yield from local and exotic varieties of cow.
Answer:

  • Milk yield from local cow : 12-15 litres/day.
  • Milk yield from exotic cow: 15-30 litres/day.

Question 10.
Are the structure and functions of the bodies of plants and animals the same?
No. The structure and functions of the bodies of plants and animals are not the same. Some differences between plants and animals:
Answer:

Plants Animals
Plants continue to grow throughout their life. Animals grow only for a specific period of life.
Growth in plants in not uniform. Growth in animals is uniform.
Plants have dividing and non-dividing tissues. Animals do not have different dividing and non­dividing tissues.
Plants are sedentary. Animals generally move from place to place in search of food, shelter and partners.
Energy needs of plants are less. Energy needs of animals are greater.
Plants can prepare their own food by photosynthesis. Animals cannot prepare their own food. They depend on plants and other animals for their food.

Class 9 Science Chapter 17 Introduction to Biotechnology Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
Protective coverings in the animal body are called tissues.
(a) meristematic
(b) muscular
(c) epithelial
(d) bone
Answer:
(c) epithelial

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Epithelial tissue is present in
(a) skin
(b) inner surface of blood vessels
(c) walls of the alveoli
(d) all of these
Answer:
(d) all of these

Question 3.
helps in selective transport of substances.
(a) Squamous epithelium
(b) Stratified epithelium
(c) Cuboidal epithelium
(d) Columnar epithelium
Answer:
(a) Squamous epithelium

Question 4.
helps in secretion of digestive juice, absorption of nutrients.
(a) Glandular epithelium
(b) Columnar epithelium
(c) Cuboidal epithelium
(d) Ciliated epithelium
Answer:
(b) Columnar epithelium

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 5.
supports internal organs
(a) Areolar tissue
(b) Adipose tissue
(c) Lymph
(d) Tendon
Answer:
(a) Areolar tissue

Question 6.
tissue helps in insulation, supply of
Answer:
(b) Adipose tissue

Question 7.
connect muscles to bones.
(a) Ligaments
(b) Tendons
(c) Cartilages
(d) Nerves
Answer:
(b) Tendons

Question 8.
join two bones to each other.
(a) Cartilages
(b) Tendons
(c) Ligaments
(d) Muscles
Answer:
(c) Ligaments

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 9.
meristem increases the length of root and stem.
(a) Intercalary
(b) Apical
(c) Lateral
(d) Permanent
Answer:
(b) Apical

Question 10.
showed that cells and tissues can be grown ex vivo.
(a) Frederick Campion Steward
(b) Gregor Johann Mendel
(c) Frederick Miescher
(d) Robert Brown
Answer:
(a) Frederick Campion Steward

Question 11.
is an example of genetically modified maize.
(a) Vaishali
(b) Vistive Gold
(c) MON 810
(d) Amflora
Answer:
(c) MON 810

Question 12.
is an exotic variety of cow.
(a) Holstein
(b) Gir
(c) Sahiwal
(d) Dangi
Answer:
(a) Holstein

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 13.
is a layer chicken.
(a) Brahma
(b) Leghorn
(c) Cochin
(d) Aseel
Answer:
(b) Leghorn

Question 14.
is a broiler chicken.
(a) Leghorn
(b) Lehman
(c) Aseel
(d) Minorca
Answer:
(c) Aseel

Question 15.
is a broiler chicken.
(a) Minroca
(b) Ancona
(c) Leghorn
(d) Cochin
Answer:
(d) Cochin

Question 16.
Cartilage is found
(a) all around the cells in body
(b) in nose, ear, larynx, trachea
(c) between the muscles and skin
(d) around the blood vessels
Answer:
(b) in nose, ear, larynx, trachea

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 17.
Which of the following is an exotic variety of COW?
(a) Plymouth Rock
(b) New Hampshire
(c) Black Rock
(d) Brown Swiss
Answer:
(d) Brown Swiss

Question 18.
epithelium is present in the inner surface of respiratory tract.
(a) Stratified
(b) Columnar
(c) Ciliated
(d) Cuboidal
Answer:
(c) Ciliated

Question 19.
Bones consist of osteocytes embedded in solid ground substance made up of
(a) calcium bicarbonate
(b) calcium carbonate
(c) calcium phosphate
(d) calcium sulphate
Answer:
(c) calcium phosphate

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 20.
helps in growth of branches, formation of leaves and flowers.
(a) Lateral meristem
(b) Apical meristem
(c) Intercalary meristem
(d) Xylem
Answer:
(c) Intercalary meristem

Question 21.
tissue is present in the hard coat of seeds, outer covering of coconut.
(a) Parenchyma
(b) Xylem
(c) Collenchyma
(d) Sclerenchyma
Answer:
(d) Sclerenchyma

Question 22.
The cells of tissue are dead.
(a) Parenchyma
(b) Collenchyma
(c) Sclerenchyma
(d) All of these
Answer:
(c) Sclerenchyma

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 23.
Cuboidal epithelium is found in
(a) inner surface of mouth, blood vessels
(b) inner layer of skin
(c) tubules of kidney, salivary gland
(d) inner surface of respiratory tract
Answer:
(c) tubules of kidney, salivary gland

Find the odd man out:

Question 1.
Cell body, muscle fibre, axon, dendrites.
Answer:
Muscle fibre. It is a muscle cell whereas the others are parts of nerve cell.

Question 2.
Tracheids, vessels, companion cells, xylem fibres.
Answer:
Companion cells. It is an element of phloem whereas the others are elements of xylem.

Question 3.
Sieve tubes, tracheids, phloem parenchyma, phloem fibres.
Answer:
Tracheids. It is an element of xylem whereas the others are elements of phloem.

Question 4.
Holstein, Brown swiss, Bombyx mori, Jersey.
Answer:
Bombyx mori. It is a variety of silkworm whereas the others are varieties of cows.

Complete the analogy:

(1) Inner surface of mouth : Squamous epithelium :: Inner surface of intestine :
(2) Glandular epithelium : Secretion of sweat, oil :: Cuboidal epithelium:
(3) Respiratory tract: Ciliated columnar epithelium :: Kidney tubules:
(4) Outer layer of skin : Stratified epithelium :: Inner layer of skin :
(5) Muscular tissue : Movement :: Nervous tissue :
(6) Tendons : Join muscles to bones :: Ligaments :
Answer:
(1) Columnar epithelium
(2) Secretion of saliva
(3) Cuboidal epithelium
(4) Glandular epithelium
(5) Conduction of excitation
(6) Join two bones to each other

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) Squamous epithelium (a) Secretion of digestive juice
(2) Cuboidal epithelium (b) Selective transport of substances
(3) Columnar epithelium (c) Protection of organs
(4) Stratified epithelium (d) Secretion of saliva

Answer:
(1 – b),
(2 – d),
(3 – a),
(4 – c)

Question 2.

Column ‘A’ Column ‘B’
(1) Brahma (a) Local variety of cow
(2) Jersey (b) Layer chicken
(3) Devin (c) Exotic variety of cow
(4) Lehman (d) Broiler chicken

Answer:
(1 – d),
(2 – C),
(3 – a),
(4 – h)

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 3.

Column ‘A’ Column ‘B’
(1) Potato (a) Vistive Gold
(2) Maize (b) Amflora
(3) Soybean (c) Vaishali
(4) Tomato (d) MON 863

Answer:
(1-b),
(2-d),
(3-a),
(4-c)

Answer the following:

Question 1.
Explain the different types of muscular tissues.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 5

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Explain the types of complex permanent tissues.
Answer:

Name of tissue Xylem Phloem
Characteristics Consists of thick-walled dead cells Consists of cytoplasm containing living cells.
Types of cells Tracheids, vessels and xylem fibres – dead cells. Xylem parenchyma – living cells. Sieve tubes, companion cells, phloem parenchyma – living cells. Phloem fibres – dead cells.
Function A structure like interconnected tubes, conduct water and minerals only in upward direction. Tubes joined to each other, conduct sugar and amino acids from leaves to various parts in upward and downward direction

State whether the following statements are true or false. Correct the false statement:

Question 1.
Lateral meristem increase girth(diameter) of the root and stem.
Answer:
True

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Due to use of seeds of GM crops, there is decrease in nutritive value and increase in loss of crops.
Answer:
False. Due to use of seeds of GM crops there is improvement in nutritive value and decrease in loss of crops.

Question 3.
In a bioreactor, cells can be grown in a more nutritive medium and protected from pathogens.
Answer:
True

Question 4.
Rare and endangered plants can be grown by tissue culture and can be protected from extinction.
Answer:
True

Question 5.
The silk thread is formed from the secretion of the sweat glands of silkworm.
Answer:
False. The silk thread is formed from the secretion of the salivary glands of silkworm.

Question 6.
Rhode Island Red, New Hampshire, Plymouth Rock, Black Rock are varieties of chicken reared for both eggs as well as meat.
Answer:
True

Question 7.
Phloem consists of thick-walled dead cells.
Answer:
False. Phloem consists of living cells containing cytoplasm.

Question 8.
Stratified squamous epithelium is present in inner surface of mouth, oesophagus, blood vessels and alveoli.
Answer:
False. Stratified squamous epithelium is present in the outer layer of skin.

Question 9.
Broiler chickens are raised for eggs.
Answer:
False. Broiler chickens are raised for meat.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 10.
Non-striated muscles bring about contraction and relaxation of heart.
Answer:
False. Cardiac muscles bring about contraction and relaxation of heart.

Name the following:

Question 1.
Tissue responsible for growth of branches, formation of leaves and flowers.
Answer:
Intercalary meristem.

Question 2.
Tissue that helps in insulation, supply of energy, storage of fats.
Answer:
Adipose tissue.

Question 3.
Tissue present in outer layer of skin.
Answer:
Glandular epithelium.

Question 4.
Tissue present in nose, ear, larynx, trachea.
Answer:
Cartilage.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 5.
Dead cells in xylem.
Answer:
Tracheids, vessels and xylem fibres.

Question 6.
Living cells in xylem.
Answer:
Xylem parenchyma.

Question 7.
Living cells in phloem.
Answer:
Sieve tubes, companion cells, phloem parenchyma.

Question 8.
Dead cells in phloem.
Answer:
Phloem fibres.

Question 9.
GM crops.
Answer:
BT cotton, Amflora, Golden Rice.

Question 10.
Local Indian varieties of cow.
Answer:
Sahiwal, Sindhi, Lai kandhari, Devni, Khillari, Dangi.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 11.
Exotic varieties of cow.
Answer:
Jersey, Brown swiss, Holstein.

Question 12.
Layers.
Answer:
Leghorn, Minorca, Ancona, Lehman.

Question 13.
Broilers.
Answer:
Brahma, Long, Cochin, Aseel.

Question 14.
Chickens reared for both eggs and meat.
Answer:
Rhode Island Red, New Hampshire, Plymouth Rock, Black Rock.

Give scientific reason.

Question 1.
Blood is a complex tissue.
Answer:

  • A complex tissue is made up of more than one type of cells.
  • In blood, cells of different types, colour and shapes are mixed together.
  • Blood contains erythrocytes (RBCs), leucocytes (WBCs) and platelets in a liquid plasma.
  • Therefore, blood is a complex tissue.

Define the following:

Question 1.
Tissue
Answer:
A group of cells having the same origin, same structure and same function is called tissue.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Differentiation
Answer:
When cells acquire a specific structure, shape and location and perform a specific function, it is called differentiation.

Question 3.
Biotechnology
Answer:
The techniques of bringing about improvements in living organisms by artificial genetic changes and by hybridization for the welfare of human beings, are together called ‘Biotechnology’.

Question 4.
Tissue culture
Answer:
Ex vivo growth of cells or tissues in an aseptic and nutrient-rich medium’ is called tissue culture.

Write short notes

Question 1.
Animal husbandry
Answer:

  • In India, animal husbandry is practised for milk production and for using the cattle as help in farming operations.
  • Example – Cows and buffaloes are raised for milk and bulls and male buffaloes for pulling the heavy loads.
  • Local Indian varieties of cows like Sahiwal, Sindhi, Gir, Lai kandhari, Devni, Khillari, Dangi, etc. and exotic varieties like Jersey, Brown swiss, Holstein, etc. are kept for their milk.
  • Proper care of cattle is necessary for a clean and high yield of milk which includes:
    (a) A balanced diet, i.e. all constituents of food should be given to cattle. It must include fibre- rich coarse food, fodder, and sufficient water.
    (b) The cattle-shed should be clean and dry with proper ventilation and a roof.
    (c) Cattle should be regularly vaccinated.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Poultry farming.
Answer:

  • Rearing of egg and meat yielding chickens is called poultry farming.
  • Chickens raised for laying eggs are called layers . e.g. Leghorn, Minorca, Ancona, Lehman.
  • Chickens raised for meat are called broilers, e.g. Brahma, Long, Cochin, Aseel.
  • Rhode Island Red, New Hampshire, Plymouth Rock, Black Rock are varieties of chicken reared for both eggs as well as meat.
  • The objectives behind development of new hybrid varieties from a cross between Indian varieties like Aseel and exotic varieties like Leghorn are to produce good quality chickens in large numbers.
  • To develop the ability to withstand high temperature, to use by-products of agriculture as poultry feed, etc.

Distinguish between:

Question 1.
Xylem and Phloem
Answer:

Xylem Phloem
(i) Consists of mostly thick-walled dead cells. (i) Consists of mostly living cells containing cytoplasm.
(ii) The types of cells include dead cells- tracheids, vessels and xylem fibres and living cells-Xylem parenchyma. (ii) The types of cells include living cells – Sieve tubes, companion cells, phloem parenchyma and dead cells – phloem fibres.
(iii) Structure like interconnected tubes, conduct water and minerals only in upward direction. (iii) Tubes joined to each other, conduct sugar and amino acids from leaves to various parts in both upward and downward direction.

Question 2.
Striated muscles and Non-striated muscles.
Answer:

Striated muscles Non-striated muscles
(i) Muscle cells are long, cylindrical, multinucleate and have no branches. (i) Muscle cells are short, spindle-shaped, uninucleate and have no branches.
(ii) There are alternate dark and light bands on these muscles and they are attached to bones. (ii) Dark and light bands are absent. Not attached to bones.
(iii) They move as per our will, hence they are called voluntary muscles. (iii) They are not under the control of our will, hence they are called involuntary muscles.
(iv) These muscles bring about movements of arms and legs, running, speaking etc. (iv) These muscles bring about movement of eyelids, passage of food through alimentary canal, contraction and relaxation of blood vessels etc.

Question 3.
Cartilage and Bone
Answer:

Cartilage Bone
(i) They are present in nose, ear, larynx, trachea. (i) They form the skeleton of the body.
(ii) They contain cells supported by fibrous, flexible, jelly-like ground substance. (ii) They contain osteocytes embedded in solid ground substance made up of calcium phosphate.
(iii) Lubricates the surface of bones, gives support and shape to organs. (iii) Supports and protects different organs, helps in movement.

Answer the following questions:

Question 1.
Close your eyes and try to identify different objects by feeling them with your hand. Why is it possible for you to identify things like a note-book, text-book, bench, compass-box, etc. only by touching them?
Answer:

  • We can identify objects just by touching them because of the memory that we retain in our brain.
  • The nervous tissue enables us to respond to the stimuli of touch.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Answer in detail:

Question 1.
Write the location, structure and function of following epithelial tissues:
(a) Squamous epithelium
(b) Stratified epithelium
(c) Glandular epithelium
(d) Columnar epithelium
(e) Ciliated epithelium
(f) Cuboidal epithelium
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 8

Question 2.
Write the structure and function of the following connective tissues:
(a) Blood
(b) Lymph
(c) Areolar tissue
(d) Adipose tissue
(e) Cartilage
(f) Bones
(g) Tendons and ligaments
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 9

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Describe the structure of nervous tissue with the help of a neat labelled diagram.
Answer:

  • Nervous tissue enables us to respond to stimuli like touch, sound, odour, colour, etc.
  • Cells of the nervous tissue are specifically made to become excited and conduct the excitation from one part of the body to another.
  • The main part of the nerve cell is the cell body which contains the cytoplasm and the nucleus.
  • Numerous, small, branched fibres called dendrites arise from the cell body.
  • One of the fibres, is extremely long and is called
    the axon.
  • The length of the nerve cell may be up to one metre.
  • Many nerve cells are bound together with the help of connective tissue to form a nerve.
  • Nervous tissue is present in the brain, spinal cord and the network of nerves spread all throughout the body.
  • In most animals, action in response to a stimulus occurs due to the integrated functioning of the nervous tissue and muscular tissue.
    Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 1

Question 3.
Explain the types of simple permanent tissues in plants.
Answer:
Types of simple permanent tissues.

Question 4.
Explain types, location and function of Meristematic tissue in tabular form:
Answer:

Types Apical Intercalary Lateral
Location At tip of the root and stem At the base of the petiole of leaves and branches At sides of root and stem
Function Increase the length of the root and stem Growth of branches, the formation of leaves and flowers Increases diameter of the root and stem

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 5.
Explain the types of simple permanent tissues in plants.
Answer:
Types of simple permanent tissues.

Name of Tissue Parenchyma Collenchyma Sclerenchyma
Struc­ture of cells Living cells with thin cell wall and intercellular spaces. Elongated living cells with thickened cell wall at corners due to cellulose and pectin. Dead and fibrous cells with tapering ends, cell wall contains lignin.
Loca­tion All parts like roots, stem, leaves, flowers and seeds. At the base of leaf petiole, branches and stem Stem, veins of leaves, hard coats of seeds, outer covering of coconut.
Func­tions Support, storage of food and filling vacant spaces. Support and flexibility to various parts. Give strength and rigidity to parts of the plants.
Sub­ types Chlorenchyma: Leaves, perform photosynthesis. Aerenchyma: Helps aquatic plants, leaves and stem to float.

Question 6.
Write down the applications of biotechnology (tissue culture) in floriculture, nurseries and forestry.
Answer:
Applications of biotechnology (tissue culture) in floriculture, nurseries and foresty:

  • Tissue culture can be used to grow those plants on a large scale which bear flowers, fruits of excellent quality.
  • Fully grown plants can be produced in shorter durations.
  • Plants can be grown on a large scale even if means of pollination or germinating seeds are not available.
  • For example, orchids or pitcher plants do not germinate but these plants can easily be produced by means of tissue culture.
  • In a bioreactor, cells can be grown in a more nutritive medium and protected from pathogens.
  • Bioreactors are useful for producing plantlets on a very large scale.
  • A large number of seedlings/plantlets can be produced in a short time using minimum resources and materials.
  • Usually, plants produced by tissue culture and genetic modification techniques are disease-free.
  • Plantlets produced by tissue culture technique of the meristem are virus-free.
  • Embryos produced using conventional hybridization technique between two or more varieties may not grow fully for some reasons. Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology
  • However, embryos produced by tissue culture technique always complete their growth.
  • Rare and endangered plants can be grown using tissue culture technique and can thus be protected from extinction.
  • Similarly, various parts and seeds of such plants can be preserved by tissue culture and those varieties can be protected.

9th Std Science Questions And Answers:

Classification of Plants Class 9 Science Chapter 6 Questions And Answers Maharashtra Board

Class 9 Science Chapter 6

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 6 Classification of Plants Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 6 Classification of Plants Question Answer Maharashtra Board

Class 9 Science Chapter 6 Classification of Plants Question Answer Maharashtra Board

1. Match the proper terms from columns A and C with the description in column B.
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 10

2. Complete the sentences by filling in the blanks and explain those statements.
(angiosperms, gymnosperms, spore, Bryophyta, thallophyta, zygote)
a. ……………….. plants have soft and fiber-like body.

b. ……………….. is called the ‘amphibian’ of the plant kingdom.
Answer:
Bryophyta plant is called the ‘amphibian’ of the plant kingdom

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

c. In pteridophytes, asexual reproduction occurs by ……………….. formation and sexual reproduction occurs by ………………..formation.
Answer:
Spore, zygote: Pteridophyta plants show alteration of generation. One generation reproduces by spore-formation and the next generation reproduces sexually by zygote formation.

d. Male and female flowers of ……………….. are borne on different sporophylls of the same plant.
Answer:
Gymnosperms bear their male and female flowers on different sporophylls of the same plant

3. Answer the following questions in your own words.

a. Write the characteristics of subkingdom Phanerogams.
Answer:

  • Plants which have special structures for reproduction and produce seeds are called Phanerogams.
  • In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food.
  • During the germination of the seed, the stored food is used for the initial growth of the embryo.
  • Depending upon whether seeds are enclosed in a fruit or not phanerogams are classified into gymnosperms and angiosperms.

b. Distinguish between monocots and dicots.
Answer:

Dicots Monocots
Seed Two cotyledons Single cotyledon
Root Well developed, primary root (Taproot) Fibrous roots
Stem Strong, hard. e.g. Banyan tree Hollow, e.g. Bamboo
False, e.g. Banana
Disc-like, e.g. Onion.
Leaf Reticulate venation Parallel venation
Flower Flowers with 4 or 5 parts or in their multiples (tetramerous or pentamerous) Flowers with 3 parts or in multiples of three (trimerous).

c. Write a paragraph in your own words about the ornamental plants called ferns.
Answer:

  • Ferns belong to the group of plants called Pteridophyta.
  • They have well-developed roots, stem and leaves but do not bear flowers and fruits.
  • They have separate tissues for the conduction of food and water.
  • They reproduce with the help of spores formed along the back or posterior surface of their leaves.
  • They reproduce asexually by spore formation and sexually by zygote formation.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

d. Sketch, label and describe the Spirogyra.

e. Write the characteristics of the plants belonging to division Bryophyta.
Answer:

  • Bryophyta group of plants are called the amphibians of the plant kingdom because they grow in moist soil but need water for reproduction.
  • These plants are thalloid, multicellular and autotrophic.
  • They reproduce by spore-formation.
  • Their plant body structure is flat, ribbon-like, long, without true roots, stem and leaves.
  • Instead, they have stem-like or leaf-like parts and root-like rhizoids.
  • They do not have specific tissues for the conduction of food and water.
  • Examples: Moss (Funaria), Anthoceros, Riccia etc.?

4. Sketch and label the figures of the following plants and explain them into brief.
Marchantia, Funaria, Fern, Spirogyra.

Question 1.
Spirogyra.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 5

  • Spirogyra belongs to the division thallophyta. They are called as algae.
  • It grows mainly in water.
  • It does not have specific parts like root-stem- leaves-flowers but are autotrophic due to the presence of chlorophyll.
  • The plant body of Spirogyra is soft and fibre-like.
  • It has spirally arranged chloroplasts in its cell.

Question 2.
Funaria and Marchantia (Bryophyta)
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 6

  • These plants are called ‘amphibians’ of the plant kingdom because they grow mostly in soil and need water for reproduction.
  • They do not have specific tissues for the conduction of food and water.
  • The plant body is fiat, ribbon-like long, without true roots, stem and leaves
  • Instead, they have stem-like or leaf-like parts and root like rhizoids.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
Fern (Pteridophyta):
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 7

  • They have well-developed roots, stem and leaves for the conduction of food and water.
  • They do not bear flowers and fruits.
  • They reproduce with the help of spores present along the back or posterior surface of the leaves.

5. Collect a monocot and dicot plant available in your area. Observe the plants carefully and describe them in scientific language.

6. Which criteria are used for the classification of plants? Explain with reasons.
Answer:
Criteria for classification of plants:

  • If plants do not bear flowers, fruits and seeds, they are non-seed bearing plants. If they bear flowers, fruit and seeds, they are seed-bearing plants.
  • Presence or absence of conducting tissues- Plants such as pteridophytes, gymnosperms and angiosperms which possess conducting tissues are included in vascular plants whereas thallophytes and bryophytes which do not possess conducting tissues are included under non-vascular plants.
  • Depending upon whether the seeds are enclosed in fruit or not, plants are classified as gymnosperms (naked-seeds) and angiosperms (seeds covered by fruit) Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants
  • Depending upon the number of cotyledons in seeds, plants are classified into dicotyledons and monocotyledons

Class 9 Science Chapter 6 Classification of Plants Intext Questions and Answers

Can you recall?

Question 1.
How have living organisms been classified?
Answer:
(i) Organisms have been classified based on the following:

  • Cell structure
  • Body Organisation
  • Mode of nutrition
  • Reproduction

(ii) Organisms are also classified at kingdom level and groups and subgroups.

Activity-based questions

Question 1.
You may have seen a lush green soft carpet on old walls, bricks and rocks in the rainy season. Scrape it gently with a small ruler, observe it under a magnifying lens and discuss.
Answer:

  • It shows considerable tissue complexity and is differentiated into two main parts: a root and a shoot.
  • They have a variety of specialized tissues within these two regions of the body.
  • Same kind of cells are seen throughout the whole body except reproductive cells.

Question 2.
You may have seen ferns among the ornamental plants in a garden. Take a leaf of a fully grown fern and observe it carefully.
Answer:

  • New leaves typically expand by the unrolling in a tight spiral manner.
  • The anatomy of fern leaves can either be simple or highly divided.
  • They show the presence of spores formed along the back or posterior surface of their leaves.

Question 3.
Observe all garden plants like Cycas, Christmas tree, Hibiscus, Lily, etc. and compare them. Note the similarities and differences among them. Which differences did you notice in gymnosperms and angiosperms?
Answer:
Cycas and Christmas tree are gymnosperms, whereas Hibiscus and lily are angiosperms.

  1. Similarities: These plants have special structures for reproduction and produce seeds. During the germination of the seed, the stored food is used for the initial growth of the embryo.
  2. Differences: In gymnosperms, reproductive organs have cones whereas in angiosperms reproductive organs have flowers.
  3. In gymnosperms, seeds are without natural coverings whereas in angiosperms seeds are enclosed in natural coverings called fruits.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 4.
Soak the seeds of corns, beans, groundnut, tamarind, mango, wheat, etc. in water for 8 to 10 hrs. After they are soaked, check each seed to see whether it divides into two equal halves or not and categorize them accordingly.
Answer:
Monocots: com, wheat (it cannot be divided into equal halves)
Dicots: beans, groundnut, tamarind and mango (it can be divided into two equal halves)

Class 9 Science Chapter 6 Classification of Plants Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
The five-kingdom classification was proposed b7
(a) Robert Whittaker
(b) Robert Hooke
(c) Eichler
(d) Louis Pasteur
Answer:
(a) Robert Whittaker

Question 2.
In 1883, classified plants into two sub-kingdoms.
(a) Robert Whittaker
(b) Alexander Fleming
(c) Eichler
(d) Robert Hooke
Answer:
(c) Eichler

Question 3.
Ulothrix, ulva, sargassum belong to
(a) Bryophyta
(b) Thallophyta
(c) Pteridophyta
(d) Gymnosperms
Answer:
(b) Thallophyta

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 4.
is a bryophyte.
(a) Ulva
(b) Nephrolepis
(c) Funaria
(d) Equisetum
Answer:
(c) Funaria

Question 5.
In the seeds are naked.
(a) Pteridophyta
(b) Angiosperms
(c) Gymnosperms
(d) Bryophyta
Answer:
(c) Gymnosperms

Question 6.
In the flowers are reproductive organs.
(a) Angiosperms
(b) Gymnosperms
(c) Pteridophyta
(d) Bryophyta
Answer:
(a) Angiosperms

Question 7.
In the flowers are tetramerous or pentamerous.
(a) Monocotyledons
(b) Dicotyledons
(c) Gymnosperms
(d) Pteridophyta
Answer:
(b) Dicotyledons

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 8.
In monocotyledonous plants, the stem is
(a) hollow
(b) false
(c) disc-like
(d) all of these.
Answer:
(d) all of these

Question 9.
Lycopodium belongs to
(a) Thallophyta
(b) Bryophyta
(c) Gymnosperms
(d) Pteridophyta
Answer:
(d) Pteridophyta

Question 10.
Leaves of show reticulate venation.
(a) Bamboo
(b) Banana
(c) Onion
(d) Banyan
Answer:
(d) Banyan

Question 11.
Various types of fungi like yeasts and moulds are included in the group
(a) Thallophyta
(b) Halophyte
(c) Xenophyta
(d) Angiosperms
Answer:
(a) Thallophyta

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 12.
Bryophytes have a root-like structure called
(a) Nodes
(b) Rhizoids
(c) Nodules
(d) Aerenchyma
Answer:
(b) Rhizoids

Question 13.
reproduce with the help of spores formed along the back or posterior surface of their leaves.
(a) Halophyta
(b) Pteridophyta
(c) Thallophyta
(d) Angiosperms
Answer:
(b) Pteridophyta

Question 14.
In ……………………….., the reproductive organs cannot be seen.
(a) Pteridophyta
(b) Cryptogams
(c) Thallophyta
(d) Angiosperms
Answer:
(b) Cryptogams

Question 15.
are mostly evergreen, perennial and woody.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(c) Gymnosperms

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 16.
Gymnosperms bear male and female flowers on different of the same plant.
(a) Branches
(b) Roots
(c) Sporophylls
(d) Flowers
Answer:
(c) Sporophylls

Question 17.
In the seeds are not enclosed by fruit.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(c) Gymnosperms

Question 18.
In the seeds are enclosed by fruit.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(d) Angiosperms

Question 19.
The plants whose seeds cannot be divided into equal parts are called
(a) Algae
(b) Fungus
(c) Dicotyledons
(d) Monocotyledons
Answer:
(d) Monocotyledons

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 20
The plants whose seeds can be divided into equal parts are called
(a) Algae
(b) Fungus
(c) Dicotyledons
(d) Monocotyledons
Answer:
(c) Dicotyledons

Find the odd one out:

Question 1.
Ulothrix, Ulva, Nephrolepis, Sargassum
Answer:
Nephrolepis: It belongs to division Pteridophyta whereas the others belong to division thallophyta.

Question 2.
Funaria, Marchantia, Anthoceros, Spirogyra
Answer:
Spirogyra:

Question 3.
Marsilea, Pteris, Lycopodium, Riccia
Answer:
Riccia:

Question 4.
Cycas, Mango, Apple, Banyan
Answer:
Cycas:

Question 5.
Onion, Papaya, Wheat, Green peas
Answer:
Green peas:

Complete the analogy:

(1) Spirogyra : Thallophyta : : Riccia :
(2) Moss : Bryophyta : : Selaginella :
(3) Nephrolepis : Pteridophyta :: Ulothrix :
(4) Pteridophyta : Roots :: Bryophyta :
(5) Gymnosperms : naked seeds : : Angiosperms :
(6) Dicotyledon : Reticulate venation : : Monocotyledon:
(7) Bamboo stem: Hollow:: Onion Stem:
(8) Monocotylendon : Tap root:: Dicotyledon :
Answer:
(1) Bryophyta
(2) Pteridophyta
(3) Thallophyta
(4) Rhizoids
(5) Covered seeds
(6) Parallel venation
(7) Disc like
(8) Fibrous roots

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Difference between:

Question 1.
Thallophyta and Bryophyta
Answer:

Thallophyta Bryophyta
These plants grow mainly in water They grow in moist soil but need water for reproduction

Question 2.
Gymnosperms and Angiosperms
Answer:

Gymnosperms Angiosperms
No natural covering on seeds Seeds are formed in fruits

Question 3.
Algae and Moss
Answer:

Algae Moss
These plants mainly grow in water. These plants need water for reproduction.

State whether the following statements are true or false. Correct the false statements:

(1) Thallophyta are called as the amphibians of the plant kingdom.
(2) Fungi like yeasts and moulds are included in division bryophyta.
(3) Moss (Funaria) belongs to division bryophyta.
(4) Bryophyta have specific tissues for conduction of food and water.
(5) Plants belonging to Thallophyta group are only unicellular.
(6) Pteridophytes have well developed roots, stems and leaves.
(7) Pteridophytes reproduce with the help of spores formed along the back or posterior surface of their leaves.
(8) Nephrolepis belongs to division Pteridophyta.
(9) Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into monocots and dicots.
(10) Gymnosperms are mostly evergreen, perennial and woody.
(11) Gymnosperms bear male and female flowers on different sporophylls of different plants.
(12) In Angiosperms, the seeds are covered by fruits.
(13) Dicotyledonous plants show reticulate venation.
(14) Moncotyledonous plants have trimerous flowers.
(15) In dicotyledonous plants, the stem is strong and hard.
Answer:
(1) False. Thallophyta plants grow mainly in water.
(2) False. Fungi like yeasts and moulds are included in division thallophyta.
(3) True
(4) False. Bryophyta do not have specialised tissuesfor conduction of food and water.
(5) False. Plants belonging to thallophyta group may be unicellular or multicellular.
(6) True
(7) True Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants
(8) True
(9) False. Depending whether seeds are enclosed in. a fruit or not, angiosperms are classified into monocots and dicots.
(10) True
(11) False. Gymnosperms bear male and female flowers on different sporophylls of the same plant.
(12) True
(13) False. Dicotyledonous plants show parallel venation.
(14) True
(15) True.

Give name

Question 1.
What are ornamental plants are called?
Answer:
Ferns

Question 2.
Plants with two cotyledons are called.
Answer:
Dicots

Question 3.
Plants with single cotyledon are called.
Answer:
Monocots

Question 4.
Type of venation showed by hibiscus plant leaves
Answer:
Reticulate venation

Question 5.
Type of venation showed by lily plant leaves
Answer:
Parallel venation

One line answers

Question 1.
Which plants are mostly evergreen, perennial and woody?
Answer:
Gymnosperms are mostly evergreen, perennial and woody.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
Which type of venation showed by dicot plants?
Answer:
Leaves of dicot plants show reticulated venation.

Question 3.
Which type of venation showed by monocot plants?
Answer:
Leaves of monocot plants show parallel venation

Question 4.
How are angiosperms classified into monocot and dicot?
Answer:
Depending whether seeds and enclosed in fruit or not, angiosperms are classified into monocot and dicot

Question 5.
In which division are fungi like moulds and yeast classified?
Answer:
Fungi like moulds and yeast classified in division thallophyta.

Question 6.
Plants belonging to which group may be unicellular or multicellular?
Answer:
Plants belonging to thallophyta group may be unicellular or multicellular

Give scientific reason

Question 1.
Thallophyta plants have thin and fibre like body
Answer:
Thallophyta: These plants grow mainly in water i.e. fresh water as well as in saline water, therefore they usually have a soft and fibre-like (filamentous) body.

Question 2.
Bryophyta plants are called the amphibian plants.
Answer:
Bryophyta: They grow in moist soil but need water for reproduction. Therefore, they are called ‘amphibians of plant kingdom’.

Question 3.
Gymnosperms bear their male and female flowers on different sporophylls of the same plant
Answer:
Gymnosperms: As these plants do not take the assistance of pollinators i.e. vectors, the male and female flowers are present on the different sporophyll of the same plant for successful fertilisation.

Write note on

Question 1.
August W. Eichler
Answer:
In 1883, Eichler, a botanist, classified the Kingdom Plantae into two subkingdoms. As a result, two subkingdoms, cryptogams and phanerogams were considered for plant classification.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
Thallophyta
Answer:
These plants grow mainly in water. This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll, is called algae. Algae show great diversity. They may be unicellular or multicellular, and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a soft and fibre-like body. Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Question 3.
Bryophyta
Answer:
This group of plants is called the amphibians’ of the plant kingdom because they grow in moist soil but need water for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by spore formation. The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves. Instead, they have stem-like or leaf-like parts and root-like rhizoids. They do not have specific tissues for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.

Question 4.
Pteridophyta
Answer:
Plants from this group have well developed roots, stem and leaves and separate tissues for conduction of food and water. But, they do not bear flowers and fruits. They reproduce with the help of spores formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce asexually by spore-formation and sexually by zygote formation. They have a well-developed conducting system.

Question 5.
Phanerogams
Answer:
Plants which have special structures for reproduction and produce seeds are called phanerogams. In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into gymnosperms and angiosperms.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 6.
Gymnosperms
Answer:
Gymnosperms are mostly evergreen, perennial and woody. Their stems are without branches. The leaves form a crown. These plants bear male and female flowers on different sporophylls of the same plant. Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.

Question 7.
Angiosperms
Answer:
The flowers these plants bear are their reproductive organs. Flowers develop into fruits and seeds are formed within fruits. Thus, these seeds are covered; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Complete the flow chart.

Question 1.
Living Organisms
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 1
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 2

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
Kingdom: Plantae
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 3
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 4

Distinguish between:

Question 1.
Bryophyta and Pteridophyta:
Answer:

Bryophyta Pteridophyta
Bryophytes grow in soil but need water for reproduction. Pteridophytes grow in soil.
Plant body is without specific parts like true roots, stem and leaves. Plant body is differentiated into true roots, stem and leaves.
Conducting tissues for food and water absent. Conducting tissues for food and water present.
Examples: Moss (Funaria), Marchantia, Anthoceros, etc. Examples: Nephrolepis, Marsilea, Pteris, Adiantum, Lycopodium etc.

Question 2.
Angiosperms and Gymnosperms.
Answer:

Angiosperms Gymnosperms
(i) In Angiosperms, the stems have branches. (i) In Gymnosperms, the stems are without branches.
(ii) Reproductive organs are flowers. (ii) Reprodcutive organs are cones.
(iii) Seeds are enclosed in natural coverings, i.e., fruits. (iii) Seeds are not enclosed in natural coverings.
(iv) Examples: Mango, Bamboo, etc. (iv) Examples: Cycas, Picea etc.

Question 3.
Cryptogams and Phanerogams.
Answer:

Cryptogams Phanerogams
(iii) Their reproductive organs are hidden. (iii) Their reproductive organs are exposed.
(iii) They reproduce by forming spores. (iii) They reproduce by forming seeds.
(iii) They are less evolved plants. (iii) They are highly evolved plants.
(iv) They are divided into Thallophyta, (iv) They are divided into Gymnosperms and
Bryophyta, Pteridophyta. Angiosperms.

Distinguish between:

Question 1.
Thallophyta

Answer:
Spirogvra, Ulothrix, Ulva, Sargassum

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
Bryophyta
Answer:
Moss (Funaria), Marchantia, Anthoceros, Riccia

Question 3.
Pteridophyta
Answer:
Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium

Question 4.
Gymnosperms
Answer:
Cycas, Picca (Christmas tree), Thuja (Morpankhi), Pinus (Deodar)

Question 5.
Angiosperms
Answer:
Tamarind, Mango, Apple, Lemon

Question 6.
Monocot plants
Answer:
Bamboo, bananas, corn, daffodils, garlic, ginger, grass, lilies, onions, orchids, rice, sugarcane, tulips, and wheat

Question 7.
Dicot plants
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

Observe the figure and answer the questions

1. Dicot Plants
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 8

Question 1.
What are the characteristics of the above plants in terms of root system?
Answer:
Well developed, primary root (Tap root)

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
What are the characteristics of the above plants in terms of flowers?
Answer:
Flowers with 4 or 5 parts or in their multiples (tetramerous or pentamerous)

Question 3.
What are the characteristics of the above plants in terms of leaf venations?
Answer:
Reticulate Venation

Question 4.
What are the characteristics of the above plants in terms of type of stem?
Answer:
Strong and hard

Question 5.
What are the characteristics of the above plants in terms of seed?
Answer:
Two cotyledons

Question 6.
Give example of the following types of plants
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

2. Monocot Plants
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 9

Question 1.
What are the characteristics of the above plants in terms of root system?
Answer:
Fibrous roots

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
What are the characteristics of the above plants in terms of flowers?
Answer:
Flowers with 3 parts or in multiples of three (trimerous).

Question 3.
What are the characteristics of the above plants in terms of leaf venations?
Answer:
Parallel Venation

Question 4.
What are the characteristics of the above plants in terms of type of stem?
Answer:
Hollow, False or Disc-like

Question 5.
What are the characteristics of the above plants in terms of seed?
Answer:
Single cotyledons

Question 6.
Give example of the following types of plants
Answer:
Bamboo, bananas, com, daffodils, garlic, ginger, grass, lilies, onions, orchids, rice, sugarcane, tulips, and wheat

3. Spirogyra
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 11

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division I Thallophyta.

Question 2.
Where does this plant grow?
Answer:
These plants grow mainly in water.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
Are these types of plants unicellular or multicellular?
Answer:
They may be unicellular or multicellular and microscopic or large.

Question 4.
Are these types of plant autotropic?
Answer:
They are autotrophic due to the presence of chlorophyll but types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Question 5.
Do these plants have a root-stem-leaves-flowers system?
Answer:
They do not have specific parts like root-stem- leaves-flowers.

Question 6.
How is the body of these types of plants?
Answer:
These plants usually have a soft and fibre-like body.

4. Funaria
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 12

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division II Bryophyta.

Question 2.
Where does this plant grow?
Answer:
They grow in moist soil but need water for reproduction.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
What are these group of plants called in the plant kingdom?
Answer:
This group of plants is called the ‘amphibians’ of the plant kingdom.

Question 4.
Are these types of plant autotropic?
Answer:
They reproduce by spore formation.

Question 5.
Do these plants have root-stem-leaves-flowers system?
Answer:
The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves.

Question 6.
What do these plants have instead of roots?
Answer:
They have root like rhizoids.

5. Fern
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 13

Question 1.
Which division of plants does this plant come? under?
Answer:
This plant come under Division III Pteridophy ta.

Question 2.
Where does this plant grow?
Answer:
They grow in soil.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
How do these plants reproduce?
Answer:
These plants reproduce asexually by spore- formation and sexually by zygote formation.

Question 4.
Do these plants produce flowers and fruits?
Answer:
They do not bear flowers and fruits.

Question 5.
Do these plants have root-stem-leaves-flowers system?
Answer:
Plants from this group have well developed roots, stem and leaves and separate tissues for conduction of food and water.

Question 6.
Where are the spores formed in the plants body?
Answer:
The spores formed along the back or posterior surface of their leaves.

6. Cycas
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 14

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division III Phanerogams Division I Gymnosperms.

Question 2.
Explain structure of these types of plants?
Answer:
Gymnosperms are mostly evergreen perennial and woody.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
How is stem and leaves of these types of plants?
Answer:
Their stems are without branches and the leaves form a crown.

Question 4.
Where are the male and female flowers located?
Answer:
These plants bear male and female flowers on different sporophylls of the same plant.?

Question 6.
Give some examples of these types of plants?
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

7. Monocot and Dicot plants
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 15

Question 1.
Which division of plants does this plant come under?
Answer:
ThisplantcomeunderDivisionlllPhanerogams Division II Angiosperms

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
How are the seeds of these types of planis?
Answer:
The seeds are formed within fruits thus these seeds are covered

Question 3.
How can we classify the plants according to their seeds in this division?
Answer:
The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Question 4.
How the venations are present on the leaves of these types of plants?
Answer:
These plants bear parallel or reticulated venations on the leaves.

Question 5.
How is the root system of these types of plants?
Answer:
The root systems of these types of plant are tap roots or fibrouš roots.

Complete the paragraph

Question 1.
Thallophyta plants grow mainly in …………….. . This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of …………….., is called algae. Algae show great diversity. They may be unicellular or …………….., and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a …………….. and fibre-like body. Various types of …………….. like yeasts and moulds which do not have …………….. are also included in this group.
Answer:
Thallophyta plants grow mainly in water. This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll, is called algae. Algae show great diversity. They may be unicellular or multicellular, and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a soft and fibre-like body. Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
…………….. group of plants is called the amphibians’ of the plant kingdom because they grow in moist soil but need …………….. for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by …………….. formation. The structure of the plant body of bryophytes is flat, ribbon?like long, without true …………….., stem and leaves. Instead, they have stem-like or leaf?like parts and root-like ……………. . They do not have specific …………….. for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.
Answer:
Bryophyta group of plants is called the ‘amphibians’ of the plant kingdom because they grow in moist soil but need water for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by spore formation. The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves. Instead, they have stem-like or leaf-like parts and root-like rhizoids. They do not have specific tissues for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.

Question 3.
Plants from Pteridophyta group have well developed roots, stem and leaves and separate …………….. for conduction of food and water. But,
they do not bear …………….. and ……………… They reproduce with the help of …………….. formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce …………….. by spore-formation and sexually by …………….. formation. They have a well-developed conducting system.
Answer:
Plants from Pteridophyta group have well developed roots, stem and leaves and separate tissues for conduction of food and water. But, they do not bear flowers and fruits. They reproduce with the help of spores formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce asexually by spore-formation and sexually by zygote formation. They have a well-developed conducting system.

Question 4.
Phanerogams plants which have special structures for …………….. and produce …………….. In these plants, after the process of reproduction, seeds are formed which contain the …………….. and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in …………….. a or not, phanerogams are classified into …………….. and ……………. .
Answer:
Phanerogams plants which have special structures for reproduction and produce seeds. In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into gymnosperms and angiosperms.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 5.
Gymnosperms are mostly …………….., perennial and woody. Their stems are without …………….. The leaves form a …………….. These plants bear male and female flowers on different …………….. of the same plant …………….. of these plants do not have natural coverings, i.e. these plants do not form …………….. and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.
Answer:
Gymnosperms are mostly evergreen, perennial • and woody. Their stems are without branches. The leaves form a crown. These plants bear male and female flowers on different sporophylls of the same plant. Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.

Question 6.
The flowers of Angiosperms plants bear are their …………….. organs Flowers develop into …………….. and seeds are formed within …………….. . Thus, these seeds are ……………..; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called …………….. plants and those whose seeds cannot be divided into equal parts are called …………….. plants.
Answer:
The flowers of Angiosperms plants bear are their reproductive orgAnswer: Flowers develop into fruits and seeds are formed within fruits. Thus, these seeds are covered; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Answer the questions in detail:

Question 1.
Write the characteristics of Thallophyta.
Answer:

  • Thallophyta plants grow mainly in water.
  • The group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll are called algae.
  • Algae show great diversity They may be unicellular or multicellular and microscopic or large.
  • Some of these are found in freshwater while some are found in saline water.
  • Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.
  • Examples: Spirogyra, Ulothrix, Ulva, etc.

Question 2.
Write the characteristics of Gymnosperms.
Answer:

  • Gymnosperms are mostly evergreen, perennial and woody.
  • Their stems are without branches.
  • The leaves form a crown.
  • These plants bear male and female flowers on different sporophylls of the same plant.
  • Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms (gmnos: naked, sperms: seeds)
  • Examples: Cycas, Picea (christmas tree), Thuja, Pinus (deodar), etc.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Make concept diagram

Question 1.
Plant classification
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 16

Question 2.
Taxonomy of carnivorous 1ant
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 17

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
Taxonomy of mango plant
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 18

9th Std Science Questions And Answers:

Useful and Harmful Microbes Class 9 Science Chapter 8 Questions And Answers Maharashtra Board

Class 9 Science Chapter 8

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 8 Useful and Harmful Microbes Question Answer Maharashtra Board

Class 9 Science Chapter 8 Useful and Harmful Microbes Question Answer Maharashtra Board

1. Complete the statements using the proper option from those given below. Explain the statements. (mycotoxins, budding, rhizobium)

a. Yeast reproduces asexually by the …………………….. method.
Answer:
The yeast cells develop small round bodies on the parent cell. These are called buds. New daughter cells develop from these buds.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

b. Toxins of fungal origin are called …………………….. .
Answer:
Mycotoxins are poisonous chemicals released into the food by fungi. This makes the food poisonous.

c. Leguminous plants can produce more proteins due to …………………….. .
Answer:

  • (i) Nitrogenous compounds are required to produce proteins.
  • (ii) Rhizobia produce nitrogenous compounds by fixing atmospheric nitrogen and make it available for their host plants like leguminous plants.

2. Write the names of microbes found in the following food materials.
yogurt, bread, root nodules of leguminous plants, idli, dosa, spoiled potato curry.
Answer:

Food materials Microbes
Yogurt Lactobacilli
Bread Yeast
Root nodules of leguminous plants Rhizobium
Idli Yeast, bacteria
Dosa Yeast, bacteria
Spoiled potato curry Clostridium

3. Identify the odd word out and say why it is the odd one?

a. Pneumonia, diphtheria, chicken pox, cholera.
Answer:
Chickenpox. It is caused by a virus, whereas others are caused by bacteria.

b. Lactobacilli, rhizobia, yeast, clostridia.
Answer:
Yeast. It is a fungus, whereas the rest are bacteria.

c. Root rot, rust (tambura), rubella, mozaic.
Answer:
Rubella. It is a disease of humans, whereas the rest are diseases of plants.

4. Give scientific reasons.

a. Foam accumulates on a the surface of ‘dal’ kept for a long time in summer.
Answer:

  • Dal is rich in proteins.
  • During summer, bacteria attack the dal and cause fermentation resulting in the production of carbon dioxide gas.
  • Therefore, foam accumulates on the surface of the ‘dal’ kept for long time in summer.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

b. Why are naphthalene balls kept with clothes to be put away.
Answer:

  • Naphthalene balls are balls of chemical pesticide and deodorant.
  • They help to kill or repel insects such as moths, cockroaches, mice etc.
  • Therefore, naphthalene balls are kept with clothes to be put away to prevent clothes from getting damaged.

5. Write down the modes of infection and the preventive measures against fungal diseases.
Answer:

  • Mode of infection: Contact with infected person or his/her belongings like clothes.
  • Preventive measure: Personal hygiene and avoid contact with infected person.

6. Match the pairs.

‘A’ group ‘B’ group
1. Rhizobium a. Food poisoning
2. Clostridium b. Nitrogen fixation
3. Penicillium c. Bakery products
4. Yeast d. Production of antibiotics
Answer:
(1 – b),
(2 – a),
(3 – d),
(4 – c)

7. Answer the following questions.

a. Which vaccines are given to infants? Why?
Answer:

  • Hepatitis A and B, DTP (Diphtheria, Tetanus, Pertussis.) Polio, MMR (Measles, Mumps, Rubella), Chicken pox, Influenza, Tetanus, BCG, Rotavirus, etc.
  • Vaccines consist of dead or weakend microbes. When these are swallowed or injected, the body produces antibodies to fight them.
  • These antibodies remain in the body and protect it from any future attack of the disease causing microbes.
  • Therefore, vaccines are given to infants for preventing diseases.

b. How is a vaccine produced?
Answer:

  • Vaccines are made using the disease causing bacteria or virus but in a form that will not harm the human beings.
  • Vaccine is made from dead or weakened microbes or their toxins.
  • Vaccine stimulates the immune system to produce antibodies which give life-long protection against the disease.
  • There are specific vaccines for specific diseases.

c. How do antibiotics cure disease?
Answer:
Antibiotics cure diseases by destroying or preventing the growth of harmful micro-organisms.

d. Are the antibiotics given to humans and animals the same? Why?
Answer:

  • Generally, antibiotics work against any harmful bacteria, whether it is attacking humans or animals.
  • But some of them are better suited to humans while some are better for animals. This is due to the adverse effects they show in different species. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
  • Also, the dosages of antibiotics for humans and animals differ.

e. Why is it necessary to safely store the pathogens of a disease against which vaccines are to be produced?
Answer:

  • Pathogens are microbes which can cause diseases in us.
  • For the preparation of a vaccine, a particular pathogen is cultured and grown in a laboratory.
  • If these pathogens are not safely stored, they many get modified due to environmental factors, resulting in decrease in the efficiency of the vaccine.
  • Also, the live pathogens may escape and cause diseases in us.

8. Answer the following questions in brief.

a. What are ‘broad-spectrum antibiotics’?
b. What is fermentation?
Answer:

  • Yeast uses sugar for food.
  • Yeast grows and multiplies rapidly due to the carbon compounds in the sugar solution.
  • In the process of obtaining nutrition, yeast cells convert the carbohydrates in the food into alcohol and carbon dioxide.
  • Also, the bacteria Lactobacilli convert lactose, the sugar in milk into lactic acid.
  • This process is called fermentation.

c. Define ‘Antibiotic’.
Answer:

  • Carbon compounds obtained from some bacteria and fungi for destroying or preventing the growth of harmful micro -organisms are called antibiotics.
  • Antibiotics, a discovery of the 20th century, have brought a revolution in the field of medicine.
  • Antibiotics mainly act against bacteria. Some antibiotics can destroy protozoa.
  • Some antibiotics are useful against a wide variety of bacteria they are called broad-spectrum antibiotics. Examples – Ampicillin, amoxicillin, tetracycline, etc.
  • When the pathogen cannot be identified even though the symptoms of the disease are visible, broad-spectrum antibiotics are used. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
  • Whenever a pathogenic micro-organism is definitely known, then narrow-spectrum antibiotics are used. Examples: Penicillin, gentamycin, erythromycin, etc.

Class 9 Science Chapter 8 Useful and Harmful Microbes Intext Questions and Answers

Can you recall?

Question 1.
What is meant by microbes? What are their characteristics?
Answer:
Microbes are tiny microscopic organisms which cannot be seen with the unaided eye.

Characteristics of Microbes.

  • They are the smallest organisms on earth.
  • They are composed of prokaryotic or eukaryotic cells.
  • They can be seen only with the help of a microscope.
  • They are found in any kind of environment ranging from coolest polar regions to hottest of deserts. Also found in soil, water and air.
  • Some of them are useful, whereas some of them are harmful micro-organisms.

Question 2.
How do you observe microbes?
Answer:
Microbes are observed using a microscope.

Answer the following.

Question 1.
Why are wineries located near Nashik in Maharashtra?
Answer:

  • Nashik in Maharashtra is the leading grape producer in the country as it has the soil suitable for the production of grapes. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
  • Glucose and fructose, the sugars present in grape juice are fermented with the help of yeast to produce wines. Therefore, wineries are located near Nashik in Maharashtra.

Question 2.
Find out the uses of fungi to plants and animals?
Answer:

  • Fungi decompose the bodies of dead animals and convert them into simple carbon compounds. These substances easily mix with air, water and soil from where they are again absorbed by plants and enter the food chain.
  • Some fungi living in symbiotic association with plants help to absorb water and inorganic compounds like nitrate and phosphate.
  • Fungi are also used to derive antibiotics like penicillin which are useful to animals.
  • Ants grow fungi in their anthill and obtain food from it.
  • Some species of wasps and insects lay their eggs in the fungal bodies growing on trees, thus ensuring a food supply for their larvae.

Question 3.
What is the structure of lichen, a condiment? Where else is it used?
Answer:

  • Lichen is a symbiotic association between a fungus and an algae (Cyanobacterium).
  • Lichens are sensitive to environmental disturbances and are used in assessing air pollution in an area.
  • Lichens are also used in making dyes, perfumes and in traditional medicines.
  • A few lichen species are eaten by insects or animals such as reindeer.

Open-ended questions

Answer the following questions:

Class 9 Science Chapter 8 Useful And Harmful Microbes Exercise Question 1.
How is yoghurt made from milk? What exactly happens in this process?
Answer:

  • Milk contains sugar called lactose which is broken down with help of Lactobacilli.
  • Lactobacilli converts lactose into Lactic Acid. This process is called fermentation. As a result, the pH of milk decreases causing coagulation of milk proteins.
  • Thus, milk proteins are separated from other constituents of milk and milk changes to yoghurt.

Useful And Harmful Microbes Class 9 Exercise Question 2.
Sometimes, you may notice a black powder or white discs floating on the pickle or murabba, when a jar is opened after a long time. What exactly is this? Why are such food items not good to eat?
Answer:

  • A black powder or white disc floating on the pickle or murabba are fungi.
  • Different fungal species depend on host (pickle and murabba) for their growth and reproduction.
  • During this process, fungi secretes mycotoxins which are poisonous chemicals; which ultimately spoil the food. Consuming such food can cause food poisoning. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
  • Therefore, such food items are not good to eat.

8 Useful And Harmful Microbes Class 9 Question 3.
How many different industries depend upon the Lactobacilli bacteria?
Answer:
Industries like milk products, cider, cocoa, pickles, pharmaceuticals depend on Lactobacilli bacteria.

Useful And Harmful Microbes Class 9 Question 4.
Which types of cottage industries and factories can be started in areas with abundant milk production?
Answer:
Cottage industries like the manufacture of milk products like ghee, cheese, paneer, curd, shrikhand, etc. and chocolate making can be started in areas with abundant milk production.

9th Class Science Chapter 8 Useful And Harmful Microbes Exercise Question 5.
Which changes do you notice in leather articles and gunny (jute) bags during the rainy season?
Answer:
In rainy season we can notice whitish-green cotton-like growth or black powder or white discs on leather articles and gunny (jute) bags during the rainy season as these articles are infected by fungus.

Class 9th Science Chapter 8 Useful And Harmful Microbes Exercise Question 6.
For how long afterwards can you use those articles?
Answer:
Those articles cannot be used for long as they wear out and do not last long.

Question 7.
Why do these articles not get spoilt during the summer or winter?
Answer:

  • Spores of fungi can germinate when there is sufficient moisture.
  • During summer or winter the weather is hot and dry and so fungus cannot grow in such weather.
  • Also microbes cannot survive extreme hot or cold temperatures of summer or winter. Therefore, these articles do not get spoilt during summer or winter.

Question 8.
Why do doctors advise you to take yoghurt or buttermilk if you have indigestion or abdominal discomfort?
Answer:

  • The Lactobacilli present in yoghurt or buttermilk help to restore the natural microbial flora in the intestine, thus helping in digestion and absorption of nutrients.
  • Also buttermilk helps to cool down the stomach and works as a laxative to ease the congestion during abdominal discomfort.

Question 9.
Sometimes, yoghurt becomes bitter and froths up. Why does this happen?
Answer:

  • Sometimes yoghurt becomes bitter due to excess fermentation by bacteria.
  • Excess amount of lactic acid is produced making the curd bitter.

Question 10.
Which different milk products are obtained at home by fermentation of the cream from the milk?
Answer:
Yoghurt, buttermilk, ghee, cheese, shrikhand, sour cream, etc.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 11.
Recently, it has been made compulsory in India and some other countries to mix 10% ethanol with fuels like petrol and diesel. What is the reason for this?
Answer:

  • Ethanol is a smokeless and high quality fuel. So it helps to reduce pollution when mixed with petrol or diesel.
  • As petrol or diesel is a fossil fuel less consumption of it will lead to resourceful use of it and making the country self-efficient by moving towards sustainable fuel like ethanol.

Question 12.
Chapattis made from wheat only swell up but bread becomes spongy, soft and easy to digest. Why is it so?
Answer:

  • The chapatti dough has water, which on heating converts into steam and tries to escape.
  • While doing so, it lifts up the upper layer of the chapatti. Therefore, the chapatti swells up.
  • Bread is made by adding yeast to the flour.
  • In the process of obtaining nutrition, the yeast cells convert the carbohydrates into alcohol and carbon dioxide.
  • When this dough is baked, the carbon dioxide escapes out making the bread spongy, soft and easy to digest.

Question 13.
Salt is applied on the inner surface of pickle jars and the pickle is covered with oil. Why is this done?
Answer:

  • Salt acts as a preservative. It prevents the growth of bacteria by forcing the microbes to lose water by osmosis. Hence, salt is applied on the inner surface of pickle jars.
  • Pickle is covered with oil as oil acts as preservative. It seals off the air from the item that is being pickled and provides an environment in which microbes cannot grow.

Question 14.
Which preservatives are mixed with ready to eat foods to prevent them from spoiling?
Answer:
Common salt, sugar, sodium benzoate, citric acid, sodium meta-bi-sulfite etc. are some 1 of the preservatives mixed with ready-to-eat foods to prevent them from spoiling.

Question 15.
Which plant and animal diseases are caused by micro-organisms and what are the 1 measures to be taken against them? Answer:
Plant diseases:

  • Citrus canker is a bacterial disease that affects 1 trees of citrus fruits.
  • Rust of wheat is a fungal disease that affects wheat crops.
  • Yellow vein mosaic is a viral disease which affects vegetables like bhindi (okra).

Preventive Measures:

  • Seeds which are healthy and disease-free should be selected for sowing.
  • Infected plants should be removed.
  • Plants should be sprayed with fungicides and germicides to prevent diseases.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Animal diseases:

  • Anthrax is a disease that affects cattle. It is caused by a bacterium.
  • Foot and mouth is a dangerous disease in cattle caused by a virus.
  • Rabies is a viral disease that affects animals.

Preventive Measures:

  • The place where animals are kept should be washed with germicides.
  • Animals should be dewormed regularly.
  • The animals should be treated with necessary antibiotics for infectious diseases.
  • They should be regularly vaccinated.
  • Take the animals to a veterinary hospital for proper treatment and vaccination.

Answer the following questions:

Question 1.
Bring ‘active dry yeast’ from the market. Mix a spoonful of yeast, two spoonfuls sugar with a sufficient quantity of lukewarm water in a bottle. Fix a colourless, transparent balloon on the mouth of that bottle.

What changes do you observe after 10 minutes? Mix limewater with the gas accumulated in the balloon. Collect that limewater in a beaker and observe it. What do you notice?
Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes 3
Answer:

  • After 10 minutes, the balloon is filled with a gas and gets inflated.
  • Lime water turns milky thus proving that the gas accumulated is carbon dioxide.

Class 9 Science Chapter 8 Useful and Harmful Microbes Additional Important Questions and Answers

Select the correct option:

Question 1.
The rod-shaped bacteria found in milk or buttermilk are called ……………………. .
(a) Rhizobium
(b) Clostridium
(c) Lactobacilli
(d) Saccharomyces
Answer:
(c) Lactobacilli

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 2.
Yoghurt has a specific sour taste due to ……………………. .
(a) lactic acid
(b) citric acid
(c) acetic acid
(d) alcohol
Answer:
(a) lactic acid

Question 3.
Bacteria found in the root nodules of leguminous plants are ……………………. .
(a) clostridium
(b) streptococcus
(c) Lactobacilli
(d) Rhizobium
Answer:
(d) Rhizobium

Question 4.
A mutually beneficial relationship is called ……………………. .
(a) symbiosis
(b) parasitism
(c) autotropism
(d) none of these
Answer:
(a) symbiosis

Question 5.
Carbon compounds obtained from bacteria and fungi for destroying or preventing the growth of harmful micro-organisms are called ……………………. .
(a) probiotics
(b) antibiotics
(c) antibodies
(d) antigens
Answer:
(b) antibiotics

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 6.
Fungi release ……………………. into the food, making the food poisonous.
(a) cyanotoxins
(b) dinotoxins
(c) mycotoxins
(d) cytotoxins
Answer:
(c) mycotoxins

Question 7.
……………………. produce bottle-shaped endospores in adverse conditions.
(a) Lactobacilli
(b) Clostridium
(c) Yeast
(d) Rhizobium
Answer:
(b) Clostridium

Question 8.
……………………. conducted important research on the toxin responsible for gas gangrene and the antitoxin responsible for treating it.
(a) Ida Bengston
(b) Van Ermengem
(c) Louis Pasteur
(d) Alexander Fleming
Answer:
(a) Ida Bengston

Question 9.
……………………. is a smokeless and high quality fuel.
(a) Methanol
(b) Ethanol
(c) Petrol
(d) Diesel
Answer:
(b) Ethanol

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 10.
Antibiotics mainly act against ……………………. .
(a) bacteria
(b) viruses
(c) algae
(d) fungi
Answer:
(a) Bacteria

Question 11.
……………………. is a broad-spectrum antibiotic.
(a) Gentamycin
(b) Penicillin
(c) Amoxicillin
(d) Erythromycin
Answer:
(c) Amoxicillin

Question 12.
……………………. is a narrow-spectrum antibiotic.
(a) Ampicillin
(b) Amoxicillin
(c) Tetracyclin
(d) Penicillin
Answer:
(d) Penicillin

Question 13.
Antibiotic penicillin was discovered by ……………………. .
(a) Louis Pasteur
(b) Alexander Fleming
(c) Ida Bengston
(d) Van Ermengem
Answer:
(b) Alexander Fleming

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 14.
……………………. proved that the anaerobic bacterium Clostridium botulinum is responsible for food poisoning.
(a) Louis Pasteur
(b) Ida Bengston
(c) Alexander Fleming
(d) Van Ermengem
Answer:
(d) Van Ermengem

Question 15.
AIDS is caused by ……………………. .
(a) virus
(b) bacteria
(c) protozoa
(d) fungi
Answer:
(a) Virus

Question 16.
Dengue is caused by ……………………. .
(a) droplets spread in air
(b) contact with infected person
(c) mosquito bite
(d) contaminated water and food
Answer:
(c)mosquitobite

Question 17.
Pneumonia is caused by ……………………. .
(a) virus
(b) bacteria
(c) fungi
(d) protozoa
Answer:
(b) bacteria

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 18.
……………………. can be prevented by vaccination.
(a) Malaria
(b) AIDS
(c) Leprosy
(d) Chicken pox
Answer:
(d) Chicken pox

Question 19.
Malaria is caused by ……………………. .
(a) protozoa
(b) bacteria
(c) fungi
(d) virus
Answer:
(a) protozoa

Question 20.
Bird flu (H7N9) and swine flu (H1N1) are caused by ……………………. .
(a) bacteria
(b) protozoa
(c) fungi
(d) virus
Answer:
(d) virus

Question 21.
The Lactobacilli convert lactose, the sugar in the milk, into ……………………. .
(a) lactic acid
(b) acetic acid
(c) alcohol
(d) citric acid
Answer:
(a) lactic acid

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 22.
The ……………………. destroys harmful microbes present in the milk.
(a) high pH
(b) neutral pH
(c) low pH
(d) none of these
Answer:
(c) low pH

Question 23.
Lactobacilli kill the harmful bacteria like ……………………. present in the alimentary canal.
(a) Rhizobium
(b) Saccharomyces
(c) Clostridium
(d) Alcanivorax
Answer:
(c) Clostridium

Question 24.
During fermentation, yeast cells convert carbohydrates into ……………………. .
(a) glucose and fructose
(b) alcohol and carbon dioxide
(c) proteins and fats
(d) fatty acids and amino acids
Answer:
(b) alcohol and carbon dioxide

Question 25.
Molasses is fermented with the help of yeast called ……………………. .
(a) Yarrowia lipolytica
(b) Alcanivorax
(c) Rhizobia
(d) Saccharomyces
Answer:
(d) Saccharomyces

Question 26.
A yeast ……………………. is used to absorb the toxins released during the production of palm oil.
(a) Yarrowia lipolytica
(b) Alcanivorax
(c) Saccharomyces cerevisiae
(d) Penicillium
Answer:
(a) Yarrowia lipolytica

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 27.
The bacteria which spoil cooked food are ……………………. .
(a) Saccharomyces
(b) Lactobacilli
(c) Clostridium
(d) Rhizobium
Answer:
(c) Clostridium

Question 28.
……………………. can grow and reproduce only in living cells.
(a) Bacteria
(b) Viruses
(c) Fungi
(d) Protozoa
Answer:
(b) Viruses

Find the odd man out:

Question 1.
AIDS, Hepatitis, Leprosy, Dengue.
Answer:
Leprosy. It is caused by bacteria, whereas the rest are caused by viruses.

Question 2.
Cholera, Leprosy, Pneumonia, Influenza.
Answer:
Influenza. It is caused by a virus, whereas the rest are caused by bacteria.

Question 3.
Ampicillin, Amoxycillin, Penicillin, Tetracycline.
Answer:
Penicillin. It is a narrow-spectrum antibiotic, whereas others are broad-spectrum antibiotics.

Question 4.
Tetracycline, Penicillin, Gentamycin, Erythromycin.
Answer:
Tetracycline. It is a broad-spectrum antibiotic, whereas others are narrow-spectrum antibiotics.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Complete the analogy:

Question 1.
(1) Dengue : Virus :: Malaria : …………………………. .
(2) Hepatitis : Virus :: Pneumonia : …………………………. .
(3) Cholera : Bacteria :: Swine flu : …………………………. .
(4) Swine flu : HJNJ : : Bird Flu : …………………………. .
(5) Measles : Virus :: Ringworm : …………………………. .
(6) Yoghurt: Lactobacilli : : Bread : …………………………. .
(7) Oil spills: Alcanivorax :: Absorption of arsenic : …………………………. .
(8) Rhizobium : Nitrogen fixation : : Clostridium : …………………………. .
Answer:
(1) Protozoa
(2) Bacteria
(3) Virus
(4) HyN9
(5) Fungi
(6) Yeast
(7) Saccharomyces cerevisiae
(8) Food poisoning.

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) Leprosy (a) Virus
(2) Ringworm (b) Fungi
(3) Influenza (c) Protozoa
(4) Malaria (d) Bacteria

Answer:
(1 – d),
(2 – b),
(3 – a),
(4 – c)

State whether the following statements are true or false. Correct the false statements:

(1) Lactobacilli are aerobic bacteria.
(2) Lactobacilli converts lactose sugar into alcohol.
(3) Yoghurt has a specific sour taste due to acetic acid.
(4) The bacteria Clostridium are present in the root . nodules of leguminous plants.
(5) Yeast cell is a prokaryotic cell.
(6) The use of Rhizobium has helped to reduce the use of chemical fertilizers.
(7) Ethanol is a smokeless and high quality fuel.
(8) A yeast, Saccharomyces cerevisiae is used for absorbing toxins released during palm oil production.
(9) Gentamycin is a narrow-spectrum antibiotic.
(10) Antibiotics mainly act against bacteria. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
(11) Oil spills in oceans are cleared with the help of Clostridium bacteria.
(12) Tetracycline is a narrow-spectrum antibiotic.
(13) Amoxicillin is a broad-spectrum antibiotic.
(14) Penicillin is a group of antibiotics obtained from a fungus Saccharomyces.
(15) Antibiotic Penicillin was discovered by Alexander Fleming.
(16) The bacteria Lactobacilli cause food-poisoning.
(17) Clostridium bacteria grow in aerobic conditions.
(18) AIDS is caused by a virus.
(19) Influenza is caused by a bacteria.
(20) Antibiotics useful to one person can be suggested to others also.
(21) Dengue is caused by a bacteria.
(22) Dandruff and ringworm are caused by fungi.
(23) Pneumonia is spread through droplets spread in air by infected person.
(24) Chicken pox spread due to contaminated food and water.
(25) Ida Bengston was honoured with the Typhoid Medal’ in 1947.
Answer:
(1) False. Lactobacilli are anaerobic bacteria.
(2) False. The Lactobacilli converts lactose sugar into lactic acid.
(3) False. Yogurt has a specific sour taste due to lactic acid.
(4) False. The bacteria Rhizobium are present in the root nodules of leguminous plants.
(5) False. Yeast cell is a eukaryotic cell.
(6) True. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
(7) True.
(8) False. A yeast, Yarrowia lipolytic is used to absorb the toxins released during the production of palm oil.
(9) True.
(10) True.
(11) False. Oil spills in oceans are cleared with the help of Alcanivorax bacteria.
(12) False. Tetracycline is a broad-spectrum antibiotic.
(13) True.
(14) False. Penicillin is a group of antibiotics obtained from a fungus Penicillium.
(15) True.
(16) False. The bacteria Clostridium cause food? poisoning.
(17) False. Clostridium bacteria grow in anaerobic conditions.
(18) True.
(19) False. Influenza is caused by a virus.
(20) False. Antibiotics useful to one person cannot be suggested to others as different diseases require different antibiotics.
(21) False. Dengue is caused by a virus.
(22) True.
(23) True.
(24) False. Chicken pox spread due to contact with infected person.
(25) False. Ida Bengston was honoured with the ‘Typhus Medal’ in 1947.

Complete the statements using the proper option from those given below. Explain the statements: (mycotoxins, budding, Rhizobium, molasses, endospores, broad-spectrum, Lactobacilli)

Question 1.
Lactobacilli bacteria are used for making yoghurt.
Answer:
The lactobacilli convert lactose, the sugar in the milk, into lactic acid. As a result, the pH of milk decreases causing a coagulation of milk proteins. Milk changes into yogurt.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 2.
The use of Rhizobium has helped to reduce the use of chemical fertilizers.
Answer:
Rhizobium bacteria are found in the root nodules of leguminous plants. They help to convert atmospheric nitrogen into nitrogen compounds and provide it to the plants. This helps to reduce the use of chemical fertilizers and their adverse effects.

Question 3.
Amoxicillin is a broad-spectrum antibiotic.
Answer:
This antibiotic is useful against a wide variety of bacteria. It is used against pathogens which cannot be identified during symptoms of a disease.

Question 4.
Ethanol is produced by the fermentation of molasses.
Answer:
Molasses is produced from sugarcane juice. It is rich in carbohydrates. When it is fermented with the help of the yeast called Saccharomyces, ethanol (C2H5OH) is produced.

Question 5.
Clostridium bacteria produce bottle-shaped endospores.
Answer:
These endospores help them to survive in adverse conditions.

Give scientific reasons:

Question 1.
Lactobacilli are used for making yoghurt from milk.
Answer:

  • Lactobacilli convert lactose, the sugar in the milk, into lactic acid. This process is called fermentation.
  • As a result, the pH of milk decreases causing coagulation of milk proteins.
  • Thus, milk proteins are separated from other constituents of milk and milk changes into yoghurt.
  • Yoghurt has a specific sour taste due to lactic acid. The low pH destroys harmful microbes present in the milk. Therefore, Lactobacilli are used for making yoghurt from milk.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 2.
Antibiotics should be taken only when prescribed by a doctor.
Answer:

  • Antibiotics are a group of medicines used to kill disease-causing bacteria and certain protozoa.
  • The doctor selects and prescribes the antibiotic best suited for our disease.
  • If taken in extra dose, they can kill the useful bacteria present in our body.
  • If the course of antibiotics is not completed, the bacteria develop resistance to that antibiotic making it ineffective.
  • Therefore, antibiotics should be taken only when prescribed by a doctor.

Question 3.
Nowadays, seeds are coated with Rhizobial solution or powder before sowing.
Answer:

  • When seeds coated with Rhizobial solution or powder are sown, Rhizobia enter the plantlets.
  • This is called Rhizobial inoculation.
  • Rhizobia can produce nitrogenous compounds from atmospheric nitrogen.
  • This experiment has helped in the supply of nitrogen to cereal and other crops, besides leguminous crops.
  • Therefore, nowadays seeds are coated with Rhizobial solution or powder before sowing.

Question 4.
Antibiotics are not effective against common cold or influenza.
Answer:

  • Antibiotics are a group of medicines used to control inflections caused by bacteria.
  • Common cold or influenza is caused by a virus.
  • Antibiotics are not effective against viruses.
  • Therefore, antibiotics are not effective against common cold or influenza.

Question 5.
Cotton fabrics, gunny bags, leather items and wooden items do not last long.
Answer:

  • Microscopic spores of fungi are present in the air.
  • If there is sufficient moisture, spores germinate on cotton fabric, gunny bags, leather, wooden items etc.
  • The fungal hyphae (fibres of the fungus) penetrate deep into the material to obtain nutrition and to reproduce.
  • This causes the materials to wear and become weak.
  • As a result, cotton fabric, gunny bags, leather and wooden items do not last long.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 6.
Food on which fungi has grown cannot be eaten.
Answer:

  • Various species of fungi grow on food items like pickles, murabba, jam, sauce, chutney etc.
  • They use the nutrients in these food items for growth and reproduction.
  • During this activity, fungi release mycotoxins, certain poisonous chemicals, into the food and thus food becomes poisonous.
  • Hence, the food on which fungi have grown cannot be eaten.

Write short notes:

Question 1.
Rhizobial inoculation.
Answer:

  • Nowadays, seeds are coated with rhizobial solution or powder before sowing.
  • After sowing, Rhizobia enter the plantlets.
  • This is called Rhizobial inoculation.
  • This experiment has helped in the supply of nitrogen to cereal and other crops, besides leguminous crops.

Question 2.
Bio-remediation.
Answer:

  • Bio-remediation is a technique that involves the use of organisms to break down environmental pollutants.
  • Generally, fungi like yeast and bacteria are used for bio-remediation.
  • A yeast, Yarrowia lipolytica is used to absorb the toxins released during the production of palm oil and the heavy metals and minerals released in some other industrial processes.
  • Saccharomyces cerevisiae is used for absorption of a pollutant, arsenic.
  • Oil spills in oceans are cleaned with the help of Alcanivorax bacteria.

Question 4.
Clostridium.
Answer:

  • Clostridium are the bacteria that spoil food.
  • Out of about 100 different species of this bacterium, some are free living in the soil whereas some live in the alimentary canals of humans and other animals.
  • These bacteria are rod-shaped and produce bottle-shaped endospores in adverse conditions.
  • One special characteristic of these bacteria is that they cannot withstand the normal oxygen level of the air because they grow in anaerobic conditions.

Write down the mode of infection and preventive measures for the following:

Question 1.
Write down the causative pathogen, mode of infection and preventive measures of AIDS.
Answer:

  • Causative Pathogen: Virus.
  • Mode of infection: Through blood and semen of infected person and milk of mother suffering from AIDS.
  • Preventive measure: Safe sexual contact, avoid resuse of needles and injections.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 2.
Write down the modes of infection and preventive measures against Bird Flu (H7N9) and Swine Flu (HjN.,).
Answer:

  • Mode of infection: Contact with infected birds and animals.
  • Preventive measure: Personal hygiene, properly cooked meat.

Question 3.
Write down the modes of infection and preventive measures against Malaria and dengue.
Answer:

  • Mode of infection: Mosquito bite, unclean surroundings.
  • Preventive measure: Cleanliness of surroundings, preventing stagnation of water, controlling mosquitoes.

Question 4.
Write down the modes of infection and preventive measures against Pneumonia.
Answer:

  • Mode of infection: Droplets spread in air by infected person.
  • Preventive measure: Vaccination, avoiding contact with infected person.

Question 5.
Write down the modes of infection and preventive measures for leprosy.
Answer:

  • Mode of infection: Long term contact with infected person.
  • Preventive measure: Avoiding contact with infected persons and their belongings.

Question 5.
What are the modes of infection and preventive measures for Hepatitis?
Answer:

  • Mode of infection: Contaminated water and food.
  • Preventive measure: Use clean and filtered water, proper storage of food.

Question 6.
What are the modes of infection and preventive measures for Influenza.
Answer:

  • Mode of infection: Contact with infected person.
  • Preventive measure: Personal hygiene and avoiding contact with infected person.

Answer the following questions:

Question 1.
How is bread made?
Answer:

  • Bread is made using flour, yeast, salt and water. The yeast uses sugar as food.
  • In the process of obtaining nutrition, yeast cells convert the carbohydrates into alcohol and carbon dioxide. This process is called fermentation. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
  • The bubbles of carbon dioxide given off cause the dough to rise.
  • This dough can be used to make bread.
  • When this dough is baked, more bubbles of carbon dioxide reformed due to heat. As the gas escapes, the bread rises and becomes soft and fluffy.

Question 2.
What is the advantage of Rhizobium to farmers?
Answer:

  • The use of Rhizobium has helped to reduce the use of chemical fertilizers and thereby their adverse effects.
  • It has also helped to reduce expenses on fertilizers and thus benefited the farmers.

Question 3.
How can we observe Lactobacilli in buttermilk?
Answer:

  • Smear a drop of fresh buttermilk on a glass slide.
  • Stain it with methylene blue and put a coverslip over it.
  • Observe the smear under the 10X objective of a compound microscope and then with the more powerful 60X objective.
  • The blue rod-shaped organisms moving about are Lactobacilli.

Question 4.
What is symbiosis? Give example.
Answer:

  • Symbiosis is a mutually beneficial relationship.
  • Example: Rhizobium living in root nodules of leguminous plants supply nitrates, nitrites and amino acids to that plant and in exchange get energy in the form of carbohydrates.

Complete the following table:

Question 1.
The spread and prevention of disease
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes 1
Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes 2

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 2.
Different species of Clostridium bacteria and the diseases caused by them.
Answer:

Species of Clostridium Diseases
Clostridium tetani Tetanus
Clostridium perfringens Food poisoning
Clostridium botulinum Botulism (Paralysis of muscles)
Clostridium difficile Colitis (Infection of the intestine)

Answer in detail:

Question 1.
How is alcohol produced?
Answer:

  • Alcohol is often produced along with sugar in sugar factories.
  • Molasses is produced from sugarcane juice. It is rich in carbohydrates.
  • Molasses is fermented with the help of the yeast Saccharomyces.
  • In this process, ethanol (C2H5OH) alcohol is produced as a primary product and ester and other alcohols are produced as secondary products.
  • Besides molasses, maize, barley and other grains are also used for industrial production of alcohol.
  • Glucose and fructose, the sugars present in grape juice are also fermented with the help of yeast to produce alcohol which is used to make wines.

Question 2.
Give the uses of Lactobacilli.
Answer:
Uses of Lactobacilli:

  • Various milk products like yoghurt, buttermilk, ghee, cheese, shrikhand, etc. can be obtained by fermentation of milk.
  • Lactobacilli fermentation is useful for large scale production of cider, cocoa, pickles of vegetables etc.
  • Lactobacilli and some other useful microbes taken together are used to treat abdominal discomfort.
  • Leavened fodder offered to domestic cattle like cows and buffaloes is fodder fermented with the help of lactobacilli.
  • The Lactobacilli fermentation process is used to make wine and some types of bread.

Question 3.
What is Penicillin? What is it used for?
Answer:

  • Penicillin is a group of antibiotics obtained from a fungus, penicillium.
  • It is used for controlling the infections caused by bacteria like Staphylococci, Clostridia, Streptococci, etc.
  • Medicines containing Penicillin are useful to treat certain bacterial infections of the ear, nose, throat and skin as well as diseases like Pneumonia and scarlet fever.

Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes

Question 4.
How was the antibiotic penicillin discovered?
Answer:

  1. Alexander Fleming, a professor of microbiology at St. Mary’s Hospital had cultured varieties of bacteria and fungi in petri dishes in his laboratory.
  2. On 3rd September 1928, while observing Staphylococci cultures, he made an interesting observation in one petri dish.
  3. In that petri dish, fungal colonies had grown but the area around those colonies was clean and clear, i.e. the bacteria had actually been destroyed.
  4. After further studies, he confirmed that the fungus growing there was Penicillium and its secretion had destroyed the bacterial colonies.
  5. Thus, the first antibiotic – penicillin had been discovered accidentlly and this formed the basis to find cures for incurable diseases.

Question 5.
What are the precautions to be followed while taking antibiotics?
Answer:

  • Antibiotics should be taken only when prescribed by a doctor.
  • Don’t purchase any antibiotic from medical stores without a prescription from a doctor.
  • Don’t consume antibiotics on your own to treat common diseases like a throat infection, common cold or influenza.
  • Even if you feel well before completing of the prescribed course of the antibiotics, you must continue and complete it.
  • Don’t suggest to others the antibiotics which were useful to you.

Question 6.
How can we observe Rhizobium bacteria in the roots of leguminous plant?
Answer:

  • Take a plantlet of fenugreek, groundnut or any other bean and sterilize it with a 3 to 5% solution of hydrogen peroxide. Maharashtra Board Class 9 Science Solutions Chapter 8 Useful and Harmful Microbes
  • Afterwards, keep it in a 70% solution of ethyl alcohol for 4 to 5 minutes.
  • Clean the roots with sterile water and take thin sections of the root nodules.
  • Select a good section and place it an a solution of saffranin for 2 to 3 minutes.
  • Place the stained section on a glass slide, cover it with a coverslip and observe it under the compound microscope. The pinkish rod-shaped organisms are the Rhizobium bacteria.

9th Std Science Questions And Answers:

Measurement of Matter Class 9 Science Chapter 4 Questions And Answers Maharashtra Board

Class 9 Science Chapter 4

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 4 Measurement of Matter Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 4 Measurement of Matter Question Answer Maharashtra Board

Class 9 Science Chapter 4 Measurement of Matter Question Answer Maharashtra Board

1. Give examples.

a. Positive radicals
Answer:
Na+– Sodium ion, K+ – Potassium ion

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

b. Basic radicals
Answer:
Na+ – Sodium ion, K+ – Potassium ion, Ag+ – Silver ion

c. Composite radicals
Answer:
\(\mathrm{SO}_{4}^{2-}, \mathrm{NH}_{4}^{+}\)

d. Metals with variable valency
Answer:
(a) Iron (Ferrum)
(i) Fe2+ – Ferrous [Iron – II]
(ii) Fe3+ – Ferric [Iron – III]

(b) Copper (Cuprum)
(i) Cu+ – Cuprous [Copper -1]
(ii) Cu2+ – Cupric [Copper – II]

(c) Mercury (Hydragyrum)
(i) Hg+ – Mercurous [Mercury -1]
(ii) Hg2+ – Mercuric [Mercury – II]

e. Bivalent acidic radicals
Answer:
O2- – Oxide, S2- – Sulphide, \(\mathrm{CO}_{3}^{2-}\) – Carbonate

f. Trivalent basic radicals
Answer:
Al3+ – Aluminium, Cr3+ – Chromium, Fe3+ – Ferric.

2. Write symbols of the following elements and the radicals obtained from them, and indicate the charge on the radicals.
Mercury, potassium, nitrogen, copper, sulphur, carbon, chlorine, oxygen
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 24

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

3. Write the steps in deducing the chemical formulae of the following compounds.
Sodium sulphate, potassium nitrate, ferric phosphate, calcium oxide, aluminium hydroxide
Answer:
In order to write the chemical formulae of compounds, it is necessary to know the symbols and valency of various radicals.

1. Sodium Sulphate:
Step – 1 : To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
\(\mathrm{Na} \quad \mathrm{SO}_{4}\)
Step – 2 : To write the valency below the respective radical.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 40
Step – 3: To cross-multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 41
Step – 4 : To write down the chemical formula of the compound.
Na2 SO4
(Sodium sulphate)

2. Potassium Nitrate:
Step -1 : To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
K NO3
Step – 2 : To write the valency below the respective radical.
\(\begin{array}{cc}
\mathrm{K} & \mathrm{NO}_{3} \\
1 & 1
\end{array}\)
Step – 3: To cross-multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 42
Step – 4 : To write down the chemical formula of the compound.
KNO3
(Potassium nitrate)

3. Ferric phosphate:
Step -1 : To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
Fe PO4
Step – 2 : To write the valency below the respective radical.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 43
Step – 3: To cross-multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 44
Step – 4 : To write down the chemical formula of the compound.
FePO4
(Ferric phosphate)

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

4. Calcium oxide:
Step – 1 : To write the symbols of the radicals (Basic radical on the left and acidic radicals on the right)
Ca O
Step – 2 : To write the valency below the respective radical.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 60
Step – 3: To cross-multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 45
Step – 4 : To write down the chemical formula of the compound.
CaO
(Calcium oxide)

5. Aluminium hydroxide:
Step – 1 : To write the symbols of the radical (Basic radical on the left and acidic radical on the right)
Al OH
Step – 2 : To write the valency below the respective radical.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 61
Step – 3: To cross-multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 46
Step – 4 : To write down the chemical formula of the compound.
Al(OH)3
(Aluminium hydroxide)

6. Calcium carbonate:
Step – 1 : To write the symbols of the radical (Basic radical on the left and acidic radicals on the right)
Ca CO3
Step – 2 : To write the valency below the respective radical.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 47
Step – 3: To cross-multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 48
Step – 4 : To write down the chemical formula of the compound.
CaCO3
(Calcium Carbonate)

7. Sodium dichromate:
Step – 1 : To write the symbols of the radicals (Basic radical on the left and acidic radical on the right)
Na Cr2O7
Step – 2 : To write the valency below the respective radical.
\(\begin{array}{cc}
\mathrm{Na} & \mathrm{Cr}_{2} \mathrm{O}_{7} \\
1 & 2
\end{array}\)
Step – 3: To cross-multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 49
Step – 4 : To write down the chemical formula of the compound.
Na2Cr2O7
(Sodium dichromate)

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

4. Write answers to the following questions and explain your answers.

a. Explain how the element sodium is monovalent.

Answer:

  1. The number of protons or electrons (atomic number) in Sodium (Na) atom is 11. Therefore the electronic configuration of sodium atom is (2, 8,1).
  2. In chemical reaction, sodium atom has the capacity to give away le_ from its outermost orbit to form Na+ ion with stable electronic configuration (2, 8).
  3. As sodium atom gives away le- and a cation of sodium is formed, hence the valency of sodium is 1 and therefore, the element sodium is monovalent.

b. M is a bivalent metal. Write down the steps to find the chemical formulae of its compounds formed with the radicals, sulphate and phosphate.
Answer:
M is a bivalent metal. Following are the steps to find the chemical formulae of its compounds formed with the radicals, sulphate and phosphate:

(i) Compound of metal ‘M’ with radical sulphate
Step – 1: To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
M SO4
Step – 2: To write the valency below the respective radical.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 21
Step – 3: To cross multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 21
Step – 4: To write down the chemical formula of the compound.
M SO4

(ii) Compound of metal ‘M’ with radical phosphate.
Step – 1: To write the symbols of the radicals (Basic radicals on the left and acidic radicals on the right)
M PO4
Step – 2: To write the valency below the respective radical.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 23
Step – 3: To cross multiply as shown by arrows the number of radicals.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 51
Step – 4: To write down the chemical formula of the compound.
M3 (PO4)2

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

c. Explain the need for a reference atom for atomic mass. Give some information about two reference atoms.
Answer:

  • The mass of an atom is concentrated in its nucleus and it is due to the protons (p) and neutrons (n) in it.
  • Since an atom is very very tiny, it was not possible to measure atomic mass accurately. Therefore, the concept of relative mass of an atom was formed.
  • To express relative mass of an atom, reference of atom is considered. The two reference atoms were as follows:

(a) Hydrogen (H) atom: The hydrogen atom is the lightest. The relative mass of a hydrogen atom is 1 which has only 1 proton in its nucleus. On this scale, the relative atomic mass of many elements comes out to be fractional. Therefore, carbon was selected as a reference atom.

(b) Carbon (C) atom: The carbon atom is selected as reference atom. In this scale, the relative mass of a carbon atom is accepted as 12.

  • The relative atomic mass of 1 hydrogen (H) atom compared to the carbon (C) atom becomes

d. What is meant by Unified Atomic Mass.
Answer:

  • During earlier time, relative mass of an atom was considered for measuring the mass of an atom directly. But since the founding of unified mass, relative mass is not accepted henceforth.
  • Unified atomic mass is the unit of atomic mass called as Dalton.
  • Its symbol is ‘u’. lu = 1.66053904 x 10-27 kg.

e. Explain with examples what is meant by a ‘mole’ of a substance.
Answer:

  • A mole is that quantity of a substance whose mass in grams is equal in magnitude to the molecular mass of that substance in Daltons.
  • For example: Atomic mass of oxygen atom (O) is 16u. Thus, the molecular mass of oxygen molecule (O2) is 16 x 2 = 32u. Therefore, 32 g of oxygen is 1 mole of oxygen.

5. Write the names of the following compounds and deduce their molecular masses.
Na2SO4, K2CO3, CO2, MgCl2, NaOH, AlPO4, NaHCO3
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 5

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

6. Two samples ‘m’ and ‘n’ of slaked lime were obtained from two different reactions. The details about their composition are as follows:
‘sample m’ mass : 7g
Mass of constituent oxygen : 2g
Mass of constituent calcium : 5g
‘sample n’ mass : 1.4g
Mass of constituent oxygen : 0.4g
Mass of constituent calcium : 1.0g

Which law of chemical combination does this prove? Explain.
Answer:
(i) The expected proportion by weight of the constituent elements of quick lime that is calcium oxide would be from its known molecular formula CaO. The atomic mass of Ca and O are 40 and 16 respectively. This means, the proportion by weight of the constituent elements Ca and O in the compound CaO is 40 :16 which is 5 : 2.

(ii) Now, for the given sample’m’ of CaO = 5 g
mass of given sample = 7 g
mass of constituent Ca in sample’m’ = 5 g
mass of constituent O in sample’m’ = 2 g

(iii) This means that 7 g of calcium oxide contairis 5 g of calcium (Ca) and 2 g of oxygen (O); apd the proportion by weight of calcium and oxygen in it is 5 : 2.

(iv) Now, for the given sample ‘n’ of CaO mass of given sample CaO = 1.4 g
Mass of constituent Ca in sample ‘n’ = 1.0 g
Mass of constituent O in sample ‘n’ = 0.4 g
This means that 1.4g of calcium oxide contains 1.0 g of calcium (Ca) and 0.4 g of oxygen (O); and the proportion by weight of calcium and oxygen in it is 5 : 2.

(v) Above samples’m’ and ‘n’ of calcium oxide (CaO) shows that the proportion by weight of the constituent elements in different samples of a compound is always constant that is the proportion by weight of calcium (Ca) and oxygen (O) in different samples of calcium oxide (CaO) is constant.

(vi) The experimental value of proportion by weight of the constituent elements matched with the expected proportion calculated by molecular mass. This proves and verifies the law of constant proportion.

The law states that ‘The proportion by weight of the constituent elements in the various samples of a compound is fixed’.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

7. Deduce the number of molecules of the following compounds in the given quantities.
32g oxygen, 90g water, 8.8g carbon dioxide, 7.1g chlorine.

Class 9 Science Chapter 4 Measurement Of Matter Notes Question 1.
32g oxygen
Answer:
Given : Mass of oxygen (O2) m = 32g
To find : Number of molecules in 32g of oxygen.
Solution : Atomic mass of oxygen (O) = 16
∴ Molecular mass of oxygen (O2) M = 16 x 2 = 32
According to the formula, Number of moles in the given O2 (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 12
1 mol of O2 contains 6.022 x 1023 molecules that is 32 g of 02 contains 6.022 * 1023 molecules of O2.
32g of oxygen contains 6.022 x 1023 molecules of oxygen.

Class 9 Science Chapter 4 Measurement Of Matter Answers Question 2.
90g water
Answer:
Given : Mass of water (H2O) m = 90g.
To find : Number of molecules in 90g of water.
Solution : Molecular mass of (H2O) M = (Atomic mass of H) x 2 + (Atomic mass of O) x 1
∴ Molecular mass of (H2O) M = 1 x 2 +16
∴ Molecular mass of (H2O) M = 18
According to the formula,
Number of moles in the given H2O (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 13
1 mol of H2O contains 6.022 x 1023 molecules.
5 mol of H2O contains 5 x 6.022 x 1023 molecules. = 30.11 x 1023 molecules, that is 90g of H2O contains 30.11 x 1023 molecules of H20.
90g of water contains 30.11 x 1023 molecules of water.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

4 Measurement Of Matter Exercise Question 3.
8.8g carbon dioxide
Answer:
Given : Mass of Carbon dioxide (CO2)m = 8.8g.
To find : Number of molecules in 8.8g of carbon dioxide.
Solution : Molecular mass of (CO2)M = (Atomic mass of C) x 1 + (Atomic mass of O) x 2
∴ Molecular mass of (CO2)M = 12 x 1 + 16 x 2 = 12 + 32
Molecular mass of (CO2)M = 44
According to the formula, Number of moles in the given CO2 (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 14
∴ 1 mol of CO2 contains 6.022 x 1023 molecules.
∴ 0.2 mol of CO2 contains 0.2 x 6.022 x 1023 molecules.
= 1.2044 x 1023 molecules,
that is 8.8g of CO2 contains 1.2044 x 1023 molecules of CO2.
8.8g of CO2 contains 1.2044 x 1023 molecules of CO2.

Class 9 Science Solutions Maharashtra Board Question 4.
7.1g chlorine
Answer:
Given : Mass of Chlorine (Cl2)m = 7.1g.
To find : Number of molecules in 7.1g of chlorine.
Solution : Atomic mass of (Cl) = 35.5
∴ Molecular mass of chlorine (Cl2)M = 35.5 x 2 = 71
According to the formula, Number of moles in the given Cl2 (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 15
∴ 1 mol of Cl2 contains 6.022 x 1023 molecules.
∴ 0.1 mol of Cl2 contains 0.1 x 6.022 x 1023 molecules.
= 0.6022 x 1023 molecules,
that is 7.1g of Cl2 contains 0.6022 x 1023 molecules of Cl2.
7.1g of Cl2 contains 0.6022 x 1023 molecules of chlorine.

8. If 0.2 mol of the following substances are required how many grams of those substances should be taken? Sodium chloride, magnesium oxide, calcium carbonate
Answer:
Given : Number of moles of sodium chloride (NaCl) n = 0.2 mol
To find : Mass in grams of 0.2 mol of NaCl
Solution:
Molecular mass of (NaCl)M = (Atomic mass of Na) x 1 + (Atomic mass of Cl) x 1
= 23 x 1 + 35.5 x 1
= 23 + 35.5
Molecular mass of (NaCl)M = 58.5
According to the formula,
Number of moles in the given NaCl (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 16
Mass of NaCl in grams (m) = 0.2 x 58.5
Mass of NaCl in grams (m) = 11.7 g
Mass of 0.2 mole of NaCl is 11.7g

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Class 9 Science Chapter 3 Current Electricity Intext Questions and Answers

Maharashtra State Board Class 9 Science Solutions Question 1.
What is the type of chemical bond in NaCl and MgCl2?
Answer:

  • The type of chemical bond in NaCl and MgCl2 is ionic bond.

9th Class Science Chapter 4 Measurement Of Matter Question 2.
Determine the valencies of H, Cl, O and Na from the molecular formulae H2, HC1, H2O and NaCl.
Answer:
(i) In the molecular formula HCl
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 3
∴ The valency of H is 1 and Cl is 1.

(ii) In the molecular formula H2O
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 2
∴ The valency of H is 1 and O is 2.

(iii) In the molecular formula NaCl
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 50
∴ The valency of Na is 1 and Cl is 1.
∴ From all the above, the valencies of the given elements are as follows : H = 1, Cl = 1, O = 2 and Na = l.

Measurement Of Matter Class 9 Exercise Answers Question 3.
How is an element indicated in Chemistry?
Answer:
In chemistry an element is indicated by its symbol.

Question 4.
Write down the symbols of the elements you know.
Answer:
Symbols of some elements are

  • Hydrogen – H
  • Helium – He
  • Boron – B
  • Carbon – C
  • Aluminium – A1

Question 5.
Write down the symbols for the following elements.
Antimony, Iron, Gold, Silver, Mercury, Lead, Sodium
Answer:
The symbols of given elements are as follows:

  • Antimony – Sb
  • Iron – Fe
  • Gold – Au
  • Silver – Ag
  • Mercury – Hg
  • Lead – Pb
  • Sodium – Na

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Following are atomic masses of a few elements in Daltons and the molecular formulae of some compounds. Deduce the molecular masses of those compounds:

Atomic masses – H(l), 0(16), N(14), C(12), K(39), S(32) Ca(40), Na(23), Cl(35.5), Mg(24), Al(27)

Question 1.
Molecular formula – NaCl
Answer:
Molecular mass of NaCl (M)
= (Atomic mass of Na) x 1 + (Atomic mass of Cl) x 1
= (23 x 1) + (35.5 x 1)
= 23 + 35.5
= 58.5
∴ Molecular mass of NaCl (M) = 58.5

Question 2.
Molecular formula – MgCl2
Answer:
Molecular mass of MgCl2 (M)
= (Atomic mass of Mg) x 1 + (Atomic mass of Cl) x 2
= (24 x 1) + (35.5 x 2)
= 24 + 71
= 95
∴ Molecular mass of MgCl2 (M) = 95?

Question 3.
Molecular formula – KNO3
Answer:
Molecular mass of KNO3 (M)
= (Atomic mass of K) x 1 + (Atomic mass of N) x 1 + (Atomic mass of O) x 3
= (39 x 1) + (14 x 1) + (16 x 3)
= 39 + 14 + 48
= 101
Molecular mass of KNO3 (M) = 101

Question 4.
Molecular formula – H2O2
Answer:
Molecular mass of H2O2 (M)
= (Atomic mass of H) x 2 + (Atomic mass of O) x 2
= (1 x 2) + (16 x 2)
= 2 + 32
= 34
∴ Molecular mass of H2O2 (M) = 34.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 5.
Molecular formula – A1C13
Answer:
Molecular mass of A1C13 (M)
= (Atomic mass of Al) x 1 + (Atomic mass of Cl) x 3
= (27 x 1) + (35.5 x 3)
= 27 + 106.5
= 133.5
∴ Molecular mass of A1C13 (M) = 133.5

Question 6.
Molecular formula – Ca(OH)2
Answer:
Molecular mass of Ca(OH)2 (M)
= (Atomic mass of Ca) x 1 + (Atomic mass of O + Atomic Mass of H) x 2
= (40 x 1) + (16 + 1) x 2
= 40 + (17 x 2)
= 40 + 34
= 74
∴ Molecular mass of Ca(OH)2 (M)
= 74

Question 7.
Molecular formula – MgO
Answer:
Molecular mass of MgO (M)
= (Atomic mass of Mg) x 1 + (Atomic mass of 0)xl
= (24 x 1) + (16 x 1)
= 24 + 16
= 40
Molecular mass of MgO (M) = 40

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 8.
Molecular formula – H2S04
Answer:
Molecular mass of H2S04 (M)
= (Atomic mass of H) x 2 + (Atomic mass of S) x 1 + (Atomic mass of O) x 4
= (1 x 2) + (32xl) + (16×4)
= 2 + 32 + 64
= 98
Molecular mass of H2S04 (M) = 98

Question 9.
Molecular formula – HN03
Answer:
Molecular mass of HN03 (M)
= (Atomic mass of H) x 1 + (Atomic mass of N) x 1 + (Atomic mass of O) x 3
= (lxl)+ (14xl)+ (16×3)
= 1 + 14 + 48
= 63
Molecular mass of HNOs (M) = 63

Question 10.
Molecular formula – NaOH
Answer:
Molecular mass of NaOH (M)
= (Atomic mass of Na) x 1 + (Atomic mass of O) x 1 + (Atomic mass of H) x 1
= (23 x 1) + (16 x 1) + (l x l)
= 23 + 16 + 1
= 40
Molecular mass of NaOH (M) = 40

Question 11.
How many molecules of water are there in 36 g water?
Answer:
Given : Mass of water (H2O) m = 36g
To find : Number of molecules in 36g of water
Solution :
Molecular mass of (H2O) M = (Atomic mass of H) x 2 + (Atomic mass of O) x 1 Molecular mass of (H2O) M
= (1 x 2) + 16 x 1
Molecular mass of (H2O) M = 18
According to the formula,
Number of moles in the given H2O (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 19
1 mol of H2O contains 6.022 x 1023 molecules.
∴ 2 mol of H2O contains 2 x 6.022 x 1023 molecules.
= 12.044 x 1023 molecules, that is 36g of H2O contains 12.044 x 1023 molecules of H2O.
36 g of water contains 12.044 x 1023 molecules of water.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 12.
How many molecules of H2S04 are there in a 49 g sample?
Answer:
Given : Mass of Sulphuric acid (H2SO4) m = 49g
To find : Number of molecules in 49g of H2SO4
Solution:
Molecular mass of (H2SO4) M = (Atomic mass of H) x 2 + (Atomic mass of S) x 1 + (Atomic mass of O) x 4
Molecular mass of (H2SO4)M = (1 x 2) + (32 x 1) + (16 x 4)
= 2 + 32 + 64
= 98.
According to the formula,
Number of moles in the given H2SO4 (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 20
∴ 1 mol of H2SO4 contains 6.022 x 1023 molecules.
∴ 0.5 mol of H2SO4 contains 0.5 x 6.022 x 1023 molecules.
= 3.011 x 1023 molecules,
that is 49g of H2SO4 contains 3.011 x 1023 molecules of H2SO4.
49 g of Sulphuric acid contains 3.011 x 1023 molecules of H2SO4.

Question 13.
Fill the following tables.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 25

Question 14.
Complete the following chart.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 27

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 15.
The relative atomic masses of some elements in the chart below are given. You have to find the relative atomic masses of the others.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 28

Question 16.
Classify the following radicals into simple radicals and composite radicals: (Use your brain power;
\(\begin{array}{l}
\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cl}^{-}, \mathrm{SO}_{4}^{2-}, \mathrm{Fe}^{2+}, \mathrm{ClO}_{3}^{-}, \mathrm{NH}_{4}^{+}, \mathrm{Br}^{-} \\
\mathrm{NO}_{3}^{-}, \mathrm{Na}^{+}, \mathrm{Cu}^{+}
\end{array}\)
Answer:

Simple radicals Composite radicals
Ag+ \(\mathrm{SO}_{4}^{2-}\)
Mg2+ \(\mathrm{ClO}_{3}^{-}\)
Cl \(\mathrm{NH}_{4}^{+}\)
Fe2+ \(\mathrm{NO}_{3}^{-}\)
Br
Na+
Cu+

Question 17.
Which are the basic radicals and which are the acidic radicals among the following?
\(\begin{array}{l}
\mathrm{Ag}^{+}, \mathrm{Cu}^{2+}, \mathrm{Cl}^{+}, \mathrm{I}, \mathrm{SO}_{4}^{2-}, \mathrm{Fe}^{3+}, \mathrm{Ca}^{2+}, \mathrm{NO}_{3} ; \mathrm{S}^{2}, \mathrm{NH}_{4}^{+} \\
\mathrm{K}^{+}, \mathrm{MnO}_{4}, \mathrm{Na}^{+}
\end{array}\)
Answer:

Basic Radical Acidic Radical
(i)Ag+ (i) Cl
(ii) Cu2+ (ii) I
(iii) Fe3+ \(\text { (iii) } \mathrm{SO}_{4}^{2-}\)
(iv) Ca2+ \(\text { (iv) } \mathrm{NO}_{3}^{-}\)
Wnh; \(\text { (v) } \mathrm{S}^{2-}\)
(vi) K+ \(\text { (vi) } \mathrm{MnO}_{4}^{-}\)
(vii) Na+

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Give examples:

Question 1.
Make a list of elements in the monoatomic and in the diatomic molecular state. (Make a list and discuss;
Answer:

  • Elements in the monoatomic molecular state are: Helium (He), Neon (Ne), Argon (Ar), Sodium (Na), Copper (Cu),
  • Elements in the diatomic molecular state are:
    Oxygen (O2), Nitrogen (N2), Hydrogen (H2), Chlorine (Cl2), Fluorine (F2).

Problem-based questions

Answer the following questions:

Question 1.
Is it possible to weigh one molecule using a weighing balance?
Answer:
No, it is not possible to weigh one molecule using a weighing balance.

Question 2.
Will the number of molecules be the same in equal weights of different substances?
Answer:
No, the number of molecules will not be the same in equal weights of different substances.

Question 3.
If we want equal number of molecules of different substances, will it work to take equal weights of those substances.
Answer:
No, if we want equal number of molecules of different substances, it will not work to take equal weights of those substances.

Answer the following:

Question 1.
What is the Dalton’s atomic theory?
Answer:
Dalton’s Atomic theory-

  • All matter is made of atoms. Atoms are indivisible and indestructible.
  • All atoms of a given element are identical in mass and properties.
  • Compounds are formed by a combination of two or more different kinds of atoms.
  • A chemical reaction is a rearrangement of atoms.

Question 2.
How are compounds formed?
Answer:
Compounds are formed by a chemical combination of two or more different kinds of atoms.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 3.
What are the molecular formulae of salt, slaked lime, water, lime, limestone?
Answer:
The molecular formulae for
Salt – Sodium chloride – NaCl
Slaked lime – Calcium hydroxide Ca(OH)2
Water – H2O
Lime – Calcium oxide – CaO
Lime stone – Calcium carbonate – CaCO3

Question 4.
From which experiments was it discovered that atoms have an internal structure? When?
Answer:

  • In 1911, Earnest Rutherford conducted a well known experiment called as ‘Gold foil experiment’.
  • From this experiment it was discovered that atoms have internal structure.

Question 5.
What are the two parts of an atom? What are they made up of?
Answer:
The two parts of atoms are nucleus and extra nuclear part. Nucleus is made up of positively charged protons and electrically neutral neutrons and the extra nuclear part is made up of negatively charged electrons revolving around the nucleus in different orbits.

Open-ended questions

Q.3. 2. Answer the following questions:

Question 1.
How will the compounds, MgCl2 and CaO be formed from their elements?
Answer:
(1) Magnesium Chloride (MgCl2)
Magnesium atom (Mg): Electronic configuration
\((2,8,2) \stackrel{-2 e^{-}}{\longrightarrow}\) Magnesium ion Mg2+ (2,8).
Chlorine atom (Cl). Electronic configuration \((2,8,7) \stackrel{+1 e^{-}}{\longrightarrow}\) Chloride ion Ch (2,8,8).
∴ Mg2+ + 2CT → MgCl2 (Magnesium Chloride)

  • A Magnesium atom gives away 2e and a cation of Magnesium (Mg2+) is formed, hence, the valency of magnesium is two.
  • Two chlorine atoms takes le each and forms two anions of chlorine (2Cl) (chloride), and thus, the valency of chlorine is one.
  • After the give and take of electrons is over, the electronic configuration of all the resulting ions has a complete octet.
  • Due to the attraction between the unit but opposite charges on all the ions, one chemical bond known as ionic bond is formed between Mg2+ and 2C1 each and the compound MgCl2 is formed.

(2) Calcium Oxide (CaO)
Calcium atom (Ca): Electronic configuration
\((2,8,8,2) \stackrel{-2 e^{-}}{\longrightarrow}\) Calcium ion Ca2+ (2,8,8).
Oxygen atom (O). Electronic configuration (2,6)
\(\stackrel{+2 e^{-}}{\longrightarrow}\) Oxygen ion O2- (2,8).
∴ Ca2+ + O2- → CaO

  • A calcium atom gives away 2er and a cation of calcium (Ca2+) is formed, hence, the valency of calcium is two.
  • An oxygen atom takes 2e and forms anions of oxygen (O2-) (oxide), and thus, the valency of oxygen is two.
  • After the give and take of electrons is over, the electronic configuration of both the resulting ions has a complete octet.
  • Due to the attraction between the unit but opposite charges on the two ions, one chemical bond known as ionic bond is formed between Ca2+ and O2- and the compound CaO is formed.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 2.

  • Take 56 g calcium oxide in a large conical flask and put 18 g water in it.
  • Observe what happens.
  • Measure the mass of the substance formed.
  • What similarity do you find? Write your inference.

Answer:
(i) When 18 g of water is added to 56 g of calcium oxide, calcium oxide combines with water to form calcium hydroxide Ca(OH)2

(ii) The mass of calcium hydroxide formed is 74 g.?

(iii) In this activity the total mass of reactants, Calcium oxide + Water = 56 g +18 g = 74 g.
It is equal to the mass of the product formed. Ca(OH)2 = 74g.

This activity verifies the Law of Conservation of Matter, i.e., in a chemical reaction, the total weight of the reactants is same as the total weight of the products formed due to the chemical reactions.

Question 3.

  • Take a solution of calcium chloride in a conical flask and a solution of sodium sulphate in a test tube.
  • Tie a thread to the test tube and insert it in the conical flask.
  • Seal the conical flask with an airtight rubber cork.
  • Weigh the conical flask using a balance.
  • Now tilt the conical flask so that the solution in the test tube gets poured in the conical flask.
  • Now weigh the conical flask again.

Answer:

  • In this activity, a white precipitate of CaS04 in NaOl is seen in the conical flask after the reaction.
  • There is no change in the weight of the flask before and after the reaction.
  • This activity verifies the Law of Conservation of Matter i.e., in a chemical reaction, the total weight of the reactants is same as the total weight of the products formed due to the chemical reactions.

Question 4.
Using the chart of ions/radicals and the cross-multiplication method, write the chemical formulae of the following compounds : Calcium carbonate, Sodium bicarbonate, Silver chloride, Calcium hydroxide, Magnesium oxide, Ammonium phosphate, Cuprous bromide, Copper sulphate, Potassium nitrate, Sodium dichromate.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 39
Answer:
Calcium carbonate – CaCO3 Sodium bicarbonate – NaHCO3 Silver chloride – AgCl, Calcium hydroxide – Ca(OH)2, Magnesium oxide – MgO, Ammonium phosphate – (NH4)3PO4, Cuprous bromide – CuBr, Copper sulphate – CuSO4, Potassium nitrate – KNO3, Sodium dichromate – Na2Cr2O7.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Class 9 Science Chapter 3 Current Electricity Additional Important Questions and Answers

(A) Select the correct option:

Question 1.
The proportion by weight of hydrogen and oxygen in water is ……………………….. .
(a) 8 : 1
(b) 2 : 1
(c) 1 : 2
(d) 1 : 8
Answer:
(d) 1: 8

Question 2.
The proportion by weight of carbon and oxygen in carbon dioxide is ……………………….. .
(a) 8 : 3
(b) 3 : 8
(c) 3 : 2
(d) 2 : 3
Answer:
(b) 3 : 8

Question 3.
A nucleus of an atom is made up of positively charged ………………………… and electrically neutral ……………………….. .
(a) protons; neutrons
(b) electrons; neutrons
(c) neutrons; protons
(d) neutrons; electrons
Answer:
(a) protons; neutrons

Question 4.
The size of an atom is determined by its ……………………….. .
Answer:
radius

Question 5.
Atomic radius is expressed in ……………………….. .
(a) milimetres
(b) centimetres
(c) nanometres
(d) picometres
Answer:
(c) nanomet res

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 6.
The atomic size depends on the number of ………………………… in the atom.
(a) protons
(b) nucleus
(c) neutrons
(d) electron orbits
Answer:
(d) electron orbits

Question 7.
The mass of an atom is concentrated in its ……………………….. .
(a) protons
(b) nucleus
(c) neutrons
(d) electrons
Answer:
(b) nucleus

Question 8.
The total number of protons and neutrons in the atomic nucleus is called the ……………………….. .
(a) atomic number
(b) electronic configuration
(c) atomic mass number
(d) valency
Answer:
(c) atomic mass number

Question 9.
A ………………………… is that quantity of a substance whose mass in grams is equal in magnitude to the molecular mass of that substance in Daltons.
(a) mole
(b) dalton
(c) dozen
(d) gross
Answer:
(a) Mole

Question 10.
Avogadro’s number is denoted by the symbol ……………………….. .
(a) NG
(b) Nv
(c) NA
(d) ND
Answer:
(c) NA

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 11.
A mole of any substance stands for ………………………… molecules.
(a) 60.22 x 1023
(b) 6.022 x 1022
(c) 6.022 x 1023
(d) 60.22 x 1022
Answer:
(a) 60.22 x 1023

Question 12.
The capacity of an element to combine is called its ……………………….. .
(a) valency
(b) electronic configuration
(c) atomic number
(d) volence electrons
Answer:
(a) valency

Question 13.
Electronic configuration of sodium atom is ……………………….. .
(a) (2, 8, 3)
(b) (2, 8, 7)
(c) (2, 8, 2)
(d) (2, 8,1)
Answer:
(d) (2,8,1)

Question 14.
Electronic configuration of chlorine atom is ……………………….. .
(a) (2, 8, 3)
(b) (2, 8, 7)
(c) (2, 8, 2)
(d) (2, 8, 1)
Answer:
(b) (2, 8, 7)

Question 15.
Positively charged ions are called as ……………………….. .
(a) cations
(b) anions
(c) nucleous
(d) protons
Answer:
(a) cations

Question 16.
Negatively charged ions are called as ……………………….. .
(a) cations
(b) anions
(c) nucleus
(d) electrons
Answer:
(b) anions

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 17.
Iron (Fe) exhibits the variable valencies ……………………….. .
(a) 1 and 2
(b) 2 and 3
(c) 3 and 4
(d) 2 and 4
Answer:
(b) 2 and 3

Question 18.
Cationic radicals are called as ………………………… radicals.
(a) basic
(b) acidic
(c) neutral
(d) mixed
Answer:
(a) basic

Question 19.
Anionic radicals are called as ………………………… radicals.
(a) basic
(b) acidic
(c) neutral
(d) mixed
Answer:
(b) acidic

Question 20.
The unit Dalton is used to express …………………………
(a) atomic mass
(b) atomic radius
(c) atomic number
(d) mass number
Answer:
(a) atomic mass

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 21.
The valency of element with electronic configuration ………………………… is 2.
(a) (2,5)
(b) (2, 4)
(c) (2, 6)
(d) (2, 7)
Answer:
(c) (2, 6)

Question 22.
The symbol of Avogadro’s number is ……………………….. .
(a) ND
(b) N0
(c) NB
(d) NA
Answer:
(d) NA

Question 23.
………………………. is bicarbonate radical.
\((a) \mathrm{HCO}_{3}^{2-} (b) \mathrm{CO}_{3}^{-}
(c) \mathrm{HCO}_{3}^{-}
(d) \mathrm{CO}_{3}^{2-}\)
Answer:
\(\text { (c) } \mathrm{HCO}_{3}^{-}\)

Question 24.
Molecular formula of sodium sulphate is ……………………….. .
(a) Na(SO4)2
(b) Na2SO4
(c)Na2(SO4)2
(d)NaSO4
Answer:
(b) Na2SO4

Question 25.
………………………… is a composite radical.
(a) Fe3+
(b) Ca2+
(c) NH4+
(d) S2-
Answer:
(c) NH

Question 26.
A mole of any substance stands for ………………………… molecules.
(a) 6.022 x 1023
(b) 6.022 x 1022
(c) 60.22 x 1023
(d) 60.22 x 1022
Answer:
(a) 6.022 x 1023

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 27.
The mass of an atom is concentrated in its ………………………… .
(a) nucleus
(b) electrons
(c) extranuclear part
(d) protons
Answer:
(a) nucleus

Question 28.
………………………… g of water make 1 mole of water.
(a) 32
(b) 33
(c) 16
(d) 18
Answer:
(d) 18

Complete the analogy:

(1) Electron : extra nuclear part:: Neutron ………………………… .
(2) Sodium: (2, 8, 1):: Chlorine:: ………………………… .
(3) K : basic radical :: Br : ………………………… .
(4) Cut: simple radical:: NH4+ : ………………………… .
(5) Sodium sulphate: Na2SO4:: Potassium Sulphate: ………………………… .
(6) Mercurous: Hg+:: Mercuric : ………………………… .
(7) Positively charged ion : cation:: Negatively charged ion : ………………………… .
(8) 12: 1 dozen :: 144 : ………………………… .
(9) Hydrogen : \(\odot\) :: Copper : ………………………… .
(10) Law of constant proportions : J. L. Proust::
Law of conservation of matter : ………………………… .
Answer:
(1) nucleus
(2) (2, 8, 7)
(3) acidic radical
(4) composite radical
(5) K2SO4
(6) Hg2+
(7) anion
(8) 1 gross
(9) ©
(10) Antoine Lavoisier.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Match the columns:

Column A’ Column ‘B’
Example Atomic radius (in metres)
(1) Water molecule
(2) Haemoglobin molecule
(3) Hydrogen atom
(a) 10-10
(b) 10-9
(c) 10-8

Answer:
(1-b),
(2- c),
(3 – a)

Column ‘A’ Column ‘B’
Element Atomic mass
(1) Neon (a) 35.5
(2) Silicon (b) 32
(3) Chlorine (c) 28
(4) Sulphur (d) 20

Answer:
(1 – d),
(2 – c),
(3 – a),
(4 – b)

Column ‘A’ Column ‘B’
Molecule Molecular mass in grams
(1) h2 (a) 32 g
(2) H2O (b) 34 g
(3) O2 (c) 2 g
(4) H2O2 (d) 18 g

Answer:
(1 – c),
(2 – d),
(3 – a),
(4 – b)

Column ‘A’ Column B’
Radicals Names
(1) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) (a) Carbonate
(2) \(\mathrm{ClO}_{3}^{-}\) (b) Chromate
(3) \(\mathrm{CO}_{3}^{2-}\) (c) Dichromate
(4) \(\mathrm{CrO}_{4}^{2-}\) (d) Chlorate

Answer:
(1 – c),
(2 – d),
(3 – a),
(4 – b)

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Answer the following in one sentence:

Question 1.
What are valence electrons?
Answer:
The electrons present in the outermost orbit of an atom are called valence electrons.

Question 2.
Give the formula to determine the number of moles of a substance.
Answer:
The formula to determine the number of moles of a substance is as given below.
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 4

Question 3.
What are basic radicals? Give examples.
Answer:
The radicals which are formed by removal of electrons from the atoms of metals are called as basic radicals, e.g., Na+, Cu2+

Question 4.
What are acidic radicals? Give examples.
Answer:
The radicals which are formed by adding electrons to the atoms of non-metals are called as acidic radicals, e.g., CT, S2-

State whether the following statement is ‘True’ or ‘False’. Correct the false statement.

(1) Molecular state of oxygen is monoatomic.
(2) The capacity of an element to combine is called its valency.
(3) Anionic radicals are basic radicals.
(4) The magnitude of charge on any radical is its atomic number.
(5) In a chemical reaction, mass of original matter and mass of matter newly formed as a result of chemical change are equal.
(6) The proportion by weight of carbon and oxygen in carbon dioxide is 3 : 5.
(7) Relative mass of hydrogen is 1.
(8) The number of molecules in a given quantity of a substance is determined by its atomic mass.
(9) Avogadro’s number is 6.022 x 1023
(10) Valency of sodium is 2.
Answer:
(1) False. Molecular state of oxygen is diatomic:
(2) True
(3) False. Anionic radicals are acidic radicals.
(4) False. Magnitude of charge on any radical is its valency.
(5) True
(6) False. The proportion by weight of carbon and oxygen in carbon dioxide is 3 : 8.
(7) True
(8) False. The number of molecules in a given quantity of a substance is determined by its molecular mass.
(9) True
(10) False. Valency of sodium is 1.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Name the following:

Question 1.
Scientist who gave Law of Conservation of Matter.
Answer:
Antoine Lavoisier

Question 2.
Scientist who gave Law of Constant Proportion.
Answer:
J. L. Proust

Question 3.
What are protons and neutrons present in nucleus together called as?
Answer:
Nucleons

Question 4.
Unit used to express atomic radius.
Answer:
Nanometre

Question 5.
The number (p + n) in the atomic nucleus is called as?
Answer:
Atomic mass number

Question 6.
Name the unit of atomic mass.
Answer:
Dalton (u)

Question 7.
Write molecular formula of two ionic compounds containing chlorine.
Answer:
NaCl, MgCl2

Question 8.
Give two monoatomic radicals.
Answer:
Na+, Cl

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 9.
Give two examples of simple radicals.
Answer:
Ag+, O2-

Give scientific reasons:

Question 1.
An atom is electrically neutral though it contains charged particles.
Answer:

  • An atom is made up of a nucleus and an extranuclear part. Protons and neutrons are present in the nucleus.
  • The nucleus is positively charged. The extranuclear part is made up of negatively charged electrons.
  • Protons are positively charged, electrons are negatively charged and neutrons are without any charge.
  • The magnitude of their charges is the same when they are equal in number.
  • Hence, the negative charge on all the extra, nuclear electrons together balances the positive charge on the
  • nucleus.
  • Therefore, an atom is electrically neutral though it contains charged particles.

Question 2.
Neon is chemically inert element.
Answer:

  • Atomic number of neon is 10, so its electronic configuration is (2, 8). There are 8 electrons in its 2nd shell, fulfilling its capacity.
  • Thus, neon has a complete octet.
  • It has a stable orbit therefore, it does not indulge in chemical reactions. Hence, neon is a chemically inert element.

Question 3.
The valency of sodium (Na) is one.
Answer:

  • The electronic configuration of sodium (Na) is (2, 8,1). It has 1 electron in its 3rd orbit.
  • It tends to give up this electron so that it is left up with (2, 8), having 8 electrons in the second orbit, with a stable state.
  • The loss of one electron leads to the formation of sodium ion (Na+) which is positively charged as it has lost one electron.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 4.
The valency of chlorine (Cl) is one.
Answer:

  • The electronic configuration of chlorine (Cl) is (2, 8, 7). It has 7 electrons in its 3rd orbit.
  • It tends to take one electron from another atom so that it has 8 electrons in the outermost orbit with electronic configuration (2,8,8) with stable state.
  • The gaining of one electron leads to formation of chloride ion (Cl) which is negatively charged as it has gained one electron.

Question 5.
The valency of Magnesium (Mg) is two.
Answer:

  • The electronic configuration of Magnesium (Mg) is (2,8,2), it has 2 electrons in its 3rd orbit.
  • It tends to give these ‘2’ electrons so that it is left up with (2, 8), having 8 electrons in the second orbit, with a stable state.
  • The loss of two electrons leads to the formation of Magnesium ion (Mg2+) which is double positively charged as it has lost two electrons.

Question 6.
Valency is always a whole number.
Answer:

  • The number of electrons that an atom of an element gives away, takes up or shares forming a bond is called the valency of that element.
  • These electrons are always in whole numbers and not in fractions.
  • Therefore, valency is always a whole number.

Question 7.
Atomic size of potassium is bigger than atomic size of sodium.
Answer:

  • The atomic size of an element depends on the number of electron orbits in the atom of that element.
  • The greater the number of orbits, the larger the size.
  • Atomic number of potassium (K) is 19. Hence, its electronic configuration is (2, 8, 8,1). While atomic number of sodium (Na) is 11. Hence its electronic configuration is (2, 8,1)
  • Number of orbits in potassium atom is 4, while that in sodium atom is 3.
  • Hence, atomic size of potassium is bigger than atomic size of sodium.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 8.
The atomic size of sodium is bigger than atomic size of Magnesium.
Answer:

  • The atomic size of an element depends on the number of electron orbits in the atom of that element.
  • If 2 atoms have the same outermost orbit, then the atom having the larger number of electrons in the outermost orbit is smaller than the one having fewer electrons in the same outermost orbit.
  • Atomic number of sodium (Na) is 11. Hence, its electronic configuration is (2, 8, 1) while atomic number of magnesum (Mg) is 12 and hence its electronic configuration is (2, 8, 2).
  • As compared to sodium atom Magnesum atom has larger number of electrons n its electronic configuration.
  • Therefore, atomic size of sodium is bigger than atomic size of Magnesium.

Write the names of the following compounds and deduce their molecular masses:

Atomic masses : H(1), 0(16), N(14), C(12), K(39), S(32), Ca(40), Na(23), C1(35.5), Mg(24), A1(27), P(31)

Question 1.
Molecular mass of K2CO3
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 6

Question 2.
Molecular mass of CO2
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 7

Question 3.
Molecular mass of MgCl2
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 8

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 4.
Molecular mass of NaOH
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 8

Question 5.
Molecular mass of AIPO4
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 10

Question 6.
Molecular mass of NaHCO3
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 11

Numerical.

Question 1.
Magnesium Oxide:
Answer:
Given : Number of moles of Magnesium oxide (MgO)n = 0.2 mol
To find : Mass in grams of 0.2 mol of MgO
Solution:
Molecular mass of (MgO)M
= (Atomic mass of Mg) x 1 + (Atomic mass of O) x 1
= 24 x 1 + 16 x 1
= 24 + 16
Molecular mass of (MgO)M = 40
According to the formula Number of moles in the given MgO (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 17
Mass of MgO in grams (m) = 0.2 x 40
Mass of MgO in grams (m) = 8 g.
Mass of 0.2 mole of MgO is 8 g

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 2.
Calcium Carbonate:
Answer:
Given : Number of moles of Calcium carbonate (CaCO3) n = 0.2 mol
To find : Mass in grams of 0.2 mol of CaCO3
Solution:
Molecular mass of (CaCO3) M
= (Atomic mass of Ca) x 1 + (Atomic mass of C) x 1 + (Atomic mass of O) x 3
= (40 x l) + (12 x 1) +(16 x 3)
= 40+ 12+ 48
Molecular mass of (CaCO3) M = 100
According to the formula Number of moles in the given CaCO3 (n)
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 18
∴ Mass of CaCO3 in grams (m) = 0.2 x 100
∴ Mass of CaCO3 in grams (m) = 20 g
Mass of 0.2 mole of CaCO3 is 20 g

State laws/Define the following:

Question 1.
Law of Conservation of Matter.
Answer:
In a chemical reaction, the total weight of the reactants is same as the total weight of the products formed due to chemical reaction.

Question 2.
Law of Constant Proportion.
Answer:
The proportion by weight of the constituent elements in the various samples of a compound is fixed.

Question 3.
Molecular Mass:
Answer:
The molecular mass of a substance is the sum of the atomic masses of all the atoms in a single molecule of that substance.

Question 4.
Mole
Answer:
A mole is that quantity of a substance whose mass in grams is equal in magnitude to the molecular mass of that substance in Daltons.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Question 5.
Valency
Answer:
The capacity of an element to combine is called its valency.

Question 6.
Electronic definition of Valency
Answer:
The number of electrons that an atom of an element gives away or takes up while forming an ionic bond is called valency of that element.

Question 7.
Radicals
Answer:
The positively or negatively charged ions that take part independently in chemical reactions are called radicals.

Question 8.
Atomic size determination
Answer:
The size of an atom is determined by its radius. The atomic radius of an isolated atom is the distance between the nucleus of an atom and its outermost orbit.

Question 9.
Atomic mass number
Answer:
The number of protons and neutrons in the atomic nucleus is called the atomic mass number.

Question 10.
Unified mass
Answer:
Unified mass is the standard unit of atomic mass that quantifies mass on an atomic or molecular scale. Its symbol is ‘u’.
1 u = 1.66053904 x 10-27 kg.

Question 11.
Molecular mass of a substance
Answer:
The molecular mass of a substance is the sum of the atomic masses of all the atoms in a single molecule of that substance. Like atomic mass, molecular mass is also expressed in the unit Dalton (u).

Answer the following questions:

Question 1.
What is variable valency?
Answer:

  • Under different conditions, the atoms of some elements give away or take up a different number of electrons.
  • In such cases, those elements exhibit more than one valency.
  • This property of elements is called variable valency.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

Complete the following table:

Question 1.
Write down the cations and anions obtained from the compounds in the following chart.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 26

Answer the following questions:

Question 1.
Using the chart of ions/radicals and the cross-multiplication method, write the chemical formulae of the following compounds:

(a) Calcium carbonate
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 29
∴ Chemical formula of Calcium carbonate is CaCO3

(b) Sodium bicarbonate
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 30
∴ Chemical formula of Sodium bicarbonate is NaHCO3

(c) Silver chloride
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 31
∴ Chemical formula of Silver chloride is AgCl

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

(d) Calcium hydroxide Answer: Symbol Ca OH
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 52
∴ Chemical formula of Calcium hydroxide is Ca(OH)2

(e) Magnesium oxide
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 53
∴ Chemical formula of Magnesium oxide is MgO

(f) Ammonium phosphate
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 54
∴ Chemical formula of Ammonium phosphate is (NH4)3PO4

(g) Cuprous bromide
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 55
∴ Chemical formula of Cuprous bromide is CuBr.

(h) Copper sulphate
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 57
∴ Chemical formula of Copper sulphate is CuSO4.

(i) Potassium nitrate
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 58
∴ Chemical formula of Potassium nitrate is KNO3.

Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter

(j) Sodium dichromate
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 4 Measurement of Matter 59
∴ Chemical formula of Sodium dichromate is Na2Cr2O7.

9th Std Science Questions And Answers: