Class 12

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 1 Rotational Dynamics Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 1 Rotational Dynamics

1. Choose the correct option.

i) When seen from below, the blades of a ceiling fan are seen to be revolving anticlockwise and their speed is decreasing. Select the correct statement about the directions of its angular velocity and angular acceleration.
(A) Angular velocity upwards, angular acceleration downwards.
(B) Angular velocity downwards, angular acceleration upwards.
(C) Both, angular velocity and angular acceleration, upwards.
(D) Both, angular velocity and angular acceleration, downwards.
Answer:
(A) Angular velocity upwards, angular acceleration downwards.

ii) A particle of mass 1 kg, tied to a 1.2 m long string is whirled to perform the vertical circular motion, under gravity. The minimum speed of a particle is 5 m/s. Consider the following statements.
P) Maximum speed must be 5 5 m/s.
Q) Difference between maximum and minimum tensions along the string is 60 N. Select the correct option.
(A) Only statement P is correct.
(B) Only statement Q is correct.
(C) Both the statements are correct.
(D) Both the statements are incorrect.
Answer:
(C) Both the statements are correct.

iii) Select the correct statement about the formula (expression) of the moment of inertia (M.I.) in terms of mass M of the object and some of its distance parameter/s, such as R, L, etc.
(A) Different objects must have different expressions for their M.I.
(B) When rotating about their central axis, a hollow right circular cone and a disc have the same expression for the M.I.
(C) Expression for the M.I. for a parallelepiped rotating about the transverse axis passing through its centre includes its depth.
(D) Expression for M.I. of a rod and that of a plane sheet is the same about a transverse axis.
Answer:
(B) When rotating about their central axis, a hollow right circular cone and a disc have the same expression for the M.I.

iv) In a certain unit, the radius of gyration of a uniform disc about its central and transverse axis is \(\sqrt{2.5}\). Its radius of gyration about a tangent in its plane (in the same unit) must be
(A) \(\sqrt{5}\)
(B) 2.5
(C) 2\(\sqrt{2.5}\)
(D) \(\sqrt{12.5}\)
Answer:
(B) 2.5

v) Consider following cases:
(P) A planet revolving in an elliptical orbit.
(Q) A planet revolving in a circular orbit.
Principle of conservation of angular momentum comes in force in which of these?
(A) Only for (P)
(B) Only for (Q)
(C) For both, (P) and (Q)
(D) Neither for (P), nor for (Q)
Answer:
(C) For both, (P) and (Q)

X) A thin walled hollow cylinder is rolling down an incline, without slipping. At any instant, the ratio ”Rotational K.E.:
Translational K.E.: Total K.E.” is
(A) 1:1:2
(B) 1:2:3
(C) 1:1:1
(D) 2:1:3
Answer:
(D) 2:1:3

2. Answer in brief.

i) Why are curved roads banked?
Answer:
A car while taking a turn performs circular motion. If the road is level (or horizontal road), the necessary centripetal force is the force of static friction between the car tyres and the road surface. The friction depends upon the nature of the surfaces in contact and the presence of oil and water on the road. If the friction is inadequate, a speeding car may skid off the road. Since the friction changes with circumstances, it cannot be relied upon to provide the necessary centripetal force. Moreover, friction results in fast wear and tear of the tyres.

To avoid the risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is tilted inward, i.e., the outer side of the road is raised above its inner side. This is called banking of road. On a banked road, the resultant of the normal reaction and the gravitational force can act as the necessary centripetal force. Thus, every car can be safely driven on such a banked curve at certain optimum speed, without depending on friction. Hence, a road should be properly banked at a bend.

The angle of banking is the angle of inclination of the surface of a banked road at a bend with the horizontal.

ii) Do we need a banked road for a two wheeler? Explain.
Answer:
When a two-wheeler takes a turn along an unbanked road, the force of friction provides the centripetal force. The two-wheeler leans inward to counteract a torque that tends to topple it outward. Firstly, friction cannot be relied upon to provide the necessary centripetal force on all road conditions. Secondly, the friction results in wear and tear of the tyres. On a banked road at a turn, any vehicle can negotiate the turn without depending on friction and without straining the tyres.

iii) On what factors does the frequency of a conical pendulum depend? Is it independent of some factors?
Answer:
The frequency of a conical pendulum, of string length L and semivertical angle θ, is
n = \(\frac{1}{2 \pi} \sqrt{\frac{g}{L \cos \theta}}\)
where g is the acceleration due to gravity at the place.
From the above expression, we can see that

1. n ∝ \(\sqrt{g}\)
2. n ∝ \(\frac{1}{\sqrt{L}}\)
3. n ∝ \(\frac{1}{\sqrt{\cos \theta}}\)
   (if θ increases, cos θ decreases and n increases)
4. The frequency is independent of the mass of the bob.

iv) Why is it useful to define radius of gyration?
Answer:
Definition : The radius of gyration of a body rotating about an axis is defined as the distance between the axis of rotation and the point at which the entire mass of the body can be supposed to be concentrated so as to give the same moment of inertia as that of the body about the given axis.

The moment of inertia (MI) of a body about a given rotation axis depends upon

1. the mass of the body and
2. the distribution of mass about the axis of rotation. These two factors can be separated by expressing the MI as the product of the mass (M) and the square of a particular distance (k) from the axis of rotation. This distance is called the radius of gyration and is defined as given above. Thus,
   

Physical significance : The radius of gyration is less if I is less, i.e., if the mass is distributed close to the axis; and it is more if I is more, i.e., if the mass is distributed away from the axis. Thus, it gives the idea about the distribution of mass about the axis of rotation.

v) A uniform disc and a hollow right circular cone have the same formula for their M.I., when rotating about their central axes. Why is it so?
Answer:
The radius of gyration of a thin ring of radius Rr about its transverse symmetry axis is
K_r = \(\sqrt{I_{\mathrm{CM}} / M_{\mathrm{r}}}\) = \(\sqrt{R_{\mathrm{r}}^{2}}\) = R_r
The radius of gyration of a thin disc of radius R_d about its transverse symmetry axis is

Question 3.
While driving along an unbanked circular road, a two-wheeler rider has to lean with the vertical. Why is it so? With what angle the rider has to lean? Derive the relevant expression. Why such a leaning is not necessary for a four wheeler?
Answer:
When a bicyclist takes a turn along an unbanked road, the force of friction \(\vec{f}_{\mathrm{s}}\) provides the centripetal force; the normal reaction of the road \(\vec{N}\) is vertically up. If the bicyclist does not lean inward, there will be an unbalanced outward torque about the centre of gravity, f_s.h, due to the friction force that will topple the bicyclist outward. The bicyclist must lean inward to counteract this torque (and not to generate a centripetal force) such that the opposite inward torque of the couple formed by \(\vec{N}\) and the weight \(\vec{g}\), mg.a = f_s.h₁

Since the force of friction provides the centripetal force,
f_s = \(\frac{m v^{2}}{r}\)
If the cyclist leans from the vertical by an angle 9, the angle between \(\vec{N}\) and \(\vec{F}\) in above figure.

Hence, the cyclist must lean by an angle
θ = tan^-1\(\left(\frac{v^{2}}{g r}\right)\)

When a car takes a turn along a level road, apart from the risk of skidding off outward, it also has a tendency to roll outward due to an outward torque about the centre of gravity due to the friction force. But a car is an extended object with four wheels. So, when the inner wheels just get lifted above the ground, it can be counterbalanced by a restoring torque of the couple formed by the normal reaction (on the outer wheels) and the weight.

Question 4.
Using the energy conservation, derive the expressions for the minimum speeds at different locations along a vertical circular motion controlled by gravity. Is zero speed possible at the uppermost point? Under what condition/s? Also prove that the difference between the extreme tensions (or normal forces) depends only upon the weight of the object.
Answer:
In a non uniform vertical circular motion, e.g., those of a small body attached to a string or the loop-the-loop manoeuvers of an aircraft or motorcycle or skateboard, the body must have some minimum speed to reach the top and complete the circle. In this case, the motion is controlled only by gravity and zero speed at the top is not possible.

However, in a controlled vertical circular motion, e.g., those of a small body attached to a rod or the giant wheel (Ferris wheel) ride, the body or the passenger seat can have zero speed at the top, i.e., the motion can be brought to a stop.

Question 5.
Discuss the necessity of radius of gyration. Define it. On what factors does it depend and it does not depend? Can you locate some similarity between the centre of mass and radius of gyration? What can you infer if a uniform ring and a uniform disc have the same radius of gyration?
Answer:
Definition : The radius of gyration of a body rotating about an axis is defined as the distance between the axis of rotation and the point at which the entire mass of the body can be supposed to be concentrated so as to give the same moment of inertia as that of the body about the given axis.

The moment of inertia (MI) of a body about a given rotation axis depends upon

1. the mass of the body and
2. the distribution of mass about the axis of rotation. These two factors can be separated by expressing the MI as the product of the mass (M) and the square of a particular distance (k) from the axis of rotation. This distance is called the radius of gyration and is defined as given above. Thus,

Physical significance : The radius of gyration is less if I is less, i.e., if the mass is distributed close to the axis; and it is more if I is more, i.e., if the mass is distributed away from the axis. Thus, it gives the idea about the distribution of mass about the axis of rotation.

The centre of mass (CM) coordinates locates a point where if the entire mass M of a system of particles or that of a rigid body can be thought to be concentrated such that the acceleration of this point mass obeys Newton’s second law of motion, viz.,
\(\vec{F}_{\mathrm{net}}\) = M\(\overrightarrow{\mathrm{a}}_{\mathrm{CM}}\), where \(\vec{F}_{\mathrm{net}}\) is the sum of all the external forces acting on the body or on the individual particles of the system of particles.

Similarly, radius of gyration locates a point from the axis of rotation where the entire mass M can be thought to be concentrated such that the angular acceleration of that point mass about the axis of rotation obeys the relation, \(\vec{\tau}_{\mathrm{net}}\) = M\(\vec{\alpha}\), where \(\vec{\tau}_{\text {net }}\) is the sum of all the external torques acting on the body or on the individual particles of the system of particles.

Question 6.
State the conditions under which the theorems of parallel axes and perpendicular axes are applicable. State the respective mathematical expressions.
Answer:
The theorem of parallel axis is applicable to any body of arbitrary shape. The moment of inertia (MI) of the body about an axis through the centre mass should be known, say, I_CM. Then, the theorem can be used to find the MI, I, of the body about an axis parallel to the above axis. If the distance between the two axes is h,
I = I_CM + Mh² …(1)
The theorem of perpendicular axes is applicable to a plane lamina only. The moment of inertia I_z of a plane lamina about an axis-the z axis- perpendicular to its plane is equal to the sum of its moments of inertia I_x and I_y about two mutually perpendicular axes x and y in its plane and through the point of intersection of the perpendicular axis and the lamina.
I_z = I_x + I_y …. (2)

Question 7.
Derive an expression that relates angular momentum with the angular velocity of a rigid body.
Answer:
Consider a rigid body rotating with a constant angular velocity \(\vec{\omega}\) about an axis through the point O and perpendicular to the plane of the figure. All the particles of the body perform uniform circular motion about the axis of rotation with the same angular velocity \(\vec{\omega}\). Suppose that the body consists of N particles of masses m₁, m₂, …, m_n, situated at perpendicular distances r₁, r₂, …, r_N, respectively from the axis of rotation.

The particle of mass m₁ revolves along a circle of radius r₁, with a linear velocity of magnitude v₁ = r₁ω. The magnitude of the linear momentum of the particle is
p₁ = m₁v₁ = m₁r₁ω
The angular momentum of the particle about the axis of rotation is by definition,
\(\vec{L}_{1}\) = \(\vec{r}_{1}\) × \(\vec{p}_{1}\)
∴ L₁ = r₁p₁ sin θ
where θ is the smaller of the two angles between \(\vec{r}_{1}\) and \(\vec{p}_{1} \text { . }\)
In this case, θ = 90° ∴ sin θ = 1
∴ L₁ = r₁p₁ = r₁m₁r₁ω = m₁r₁²ω
Similarly L₂ = m₂r₂²ω, L₃ = m₃r₃²ω, etc.
The angular momentum of the body about the given axis is
L = L₁ + L₂ + … + L_N

where I = \(\sum_{i=1}^{N} m_{i} r_{i}^{2}\) = moment of inertia of the body about the given axis.
In vector form, \(\vec{L}\) = \(I \vec{\omega}\)
Thus, angular momentum = moment of inertia × angular velocity.
[Note : Angular momentum is a vector quantity. It has the same direction as \(\vec{\omega}\).]

Question 8.
Obtain an expression relating the torque with angular acceleration for a rigid body.
Answer:
A torque acting on a body produces angular acceleration. Consider a rigid body rotating about an axis passing through the point O and perpendicular to the plane of the figure. Suppose that a torque \(\vec{\tau}\) on

the body produces uniform angular acceleration \(\vec{\alpha}\) along the axis of rotation.
The body can be considered as made up of N particles with masses m₁, m₂, …, m_N situated at perpendicular distances r₁, r₂, …, r_N respectively from the axis of rotation, \(\vec{\alpha}\) is the same for all the particles as the body is rigid. Let \(\vec{F}_{1}\), \(\vec{F}_{2}\), …, \(\vec{F}_{N}\) be the external forces on the particles.
The torque \(\vec{\tau}_{1}\), on the particle of mass m₁, is
\(\vec{\tau}_{1}\) = \(\vec{r}_{1}\) × \(\vec{F}_{1}\)
∴ τ₁ = r₁F₁ sin θ
where θ is the smaller of the two angles between \(\vec{r}_{1}\) and \(\vec{F}_{1} .\)

where

is the moment of inertia of the body about the axis of rotation.
In vector form, \(\vec{\tau}\) = I\(\vec{\alpha}\)
This gives the required relation.
Angular acceleration \(\vec{\alpha}\) has the same direction as the torque \(\vec{\tau}\) and both of them are axial vectors along the rotation axis.

Question 9.
State and explain the principle of conservation of angular momentum. Use a suitable illustration. Do we use it in our daily life? When?
Answer:
Law (or principle) of conservation of angular momentum : The angular momentum of a body is conserved if the resultant external torque on the body is zero.
Explanation : This law (or principle) is used by a figure skater or a ballerina to increase their speed of rotation for a spin by reducing the body’s moment of inertia. A diver too uses it during a somersault for the same reason.

(1) Ice dance :
Twizzle and spin are elements of the sport of figure skating. In a twizzle a skater turns several revolutions while travelling on the ice. In a dance spin, the skater rotates on the ice skate and centred on a single point on the ice. The torque due to friction between the ice skate and the ice is small. Consequently, the angular momentum of a figure skater remains nearly constant.

For a twizzle of smaller radius, a figure skater draws her limbs close to her body to reduce moment of inertia and increase frequency of rotation. For larger rounds, she stretches out her limbs to increase moment of inertia which reduces the angular and linear speeds.

A figure skater usually starts a dance spin in a crouch, rotating on one skate with the other leg and both arms extended. She rotates relatively slowly because her moment of inertia is large. She then slowly stands up, pulling the extended leg and arms to her body. As she does so, her moment of inertia about the axis of rotation decreases considerably,and thereby her angular velocity substantially increases to conserve angular momentum.

(2) Diving :
Take-off from a springboard or diving platform determines the diver’s trajectory and the magnitude of angular momentum. A diver must generate angular momentum at take-off by moving the position of the arms and by a slight hollowing of the back. This allows the diver to change angular speeds for twists and somersaults in flight by controlling her/his moment of inertia. A compact tucked shape of the body lowers the moment of inertia for rotation of smaller radius and increased angular speed. The opening of the body for the vertical entry into water does not stop the rotation, but merely slows it down. The angular momentum remains constant throughout the flight.

Question 10.
Discuss the interlink between translational, rotational and total kinetic energies of a rigid object that rolls without slipping.
Answer:
Consider a symmetric rigid body, like a sphere or a wheel or a disc, rolling on a plane surface with friction along a straight path. Its centre of mass (CM) moves in a straight line and, if the frictional force on the body is large enough, the body rolls without slipping. Thus, the rolling motion of the body can be treated as translation of the CM and rotation about an axis through the CM. Hence, the kinetic energy of a rolling body is
E = E_tran + E_rot ……. (1)

where E_tran and E_rot are the kinetic energies associated with translation of the CM and rotation about an axis through the CM, respectively.

Let M and R be the mass and radius of the body. Let ω, k and i be the angular speed, radius of gyration and moment of inertia for rotation about an axis through its centre, and v be the translational speed of the centre of mass.

Question 11.
A rigid object is rolling down an inclined plane. Derive expressions for the acceleration along the track and the speed after falling through a certain vertical distance.
Answer:
Consider a circularly symmetric rigid body, like a sphere or a wheel or a disc, rolling with friction down a plane inclined at an angle 9 to the horizontal. If the frictional force on the body is large enough, the body rolls without slipping.
Let M and R be the mass and radius of the body. Let I be the moment of inertia of the body for rotation about an axis through its centre. Let the body start from rest at the top of the incline at a

height h. Let v be the translational speed of the centre of mass at the bottom of the incline. Then, its kinetic energy at the bottom of the incline is

Let a be the acceleration of the centre of mass of the body along the inclined plane. Since the body starts from rest,

Starting from rest, if t is the time taken to travel the distance L,

[Note : For rolling without slipping, the contact point of the rigid body is instantaneously at rest relative to the surface of the inclined plane. Hence, the force of friction is static rather than kinetic, and does no work on the body. Thus, the force of static friction causes no decrease in the mechanical energy of the body and we can use the principle of conservation of energy.]

Question 12.
Somehow, an ant is stuck to the rim of a bicycle wheel of diameter 1 m. While the bicycle is on a central stand, the wheel
is set into rotation and it attains the frequency of 2 rev/s in 10 seconds, with uniform angular acceleration. Calculate
(i) Number of revolutions completed by the ant in these 10 seconds.
(ii) Time taken by it for first complete revolution and the last complete revolution.
[Ans:10 rev., t_first = \(\sqrt{10}\)s, t_last = 0.5132s]
Answer:
Data : r = 0.5 m, ω₀ = 0, ω = 2 rps, t = 10 s

(i) Angular acceleration (α) being constant, the average angular speed,
ω_av = \(\frac{\omega_{\mathrm{o}}+\omega}{2}\) = \(\frac{0+2}{2}\) = 1 rps
∴ The angular displacement of the wheel in time t,
θ = ω_av ∙ t = 1 × 10 = 10 revolutions

(ii)

The time for the last, i.e., the 10th, revolution is t₁ – t₂ = 10 – 9.486 = 0.514 s

Question 13.
Coefficient of static friction between a coin and a gramophone disc is 0.5. Radius of the disc is 8 cm. Initially the
centre of the coin is 2 cm away from the centre of the disc. At what minimum frequency will it start slipping from there? By what factor will the answer change if the coin is almost at the rim?
(use g = π² m/s²)
[Ans: 2.5 rev/s, n₂ = \(\frac{1}{2}\)n₁]
Answer:
Data : µ_s = 0.5, r₁ = π cm = π × 10^-2 m, r₂ = 8 cm = 8 × 10^-2 m, g = π² m/s²
To revolve with the disc without slipping, the necessary centripetal force must be less than or equal to the limiting force of static friction.

The coin will start slipping when the frequency is

The minimum frequency in the second case will be \(\sqrt{\frac{\pi}{8}}\) times that in the first case.
[ Note The answers given in the textbook are for r₁ = 2 cm.]

Question 14.
Part of a racing track is to be designed for a radius of curvature of 72 m. We are not recommending the vehicles to drive faster than 216 kmph. With what angle should the road be tilted? At what height will its outer edge be, with respect to the inner edge if the track is 10 m wide?
[Ans: θ = tan^-1 (5) = 78.69°, h = 9.8 m]
Answer:
Data : r = 72 m, v₀ = 216 km/h, = 216 × \(\frac{5}{18}\)
= 60 m/s, w = 10 m, g = 10 m/s²
tan θ = \(\frac{v_{\mathrm{o}}^{2}}{r g}\) = \(\frac{(60)^{2}}{72 \times 10}\) = \(\frac{3600}{720}\) = 5
∴ θ = tan^-1 5 = 78°4′
This is the required angle of banking.
sin θ = \(\frac{h}{w}\)
∴ h = w sin θ = (10) sin 78°4′ = 10 × 0.9805
= 9.805 m
This gives the height of the outer edge of the track relative to the inner edge.

Question 15.
The road in the example 14 above is constructed as per the requirements. The coefficient of static friction between the tyres of a vehicle on this road is 0.8, will there be any lower speed limit? By how much can the upper speed limit exceed in this case?
[Ans: v_min ≅ 88 kmph, no upper limit as the road is banked for θ > 45°]
Answer:
Data : r = 72 m, θ = 78 °4′, µ_s = 0.8, g = 10 m/s² tan θ = tan 78°4′ = 5

= 24.588 m/s = 88.52 km/h
This will be the lower limit or minimum speed on this track.
Since the track is heavily banked, θ > 45 °, there is no upper limit or maximum speed on this track.

Question 16.
During a stunt, a cyclist (considered to be a particle) is undertaking horizontal circles inside a cylindrical well of radius 6.05 m. If the necessary friction coefficient is 0.5, how much minimum speed should the stunt artist maintain? Mass of the artist is 50 kg. If she/he increases the speed by 20%, how much will the force of friction be?
[Ans: v_min = 11 m/s, f_s = mg = 500 N]
Answer:
Data : r = 6.05 m, µ_s = 0.5, g = 10 m/s², m = 50 kg, ∆v = 20%

This is the required minimum speed. So long as the cyclist is not sliding, at every instant, the force of static friction is fs = mg = (50)(10) = 500 N

Question 17.
A pendulum consisting of a massless string of length 20 cm and a tiny bob of mass 100 g is set up as a conical pendulum. Its bob now performs 75 rpm. Calculate kinetic energy and increase in the gravitational potential energy of the bob. (Use π ² = 10 )
[Ans: cos θ = 0.8, K.E. = 0.45 J, ∆(P E.) = 0.04 J]
Answer:
Data : L = 0.2 m, m = 0.1 kg, n = \(\frac{75}{60}\) = \(\frac{5}{4}\) rps,
g = 10 m/s², π² = 10,

The increase in gravitational PE,
∆PE = mg(L – h)
= (0.1) (10) (0.2 – 0.16)
= 0.04 J

Question 18.
A motorcyclist (as a particle) is undergoing vertical circles inside a sphere of death. The speed of the motorcycle varies between 6 m/s and 10 m/s. Calculate diameter of the sphere of death. What are the minimum values are possible for these two speeds?
[Ans: Diameter = 3.2 m, (v₁)_min = 4 m/s, (v₂)_min = 4 \(\sqrt{5}\)/m s ]
Answer:

= 1.6 m
The diameter of the sphere of death = 3.2 m.

The required minimum values of the speeds are 4 m/s and 4\(\sqrt{5}\) m/s.

Question 19.
A metallic ring of mass 1 kg has moment of inertia 1 kg m² when rotating about one of its diameters. It is molten and remoulded into a thin uniform disc of the same radius. How much will its moment of inertia be, when rotated about its own axis.
[Ans: 1 kg m²]
Answer:
The MI of the thin ring about its diameter,
I_ring = \(\frac{1}{2}\)MR² = 1 kg.m²
Since the ring is melted and recast into a thin disc of same radius R, the mass of the disc equals the mass of the ring = M.
The MI of the thin disc about its own axis (i.e., transverse symmetry axis) is
I_disc = \(\frac{1}{2}\)MR² = I_ring
∴ I_disc = 1 kg.m²

Question 20.
A big dumb-bell is prepared by using a uniform rod of mass 60 g and length 20 cm. Two identical solid thermocol spheres of mass 25 g and radius 10 cm each are at the two ends of the rod. Calculate moment of inertia of the dumbbell when rotated about an axis passing through its centre and perpendicular to the length.
[Ans: 24000 g cm^-2]
Answer:
Data : M_sph = 50 g, R_sph = 10 cm, M_rod = 60 g, L_rod = 20 cm
The MI of a solid sphere about its diameter is
I_sph,CM = \(\frac{2}{5}\)M_sphR_sph
The distance of the rotation axis (transverse symmetry axis of the dumbbell) from the centre of sphere, h = 30 cm.
The MI of a solid sphere about the rotation axis, I_sph = I_{sph, CM} + M_sphh²
For the rod, the rotation axis is its transverse symmetry axis through CM.
The MI of a rod about this axis,
I_rod = \(\frac{1}{12}\) M_rodL²_rod
Since there are two solid spheres, the MI of the dumbbell about the rotation axis is

Question 21.
A flywheel used to prepare earthenware pots is set into rotation at 100 rpm. It is in the form of a disc of mass 10 kg and
radius 0.4 m. A lump of clay (to be taken equivalent to a particle) of mass 1.6 kg falls on it and adheres to it at a certain
distance x from the centre. Calculate x if the wheel now rotates at 80 rpm.
[Ans: x = \(\frac{1}{\sqrt{8}}\)m = 0.35 m]
Answer:
Data : f₁ = 60 rpm = 60/60 rot/s = 1 rot/s,
f₂ = 30 rpm = 30/60 rot/s = \(\frac{1}{2}\) rot/s, ∆E = — 100 J

(i)

This gives the MI of the flywheel about the given axis.

(ii) Angular momentum, L = Iω = I(2πf) = 2πIf
The change in angular momentum, ∆L
= L₂ – L₁ = 2πI(f₂ – f₁)
= 2 × 3.142 × 6.753(\(\frac{1}{2}\) – 1)
= -3.142 × 6.753 = -21.22 kg.m²/s

Question 22.
Starting from rest, an object rolls down along an incline that rises by 3 units in every 5 units (along it). The object gains a speed of \(\sqrt{10}\) m/s as it travels a distance of \(\frac{5}{3}\)m along the incline. What can be the possible shape/s of the object?
[Ans: \(\frac{K^{2}}{R^{2}}\) = 1. Thus, a ring or a hollow cylinder]
Answer:
Data : sin θ = \(\frac{3}{5}\), u = 0, v = \(\sqrt{10}\) m/s, L = \(\frac{5}{3}\)m, g = 10 m/s²


Therefore, the body rolling down is either a ring or a cylindrical shell.

12th Physics Digest Chapter 1 Rotational Dynamics Intext Questions and Answers

Activity (Textbook Page No. 3)

Question 1.
Attach a body of suitable mass to a spring balance so that it stretches by about half its capacity. Now whirl the spring balance so that the body performs a horizontal circular motion. You will notice that the balance now reads more for the same body. Can you explain this ?
Answer:
Due to outward centrifugal force.

Use your brain power (Textbook Page No. 4)

Question 1.
Obtain the condition for not toppling (rollover) for a four-wheeler. On what factors does it depend and how?
Answer:
Consider a car of mass m taking a turn of radius r along a level road. As seen from an inertial frame of reference, the forces acting on the car are :

1. the lateral limiting force of static friction \(\overrightarrow{f_{\mathrm{s}}}\) on the wheels-acting along the axis of the wheels and towards the centre of the circular path- which provides the necessary centripetal force,
2. the weight \(m \vec{g}\) acting vertically downwards at the centre of gravity (C.G.)
3. the normal reaction \(\vec{N}\) of the road on the wheels, acting vertically upwards effectively at the C.G. Since maximum centripetal force = limiting force of static friction,
   ma_r = \(\frac{m v^{2}}{r}\) = f_s…. (1)

In a simplified rigid-body vehicle model, we consider only two parameters-the height h of the C.G. above the ground and the average distance b between the left and right wheels called the track width.

The friction force \(\overrightarrow{f_{s}}\) on the wheels produces a torque \(\tau_{\mathrm{t}}\) that tends to overturn/rollover the car about the outer wheel. Rotation about the front-to-back axis is called roll.
\(\tau_{\mathrm{t}}\) = f_s.h = \(\left(\frac{m v^{2}}{r}\right)\)h … (2)

When the inner wheel just gets lifted above the ground, the normal reaction \(\vec{N}\) of the road acts on the outer wheels but the weight continues to act at the C.G. Then, the couple formed by the normal reaction and the weight produces a opposite torque \(\tau_{\mathrm{r}}\) which tends to restore the car back on all four wheels
\(\tau_{\mathrm{r}}\) = mg.\(\frac{b}{2}\) …. (3)
The car does not topple as long as the restoring torque \(\tau_{\mathrm{r}}\) counterbalances the toppling torque \(\tau_{\mathrm{t}}\). Thus, to avoid the risk of rollover, the maximum speed that the car can have is given by
\(\left(\frac{m v^{2}}{r}\right)\)h = mg.\(\frac{b}{2}\) ∴ v_max = \(\sqrt{\frac{r b g}{2 h}}\) … (4)

Thus, vehicle tends to roll when the radial acceleration reaches a point where inner wheels of the four-wheeler are lifted off of the ground and the vehicle is rotated outward. A rollover occurs when the gravitational force \(m \vec{g}\) passes through the pivot point of the outer wheels, i.e., the C.G. is above the line of contact of the outer wheels. Equation (3) shows that this maximum speed is high for a car with larger track width and lower centre of gravity.

There will be rollover (before skidding) if \(\tau_{t}\) ≥ \(\tau_{\mathrm{r}}\), that is if

The vehicle parameter ratio, \(\frac{b}{2 h}\), is called the static stability factor (SSF). Thus, the risk of a rollover is low if SSF ≤ µ_s. A vehicle will most likely skid out rather than roll if µ_s is too low, as on a wet or icy road.

Question 2.
Think about the normal reactions. Where are those and how much are those?
Answer:

In a simplified vehicle model, we assume the normal reactions to act equally on all the four wheels, i.e., mg/4 on each wheel. However, the C.G. is not at the geometric centre of a vehicle and the wheelbase (i.e., the distance L between its front and rear wheels) affects the weight distribution of the vehicle. When a vehicle is not accelerating, the normal reactions on each pair of front and rear wheels are, respectively,
N_f = \(\frac{d_{\mathrm{r}}}{L}\)mg and N_r = \(\frac{d_{\mathrm{f}}}{L}\) mg
where d_r and d_f are the distances of the rear and front axles from the C.G. [When a vehicle accelerates, additional torque acts on the axles and the normal reactions on the wheels change. So, as is common experience, a car pitches back (i.e., rear sinks and front rises) when it accelerates, and a car pitches ahead (i.e., front noses down). Rotation about the lateral axis is called pitch.]

Question 3.
What is the recommendations on loading a vehicle for not toppling easily?
Answer:
Overloading (or improper load distribution) or any load placed on the roof raises a vehicle’s centre of gravity, and increases the vehicle’s likelihood of rolling over. A roof rack should be fitted by considering weight limits.

Road accidents involving rollovers show that vehicles with higher h (such as SUVs, pickup vans and trucks) topple more easily than cars. Untripped rollovers normally occur when a top-heavy vehicle attempts to perform a panic manoeuver that it physically cannot handle.

Question 4.
If a vehicle topples while turning, which wheels leave the contact with the road? Why?
Answer:
Inner wheels.
Consider a car of mass m taking a turn of radius r along a level road. As seen from an inertial frame of reference, the forces acting on the car are :

1. the lateral limiting force of static friction \(\overrightarrow{f_{\mathrm{s}}}\) on the wheels-acting along the axis of the wheels and towards the centre of the circular path- which provides the necessary centripetal force,
2. the weight \(m \vec{g}\) acting vertically downwards at the centre of gravity (C.G.)
3. the normal reaction \(\vec{N}\) of the road on the wheels, acting vertically upwards effectively at the C.G. Since maximum centripetal force = limiting force of static friction,
   ma_r = \(\frac{m v^{2}}{r}\) = f_s…. (1)

In a simplified rigid-body vehicle model, we consider only two parameters-the height h of the C.G. above the ground and the average distance b between the left and right wheels called the track width.

The friction force \(\overrightarrow{f_{s}}\) on the wheels produces a torque \(\tau_{\mathrm{t}}\) that tends to overturn/rollover the car about the outer wheel. Rotation about the front-to-back axis is called roll.
\(\tau_{\mathrm{t}}\) = f_s.h = \(\left(\frac{m v^{2}}{r}\right)\)h … (2)

When the inner wheel just gets lifted above the ground, the normal reaction \(\vec{N}\) of the road acts on the outer wheels but the weight continues to act at the C.G. Then, the couple formed by the normal reaction and the weight produces a opposite torque \(\tau_{\mathrm{r}}\) which tends to restore the car back on all four wheels
\(\tau_{\mathrm{r}}\) = mg.\(\frac{b}{2}\) …. (3)
The car does not topple as long as the restoring torque \(\tau_{\mathrm{r}}\) counterbalances the toppling torque \(\tau_{\mathrm{t}}\). Thus, to avoid the risk of rollover, the maximum speed that the car can have is given by
\(\left(\frac{m v^{2}}{r}\right)\)h = mg.\(\frac{b}{2}\) ∴ v_max = \(\sqrt{\frac{r b g}{2 h}}\) … (4)

Thus, vehicle tends to roll when the radial acceleration reaches a point where inner wheels of the four-wheeler are lifted off of the ground and the vehicle is rotated outward. A rollover occurs when the gravitational force \(m \vec{g}\) passes through the pivot point of the outer wheels, i.e., the C.G. is above the line of contact of the outer wheels. Equation (3) shows that this maximum speed is high for a car with larger track width and lower centre of gravity.

Question 5.
How does [tendency to] toppling affect the tyres?
Answer:
While turning, shear stress acts on the tyre-road contact area. Due to this, the treads and side wall of a tyre deform. Apart from less control, this contributes to increased and uneven wear of the shoulder of the tyres.

Each wheel is placed under a small inward angle (called camber) in the vertical plane. Under severe lateral acceleration, when the car rolls, the camber angle ensures the complete contact area is in contact with the road and the wheels are now in vertical position. This improves the cornering behavior of the car. Improperly inflated and worn tyres can be especially dangerous because they inhibit the ability to maintain vehicle control. Worn tires may cause the vehicle to slide
sideways on wet or slippery pavement, sliding the vehicle off the road and increasing its risk of rolling over.

Question 6.
What is the recommendation for this?
Answer:
Because of uneven wear of the tyre shoulders, tyres should be rotated every 10000 km-12000 km. To avoid skidding, rollover and tyre-wear, the force of friction should not be relied upon to provide the necessary centripetal force during cornering. Instead, the road surface at a bend should be banked, i.e., tilted inward.

A car while taking a turn performs circular motion. If the road is level (or horizontal road), the necessary centripetal force is the force of static friction between the car tyres and the road surface. The friction depends upon the nature of the surfaces in contact and the presence of oil and water on the road. If the friction is inadequate, a speeding car may skid off the road. Since the friction changes with circumstances, it cannot be relied upon to provide the necessary centripetal force. Moreover, friction results in fast wear and tear of the tyres.

To avoid the risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is tilted inward, i.e., the outer side of the road is raised above its inner side. This is called banking of road. On a banked road, the resultant of the normal reaction and the gravitational force can act as the necessary centripetal force. Thus, every car can be safely driven on such a banked curve at certain optimum speed, without depending on friction. Hence, a road should be properly banked at a bend.

The angle of banking is the angle of inclination of the surface of a banked road at a bend with the horizontal.

When a two-wheeler takes a turn along an unbanked road, the force of friction provides the centripetal force. The two-wheeler leans inward to counteract a torque that tends to topple it outward. Firstly, friction cannot be relied upon to provide the necessary centripetal force on all road conditions. Secondly, the friction results in wear and tear of the tyres. On a banked road at a turn, any vehicle can negotiate the turn without depending on friction and without straining the tyres.

Question 7.
Determine the angle to be made with the vertical by a two-wheeler while turning on a horizontal track?
Answer:

When a bicyclist takes a turn along an unbanked road, the force of friction \(\vec{f}_{\mathrm{s}}\) provides the centripetal force; the normal reaction of the road \(\vec{N}\) is vertically up. If the bicyclist does not lean inward, there will be an unbalanced outward torque about the centre of gravity, f_s.h, due to the friction force that will topple the bicyclist outward. The bicyclist must lean inward to counteract this torque (and not to generate a centripetal force) such that the opposite inward torque of the couple formed by \(\vec{N}\) and the weight \(\vec{g}\), mg.a = f_s.h₁

Since the force of friction provides the centripetal force,
f_s = \(\frac{m v^{2}}{r}\)
If the cyclist leans from the vertical by an angle 9, the angle between \(\vec{N}\) and \(\vec{F}\) in above figure.

Hence, the cyclist must lean by an angle
θ = tan^-1\(\left(\frac{v^{2}}{g r}\right)\)

When a car takes a turn along a level road, apart from the risk of skidding off outward, it also has a tendency to roll outward due to an outward torque about the centre of gravity due to the friction force. But a car is an extended object with four wheels. So, when the inner wheels just get lifted above the ground, it can be counterbalanced by a restoring torque of the couple formed by the normal reaction (on the outer wheels) and the weight.

Question 8.
We have mentioned about ‘static friction’ between road and tyres. Why is it static friction? What about kinetic friction between road and tyres?
Answer:
When a car takes a turn on a level road, the point of contact of the wheel with the surface is instantaneously stationary if there is no slipping. Hence, the lateral force on the car is the limiting force of static friction between the tyres and road. Lateral forces allow the car to turn. As long as the wheels are rolling, there is lateral force of static friction and longitudinal force of rolling friction. Longtitudinal forces, which act in the direction of motion of the car body (or in the exact opposite direction), control the acceleration or deceleration of the car and therefore the speed of the car. These are the wheel force, rolling friction, braking force and air drag. If the car skids, the friction force is kinetic friction; more importantly, the direction of the friction force then changes abruptly from lateral to that opposite the velocity of skidding and not towards the centre of the curve, so that the car cannot continue in its curved path.

Question 9.
What do you do if your vehicle is trapped on a slippery or sandy road? What is the physics involved?
Answer:
Driving on a country road should be attempted only with a four-wheel drive. However, if you do get stuck in deep sand or mud, avoid unnecessary panic and temptation to drive your way out of the mud or sand because excessive spinning of your tyres will most likely just dig you into a deeper hole. Momentum is the key to getting unstuck from sand or mud. One method is the rocking method-rocking your car backwards and forwards to gain momentum. Your best option is usually to gain traction and momentum by wedging a car mat (or sticks, leaves, gravel or rocks) in front and under your drive wheels. Once you start moving, keep the momentum going until you are on more solid terrain.

Use your brain power (Textbook Page No. 6)

Question 1.
As a civil engineer, you are to construct a curved road in a ghat. In order to calculate the banking angle 0, you need to decide the speed limit. How will you decide the values of speed and radius of curvature at the bend ?
Answer:
For Indian roads, Indian Road Congress (IRC), [IRC-73-1980, Table 2, p.4], specifies the design speed depending on the classification of roads (such as national and state highways, district roads and village roads) and terrain. It is the basic design parameter which determines further geometric design features. For the radius of curvature at a bend, IRC [ibid., Table 16, p.24] specifies the absolute minimum values based on the minimum design speed. However, on new roads, curves should be designed to have the largest practicable radius, generally more than the minimum values specified, to allow for ‘sight distance’ and ‘driver comfort’. To consider the motorist driving within the innermost travel lane, the radius used to design horizontal curves should be measured to the inside edge of the innermost travel lane, particularly for wide road-ways with sharp horizontal curvature.

A civil engineer refers to banking as superelevation e;e = tan θ. IRC fixes e_max = 0.07 for a non-urban road and the coefficient of lateral static friction, µ = 0.15, the friction between the vehicle tyres and the road being incredibly variable. Ignoring the product eµ, from Eq. (6)
e + µ = \(\frac{v^{2}}{g r}\) (where both v and r are in SI units)
= \(\frac{V^{2}}{127 r}\)(where V is in km / h and r is in metre) …. (1)
The sequence of design usually goes like this :

1. Knowing the design speed V and radius r, calculate the superelevation for 75% of design speed ignormg friction : e = \(\frac{(0.75 \mathrm{~V})^{2}}{127 r}\) = \(\frac{V^{2}}{225 r}\)
2. If e < 0.07, consider this calculated value of e in subsequent calculations. If e > 0.07, then take e = e_max = 0.07.
3. Use Eq. (1) above to check the value of µ for e_max = 0.07 at the full value of the design speed V : µ = \(\frac{V^{2}}{127 r}\) – 0.07
   If µ < 0.15, then e = 0.07 is safe. Otherwise, calculate the allowable speed Va as in step 4.
4. \(\frac{V_{\mathrm{a}}^{2}}{127 r}\) = e + p = 0.07 + 0.15
   If V_a > V, then the design speed V is adequate.
   If V_a < V, then speed is limited to V_a with appropriate warning sign.

Use your brain power (Textbook Page No. 7)

Question 1.
If friction is zero, can a vehicle move on the road? Why are we not considering the friction in deriving the expression for the banking angle?
Answer:
Friction is necessary for any form of locomotion. Without friction, a vehicle cannot move. The banking angle for a road at a bend is calculated for optimum speed at which every vehicle can negotiate the bend without depending on friction to provide the necessary lateral centripetal force.

Question 2.
What about the kinetic friction between the road and the lyres?
Answer:
When a car takes a turn on a level road, the point of contact of the wheel with the surface is instantaneously stationary if there is no slipping. Hence, the lateral force on the car is the limiting force of static friction between the tyres and road. Lateral forces allow the car to turn. As long as the wheels are rolling, there is lateral force of static friction and longitudinal force of rolling friction. Longtitudinal forces, which act in the direction of motion of the car body (or in the exact opposite direction), control the acceleration or deceleration of the car and therefore the speed of the car. These are the wheel force, rolling friction, braking force and air drag. If the car skids, the friction force is kinetic friction; more importantly, the direction of the friction force then changes abruptly from lateral to that opposite the velocity of skidding and not towards the centre of the curve, so that the car cannot continue in its curved path.

Use your brain power (Textbook Page No. 12)

Question 1.
What is expected to happen if one travels fast over a speed breaker? Why?
Answer:
The maximum speed with which a car can travel over a road surface, which is in the form of a convex arc of radius r, is \(\sqrt{r g}\) where g is the acceleration due to gravity. For a speed breaker, r is very small (of the order of 1 m). Hence, one must slow down considerably while going over a speed breaker. Otherwise, the car will lose contact with the road and land with a thud.

Question 2.
How does the normal force on a concave suspension bridge change when a vehicle is travelling on it with a constant speed ?
Answer:
At the lowest point, N-mg provides the centripetal force. Therefore, N-mg = \(\frac{m v^{2}}{r}\), so that N = m(g + \(\frac{v^{2}}{r}\)).
Therefore, N increases with increasing v.

Use your brain power (Textbook Page No. 15)

Question 1.
For the point P in above, we had to extend OC to Q to meet the perpendicular PQ. What will happen to the expression for I if the point P lies on OC?
Answer:
There will be no change in the expression for the MI (I) about the parallel axis through O.

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 6 Plant Water Relation Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 6 Plant Water Relation

1. Multiple Choice Questions

Question 1.
In soil, water available for absorption by root is ……………..
(a) gravitational water
(b) capillary water
(c) hygroscopic water
(d) combined water
Answer:
(b) capillary water

Question 2.
The most widely accepted theory for ascent of sap is ……………..
(a) capillarity theory
(b) root pressure theory
(c) diffusion
(d) transpiration pull theory
Answer:
(d) transpiration pull theory

Question 3.
Water movement between the cells is due to ……………..
(a) T.E
(b) W.P
(c) D.P.D.
(d) incipient plasmolysis
Answer:
(c) D.P.D.

Question 4.
In guard cells, when sugar is converted into starch, the stomata pore ……………..
(a) closes almost completely
(b) opens partially
(c) opens fully
(d) remains unchanged
Answer:
(a) closes almost completely

Question 5.
Surface tension is due to ……………..
(a) diffusion
(b) osmosis
(c) gravitational force
(d) cohesion
Answer:
(d) cohesion

Question 6.
Which of the following type of solution has lower level of solutes than the solution?
(a) Isotonic
(b) Hypotonic
(c) Hypertonic
(d) Anisotonic
Answer:
(b) Hypotonie

Question 7.
During rainy season wooden doors warp and become difficult to open or to close because of ……………..
(a) plasmolysis
(b) imbibition
(c) osmosis
(d) diffusion
Answer:
(b) imbibition

Question 8.
Water absorption takes place through ……………..
(a) lateral root
(b) root cap
(c) root hair
(d) primary root
Answer:
(c) root hair

Question 9.
Due to low atmospheric pressure the rate of transpiration will ……………..
(a) increase
(b) decrease rapidly
(c) decrease slowly
(d) remain unaffected
Answer:
(a) increase

Question 10.
Osmosis is a property of ……………..
(a) solute
(b) solvent
(c) solution
(d) membrane
Answer:
(c) solution

2. Very short answer question

Question 1.
What is osmotic pressure?
Answer:
The pressure exerted due to osmosis is osmotic pressure.

Question 2.
Name the condition in which protoplasm of the plant cell shrinks.
Answer:
Plasmolysis

Question 3.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer:
When a pressure greater than the atmospheric pressure is applied to pure water or a solution then water potential of pure water or solution increases.

Question 4.
Which type of solution will bring about deplasmolysis ?
Answer:
Placing a plasmolysed cell in hypotonic solution will bring about deplasmolysis.

Question 5.
Which type of plants have negative root pressure?
Answer:
Plants showing excessive transpiration have negative root pressure.

Question 6.
In which conditions transpiration pull will be affected?
Answer:
Due to temperature fluctuations during day and night gas bubbles may be formed which affects transpiration pull.

Question 7.
Mention the shape of guard cells in Cyperus.
Answer:
Kidney shaped and dumbbell shaped guard cells are seen.

Question 8.
Why do diurnal changes occur in osmotic potential of guard cells?
Answer:
Enzyme activity of phosphorylase converts starch into sugar during daytime and sugar is converted to starch during night. This causes changes in osmotic potential of guard cells.

Question 9.
What is symplast pathway?
Answer:
When water is absorbed by root hair it passes across from one living cell to other living cell through the plasmodesmatal connections between them, then it is called symplast pathway across the root.

3. Answer the Following Questions

Question 1.
Describe mechanism of absorption of water.
Answer:

1. The absorption of water takes place by two modes, i.e. active absorption and passive absorption.
2. Passive absorption is the chief method of absorption (98%).
3. There is no expenditure of energy in passive absorption.
4. Transpiration pull is a driving force and water moves depending upon concentration gradient. Water is pulled upwards.
5. It occurs during daytime when there is active transpiration.
6. Active absorption occurs usually during night time as due to closure of stomata transpiration stops.
7. Water absorption is against D.ED. gradient, A.T.R energy is required which is available from respiration.
8. Active absorption may be osmotic or non- osmotic type.
9. For osmotic absorption root pressure has a role.

Question 2.
Discuss theories of water translocation.
Answer:

1. Translocation of water is transport of water along with dissolved minerals from roots to aerial parts.
2. The movement is against the gravity and described as ascent of sap.
3. The translocation occurs through lumen of water conducting tissue xylem mainly vessels and tracheids.
4. Different theories have been discussed for translocation mechanism like vital force theory (Root pressure), relay pump, physical force (capillary), etc.
5. Cohesion tension theory or transpiration pull theory is most widely accepted theory.

Question 3.
What is transpiration? Describe mechanism of opening and closing of stomata.
Answer:

1. The loss of water in the form of vapour is called transpiration.
2. Stomatal transpiration is a main type of transpiration where minute pores are concerned with it.
3. Stomata are bounded by two guard cells which in turn are surrounded by accessory cells.
4. Opening and closing of stomata is controlled by turgidity of guard cells.
5. When guard cells become turgid due to endosmosis their lateral thin and elastic wall bulges or stretch out.
6. The inner thick and inelastic wall is pulled apart, thus the stoma opens during daytime.
7. At night when guard cells become flaccid due to exosmosis the wall relaxes and stoma closes.
8. Endosmosis and exosmosis takes place due to changes in osmotic potential of guard cells.

Question 4.
What is transpiration? Explain role of transpiration.
Answer:
Transpiration : The loss of water from plant body in the form of vapour is called transpiration.

Role of transpiration:

1. Removal of excess of water
2. Helps in passive absorption of water and minerals
3. Helps in ascent of sap – transpiration pull
4. Maintains turgor of cells
5. Imparts cooling effect by reducing temperature 90% – 93% is stomatal transpiration and hence when stomata are open gaseous exchange takes place.

Question 5.
Explain root pressure theory and its limitations.
Answer:

1. Root pressure theory is proposed by J. Pristley.
2. For translocation of water, activity of living cells of root is responsible.
3. Absorption of water by root hair is a constant and continuous process and due to this a hydrostatic pressure is developed in cortical cells.
4. Owing to this hydrostatic pressure i.e. root pressure, water is forced into xylem and further conducted upwards.
5. Root pressure is an osmotic phenomenon.

Limitation of this theory:

1. Not applicable to tall plants above 20 metres.
2. Even in absence of root pressure ascent of sap is noticed.
3. In actively transpiring plants, root pressure is not developed.
4. In taller gymnosperms, root pressure is zero.
5. Xylem sap is under tension and shows negative hydrostatic pressure.

Question 6.
Explain capillarity theory of water translocation.
Answer:

1. Capillarity theory of water translocation is proposed by Bohem.
2. Capillarity is because of surface tension and cohesive forces and adhesive forces of water molecules.
3. Xylem vessels and tracheids are tubular elements having their lumen.
4. In these elements water column exists due to combined action of cohesive and adhesive forces of water and lignified wall.
5. As a result of this capillarity water is raised upwards.

Question 7.
Why is transpiration called ‘a necessary evil’?
Answer:

1. The loss of water in the form of water vapour is called transpiration.
2. About 90 – 93% of transpiration occurs through stomata, small apertures located in the epidermis of leaves.
3. For this process stomata must remain open and then only gaseous exchange by diffusion takes places.
4. Gaseous exchange is necessary for respiration and photosynthesis. If stomata remain closed then it will affect productivity of plant.
5. The process is necessary evil because water which is important for plant is lost in the process.
6. At the same time it helps in absorption of water and its translocation. Hence it cannot be avoided.
   So Curtis has rightly called it as necessary evil.

Question 8.
Explain movement of water in the root.
Answer:

1. Root hairs absorb water by imbibition then diffusion which is followed by osmosis.
2. As water is taken inside the root hair cell it becomes turgid i.e. increase in turgor pressure (T.E)
3. Root hair cell has less D.ED. but adjacent cortical cell has more D.PD.
4. The inner cortical cell has more osmotic potential so it will suck water from root hair cell.
5. Root hair cell becomes flaccid and ready to absorb soil water.
6. Water is passed on similarly in inner cortical cells.
7. Water moves rapidly through loose cortical cells up to endodermis and through passage cells in pericycle.
8. From pericycle due to hydrostatic pressure developed it is forced into protoxylem.

Question 9.
(i) Osmosis
Answer:
It is a special type of diffusion of solvent through a semipermeable membrane.

(ii) Diffusion
Answer:
It is the movement of ions/ atoms/molecules of a substance from the region of higher concentration to the region of their lower concentration.

(iii) Plasmolysis
Answer:
Exo-osmosis in a living cell when placed in hypertonic solution is called plasmolysis.

(iv) Imbibition
Answer:
It is swelling up of hydrophilic colloids due to adsorption of water.

(v) Guttation
Answer:
The loss of water in the form of liquid is called guttation.

(vi) Transpiration
Answer:
The loss of water from plant body in the form of vapour is called transpiration.

(vii) Ascent of sap
Answer:
The transport of water with dissolved minerals in it from root to other aerial parts of plant against the gravity is called ascent of sap.

(viii) Active absorption
Answer:
Water absorption by activity of root which is against the D.PD. gradient along with expenditure of A.T.E energy generated by respiration is the process of active absorption.

(ix) Diffusion Pressure Deficit (D.P.D.)
Answer:
The difference in the diffusion pressures of pure solvent and the solvent in a solution is called diffusion pressure deficit.

(x) Turgor pressure
Answer:
It is the pressure exerted by turgid cell sap on to the cell membrane and cell wall.

(xi) Water potential
Answer:
Chemical potential of water is called water potential.

(xii) Wall pressure
Answer:
Thick and rigid cell wall exerts a counter pressure to turgor pressure developed on the cell sap is called wall pressure that operates in opposite direction.

(xiii) Root pressure
Answer:
As absorption of water by root hair being a continuous process, a sort of hydrostatic pressure is developed in living cells of root, this is called root pressure.

Question 10.
Osmotic Pressure (O.P) and Turgor Pressure (T.P)
Answer:

Osmotic Pressure (O.R)	Turgor Pressure (T.P.)
1. The pressure exerted due to osmosis is called osmotic pressure.	1. The pressure exerted by turgid cell sap on cell membrane and cell wall, is called turgor pressure.
2. It is pressure caused by water when it moves by osmosis.	2. It is pressure caused by content of cell (cell sap).
3. It is generated by the osmotic flow of water through a semipermeable membrane.	3. It is maintained by osmosis.

Question 11.
How are the minerals absorbed by the plants ?
Answer:

1. Soil is the chief source of minerals for the plants.
2. Minerals get dissolved in the soil water.
3. Minerals are absorbed by the plants in the ionic form mainly through roots.
4. Absorption of minerals is independent of water.
5. Absorbed minerals are pulled upwards along with xylem sap.
6. Mineral ions can be remobilized in the plant body form older parts to young plants E.g. Ions of S, P and N.

4. Long answer questions

Question 1.
Describe structure of root hair.
Answer:

1. Water from soil is absorbed by plants with the help of root hairs.
2. Root hairs are present in zone of absorption.
3. Epidermal cells form unicellular extensions which are short lived (ephemeral) structures i.e. root hairs.
4. Root hairs are nothing but cytoplasmic extensions of epiblema cell.
5. Root hairs are long tube like structures of about 1 to 10 mm.
6. They are colourless, unbranched and very delicate structures.
7. A large central vacuole is surrounded by thin layer of cytoplasm, plasma membrane and outer cell wall.
8. The cell wall of root hair is thin and double layered with outer layer of pectin and inner layer of cellulose which is freely permeable.

Question 2.
Write on journey of water from soil to xylem in roots.
Answer:

1. Unicellular root hairs which are tubular extensions of epiblema cells absorb readily available capillary water from soil.
2. The three physical processes imbibition, diffusion and osmosis are concerned with absorption of water.
3. Water molecules get adsorbed on cell wall of root hair (imbibition).
4. They enter the root hair cell by diffusion through cell wall which is freely permeable.
5. By process of osmosis they enter through plasma membrane which is semipermeable.
6. The root hair cell becomes turgid and hence its turgor pressure increases and D.ED. value decreases.
7. The adjacent cell of cortex has more D.ED. value as its osmotic potential is more.
8. The cortical cell thus takes water from epidermal cell which is turgid. This process goes on due to gradient of suction pressure developed from cell to cell till thin walled passage cells of endodermis.
9. From endodermis it will enter pericycle and then due to hydrostatic pressure it is forced in protoxylem cell.
10. The pathway of water is by apoplast and symplast.
11. When water passes through cell wall and intercellular spaces of cortex it is apoplast pathway.
12. When water passes across living cells through their plasmodesmatal connections it is symplast pathway.

Question 3.
Explain cohesion theory of translocation of water.
Answer:

1. This is very widely accepted theory of ascent of sap proposed by Dixon and Joly.
2. It is based on principles of adhesion and cohesion of water molecules and transpiration by plants.
3. A strong force of attraction existing between water molecules is cohesion and the force of attraction between water molecules and lignified walls of xylem elements is adhesion.
4. Ascent of sap occurs through lumen of xylem elements.
5. Owing to cohesive and adhesive forces a continuous water column is maintained in xylem from root to aerial parts i.e. leaves.
6. Transpiration occurs through stomata and transpiration pull is developed in leaf vessels.
7. This tension or pull is transmitted downwards through vein to roots which triggers ascent of sap.
8. In transpiration, water is lost in vapour form and this increases D.PD. of mesophyll cells that are near guard cells.
9. Mesophyll cells absorb water from xylem in leaf and a gradient of D.PD. or suction pressure (S. E) is set.
10. Owing to this gradient from guard cell to xylem in leaf, a transpiration pull or tension is created in xylem.
11. Hence water column is pulled upward passively against gravity.

Question 4.
Write on mechanism of opening and closing of stomata.
Answer:

1. Transpiration takes place through stomata. Turgidity of guard cells controls opening and closing of stomata
2. Turgor pressure exerted on unevenly thickened wall of guard cell is responsible for the movement.
3. The outer thin wall which is elastic is stretched out which pulls inner thick inelastic wall and thus stomata open.
4. When guard cells are flaccid that results in closure of stomata.
5. According to starch-sugar in ter conversion theory enzyme phosphorylase converts starch to sugar during daytime.
6. Sugar being osmotically active, the O.E of guard cells is increased. The water is absorbed from subsidiary cells. Due to turgidity walls are stretched and stoma opens.
7. During night-time sugar is converted to starch and hence guard cells loose water and become flaccid. Hence there is closure of stomata.
8. According to proton transport theory, the movement is due to transport of H⁺ and K⁺ ions.
9. Subsidiary cells are reservoirs of K⁺ ions. Starch is converted to malic acid which dissociate into malate and proton (H⁺) during day.
10. Proton transported to subsidiary cells and K⁺ ions are taken from it. This forms potassium malate in guard cells.
11. Potassium malate increases osmotic potential and endo osmosis occurs hence turgidity of guard cells. → stomata opens,
12. The uptake of K⁺ and Cl^– ions is stopped by abscissic acid formed during night. This changes permeability. Guard cells become hypotonic and loose water as they become flaccid stomata close.

Question 5.
What is hydroponics? How is it useful in identifying the role of nutrients?
Answer:
(1) Growing plants in aqueous (soilless) medium is known as hydroponics. [Greek word hudor = water and ponos = work]

(2) It is technique of growing plants by supplying all necessary nutrients in the water supply given to plant.

(3) A nutrient medium is prepared by dissolving necessary salts of micronutrients and macronutricnts In desired quantity and roots of plants are suspended in this liquid with appropriate support.

(4) Hydroponics is of great use in studying the deficiency symptoms of different mineral nutrients.

(5) The plants uptake mineral nutrients in the form of dissolved ions with the help of root hairs from the surrounding medium or nutrient solution supplied.

(6) While preparing the required nutricnt medium particular nutrient can be totally avoided and then the effect of lack of that nutrient can be studied in variation of plant growth.

(7) Any visible change noticed from normal structure and function of the plant is the symptom or hunger sign considered.

(8) For e.g. Yellowing of leaf is observed due to loss of chlorophyll pigments or Chiorosis is noticed if Magnesium is lacking as it is a structural componen of chlorophyll pigment.

Question 6.
Explain the active absorption of minerals.
Answer:

1. Plants absorb minerals from the soil with their root system.
2. MInerals are absorbed from the soil In the form of charged particles, positively charged cations and negatively charged anions.
3. The absorption of minerals against the concentration gradient which requires expenditure of metabolic energy is called active absorption.
4. The ATP energy derived from resp’ration in root cells Is utilized for active absrption.
5. Ions get accumulated in the root hair against the concentration gradient.
6. These ions pass into cortical cells and finally reach xylem of roots.
7. Along with the water these minerals are carried to other parts of plant.

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 5 Origin and Evolution of Life Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 5 Origin and Evolution of Life

1. Multiple Choice Questions

Question 1.
Who proposed that the first form of life could have come from pre-existing non-living organic molecules?
(a) Alfred Wallace
(b) Oparin and Haldane
(c) Charles Darwin
(d) Louis Pasteur
Answer:
(b) Oparin and Haldane

Question 2.
The sequence of origin of life may be
(a) Organic materials – inorganic materials – Eobiont – colloidal aggregates – cell.
(b) Inorganic materials – organic materials – colloidal aggregates – Eobiont – cell.
(c) Organic materials – inorganic materials – colloidal aggregates – cell.
(d) Inorganic materials – organic materials – Eobiont – colloidal aggregates – cell.
Answer:
(b) Inorganic materials – organic materials- colloidal aggregates – Eobiont – cell.

Question 3.
In Hardy-Weinberg equation, the frequency of homozygous recessive individual is represented by-
(a) p²
(b) pq
(c) q²
(d) 2pq
Answer:
(c) q²

Question 4.
Select the analogous organs.
(a) Forelimbs of whale and bat
(b) Flippers of dolphins and penguin
(c) Thorn and tendrils of bougainvillea and Cucurbita
(d) Vertebrate hearts or brains
Answer:
(b) Flippers of dolphins and penguin

Question 5.
Archaeopteryx is known as missing link because it is a fossil and share characters of both
(a) Fishes and amphibians
(b) Annelida and Arthropoda
(c) Reptiles and birds
(d) Chordates and non-chordates
Answer:
(c) Reptiles and birds

Question 6.
Identify the wrong statement regarding evolution.
(a) Darwin’s variations are small and directional.
(b) Mutations are random and non- directional.
(c) Adaptive radiations leads to divergent evolution.
(d) Mutations are non-random and directional.
Answer:
(d) Mutations are non-random and directional

Question 7.
Gene frequency in a population remain constant due to ……………….
(a) Mutation
(b) Migration
(c) Random mating
(d) Non-random mating
Answer:
(c) Random mating

Question 8.
Which of the following characteristic is not : shown by the ape?
(a) Prognathous face
(b) Tail is present
(c) Chin is absent
(d) Forelimbs are longer than hind limbs
Answer:
(b) Tail is present

Question 9.
………………. can be considered as connecting link between ape and man.
(a) Australopithecus
(b) Homo habilis
(c) Homo erectus
(d) Neanderthal man
Answer:
(a) Australopithecus

Question 10.
The Cranial capacity of Neanderthal man was ……………….
(a) 600 cc
(b) 940 cc
(c) 1400 cc
(d) 1600 cc
Answer:
(c) 1400 cc

2. Very short answer questions

Question 1.
Define
(i) Gene pool
Answer:
The sum total of genes of all individuals of interbreeding population or Mendelian population is called gene pool.

(ii) Gene frequency
Answer:
The proportion of an allele in the gene pool as compared with other alleles at the same locus is termed as gene frequency.

(iii) Organic evolution
Answer:
Organic evolution can be defined as slow, gradual, continuous and irreversible changes through which the present-day complex forms of the life developed (or evolved) from their simple pre-existing forms.

(iv) Population
Answer:
All individuals of the same species form a group which is called a population.

(v) Speciation
Answer:
Formation of new species from the pre-existing single group of organisms is called speciation.

Question 2.
What is adaptive radiation?
Answer:
The process of evolution which results in transformation of original species to many different varieties is called adaptive radiation.

Question 3.
If the variation occurs in population by chance alone and not by natural selection and bring change in frequencies of an allele, what is it called?
Answer:
If the variation occurs in population by chance alone and not by natural selection to bring change in frequencies of an allele, it is called genetic drift.

Question 4.
State the Hardy-Weinberg equilibrium law.
Answer:
The Hardy-Weinberg equilibrium law states that at equilibrium point both the allelic frequency and genotypic frequency remain constant from generation to generation, in the diploid, sexually reproducing, large, free interbreeding population in which mating is random and there is absence of any other factors that change the allele frequency.

Question 5.
What is homologous organs?
Answer:
Homologous organs are those organs, which are structurally similar but perform different functions.

Question 6.
What is vestigial organ?
Answer:
Vestigial organs are imperfectly developed and non-functional organs which are in degenerate form, they may be functional in some related and other animals or in ancestor.

Question 7.
What is the scientific name of modern man?
Answer:
Homo sapiens sapiens is the scientific name of modern man.

Question 8.
What is coacervate?
Answer:
Coacervates are colloidal aggregations of hydrophobic proteins and lipids which grew in size by taking up material from surrounding aqueous medium.

Question 9.
Which period is known as ‘age of Reptilia’?
Answer:
Jurassic period from Mesozoic era is known as age of Reptilia.

Question 10.
Name the ancestor of human which is described as man with ape brain.
Answer:
Australopithecus, the ancestor of human which is described as man with ape brain.

Short Answer Questions

Question 1.
Genetic drift.
Answer:

1. Genetic drift is random, directionless fluctuation that takes place in allele frequency.
2. It occurs by pure chance, in small sized population.
3. Genetic drift becomes an evolutional factor as it can change the gene frequency.
4. Sewall wright has given this concept and hence it is also known as Sewall wright effect.
5. Due to genetic drift, some alleles of a population are lost or reduced by chance and some others may be increased.
6. Some time, a few individuals become isolated from the large population and they produce new population in new geographical area.
7. Genetic drift is also called founders’ effect because original drifted population becomes ‘founders’ in the new area.
   E.g. Non-adaptive character of huge horns in Antelope is fixed due to genetic drift.

Question 2.
Enlist the different factors that are responsible for changing gene frequency.
Answer:
Gene flow, genetic drift, gene mutations, chromosomal aberrations such as deletion, duplication, inversion and translocation, genetic recombinations, natural selection, isolation are some of the factors which are responsible for changing the gene frequency.

Question 3.
Draw a graph to show that natural selection leads to disruptive change.
Answer:

Question 4.
Significance of fossils
Answer:

1. Fossils are studied under palaeontology. They are used in reconstruction of phylogeny.
2. Fossil study helps in studying various forms and structures of extinct animals.
3. By understanding the structure of fossil, record of missing link between two groups of organisms can be deduced.
4. By studying fossils various body forms and their evolution can be understood. They also help to understand the habit and habitat.
5. Some fossils provide the evolutionary evidences such a connecting links.

Question 5.
Write the objections to Mutation theory of Hugo de Vries.
Answer:
Objections to Mutation Theory:

1. Hugo de Vries observed the large and discontinuous variation. But these were chromosomal aberrations. Only gene mutations usually bring about minor changes.
2. Rate by which mutations take place is very slow as compared to the requirement of evolution.
3. Chromosomal aberrations are very unstable.
4. The organisms with chromosomal aberration are usually sterile and thus chromosomal aberrations have little significance in evolution.

Question 6.
What is disruptive selection? Give example.
Answer:
Disruptive selection:

1. The natural selection that disrupts the mean characters of the population, is called disruptive selection.
2. Greater number of individuals acquire peripheral character value at both ends of the distribution curve. E.g. Finches with large size or small size, both will be selected.
3. Extreme phenotypes are selected in evolutionary process and intermediate forms are eliminated.
4. When distribution curve is plotted it shows two peaks for two extremes.
5. Disruptive selection is rare because, nature always tries to balance the characters.
6. It ensures the effect on the entire gene pool of a population, considering all mating types or systems.

Example of disruptive selection:
African seed cracker finches are types of seed-feeder birds which have different sizes of beak. The seeds available to them were of small and large sized. Large beak sized birds feeds on large seeds while small beak sized birds feed on small seeds.

Such large and small birds thus thrive well. However, intermediate beak sized birds are unable to feed on either type of seeds so they starve and their population was decreased gradually. Natural selection eliminated them and thus the population of finches appear disrupted.

4. Match the columns

Question 1.

Column I	Column II
(1) August Weismann	(a) Mutation theory
(2) Hugo de Vries	(b) Germplasm theory
(3) Charles Darwin	(c) Theory of acquired characters
(4) Lamarck	(d) Theory of natural selection

Answer:

Column I	Column II
(1) August Weismann	(b) Germplasm theory
(2) Hugo de Vries	(a) Mutation theory
(3) Charles Darwin	(d) Theory of natural selection
(4) Lamarck	(c) Theory of acquired characters

5. Long Answer Questions

Question 1.
Would you consider wings of butterfly and bat as homologous or analogous and why?
Answer:
Wings of butterfly are made up of chitin. They neither have bones, nor muscles in the wings. The bat’s wings are actually patagium. They have muscles and bones just as those seen in all vertebrate limb series. Therefore, these two examples cannot be homologous. However, both the animals use the wings for flight. This is an indication that their function is similar but structure is different, hence they are analogous organs.

Question 2.
What is adaptive radiation? Explain with suitable example.
Answer:

1. Adaptive radiation is the process of evolution which results in transformation of original species to many different varieties.
2. The well-known example of adaptive radiation is Darwin’s Finches. When Charles Darwin went on his voyage to Galapagos islands, he noticed finches which is a variety of small birds.
3. According to Darwin’s observations, the American main land species of finches was the original one which must have migrated to the different islands of Galapagos.
4. Since environmental conditions here were different, they adapted in various ways to the differing environmental conditions of these islands.
5. Original bird had a beak suited for eating seeds, but the changed feeding pattern has changed the shape of beaks too. Some birds also show altered beaks for insectivorous mode. Thus, this demonstrated adaptive radiation.
6. Adaptive radiation in Australian Marsupials is also well studied. In Australia, there are many marsupial mammals who evolved from common ancestor.
7. Adaptive radiation leads to divergent evolution.

Question 3.
By talking industrial melanism as one example, explain the concept of natural selection.
OR
Explain natural selection in action by quoting the example of industrial melanism.
Answer:
1. Industrial melanism is the best example of natural selection which was studied by Kettlewell. In U.K. there are two varieties of peppered moths, Biston betularia and Biston carbonaria.

2. Before industrialisation, in Great Britain, Biston betularia were more in number than Biston carbonaria. B. Betularia is greyish white while B.carbonaria is melanic form.

3. These nocturnal moths rest on tree trunk during day. White-winged moth can camouflage well with the lichen covered whitish barks of trees. They thus escaped the attention of the predatory birds. But at the same time melanic forms were visible due to white barks of the trees. Their number was thus reduced as they were preyed upon by birds.

4. Later there was an industrial revolution, which ultimately resulted in air pollution causing dark soot to settle on the barks of the trees. Lichens too were destroyed and the melanic forms were now at advantage. Melanic forms could camouflage with black tree trunks and their number increased. White-winged moth become clearly seen in changed colours of the trees and thus they were easily caught by predatory birds. This caused decrease in their number.

5. Natural selection thus acted in changed environmental conditions and helped in the establishment of a phenotypic traits. The changed traits were more adaptive and hence were selected. Natural selection encourages those genes or traits that assure highest degree of adaptive efficiency between population and its environment.

Question 4.
Describe the Urey and Miller’s experiment.
Answer:

1. Urey and Miller performed an experiment to prove Oparin’s theory of chemical evolution.

2. They selected a spark discharge apparatus that consisted of closed system of glass having tungsten electrodes, flask for water boiling, a side tube connected to a vacuum pump, a cooling jacket and U-shaped trap.

3. The entire apparatus was first evacuated and made sterile and pre-biotic atmosphere was created in it.

4. The flask was filled with some water and mixture of methane, ammonia and hydrogen in the ratio of 1 : 2 : 2 were slowly passed through the stopcock, without allowing air.

5. Heat was supplied to the flask at very low temperature causing water to boil. The flask simulated the ocean present on primitive earth. Process of evaporation and precipitation was simulated by using heating mantle and condenser respectively.

6. Water vapours along with other gases were circulated continuously through continuous electric sparks. These sparks were given to the mixture for several days causing the gases to interact. This too simulated lightning.

7. Mixture of CH₄, NH₃ and H₂ gases passed through a condenser and was condensed to liquid.

8. The liquefied mixture was collected in the U-shaped trap, present at the bottom of the apparatus. It was found that variety of simple organic compounds (urea, amino acids, lactic acid and sugars) were formed in the apparatus.

This experiment provides the evidence in support to the fact that simple molecules present in the earth’s early atmosphere combined to form the organic building blocks of life.

Question 5.
What is Isolation? Describe the different types of reproductive isolations.
Answer:
1. Isolation means separation of the population of a particular species into smaller units. The organisms belonging to these subunits are prevented from interbreeding due to some barrier. These barriers are called isolating mechanisms.

2. They prevent the genetic exchange and gene flow.

3. Due to isolating mechanisms in nature the divergence among organisms takes place gradually leading to speciation. The isolating mechanisms are of two types namely, geographical isolation and reproductive isolation.

I. Geographical Isolation : The barrier in the form of physical distance or geographical barrier is called geographical isolation. The original population gets divided into two or more groups by geographical barriers such as river, ocean, mountain, glacier, etc. Organisms cannot cross the barriers on their own and hence interbreeding is prevented between isolated groups.

The separated groups experience different environmental factors and they acquire new traits by mutations. The separated populations develop distinct gene pool and they do not interbreed. Each subgroup then evolves differently which results into formation of new species. E.g. Darwin’s Finches, African elephant, Loxodonta and Indian elephant, Elephas.

II. Reproductive Isolation : Two populations may be occupying the same area, they may not be separated by geographical barrier, but then also they are reproductively isolated. Such reproductive isolation occurs due to change in genetic material, gene pool and structure of genital organs. Such differences prevent interbreeding between population. Such isolation later leads to speciation.

III. Different types of reproductive isolations : Reproductive isolation is of two types, viz. pre-zygotic and post-zygotic isolating mechanisms.

1. Pre-zygotic or pre-mating isolating mechanisms do not allow individuals to mate with each other at all.
2. By various mechanisms the two groups remain isolated.
3. In post-zygotic or post-mating isolating mechanisms, the two individuals can mate but the result of mating is not favourable.
4. Thus the populations remain isolated without the actual genetic exchange.

Question 6.
What is Genetic variations? Explain the different factors responsible for genetic variations.
Answer:
Genetic variations : The change in gene and gene frequencies is known as genetic variation. Genetic variations are caused by following factors:
(i) Mutations : Sudden permanent heritable change is called mutation. Mutation can occur in the gene, in the chromosome structure and in chromosome number. Mutation that occurs within the single gene is called point mutation or gene mutation. This leads to the change in the phenotype of the organism, causing variations.

(ii) Genetic recombination : In sexually reproducing organisms, during gamete formation, exchange of genetic material occurs between non-sister chromatids of homologous chromosomes. This is called crossing over. It produces new genetic combinations which result in variation. Fertilization between opposite mating gametes leads to various recombinations resulting into the phenotypic variations. These result in change in the frequencies of alleles.

(iii) Gene flow : Gene flow is movement of genes into or out of a population. Gene movement may be in the form of migration of organism, or gametes (dispersal of pollens) or segments of DNA (transformation). Gene flow also alters gene frequency causing evolutionary changes.

(iv) Genetic drift : Any random fluctuation (alteration) in allele frequency, occurring in the natural population by pure chance, is called genetic drift. For example, when the size of a population is severely reduced due to natural disasters like earthquakes, floods, fires, etc. elimination of particular alleles from a population becomes possible. Smaller populations have greater chances for genetic drift. It results in the change in the gene frequency. Genetic drift is also an important factor for evolutionary change.

(v) Chromosomal aberrations : The structural, morphological change in chromosome due to rearrangement of genes is called chromosomal aberrations. Due to changes in the gene arrangement or gene sequence variations are caused.

6. Complete the chart

Era	Dominating group of animals
1. Coenozoic	————–
2. ————-	Reptiles
3. Palaeozoic	————-
4. ————	Lower Invertebrates

Answer:

Era	Dominating group of animals
1. Coenozoic	Mammals
2. Mesozoic	Reptiles
3. Palaeozoic	Insects, Fishes, Amphibians
4. Proterozoic	Lower Invertebrates

 

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 14 Biomolecules Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 14 Biomolecules

Question 1.
What are biomolecules? Give examples of biomolecules.
Answer:
Biomolecules: The lifeless, complex organic molecules which combine in a specific manner to produce life or control biological reactions are called biomolecules.

Examples: Carbohydrates, lipids (fats and oils), nucleic acids, enzymes.

Question 2.
What is the importance of biomolecules?
Answer:
Biomolecules are organic molecules which combine in a particular fashion to give complex substances which help to sustain life and produce identical daughter cells and play an important role in the actions of an organism.

• Carbohydrates are the major constituents of food and source of energy.
• Proteins help in proper functioning of living beings. They are important constituents of skin, hair, muscles. Enzymes which catalyse chemical reactions that take place in cells are proteins.
• Lipids (fats and oils) function as the storehouses of energy.
• Nucleic acids, the ribonucleic acid (RNA), and deoxyribonucleic acid (DNA) are responsible for genetic characteristics and synthesis of proteins.

Question 3.
What are carbohydrates?
OR
Define the term : Carbohydrates.
Answer:
Carbohydrates : Carbohydrates are optically active polyhydroxy aldehydes or polyhydroxy ketones, or the compounds which on hydrolysis produce polyhydroxy aldehydes or polyhydroxy ketones.

Examples : Glucose, sucrose, fructose.

Question 4.
What is monosaccharide?
Answer:
The basic unit of all carbohydrates which is a simple carbohydrate and cannot be hydrolysed further is known as monosaccharide. The monosaccharide is crystalline and soluble in water. E.g. Glucose, fructose, ribose.

Question 5.
Mention the names of monosaccharides or simple carbohydrates.
Answer:
Monosaccharides are (1) glucose (2) fructose (3) ribose.

Question 6.
State the basic unit of all carbohydrates.
Answer:
The basic unit of all carbohydrates which is a simple carbohydrate and cannot be hydrolysed further is known as monosaccharide.

Question 7.
How are carbohydrates classified?
OR
Classification of carbohydrates with examples.
Answer:
Carbohydrates are classified as monosaccharides oligosaccharides and polysaccharides.
(1) Monosaccharides : These carbohydrates cannot be further hydrolysed into smaller units. They are basic units of all carbohydrates, and are called monosaccharides.

Examples : Glucose, fructose, ribose

(2) Oligosaccharides : An oligosaccharide is a carbohydrate (sugar) which on hydrolysis gives two to ten monosaccharide units.
Depending on the number of monosaccharides produced on hydrolysis, oligosaccharides are further classified as :

Oligosaccharide is homogeneous. In this, each molecule of oligosaccharide contains the same number of monosaccharide units joined together in the same order as every other molecule of the same oligosaccharide.

(3) Polysaccharides : These are carbohydrates which on hydrolysis give a large number of monosaccharides.

Polysaccharides are tasteless, amorphous, insoluble in water. They are long chain, naturally αcurring polymers of carbohydrates.

Example : Cellulose, starch, glycogen.

Question 8.
Classify the following carbohydrates into Monosaccharide, Disaccharide, Oligosaccharide, Polysaccharide.
(1) Glucose
(2) Starch
(3) Sucrose
(4) Maltose
(5) Galactose
(6) Lactose
(7) Ribose.
Answer:

Carbohydrates	Class
(1) Glucose	Monosaccharide
(2) Starch	Polysaccharide
(3) Sucrose	Disaccharide
(4) Maltose	Disaccharide
(5) Galactose	Monosaccharide
(6) Lactose	Disaccharide
(7) Ribose	Monosaccharide

Question 9.
Classify the following carbohydrates.
(1) Cellulose,
(2) Maltose,
(3) Raffinose,
(4) Fructose.
Answer:

Carbohydrates	Class
(1)     Cellulose (2)     Maltose (3)     Raffinose (4)     Fructose	Polysaccharide Disaccharide Trisaccharide Monosaccharide

Question 10.
Classify the following into monosaccharides, oligosaccharides and polysaccharides.
(1) Starch
(2) Glucose
(3) Stachyose
(4) Maltose
(5) Raffinose
(6) Cellulose
(7) Sucrose
(8) Lactose.
Answer:

Monosaccharides	Glucose
Oligosaccharides	Stachyose, maltose, raffinose, sucrose, lactose
Polysaccharides	Starch, cellulose

Question 11.
Classify the following into monosaccharides and disaccharides.
Ribose, maltose, galactose, fructose and lactose (~2 mark each)
Answer:

Monosaccharides	Ribose, galactose, fructose
Disaccharides	Maltose, lactose

Question 12.
Give the preparation of glucose from sucrose or cane sugar.
OR
Describe the laboratory method of preparation of glucose.
Answer:
Preparation of glucose from sucrose (cane sugar) : Laboratory method.

Glucose is prepared in the laboratory by hydrolysis of sucrose by boiling it with dilute hydrαhloric acid or dilute sulphuric acid for about two hours. On hydrolysis, sucrose gives one molecule of glucose and one molecule of fructose.

Alcohol is added during cooling to separate glucose and fructose since, glucose is almost insoluble in alcohol, hence it crystallizes out first. Fructose remains in the solution as it is more soluble than glucose.

Crystals of glucose are separated out by filtration and purified by recrystallization.

Question 13.
Give the preparation of glucose from starch.
OR
How is glucose prepared on commercial scale?
Answer:
Commercially, on a large scale, glucose is prepared by hydrolysis of starch with dilute sulphuric acid. Starchy material is mixed with water and dilute sulphuric acid and heated at 393 K under 2 to 3-atm pressure. Starch is hydrolysed to give glucose.

Question 14.
Explain the structure of glucose.
Answer:
Molecular formula of glucose is C₆H₁₂O₆.

Glucose has an aldohexose structure. In other words, glucose molecule contains one aldehydic, that is, formyl group and the remaining five carbons carry one hydroxyl group (-OH) each. The six carbons in glucose form one straight chain.

Question 15.
Describe the action of following reagents on glucose :
(1) HI
(2) Hydroxyl amine (NH₂OH)
(3) Hydrogen cyanide
(4) Bromine water
(5) dil. Nitric acid
(6) Acetic anhydride.
Answer:
(1) Action of HI : Glucose on prolonged heating with HI gives n-hexane, indicates that all the six carbon atoms are linked in straight chain.

(2) Action of hydroxyl amine : Glucose reacts with hydroxyl amine in an aqueous solution to form glucose oxime. This indicates the presence of CHO group in glucose.

(3) Action of hydrogen cyanide : Glucose reacts with hydrogen cyanide to form glucose cyanohydrin.

(4) Action of bromine water : Glucose on oxidation with mild oxidising agent like bromine water gives gluconic acid, which shows that the carbonyl group in glucose is aldehyde group.

(5) Action of dll. nitric acid : Glucose on oxidation with dilute nitric acid forms dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic group (-CH₂OH) in glucose.

(6) Action of acetic anhydride : When glucose is heated with acetic anhydride in the presence of catalyst pyridine, glucose penta acetate is formed. It indicates that glucose is a stable compound and contains five hydroxyl groups.

Question 16.
Write Fischer projection formulae for
(1) Glucose
(2) Gluconic acid
(3) Saccharic acid.
Answer:
Fischer projection formulae of glucose, gluconic acid and saccharic acid :

Question 17.
Explain D and L configuration in sugars.
Answer:
The simplest carbohydrates glyceraldehyde is chosen as the standard, to assign D and L configuration to monosaccharides. Glyceraldehyde contains one asymmetric carbon atom and exist in two enantiomeric forms

The dextro entantiomer is represented as (+) glyceraldehyde and it is referred as D-configuration i.e., D-glyceraldehyde. The laevo enantiomer of glyceraldehyde is represented as ( -) glyceraldehyde and it corelated as L-configuration i.e., L-glyceraldehyde.

In Fischer projection formula, a monosaccharide is assigned D-configuration if the (- OH) hydroxyl group at the last chiral carbon and lies towards right hand side. On the other hand it is assigned L-configuration if the – OH group on the last chiral carbon atom and lies on the left hand side. In monosaccharides, the most oxidised carbon (i.e., -CHO) is at the top.

Examples :

Question 18.
Write Fischer projection formulae for (a) L-( + )-erythrose (b) L-( +) ribulose.
Answer:

Question 19.
Is the following sugar, D-sugar or L-sugar?

Answer:
The compound is L-sugar.
The compound is L-sugar.

Question 20.
Assign D/L configuration to the following monosaccharides.


Answer:

Question 21.
Explain ring structure of glucose.
Answer:

Glucose has two cyclic structures (II and III) which are in equilibrium with each other through the open chain structure (I) in aqueous solution.

The ring structure of glucose is formed by reaction between the formyl ( – CHO) group and the alcoholic (- OH) group at C – 5. Thus, the ring structure is called a hemiacetal. The two hemiacetal structures (II and III) differ only in the configuration of C – I (Fig.), the additional chiral centre resulting from ring closure. The two ring structures are called α- and β- anomers of glucose and C-l is called the anomeric carbon. The ring of the cyclic structure of glucose contains five carbons and one oxygen. Thus, it is a six membered ring. It is called pyranose structure, in analogy with the six membered heterαyclic compound pyran (IV). Hence glucose is also called glucopyranose.

Question 22.
Write the structures of α-D-( + )-glucopyranose and β-D-( +) glucopyranose.
Answer:

Question 23.
Explain Haworth formula of glycopyranose.
Answer:

In the Haworth formula the pyranose ring is considered to be in a perpendicular plane with respect to the plane of paper. The carbons and oxygen in the ring are in the places as they appear in figure. The lower side of the ring is called α-side and the upper side is the β-side. The α-anomer has its anomeric hydroxyl (- OH) group (at C-l) on the α-side, whereas the β-anomer has its anomeric hydroxyl (- OH) group (at C-l) on the β-side. The groups which appear on right side in the Fischer projection formula appear on α-side in the Haworth formula, and the groups which appear on left side in the fischer projection formula appear on a β-side in the Haworth formula.

Question 24.
Explain the structure of fructose.
Answer:
Fructose has molecular formula C₆H₁₂O₆. It contains ketonic functional group at carbon number 2 and six carbon atoms in straight chain. It belongs to D-series and is a laevo rotatory compound. It is written as D-( – )-fructose. Being an α-hydroxy keto compound fructose is a reducing sugar.

Question 25.
Draw mirror images of glucose and fructose.
Answer:
(1) Glucose

(2) Fructose

Question 26.
Write the two cyclic structures of α-D-( – )-fructofuranose and β-D-( – )-fructofuranose exist in equilibrium with open chain structure.
Answer:

Question 27.
Write the Haworth projection formulae for α -D-( -) – Fructofuranose and β – D – ( -) – Fructo- furanose.
Answer:

Question 28.
Explain the structure of sucrose.
Answer:
Sucrose is a hexasaccharide and has molecular formula C₁₂H₂₂O₁₁. The structure of shcrose contains glycosidic linkage between C – 1 of α-glucose and C – 2 of β-fructose. Since aldehyde and ketone groups of both monosaccharide units are involved in the formation of glycosidic bond, sucrose is a nonreducing sugar.

Sucrose is dextrorotatory, on hydrolysis with dilute acid or an enzyme invertase gives equimolar mixture of dextrorotatory glucose and laevorotatory fructose.

The solution is laevorotatory because laevo rotation of fructose (- 92.4°) is more than dextrorotation of glucose ( + 52.50), hence the sign of rotation is changed from (+) to (-) after hydrolysis, the product is called invert sugar.

Question 29.
Explain the structure of maltose.
Answer:
Maltose is another disaccharide obtained by partial hydrolysis of starch or made of two units of D-glucose. In maltose, C-l of one α-D-glucose is linked to C-4 of another α-D-glucose molecule by glycosidic linkage. The glucose ring which uses its hydroxyl group at C-1 is α – 1 → 4 glycosidic linkage. It is a reducing sugar because a free aldehyde group can be produced at C₁ of second glucose molecule. Maltose on hydrolysis with dilute acids gives glucose.

Question 30.
Draw a neat diagram for Haworth formula of maltose.

Question 31.
Explain the structure of lactose.
Answer:

Lactose (C₁₂H₂₂O₁₁) is a disaccharide. It is found in milk, therefore, it is also known as milk sugar. It is formed from two monosaccharide units, namely D – galactose and D – glucose. The glycosidic linkage is formed between C-l of β-D-galactose and C -4 of glucose. Therefore the linkage in lactose is called β – 1,4 – glycosidic linkage. The hemiacetal group at C-l of the glucose unit is not involved in glycosidic linkage but is free. Hence lactose is a reducing sugar. The above figure shows Haworth formula of lactose.

Question 32.
What are the hydrolysis products of (1) lactose (2) sucrose?
Answer:
(1) Lactose on hydrolysis in presence of an acid or enzyme lactase gives one molecule each of glucose and galactose

(2) Sucrose on hydrolysis in the presence of dii. acid or the enzyme invertase gives one molecule each of glucose and fructose.

Question 33.
Explain the structure of starch.
Answer:
Starch is found in cereal grains, roots, tubers, potatoes, etc. It is a polymer of α-D-glucose and consists of two components, amylose and amylopectin.

Amylose is water soluble component forms blue coloured complex with iodine. It constitutes about 20 % of starch. Amylose contains 200 to 1000 α-D-glucose units linked together by glycosidic linkage between C-l of one unit and C-4 of another unit. i.e. α-1, 4 glycosidic linkages.

Amylopectin is insoluble in water and constitutes about 80 % starch which forms blue-violet coloured complex with iodine. It is a branched chain polymer. In amylopectin, α-D-glucose molecules are linked together by glycosidic linkage between C₁ – of one unit and C-4 of another unit to form long chain and branching αcurs by glycosidic linkage between C-l and C6 glycosidic linkage.

Question 34.
What are polysaccharides?
Answer:
A large number of same or different monosaccharides are joined together by glycosidic linkages are called polysaccharides. They have general formula (C₆H₁₀O₅)_n.

Question 35.
Explain the structure of cellulose.
Answer:

Cellulose mainly αcurs in plants. Cell wall of plant cells is made up of cellulose. It is a long chain polymer. In cellulose, β-D-glucose units are linked by glycosidic linkage between C₁-of one unit of glucose and C₄ of another glucose unit. Thus cellulose contains 1 → 4β glycosidic linkages like those in cellobiose.

Question 36.
Explain the structure of glycogen.
Answer:
The glucose is stored in animal body in the form of glycogen. It is also known as animal starch because its structure is similar to amylopectin. Glycogen is highly branched. Whenever the body is required glucose, enzymes breaks the glycogen to glucose.

Question 37.
How is glycogen different from starch?
Answer:
Starch is the main storage molecules of plants whereas glycogen is the main storage molecule of animals. Starch is found in cereals, roots, tubers, etc. Glycogen is present in liver, muscles and brain.

Question 38.
What do you understand by the term glycosidic linkage?
Answer:
The linkage between two monosaccharide units through oxygen atom is called glycosidic linkage.

Question 39.
What is the basic structural difference between starch and cellulose?
Answer:
Starch is a polymer of a-glucose and consists of two components-amylose and amylopectin. In amylose α-D-D-( + )-glucose units held by C,-C4 glycosidic linkage and in amylopectin, α-D-glucose units held by C1-C4 glycosidic linkage whereas branching αcurs by C₁-C₆ glycosidic linkage. [Refer Question 35 (i) (ii) Fig.] Cellulose is a straight chain polysaccharide composed only of β-D-glucose units held by C₁-C₄ glycosidic linkage. (Refer Question 37 Fig.)

Question 40.
Define the term : Protein OR What are proteins?
Answer:
Chemically proteins are polyamides which are high molecular weight polymers of the monomer units i.e. α-amino acids. OR It can also be defined as Proteins are the biopolymers of a large number of a-amino acids and they are naturally occurring polymeric nitrogenous organic compounds containing 16% nitrogen and peptide linkages (-CO-NH-).

Question 41.
Write the common sources of protein.
Answer:
Common sources of proteins are milk, pulses, peanuts, eggs, fishes, cheese, cereals, etc. They are also the principal materials of muscle, nerves, tendons, skin, blood, enzymes, many hormones and antibiotics.

Question 42.
What are the products of hydrolysis of proteins?
Answer:
On hydrolysis, proteins give a mixture of α-anlino acids.

The α-carbon in α-amino acids ohtained by hydrolysis of proteins has ‘L’ configuration.

Question 43.
What are the a-amino acids?
Answer:
α-Amino acids are carboxylic acids having an amino (- NH₂) group bonded to the α-carbon, i.e. the carbon next to the carboxyl (- COOH) group.

α-amino acids are derivatives of carboxylic acids, obtained by replacing – H atom by amino group. They are bifunctional compounds containing acidic and basic – NH₂ groups.
Example : (where R is an alkyl group or aryl group).

The amino acids are colourless, crystalline, water soluble, high melting solids. These acids in their aqueous solutions behave like salts due to presence of both acidic, and basic. (- NH₂) groups in the same molecule.

Such a doubly charged ion is known as zwitter ion. Example :

Question 44.
What are the final products of hydrolysis of proteins?
Answer:
Proteins on hydrolysis with dilute solution of acids, alkalies or enzymes give a mixture of large number of a-amino acids as final products.

For example :

Question 45.
Write the classification of amino acids, giving examples.
Answer:
The amino acids are of three types : acidic, basic and neutral. The symbol ‘R’ in the structure of a-amino acids represents side chain and may contain additional functional groups.

(1) Acidic amino acids : If ‘R’ contains a carboxyl (- COOH) group the amino acid is acidic amino acid, i.e. If carboxyl groups are more in number than amino groups, then amino acids are acidic in nature.

Examples : Glutamic acid HOOC-CH₂-CH₂-; Aspartic acid HOO-CH₂–

(2) Basic amino acids : If ‘R’ contains an amino (1°, 2°, or 3°) group, it is called basic amino acid i.e. If amino groups are more in number than carboxyl groups then amino acids are basic in nature.

Examples : Arginine

(3) Neutral amino acids : The other amino acids having neutral or no functional group in ‘R’ are called neutral amino acids, i.e. The amino acids having equal number of amino and carboxyl groups are neutral amino acids.

Examples : Alanine CH₃-; Valine (CH₃)₂-CH

Question 46.
What are essential and non-essential amino acids? Give two examples of each.
Answer:
The amino acids, which cannot be synthesised in the body and are supplied through diet are called essential amino acids. Examples : Lysine H₂N-(CH₂)₄-; Valine (CH₃)₂CH- The amino acids which are synthesized in the body are called non-essential amino acids.

Examples : Glutamic acid HOO-CH₂-CH₂-; Serine HO-CH₂–

Question 47.
What is meant by Zwitter ion?
Answer:
An a-amino acid molecule contains both acidic carboxyl ( – COOH) group as well as basic amino (- NH₂) group. Proton transfer from acidic group to basic group of amino acid forms a salt, which is a dipolar ion called a zwitterion.

Question 48.
Draw zwitter ion of alanine and other two forms.
Answer:

Question 49.
What is a peptide bond (peptide linkage)?
OR
Define peptide bond.
Answer:
Proteins are the polymers of a-amino acids and they are connected to each other. The bond that connects a-amino acids to each other is called peptide bond (peptide linkage, – CONH -).

Question 50.
How is peptide linkage (dipeptide linkage) formed in proteins? How is tripeptide formed?
Answer:
Peptide linkage is formed by condensation of acidic group of one molecule of a-amino acid and basic -NH₂ group of other molecule of α-amino acid with elimination of water.

When one more molecule of amino acid combines with dipeptide, it forms tripeptide.

Thus, it forms tetra, penta and finally a polypeptide chain i.e. proteins. Hence, proteins are basically polypeptides.

Question 51.
Write the structures of all possible dipeptides which can be obtained from glycine and alanine.
Answer:
(1) Dipeptide from glycine :
Carboxylic group of glycine reacting with amino group another molecule of glycine to form dipeptide

(2) Dipeptide from alanine :
Carboxylic goup of alanine reaction with amino goup of another molecule of alamine to form dipeptide

(3) Dipeptide from glycine and alanine :
Carboxylic group of glycine reacting with amino group another molecule of alanine to form dipeptide

Question 52.
How are proteins classified on the basis of molecular shapes?
Answer:
On the basis of their molecular shapes proteins are classified as :
(1) Fibrous proteins : The proteins in which the polypeptide chains lie parallel (side by side) to form fibre-like structure, are called fibrous proteins. The polypeptide chains held together by hydrogen bonds. These proteins are insoluble in water.

The fibrous proteins are tough and insoluble in water, and dilute acids or bases.

Example : myαin (in muscles), keratin (in hair, nails, skin), fibroin (in silk), collagen (in tendons), etc.

(2) Globular proteins : The proteins have spherical shape. This shape results from coiling around of the polypeptide chain of protein, and have intramolecular hydrogen bonding are called globular proteins.

They are soluble in water and dilute acids or bases.

Example : Haemoglobin (in blood), albumin (in eggs), insulin (in pancreas), etc.

Question 53.
Distinguish between globular and fibrous proteins.
Answer:

Globular proteins	Fibrous proteins
(1) The chains of polypeptides of protein coil around to give a spherical shape. (2) Globular proteins are soluble in water. (3) They are sensitive to small changes of temperature and pH. (4) They possess biological activity.	(1) The proteins in which the polypeptide chains lie parallel to form fibre like structure. (2) Fibrous proteins are insoluble in water. (3) They are stable to moderate changes of temperature and pH. (4) They do not possess biological activity.

Question 54.
Draw a neat labelled diagram for the secondary structure of protein.
Answer:
Secondary structure of proteins : The three-dimensional arrangement of lαalized regions of a long polypeptide chain is called the secondary structure of protein. Hydrogen bonding between N-H proton of one amide linkage and C = O oxygen of another gives rise to the secondary structure. There are two different types of secondary structures i.e. α-helix and β-pleated sheet.

α-Helix : In a-helix structure, a polypeptide chain gets coiled by twisting into a right handed or clαkwise spiral known as a-helixn. The characteristic features of α-helical structure of protein are :

(1) Each turn of the helix has 3.6 amino acids.
(2) A C = O group of one amino acid is hydrogen bonded to N – H group of the fourth amino acid along the chain.
(3) Hydrogen bonds are parallel to the axis of helix while R groups extend outward from the helix core.
Myosin in muscle and a-keratin in hair are proteins with almost entire a-helical secondary structure.

β-Pleated sheet : In secondary structure, when two or more polypeptide chains (strands) line up side-by-side is called β-pleated sheets. The β-picate sheet structure of protein consists of extended strands of polypeptide chains held together by intermolecular hydrogen bonding. The characteristics of β-pleated sheet structure are :

• The C = O and N – H bonds lie in the planes of the sheet.
• Hydrogen bonding occurs between the N – H and C = O groups of nearby amino acid residues in the neighbouring chains.
• The R groups are oriented above and below the plane of the sheet.

The β-pleated sheet arrangement is favoured by amino acids with small R groups.

Question 55.
What is denaturation of proteins? How is denaturation brought about?
OR
What is the effect of denaturation on the structure of proteins?
Answer:
The prαess by which the molecular shape of protein changes without breaking the amide / peptide bonds that form the primary structure is called denaturation. OR Proteins gets easily precipitated. It is an irreversible change and the prαess is called denaturation of proteins.

Denaturation uncoils the protein and destroys the shape and thus loses their characteristic biological activity. Denaturation is brought about by heating the protein with alcohol, concentrated inorganic acids or by salts of heavy metals. During denaturation secondary and tertiary and quternary structures are destroyed but primary structure remains intact.

Example : Boiling of egg to coagulate egg white, conversion of milk to curd.

Question 56.
Define : Enzymes
Answer:
All biological reactions are catalysed by bio-catalyst in living organisms called enzymes.

Question 57.
What are enzymes? Explain with suitable example.
Answer:
All biological or bio-catalysts which catalyse the reactions in living organisms are called enzymes. Chemically all enzymes are proteins. They are required in very small quantities as they are catalyst also they reduce the activation energy for a particular reaction.

Example : Enzyme maltase converts maltose to glucose.

Question 58.
Explain the catalytic action of enzymes.

Answer:
Mechanism of enzyme catalysis : Action of an enzyme on a substrate is known as lock-and-key mechanism.

Accordingly, the enzyme has active site on its surface. A substrate molecule can attach to this active site only if it has the right size and shape. Once in the active site, the substrate is held in the correct orientation, enzymes provide functional group which will attack the substrate and forms the products of reaction. The products leave the active site and the enzyme is ready to act as catalyst again.

Question 59.
Give examples of industrial application of enzyme catalysis.
Answer:

• Glucose Isomerase (enzyme) is used in conversion of glucose to sweet-tasting fructose.
• New antibiotics are manufactured using penicillin acylase (enzyme).
• Laundry detergentts are manufactured using proteases (enzyme).
• Esters used in cosmetics are manufactured using genetically engineered enzyme.

Question 60.
Draw a neat diagram for enzyme catalysis.
Answer:

Question 61.
State the main functions of enzymes.
Answer:
Enzymes are biological catalyst and they are highly specific in nature. The two main functions are as follows :
(1) They lower the requirement of activation energy.
(2) They speed up the rate of reaction.
E.g. Enzyme maltase catalyses maltose to glucose.
\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \stackrel{\text { Maltase }}{\longrightarrow} 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

Question 62.
What are nucleic acids?
Answer:
Nucleic acids are unbranched polymers of repeating monomers i.e. nucleotides. In other words, nucleic acids have a polynucleotide structure which in turn consists of a base, a pentose sugar and phosphate moiety.

Nucleic acids are biomolecules which are found in the nuclei of all living cells in the form of nucleoproteins or chromosomes.

(Nucleoproteins = Proteins + Nucleic acid)
(prosthetic group)

Question 63.
State the types of nucleic acids.
Answer:
The types of nucleic acids are : Ribonucleic acid (RNA) and deoxy ribonucleic acid (DNA). DNA molecules contain several million nucleotides while RNA molecules contain a few thousand nucleotides.

Question 64.
Explain chemical composition of nucleic acids.
Answer:
Nucleic acids have a polynucleotide structure. Nucleic acids (RNA and DNA) consists of three components :
(1) monosaccharide (sugar)
(2) nitrogen containing base and
(3) phosphate group.

(1) Monosaccharides : Nucleotides of both RNA consist of five membered monosaccharide ring (furanose), called as simply sugar component.

In RNA, the sugar component of nucleotide units is D-ribose and in DNA 2-deoxy-D-ribose.
2 – deoxy means no – OH group at C₂ position.

(2) Nitrogen containing base : Total five nitrogen – containing bases are present in nucleic acids. Three bases with one ring (cytosine, uracil and thymine) are derived from the parent compound pyrimidine. Two bases with two rings (adenine and guanine) are derived from the parent compound purine. Each base in designated by a one-letter symbol. Uracil (U) αcurs only in RNA while thymine (T) ocurs only in DNA.

(3) Phosphate group : The sugar units are joined to phosphate through C₃ and C₅ hydroxyl groups.

Question 65.
What is meant by nucleosides?
OR
Write the structure of nucleoside. Give examples.
Answer:

A nucleoside contains two basic components of nucleic acids i.e. a pentose sugar and a nitrogenous base.

A nucleoside is formed when 1 -position of a pyrimidine (cytosine, thymine or uracil) or 9-position of guanine or adenine base is attached to C- l of sugar by β-linkage.

Examples: Formation of nucleoside:

Question 66.
What is meant by nucleotide?
OR
Write the structure of nucleotide. Give example.
Answer:

A nucleotide contains all three basic components of nucleic acids i.e., a pentose sugar, a phosphoric acid and a nitrogenous base. These are obtained by esterification of \(\mathrm{C}_{5}^{1}-\mathrm{OH}\) group of the pentose sugar by phosphoric acid. Nucleotides are joined together through phosphate ester linkage. Thus, nucleotides are monophosphates of nucleosides. Abridged names of some nucleotides are AMP, dAMP, UMP, dTMP and so on. Here, the first capital letter is derived from the corresponding base. MP stands for monophosphate. Small letter ‘d’ in the beginning indicates deoxyribose in the nucleotide.

Example :

 

Question 67.
Write the structure of nucleic acids.

Answer:
Nucleic acids, both DNA and RNA, are polymers of nucleotides, formed by joining the 3′ – OH group of one nucleotide with 5′ – phosphate of another nucleotide. Two ends of polynucleotide chain are distinct from each other. One end having free phosphate group of 5′ position is called 5′ end. The other end is 3′ end and has free OH – group at 3′ position.

Question 68.
Draw a schematic representation of polynucleotide structure of nucleic acids.
Answer:

Question 69.
Explain double helix.
OR
State the salient features of the Watson and Crick mode of DNA.
Answer:

The Salient features are :

1. DNA is made of two polynucleotide strands that wind into a right-handed double helix.
2. The two strands run in opposite directions: one from the Y end to the 3’ end, while the other from the 3’ end to the Y end.
3. Pcrpcndicular to the axis of the helix, the sugar – phosphate backbone lies on the outside of the helix and the bases lic on the inside.
4. The hydrogen bonding between the hases of the two DNA strands stabilizes the double helix. This gives rise to a ladder-like structure of DNA double helix.
5. Adenine always forms two hydrogen bonds with thymine and guanine forms three hydrogen bonds with cytosinc. Thus A – T arid C – G arc complementary hase pairs and the Two strands of the double helix arc complementary to each other.

Question 70.
Give scientific reasons :
1. In the preparation of glucose from sucrose, ethyl alcohol is added at the time of cooling.
Answer:
Hydrolysis of sucrose with dilute hydrαhloric acid gives glucose along with fructose.

Ethyl alcohol is added at the time of cooling in the preparation of glucose, to separate glucose from fructose. Glucose being insoluble in alcohol, crystallizes out first, while fructose being more soluble in alcohol, remains in the solution.

Question 71.
Answer in one sentence :

(1) How is glucose stored in the animal body?
Answer:
Glucose is stored in the form of glycogen in the animal body.

(2) Write other term used for carbohydrates.
Answer:
Carbohydrates are often termed as saccharides or sugars.

(3) How many moles of acetic anhydride will be required to form glucose penta acetate from 1 mole of glucose?
Answer:
10 moles of acetic anhydride.

(4) What are reducing sugars?
Answer:
Reducing sugars : Carbohydrates which reduce Fehling solution to red ppt of Cu₂0 or Tollen’s reagent to shining metallic silver are called reducing sugars. All monosaccharides and oligosaccharides except sucrose are reducing sugars.

(5) What are non-reducing sugars?
Answer:
Non-reducing sugars : Carbohydrates which do not reduce Fehling solution and Tollen’s reagent are called non-reducing sugars. E.g. sucrose.

(6) Give an example each of reducing and non-reducing sugars.
Answer:
Reducing sugars : Maltose or lactose
Non-reducing sugars : Sucrose.

(7) Name the linkage which joins two monosaccharide units through oxygen atom.
Answer:
The linkage which joins two monosaccharide units through oxygen atom is called glycosidic linkage.

(8) Name the sugar present in DNA.
Answer:
The sugar present in DNA is deoxyribose.

(9) A nucleotide from DNA containing thymine is hydrolysed. What are the products formed?
Answer:
When nucleotide from DNA containing thymine is hydrolysed, 2-deoxy-D-ribose, thymine and phosphoric acid is obtained.

(10) How is zwitterion formed?
Answer:
In aqueous solution, the carboxyl group loses a proton while the amino group accepts it, as a result, a dipolar or zwitter ion is formed.

(11) Name the amino acids which are synthesized in the body.
Answer:
The amino acids which are synthesized in the body are called non-essential amino acids. Examples : Glutamic acid, serine.

(12) Name the four bases present in DNA which of these is not present in RNA.
Answer:
Purines-adenine (A) and guanine (G); Pyrimidines-thymine (T) and cytosine (C), these four bases are present in DNA. Out of these, thymine (T) is not present in RNA.

(13) What are different types of RNA which are found in the cell?
Answer:
There are three different types of RNA found in the cell. (1) The messenger RNA which carries the message to the ribosome (2) Ribosomal RNA where synthesis of protein takes place (3) The transport RNA.

(14) State the functions of RNA and DNA.
Answer:
RNA and DNA are responsible for generic characteristics : DNA preserves the information and uses it by producing duplicate identical DNA molecules. RNA carries messages and transports them.

Multiple Choice Questions

Question 72.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. Which of the following is not sugar?
(a) Sucrose
(b) Starch
(c) Fructose
(d) Glucose
Answer:
(b) Starch

2. Which of the following is the example of disaccharide?
(a) Glucose
(b) Raffinose
(c) Cellulose
(d) Sucrose
Answer:
(d) Sucrose

3. Fructose is
(a) aldopentose
(b) aldohexose
(c) ketopentose
(d) ketohexose
Answer:
(d) ketohexose

4. Oxidation product of glucose with bromine water is
(a) sorbitol
(b) gluconic acid
(c) glutamic acid
(d) saccharic acid
Answer:
(b) gluconic acid

5. The general formula of carbohydrates is
(a) C(H₂O)
(b) C_x(H₂O)_y
(c) C_x(H₂O)
(d) C_x(H₂O)_x
Answer:
(b) C_x(H₂O)_y

6. Monosaccharides containing aldehyde group are called
(a) aldoses
(b) ketoses
(c) polysaccharides
(d) disaccharides
Answer:
(a) aldoses

7. Which of the following sugars can be used to prepare glucose on a large scale?
(a) Cellulose
(b) Cane sugar
(c) Galactose
(d) Starch
Answer:
(d) Starch

8. Which of the following carbohydrates cannot undergo hydrolysis?
(a) Glucose
(b) Sucrose
(c) Cellulose
(d) Maltose
Answer:
(a) Glucose

9. Glucose differs from fructose in
(a) the functional group
(b) the number of chiral carbon atoms
(c) the number of carbon atoms
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

10. The example of aldopentose is
(a) arabinose
(b) glucose
(c) fructose
(d) sucrose
Answer:
(a) arabinose

11. Dextrose, grape sugar and blood sugar αcurs in
(a) fructose
(b) glucose
(c) sucrose
(d) starch
Answer:
(b) glucose

12. The example of ketopentose is
(a) galactose
(b) ribose
(c) raffinose
(d) maltose
Answer:
(b) ribose

13. Cane sugar on hydrolysis gives
(a) glucose and maltose
(b) glucose and lactose
(c) glucose and fructose
(d) only glucose
Answer:
(c) glucose and fructose

14. On commerical scale, glucose is prepared from
(a) starch
(b) potato pulp
(c) sucrose
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

15. The number of monosaccharide units formed on hydrolysis of glucose are
(a) zero
(b) one
(c) two
(d) three
Answer:
(a) zero

16. Which of the following is NOT TRUE about glucose?
(a) It is monosaccharide
(b) It is a polyhydroxy aldehyde
(c) It is polyhydroxy ketone
(d) It contains six carbon atoms
Answer:
(c) It is polyhydroxy ketone

17. Final hydrolysis product of simple protein is
(a) carboxylic acid
(b) α-amino acid
(c) mineral acid
(d) acetic acid
Answer:
(b) α-amino acid

18. Haemoglobin is the example of-
(a) simple protein
(b) derived protein
(c) fibrous protein
(d) conjugated protein
Answer:
(d) conjugated protein

19. Protein are also called
(a) polysaccharides
(b) polypeptides
(c) polyglycerides
(d) polyster
Answer:
(b) polypeptides

20. The simplest amino acid is
(a) glycine
(b) oxalic acid
(c) adipic acid
(d) caprolactam
Answer:
(a) glycine

21. Amino acids usually exist in the form of zwitter ion which consist of
(a) the basic group-NH₂ and the acidic group -COOH
(b) the acidic group -N⁺H₃ and the basic group COO^–
(c) the acidic group -COO⁺ and the acidic group NH^3-
(d) acidic or basic group
Answer:
(b) the acidic group -N⁺H₃ and the basic group COO-

22. The water insoluble protein is
(a) casein of milk
(b) albumin
(c) serum albumin
(d) keratin of hair
Answer:
(d) keratin of hair

23. The main structural feature of a protein molecule is the presence of
(a) an ester linkage
(b) an ether linkage
(c) a peptide linkage
(d) all of these
Answer:
(c) a peptide linkage

24. Milk sugar is
(a) sucrose
(b) lactose
(c) maltose
(d) glucose
Answer:
(b) lactose

25. The carbohydrates used for silvering of mirror is
(a) fructose
(b) starch
(c) glucose
(d) cellulose
Answer:
(c) glucose

26. Which one of the following is NOT produced by human body?
(a) DNA
(b) Hormones
(c) Enzymes
(d) Vitamins
Answer:
(c) Enzymes

27. A biological catalyst is essentially
(a) an amino acid
(b) an enzyme
(c) a nitrogen molecule
(d) a carbohydrate
Answer:
(d) a carbohydrate

28. Which one of the following is not a constituent of RNA?
(a) Ribose
(b) Uracil
(c) Thymine
(d) Phosphate
Answer:
(b) Uracil

29. DNA is a polymer of units of
(a) sugars
(b) ribose
(c) amino acids
(d) nucleotides
Answer:
(c) amino acids

30. Which one of the following molecules will form zwitter ion?
(a) CH₃COOH
(b) CH₃CH₂NH₂
(c) CCl₃NO₂
(d) NH₂CH₂COOH
Answer:
(d) NH₂CH₂COOH

31. In metabolic prαess the maximum energy is given by
(a) carbohydrates
(b) proteins
(c) vitamins
(d) fats
Answer:
(d) fats

32. DNA has a structure of helix was reported by
(a) Herman Fischer
(b) Fedrick Sauger
(c) Andreas Marggraf
(d) Watson and Crick
Answer:
(d) Watson and Crick

33. The secondary structure of a protein is determined by
(a) co-ordinate bond
(b) covalent bond
(c) ionic bond
(d) hydrogen bond
Answer:
(d) hydrogen bond

34. In maltose, glycosidic linkage is present between the two glucose units at positions
(a) 1, 2
(b) 1, 1
(c) 1, 3
(d) 1, 4
Answer:
(d) 1, 4

35. Which of the following amino acids is basic in nature?
(a) Valine
(b) Tyrosine
(c) Arginine
(d) Luecine
Answer:
(c) Arginine

36. Sucrose molecules consists of
(a) a glucofuranose and a fructopyranose
(b) a glucofuranose and a fructofuranose
(c) a glucopyranose and a fructopyranose
(d) a glucopyranose and a’ fructofuranose
Answer:
(d) a glucopyranose and a’ fructofuranose

37. Which one of the following statements is not correct about DNA molecule?
(a) It has double helix structure
(b) It serves as hereditary material
(c) The two DNA strands are exactly similar
(d) Its replication is called semi-conservative mode of replication
Answer:
(c) The two DNA strands are exactly similar

38. Glycine on heating forms


Answer:
(a)

39. Acidic amino acid is
(a) Glutamine
(b) Glutamic acid
(c) Tyrosine
(d) Lysine
Answer:
(b) Glutamic acid

40. Basic amino acid is
(a) Lysine
(b) Glycine
(c) Cystine
(d) Alanine
Answer:
(a) Lysine

41. Precipitation of protein is referred to as
(a) destruction of proteins
(b) separation of proteins
(c) denaturation of proteins
(d) fragmentation of proteins
Answer:
(c) denaturation of proteins

42. An amino acid containing sulphur is
(a) serine
(b) cysteine
(c) valine
(d) asparagine
Answer:
(b) cysteine

43. Rhamnose is a
(a) carbohydrate
(b) protein
(c) lipid
(d) vitamin
Answer:
(a) carbohydrate

44. Lactose on hydrolysis gives
(a) glucose + glucose
(b) glucose + fructose
(c) glucose + galactose
(d) fructose + galactose
Answer:
(c) glucose + galactose

45. Raffinose on hydrolysis gives
(a) glucose + glucose + galactose
(b) glucose + fructose + galactose
(c) glucose + galactose + galactose
(d) fructose + galactose + galactose
Answer:
(b) glucose + fructose + galactose

46. Naturally αcurring glucose is
(a) dextro rotatory
(b) laevo rotatory
(c) racemic mixture
(d) all of these
Answer:
(a) dextro rotatory

47. Amylopectin is
(a) soluble in water and constitutes about 80% of starch
(b) insoluble in water and constitutes about 80% of starch
(c) Soluble in alcohol and constitutes about 60% of starch
(d) in soluble in alcohol and constitutes about 60% of starch
Answer:
(b) insoluble in water and constitutes about 80% of starch

48. Insulin contains
(a) 51 amino acids
(b) 151 amino acids
(c) 15 amino acids
(d) 115 amino acids
Answer:
(a) 51 amino acids

49. Pyranose structure of glucose is
(a) an open chain structure of glucose
(b) a structure of reduction product of glucose
(c) a cyclic six-membered structure of glucose
(d) a four-membered cyclic form of glucose
Answer:
(c) a cyclic six-membered structure of glucose

50. The number of – OH groups present in ribulose is
(a) 3
(b) 4
(c) 6
(d) 5
Answer:
(b) 4

51. Peptide linkage is

Answer:
(d)

52. Stachyose is an example of
(a) monosaccharides
(b) disaccharides
(c) trisaccharides
(d) tetrasaccharides
Answer:
(d) tetrasaccharides

53. How many moles of (CH₃CO)₂O will be required to form glucose pentaacetate form 2 moles of glucose?
(a) 2
(b) 5
(c) 10
(d) 2.5
Answer:
(c) 10

54. Which of the following NOT present in DNA?
(a) Adenine
(b) Guanine
(c) Thymine
(d) Uracil
Answer:
(d) Uracil

55. Maltose is a
(a) polysaccharide
(b) disaccharide
(c) trisaccharide
(d) monosaccharide
Answer:
(b) disaccharide

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 14 Biomolecules Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 14 Biomolecules

1. Select the most correct choice.

Question i.
CH₂OH-CO-(CHOH)₄-CH₂OH is an example of
a. Aldohexose
b. Aldoheptose
c. Ketotetrose
d. Ketoheptose
Answer:
(d) Ketoheptose

Question ii.
Open chain formula of glucose does not contain
a. Formyl group
b. Anomeric hydroxyl group
c. Primary hydroxyl group
d. Secondary hydroxyl group
Answer:
(b) Anomeric hydroxyl group

Question iii.
Which of the following does not apply to CH₂NH₂ – COOH
a. Neutral amino acid
b. L – amino acid
c. Exists as zwitterion
d. Natural amino acid
Answer:
(d) Natural amino acid

Question iv.
Tryptophan is called essential amino acid because
a. It contains an aromatic nucleus.
b. It is present in all the human proteins.
c. It cannot be synthesized by the human body.
d. It is an essential constituent of enzymes.
Answer:
(c) It cannot be synthesised by human body.

Question v.
A disulfide link gives rise to the following structure of protein.
a. Primary
b. Secondary
c. Tertiary
d. Quaternary
Answer:
(c) Tertiary

Question vi.
RNA has
a. A – U base pairing
b. P – S – P – S backbone
c. double helix
d. G – C base pairing
Answer:
(a) A – U base pairing

2. Give scientific reasons :

Question i.
The disaccharide sucrose gives negative Tollens test while the disaccharide maltose gives positive Tollens test.
Answer:
(1) In disaccharide sucrose, the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a nonreducing sugar. As there is no free aldehyde group, it does not reduce Tollen’s reagent to metallic silver. Hence, sucrose gives negative Tollen’s test.

(2) While the disaccharide maltose is a reducing sugar because a free aldehyde group can be produced at C₁ of second sugar molecule. It is a reducing sugar. It reduces Tollen’s reagent to shining silver mirror. Hence, Maltose gives positive Tollen’s test.

Question ii.
On complete hydrolysis DNA gives equimolar quantities of adenine and thymine.
Answer:
On complete hydrolysis DNA yields 2-deoxy-D-ribose, adenine, thymine, guanine, cystosine and phosphoric acid. Since adenine always forms two hydrogen bonds with thymine, the hydrolysis of DNA gives equimolar quantities of adenine and thymine.

Question iii.
α – Amino acids have high melting points compared to the corresponding amines or carboxylic acids of comparable molecular mass.
Answer:

α-Amino acids have high melting points compared to the corresponding amines or carboxylic acids of comparable molecular mass due to the presence of both acidic (carboxylic group) and basic (amino group) groups in the same molecule. In aqueous solution, proton transfer from acidic group to amino (basic) group of amino acid forms a salt, which is a dipolar ion called zwitter ion.

Question iv.
Hydrolysis of sucrose is called inversion.
Answer:

Sucrose is dextro rotatory. On hydrolysis it gives equimolar mixture of D – ( + ) glucose and D – ( -) fructose. Since the laevorotation of fructose (- 92.4°) is more than dextrorotation of glucose ( + 52.7°), the hydrolysis product has net laevorotation. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro ( + ) to laevo (-) and the product is called as invert sugar and so the hydrolysis of sucrose is called inversion.

Question v.
On boiling, egg albumin becomes opaque white.
Answer:
Upon boiling the egg, denaturation αcurs. During denaturation, secondary and tertiary structures are destroyed, but primary structure remains intact. Egg contains soluble globular proteins, which forms insoluble fibrous proteins (opque) on boiling egg.

3. Answer the following

Question i.
Some of the following statements apply to DNA only, some to RNA only and some to both. Lable them accordingly.
a. The polynucleotide is double stranded. ( …………… )
b. The polynucleotide contains uracil. ( …………… )
c. The polynucleotide contains D-ribose ( …………… ).
d. The polynucleotide contains Guanine ( …………… ).
Answer:
(1) The polynucleotide is double stranded. (DNA)
(2) The polynucleotide contains uracil. (RNA)
(3) The polynucleotide contain D-ribose (RNA)
(4) Thc polynucleotide contains Guanine (DNA, RNA)

Question ii.
Write the sequence of the complementary strand for the following segments of a DNA molecule.
a. 5′ – CGTTTAAG – 3′
b. 5′ – CCGGTTAATACGGC – 3′
Answer:
(1) DNA molecule : 5′ – CGTTTAAG – 3′
The complementary strand runs in opposite direction from the 3′ end to the 5′ end. It has the base sequence decided by complementary base pairs A – T and C – G.

(2) DNA molecule : 5′ – CCGGTTAATACGGC – 3′
The complementary strand runs in opposite direction from the 3′ end to the 5′ end. It has the base sequence decided by complementary base pairs A – T and C – G.

Question iii.
Write the names and schematic representations of all the possible dipeptides formed from alanine, glycine and tyrosine.
Answer:
(1) Dipeptide formed from alanine and glycine

(2) Dipeptide formed from alanine and tyrosine

(3) Dipeptide formed from glycine and tyrosine.

Question iv.
Give two pieces of evidence for the presence of the formyl group in glucose.
Answer:
(1) Glucose reacts with hydroxyl amine in an aqueous solution to form glucose oxime. This indicates the presence of – CHO (formyl group) in glucose.

(2) Glucose on oxidation with mild oxidising agent like bromine water gives gluconic acid which shows carbonyl group in glucose is aldehyde (formyl group) group.

4. Draw a neat diagram for the following:

Question i.
Haworth formula of glucopyranose
Answer:

Question ii.
Zwitter ion
Answer:

Question iii.
Haworth formula of maltose
Answer:

Question iv.
Secondary structure of the protein

Answer:
The structure of proteins can be studied at four different levels i.e. primary, secondary, tertiary and quaternary levels. Each level is more complex than the previous one.
(1) Primary structure of proteins :
(a) Representation by structural formula

(b) Representation with amino acid symbols

Primary structure of proteins is the sequence of constituent a-amino acid residues linked by peptide bonds. Any change in the sequence of amino acid residue creates different protein molecule. Primary structure of proteins is represented by writing the three letter symbols of amino acid residues as per their sequence in the concerned protein. The symbols are separated by dashes. According to the convention, the N-terminal amino acid residue as written at the left end and the C-terminal amino acid residue at the right end.

(2) Secondary structure of proteins : The three-dimensional arrangement of lαalized regions of a long polypeptide chain is called the secondary structure of protein. Hydrogen bonding between N-H proton of one amide linkage and C = O oxygen of another gives rise to the secondary structure. There are two different types of secondary structures i.e. α-helix and β-pleated sheet.

α-Helix : In a-helix structure, a polypeptide chain gets coiled by twisting into a right handed or clαkwise spiral known as a-helixn. The characteristic features of α-helical structure of protein are :

(1) Each turn of the helix has 3.6 amino acids.
(2) A C = O group of one amino acid is hydrogen bonded to N – H group of the fourth amino acid along the chain.
(3) Hydrogen bonds are parallel to the axis of helix while R groups extend outward from the helix core.
Myosin in muscle and a-keratin in hair are proteins with almost entire a-helical secondary structure.

β-Pleated sheet : In secondary structure, when two or more polypeptide chains (strands) line up side-by-side is called β-pleated sheets. The β-picate sheet structure of protein consists of extended strands of polypeptide chains held together by intermolecular hydrogen bonding. The characteristics of β-pleated sheet structure are :

• The C = O and N – H bonds lie in the planes of the sheet.
• Hydrogen bonding occurs between the N – H and C = O groups of nearby amino acid residues in the neighbouring chains.
• The R groups are oriented above and below the plane of the sheet.

The β-pleated sheet arrangement is favoured by amino acids with small R groups.

(3) Tertiary structure of proteins :

The three-dimensional shape acquired by the entire polypeptide chain of a protein is called its tertiary structure. The structure is stabilized and has attractive interaction with the aqueous environment of the cell due to the folding of the chain in a particular manner. Tertiary structure gives rise to two major molecular shapes i.e. globular and fibrous proteins. The main forces which stabilize a particular tertiary structure include hydrogen bonding, dipole-dipole attraction (due to polar bonds in the side chains), electrostatic attraction (due to the ionic groups like -COO^–, \(\mathrm{NH}_{3}^{+}\) in the side chain) and also London dispersion forces. Finally, disulfide bonds formed by oxidation of nearby – SH groups (in cysteine residues) are the covalent bonds which stabilize the tertiary structure.

(4) Quaternary structure of proteins The two or more polypeptide chains with folded tertiary structures forms complex protein. The spatial arrangements of these polypeptide chains with respect to each other is known as quaternary structure. Each individual polypeptide chain is called a subunit of the overall protein. For example: Haemoglobin consists of four subunits called haeme held together by intermolecular forces in a compact three dimensional shape. Haemoglobin can do its function of oxygen transport only when all the four subunits are together.

Question v.
AMP
Answer:

Question vi.
dAMP
Answer:

Question vii.
One purine base from nucleic acid
Answer:

Question viii.
Enzyme catalysis
Answer:

Activity :

• Draw the structure of a segment of DNA comprising at least ten nucleotides on a big chart paper.
• Make a model of DNA double stranded structure as group activity.

12th Chemistry Digest Chapter 14 Biomolecules Intext Questions and Answers

Try ….. this (Textbook Page No 298)

Question 1.
Observe the following structural formulae carefully and answer the questions.

(1) How many OH groups are present in glucose, fructose and ribose respectively?
(2) Which other functional groups are present in these three compounds?
Answer:
(1) Glucose contains five hydroxyl (- OH) groups.
Fructose contains five hydroxyl ( – OH) groups.
Ribose contains four hydroxyl ( – OH) groups.

(2) Glucose contains aldehyde ( – CHO) as other functional group.
Fructose contains ketonic group as other functional group.
Ribose contains aldehyde ( – CHO) as other functional group.

Use your brain power! (Textbook Page No 299)

Question 1.
Give IUPAC names to the following monosaccharides.

Answer:
(1) Aldotriose
(2) Aldopentose
(3) Ketoheptose

Problem 14.1 : (Textbook Page No 300)

Question 1.
An alcoholic compound was found to have molecular mass of 90 u. It was acetylated. Molecular mass of the acetyl derivative was found to be 174 u. How many alcoholic (- OH) groups must be present in the original compound?
Solution :
In acetylation reaction H atom of an (- OH) group is replaced by an acetyl group (- COH₃).

This results in an increase in molecular mass by [(12 + 16 + 12 + 3 x 1) – 1] that has, 42 u. In the given alcohol, increase in molecular mass = 174 u – 90 u = 84 u
∴ Number of – OH groups \(=\frac{84 \mathrm{u}}{42 \mathrm{u}}=2\)

Use your brain power! (Textbook Page No 301)

(1) Write structural formula of glucose showing all the bonds in the molecule.
(2) Number all the carbons in the molecules giving number 1 to the ( – CHO) carbon.
(3) Mark the chiral carbons in the molecule with asterisk (*).
(4) How many chiral carbons are present in glucose?
Answer:
Refer structural formula of glucose for (1) (2) and (3).

(4) There are 4 chiral carbon atoms present in glucose.

Use your brain power! (Textbook Page No 306)

Question 1.
(1) Is galactose an aldohexose or a ketohexose?
(2) Which carbon in galactose has different configuration compared to glucose?
(3) Draw Haworth formulae of α-D-galactose and β-D-galactose.
(4) Which disaccharides among sucrose, maltose and lactose is/are expected to give positive Fehling test?
(5) What are the expected products of hydrolysis of lactose?
Answer:

1. Galactose is an aldohexose.
2. Fourth carbon in galactose has different configuration compared to glucose.
3. 
4. Maltose and lactose are expected to give positive Fehling solution test.
5. The expected products of hydrolysis of lactose are D – ( +) glucose and D – ( +) galactose.

Can you think? (Textbook Page No 307)

Question 1.
When you chew plain bread, chapati or bhaakari for long time, it tastes sweet. What could be the reason?
Answer:
When chapati, bread or bhakari are chewed for long time the pulp mixes with saliva and carbohydrate component in them diseminates and gives the sweet taste.

Use your brain power! (Textbook Page No 309)

Question 1.
Tryptophan and histidine have the structures (I) and (II) respectively. Classify them into neutral? acidic/basic &amino acids and justify your answer. (Hint: Consider învolvement of lone pair in resonance).

Answer:

In tryptophan, nitrogen atom present in cyclic structure cannot donate pair of electrons as it is stabilized by resonance. The other amino group and carboxylic group present in the side chain neutralize each other. Tryptophan has equal number of amino and carboxylic groups. Hence, tryptophan is a neutral amino acid.

In histidine, amino groups are more in number than carboxyl groups therefore histidine ¡s basic in nature.

Can you think? (Textbook Page No 309)

Question 1.
Compare the molecular masses of the following compounds and explain the observed melting points.

Answer:
Above compounds have same molecular masses but they have different melting points, a-amino acids have higher melting points compared to the corresponding amines or carboxylic acids of comparable masses. This property is due to the presence of both carboxylic group (acidic) and amino group (basic) in the molecule. In aqueous solution, protons transfer from acidic group to amino (basic) group of amino acid forms a salt, which is a dipolar ion called – Zwitter ion.

Use your brain power! (Textbook Page No 310)

Question 1.
(1) Write the structural formula of dipeptide formed by combination of carboxyl group of alanine and amino group of glycine.
(2) Name the resulting dipeptide.
(3) Is this dipeptide same as glycyalanine or its structural isomer?
Answer:

(2) ala-glycine. OR ala-gly
(3) It is a structural isomer.

Question 54.
Write the names and schematic representations of all the possible dipeptides formed from alanine, glycine and tyrosine.

Problem 14.3 : (Textbook Page No 311)

Question 1.
Chymotrypsin is a digestive enzyme that hydrolyzes those amide bonds for which the carbonyl group comes from phenylalanine, tyrosine or tryptophan. Write the symbols of the amino acids and peptides smaller than pentapeptide formed by hydrolysis of the following hexapeptide with chymotrypsin. Gly-Tyr-Gly-Ala-Phe-Val
Solution :
In the given hexapeptide hydroylsis by chymotripsin can take place at two points, namely, Phe and Tyr. The carbonyl group of these residues is towards the right side, that is, toward the C-terminal. Therefore the hydrolysis products in required range will be :

Problem 14.4 : (Textbook Page No 311)

Question 1.
Write down the structures of amino acids constituting the following peptide.

Solution :
The given peptide has two amide bonds linking three amino acids. The structures of these amino acids are obtained by adding one H₂O molecule across the amide bond as follows :

Use your brain power! (Textbook Page No 313)

A protein chain has the following amino acid residues. Show and label the interactions that can be present in various pairs from these giving rise to tertiary level structure of protein.

Answer:
Tertiary level structure from amino residues.

Can you tell? (Textbook Page No 313)

Question 1.
What is the physical change observed when (a) egg is boiled, (b) milk gets curdled on adding lemon juice?
Answer:
(a) When egg is boiled, coagulation of eggwhite (insoluble fibrous proteins) takes place. This is a common example of denaturation.
(b) When lemon juice is added to milk, it gets curdled due to the formation of lactic acid. This is another example of denaturation.

Can you tell? (Textbook Page No 315)

Question 1.
What is the single term that answers all the following questions?
(1) What decides whether you are blue eyed or brown eyed?
(2) Why does wheat grain germinate to produce wheat plant and not rice plant?
(3) Which acid molecules are present in nuclei of living cells?
Answer:
(1) Nucleic acid (DNA)
(2) Nucleic acid (DNA)
(3) Nucleic acid (DNA + RNA)

Use your brain power! (Textbook Page No 317)

Question 1.
Draw structural formulae of nucleosides formed from the following sugars and bases.
(1) D – ribose and guanine
(2) D – 2 – deoxyribose and thymine
Answer:
(1) D-ribose and guanine

(2) D – 2 – deoxyribose and thymine

Problem 14.5 (Textbook Page No 318)

Queston 1.
Draw a schematic representaion of trinucicotide segment ACT of a DNA molecule.
Solution :
In DNA molecule sugar is deoxyribose. The base ‘A’ in the given segment is at 5 end while the base T at the 3’ end. I-fence the schematic representation of the given segment of DNA is

Problem 14.6 : (Textbook Page No 320)

Question 1.
Write the sequence of the complementary strand of the following portion of a DNA molecule : 5 -ACGTAC-3
Solution :
The complementary strand runs in opposite direction from the 3′ end to the 5′ end. It has the base sequence decided by complementary base pairs A – T and C – G.

Problem 14.2 : (Text Page No 303)

Question 1.
Assign D/L configuration to the following monosaccharides.

Solution :
D/L configuration is assigned to Fischer projection formula of monosaccharide on the basis of the lowest chiral carbon.

Threose has two chiral carbons C-2 and C-3. The given Fischer projection formula of threose has OH groups at the lowest C -3 chiral carbon on the right side.
∴ It is D-threose.

Ribose has three chiral carbons C – 2, C – 3 and C -4.
The given Fischer projection formula of ribose has – OH group at the lowest C -4 chiral carbon on the left side.
∴ It is L-ribose

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 4 Chemical Thermodynamics Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

1. Select the most apropriate option.

Question 1.
The correct thermodynamic conditions for the spontaneous reaction at all temperatures are
(a) ΔH < 0 and ΔS > 0
(b) ΔH > 0 and ΔS < 0
(c) ΔH < 0 and ΔS < 0
(d) ΔH < 0 and ΔS = 0
Answer:
(a) ΔH < 0 and ΔS > 0

Question ii.
A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 bar from an initial volume of 2.5 L to a final volume of 4.5 L. The change in internal energy, ΔU of the gas will be
(a) -500 J
(b) +500J
(c) -1013 J
(d) +1013 J
Answer:
(a) -500 J

Question iii.
In which of the following, entropy of the system decreases ?
(a) Crystallisation of liquid into solid
(b) Temperature of crystalline solid is increased from 0 K to 115 K
(c) H_2(g) → 2H_(g)
(d) 2NaHCO_3(s) → Na₂CO_3(s) + CO_2(g) + H₂O_(g)
Answer:
(a) Crystallisation of liquid into solid

Question iv.
The enthalpy of formation for all elements in their standard states is
(a) unity
(b) zero
(c) less than zero
(d) different elements
Answer:
(b) zero

Question v.
Which of the following reactions is exothermic ?
(a) H_2(g) → 2H_(g)
(b) C(s) → C_(g)
(c) 2Cl_(g) → Cl_2(g)
(d) H₂O_(s) → H₂O_(l)
Answer:
(c) 2Cl_(g) → Cl_2(g)

Question vi.
6.24 g of ethanol are vaporized by supplying 5.89 kJ of heat. Enthalpy of vaporization of ethanol will be
(a) 43.4 kJ mol^-1
(b) 60.2 kJ mol^-1
(c) 38.9 kJ mol^-1
(d) 20.4 kJ mol^-1
Answer:
(a) 43.4 kJ mol^-1

Question vii.
If the standard enthalpy of formation of methanol is -238.9 kJ mol^-1 then entropy change of the surroundings will be
(a) -801.7 JK^-1
(b) 801.7 JK^-1
(c) 0.8017 JK^-1
(d) -0.8017 JK^-1
Answer:
(b) 801.7 JK^-1

Question viii.
Which of the following are not state functions ?
1. Q + W 2. Q 3. W 4. H-TS
(a) 1, 2 and 3
(b) 2 and 3
(c) 1 and 4
(d) 2, 3 and 4
Answer:
(b) 2 and 3

Question ix.
For vaporization of water at 1 bar, ΔH = 40.63 kJ mol^-1 and ΔS =108.8 JK^-1 mol^-1. At what temperature, ΔG = 0?
(a) 273.4 K
(b) 393.4 K
(c) 373.4 K
(d) 293.4 K
Answer:
(c) 373.4 K

Question x.
Bond enthalpies of H – H, Cl – Cl and H – Cl bonds are 434 kJ mol^-1, 242 kJ mol^-2 and 431 kJ mol^-1, respectively. Enthalpy of formation of HCl is
(a) 245 kJ mol^-1
(b) -93 kJ mol^-1
(c) -245 kJ mol^-1
(d) 93 kJ mol^-1
Answer:
(b) -93 kJ mol^-1

2. Answer the following in one or two sentences.

Question i.
Comment on the statement: No work is involved in an expansion of a gas in vacuum.
Answer:
(1) When a gas expands against an external pressure Pex, changing the volume from V₁ to V₂, the work obtained is given by
W = -P_ex (V₂ – V₁).
(2) Hence the work is performed by the system when it experiences the opposing force or pressure.
(3) Greater the opposing force, more is the work.
(4) In free expansion, the gas expands in vaccum where it does not experience opposing force, (P = 0). Since external pressure is zero, no work is obtained.
∴ W = -P_ex (V₂ – V₁)
= -0 × (V₂ – V₁)
= 0
(5) Since during expansion in vacuum no energy is expended, it is called free expansion.

Question ii.
State the first law of thermodynamics.
Answer:
The first law of thermodynamics is based on the principle of conservation of energy and can be stated in different ways as follows :

1. Energy can neither be created nor destroyed, however, it may be converted from one form into another.
2. Whenever, a quantity of one kind of energy is consumed or disappears, an equivalent amount of another kind of energy appears.
3. The total mass and energy of an isolated system remain constant, although there may be interconservation of energy from one form to another.
4. The total energy of the universe remains constant.

Question iii.
What is enthalpy of fusion?
Answer:
Enthalpy of fusion (Δ_fusH) : The enthalpy change that accompanies the fusion of one mole of a solid into a liquid at constant temperature and pressure is called enthalpy of fusion.
For example,

This equation describes that when one mole of ice melts (fuses) at 0 °C (273 K) and 1 atmosphere, 6.1 kJ of heat will be absorbed.

Question iv.
What is standard state of a substance?
Answer:
The thermodynamic standard state of a substance (compound) is the most stable physical state of it at 298 K and 1 atmosphere (or 1 bar). The enthalpy of the substance in the standard state is represented as Δ_fH⁰.

Question v.
State whether ∆S is positive, negative or zero for the reaction 2H_(g) → H_2(g). Explain.
Answer:
(i) The given reaction, 2H_(g) → H_2(g) is the formation of H_2(g) from free atoms.
(ii) Since two H atoms form one H₂ molecule, ∆n = 1 – 2= -1 and disorder is decreased. Hence entropy change ∆S < 0 (or negative).

Question vi.
State second law of thermodynamics in terms of entropy.
Answer:
The second law of thermodynamics states that the total entropy of the system and its surroundings (universe) increases in a spontaneous process.
OR
Since all the natural processes are spontaneous, the entropy of the universe increases.
It is expressed mathematically as
∆ S_Total = ∆ S_system + ∆S_surr > 0
∆ S_Universe = ∆ S_system + ∆ S_surr > 0

Question vii.
If the enthalpy change of a reaction is ∆H how will you calculate entropy of surroundings?
Answer:
(i) For endothermic reaction, ∆H > 0. This shows the system absorbs heat from surroundings.
∴ ∆_surr H < 0.
∵ Entropy change = ∆_surr S = \(\frac{-\Delta_{\text {surr }} H}{T}\)
There is decrease in entropy of surroundings.
(ii) For exothermic reaction, ∆H < 0, hence for surroundings, ∆_surr H > 0

∴ ∆_surr > 0.

Question viii.
Comment on spontaneity of reactions for which ∆H is positive and ∆S is negative.
Answer:
Since ∆H is +ve and ∆S is -ve, ∆G will be +ve at all temperatures. Hence reactions will be non-spontaneous at all temperatures.

3. Answer in brief.

Question i.
Obtain the relationship between ∆G° of a reaction and the equilibrium constant.
Answer:
Consider following reversible reaction, aA + bB ⇌ cC + dD
The reaction quotient Q is,
Q = \(\frac{[\mathrm{C}]^{c} \times[\mathrm{D}]^{d}}{[\mathrm{~A}]^{a} \times[\mathrm{B}]^{b}}\)
The free energy change ∆G for the reaction is ∆G = ∆G° + RT in Q
Where ∆G° is the standard free energy change.
At equilibrium
Q = \(\frac{[\mathrm{C}]_{e}^{c} \times[\mathrm{D}]_{e}^{d}}{[\mathrm{~A}]_{e}^{a} \times[\mathrm{B}]_{e}^{b}}=\mathrm{K}\)
∴ ∆G = ∆G° + RT In K
∵ at equilibrium ∆G = 0
∴ 0 = AG° + RT In K
∴ ∆G° = -RT In K
∴ ∆G°= -2.303 RT log₁₀K.

Question ii.
What is entropy? Give its units.
Answer:
(i) Entropy : Being a state function and thermodynamic function it is defined as entropy change (∆S) of a system in a process which is equal to the amount of heat transferred in a reversible manner (Q_rev) divided by the absolute temperature (T), at which the heat is absorbed. Thus,

(ii) Units of entropy are JK^-1 in SI unit and cal K^-1 in c.g.s. units. It is also expressed in terms of entropy unit (e.u.). Hence 1 e.u. = 1 JK^-1.
(iii) Entropy is a measure of disorder in the system. Higher the disorder, more is entropy of the system.

Question iii.
How will you calculate reaction enthalpy from data on bond enthalpies?
Answer:
(i) In chemical reactions, bonds are broken in the reactant molecules and bonds are formed in the product molecules.
(ii) Energy is always required to break a chemical bond while energy is always released in the formation of the bond.
(iii) The enthalpy change of a gaseous reactions (Δ_fH⁰) involving substances with covalent bonds can be calculated with the help of bond enthalpies of reactants and products. (In case of solids we need lattice energy or heat of sublimation while in case of liquids we need heat of evaporation.)

If the energy required to break the bonds of reacting molecules is more than the energy released in the bond formation of the products, then the reaction will be endothermic and ∆H⁰ reaction will be positive. On the other hand if the energy released in the bond formation of the products is more than the energy required to break the bonds of reacting molecules then the reaction will be exothermic and ∆H⁰ reaction will be negative.

Question iv.
What is the standard enthalpy of combustion ? Give an example.
Answer:
Standard enthalpy of combustion or standard heat of combustion : it is defined as the enthalpy change when one mole of a substance in the standard state undergoes complete combustion in a sufficient amount of oxygen at constant temperature (298 K) and pressure (1 atmosphere or 1 bar). It is denoted by ∆_cH⁰.
E.g. CH₃OH_(l) + \(\frac {3}{2}\) O_2(g) = CO_2(g) + 2H₂O
∆_cH⁰= -726 kJ mol^-1
(∆_cH⁰ is always negative.)
[Note : Calorific value : It is the enthalpy change or amount of heat liberated when one gram of a substance undergoes combustion.

Question v.
What is the enthalpy of atomization? Give an example.
Answer:
Enthalpy of atomisation (∆_atomH) : it is the enthalpy change accompanying the dissociation of one mole of gaseous substance into its atoms at constant temperature and pressure.
For example : CH_4(g) → C_(g) + 4H_(g) ∆_atomH = 1660 kJ mol^-1

Question vi.
Obtain the expression for work done in chemical reaction.
Answer:
Consider n₁ moles of gaseous reactants A of volume V₁ change to n₂ moles of gaseous products B of volume V₂ at temperature T and pressure P.

In the initial state, PV₁ = n₁RT
In the final state, PV₂ = n₂RT
PV₂ – PV₁ = n₂RT – n₁RT = (n₂ – n₁)RT = ∆nRT
where ∆n is the change in number of moles of gaseous products and gaseous reactants.
Due to net changes in gaseous moles, there arises change in volume against constant pressure resulting in mechanical work, -P∆V.
∴ W = -P∆V = -P(V₂ – V₁) = – ∆nRT
(i) If n₁ – n₂, ∆n = 0, W = 0. No work is performed.
(ii) If n₂ > n₁, ∆n > 0, there is a work of expansion by the system and W is negative.
(iii) If n₂ < n₁, ∆n < 0, there is a work of compression, hence work is done on the system and W is positive.

Question vii.
Derive the expression for PV work.
Answer:
Consider a certain amount of an ideal gas enclosed in an ideal cylinder fitted with massless, frictionless rigid movable piston at pressure P, occupying volume V₁ at temperature T.

Fig. 4.8 : Work of expansion
As the gas expands, it pushes the piston upward through a distance d against external force F, pushing the surroundings.
The work done by the gas is,
W = opposing force × distance = -F × d
-ve sign indicates the lowering of energy of the system during expansion.
If a is the cross section area of the cylinder or piston, then,
W = \(-\frac{F}{a}\) × d × a
Now the pressure is P_ex = \(\frac{F}{a}\)
while volume change is, ΔV = d × a
∴ W = -P_ex × ΔV
If during the expansion, the volume changes from V₁ and V₂ then, ΔV = V₂ – V₁
∴ W= -P_ex(V₂ – V₁)
During compression, the work W is +ve, since the energy of the system is increased,
W = +P_ex(V₂ – V₁)

Question viii.
What are intensive properties? Explain why density is intensive property.
Answer:
(A) Intensive property : It is defined as a property of a system whose magnitude is independent of the amount of matter present in the system.
Explanation :

1. Intensive property is characteristic of the system, e.g., refractive index, density, viscosity, temperature, pressure, boiling point, melting point, freezing point of a pure liquid, surface tension, etc.
2. The intensive properties are not additive.

(B) Density is a ratio of two extensive properties namely, mass and volume. Since the ratio of two extensive properties represents an intensive property, density is an intensive property. It does not depend on the amount of a substance.

Question ix.
How much heat is evolved when 12 g of CO reacts with NO₂ ? The reaction is :
4 CO_(g) + 2 NO_2(g) → 4CO_2(g) + N_2(g), ∆H⁰ = -1200 kJ

4. Answer the following questions.

Question i.
Derive the expression for the maximum work.
Answer:

Consider n moles of an ideal gas enclosed in an ideal cylinder fitted with a massless and frictionless movable rigid piston. Let V be the volume of the gas at a pressure P and a temperature T.
If in an infinitesimal change pressure changes from P to P – dP and volume increases from V to V + dV. Then the work obtained is, dW = -(P-dP) dV
= -PdV + dPdV
Since dP.dV is negligibly small relative to PdV
dW= -PdV
Let the state of the system change from A(P₁, L₁) to B (P₂, V₂) isothermally and reversibly, at temperature T involving number of infinitesimal steps.

Then the total work or maximum work in the process is obtained by integrating above equation.

At constant temperature,

where n, P, V and T represent number of moles, pressure, volume and temperature respectively. For the process,
ΔU = 0, ΔH = 0.
The heat absorbed in reversible manner
Q_rev, is completely converted into work.
Q_rev = -W_max.
Hence work obtained is maximum.

Question ii.
Obtain the relatioship between ∆H and ∆U for gas phase reactions.
Answer:
Consider a reaction in which n₁ moles of gaseous reactant in initial state change to n₂ moles of gaseous product in the final state.
Let H₁, U₁, P₁, V₁ and H₂, U₂, P₂, V₂ represent enthalpies, internal energies, pressures and volumes in the initial and final states respectively then,

The heat of reaction is given by enthalpy change ΔH as,
ΔH = H₂ – H₁
By definition, H = U + PV
∴ H₁ = U₁ + P₁V₁ and H₂ = U₂+ P₂V₂
∴ ΔH = (U2 + P₂V₂) – (U₁ + P₁V₁)
= (U2 – U₁) + (P₂V₂ – P₁V₁)
Now, ΔU = U₂ – U₁
Since PV = nRT,
For initial state, P₁V₁= n₁RT
For final state, P₂V₂ = n₂RT
∴ P₂V₂ – P₁V₁ = n₂RT – n₁RT
= (n₂ – n₁) RT
= ΔnRT
where Δn

∴ ΔH = ΔU + ΔnRT
If Q_P and Q_V are the heats involved in the reaction at constant pressure and constant volume respectively, then since Q_P = ΔH and Q_V = ΔU.
∴ Q_P = Q_V = ΔnRT

Question iii.
State Hess’s law of constant heat summation. Illustrate with an example. State its applications.
Answer:
Statement of law of constant heat summation : It states that, the heat of a reaction or the enthalpy change in a chemical reaction depends upon initial state of reactants and final state of products and independent of the path by which the reaction is brought about (i.e. in single step or in series of steps).
OR
Heat of reaction is same whether it is carried out in one step or in several steps.
Explanation :
Consider the formation of CO_2(g).

Hess’s law treats thermochemical equations mathematically i.e., they can be added, subtracted or multiplied by numerical factors like algebraic equations.

Applications : Hess’s law is used for :

1. To calculate heat of formation, combustion, neutralisation, ionization, etc.
2. To calculate the heat of reactions which may not take place normally or directly.
3. To calculate heats of extremely slow or fast reactions.
4. To calculate enthalpies of reactants and products.

Question iv.
Although ∆S for the formation of two moles of water from H₂ and O₂ is -327 JK^-1, it is spontaneous. Explain. (Given ∆H for the reaction is -572 kJ).
Answer:
Given : ΔS= -327 JK^-1; ΔH = -572 kJ
ΔG = ΔH – TΔS, and ΔH << ΔS
∴ ΔG < 0, and hence the formation of H₂O_(l) is spontaneous.

Question v.
Obtain the relation between ∆G and ∆S_Total. Comment on spontaneity of the reaction.
Answer:
(i) Gibbs free energy, G is defined as,
G = H – TS
where H is the enthalpy, S is the entropy of the system at absolute temperature T.
Since H, T and S are state functions, G is a state function and a thermodynamic function.

(ii) At constant temperature and pressure, change in free energy ΔG for the system is represented as, ∆G = ∆H – T∆S

This is called Gibbs free energy equation for ∆G. In this ∆S is total entropy change, i.e., ∆S_Total.

(iii) The SI units of ∆G are J or kJ (or Jmol^-1 or kJmol^-1).
The c.g.s. units of ∆G are cal or kcal (or cal mol^-1 or kcal mol^-1.)

The second law explains the conditions of spontaneity as below :
(i) ∆S_total > 0 and ∆G < 0, the process is spontaneous.
(ii) ∆S_total < 0 and ∆G > 0, the process is nonspontaneous.
(iii) ∆S_total = 0 and ∆G = 0, the process is at equilibrium.

Question vi.
One mole of an ideal gas is compressed from 500 cm³ against a constant external pressure of 1.2 × 10⁵ Pa. The work involved in the process is 36.0 J. Calculate the final volume.
Answer:
Given : V₁ = 500 cm³ = 0.5 dm³;
P_ex = 1.2 × 10⁵ Pa = 1.2 bar; W= 36 J;
1 dm³ bar = 100 J; V₂ = ?
W = -P_ex (V₂ – V₁)
36 J = – 1.2 (V₂ – 0.5) dm³ bar
= -1.2 × 100 (V₂ – 0.5) J
∴ V₂ – 0.5 = \(\frac{-36}{1.2 \times 100}=-0.3\)
∴ V₂ = 0.5 -0.3 = 0.2 dm³ = 200 cm³
Ans. Final volume = 200 cm³.

Question vii.
Calculate the maximum work when 24g of O₂ are expanded isothermally and reversibly from the pressure of 1.6 bar to 1 bar at 298 K.
Answer:
Given : W₀₂ = 24 g, P₁ = 1.6 bar, P₂ = 1 bar
T = 298 K, W_max = ?

Question viii.
Calculate the work done in the decomposition of 132 g of NH₄NO₃ at 100 °C.
NH₄NO_3(s) → N₂O_(g) + 2 H₂O_(g)
State whether work is done on the system or by the system.
Answer:
NH4₄NO_3(s) → N₂O_(g) + 2 H₂O_(g)
m_NH4NO3 = 132 g; M_NH4NO3 = 80 g mol^-1
T = 273 + 100 = 373 K; Δn = ?
For the reaction,
Δn = Σn_{2 gaseous products} – Σn_{1 gaseous reactants}
= 3 – 0 = 3 mol
Since there is an increase in number of gaseous moles, the work is done by the system.
n_NH4NO3 = \(\frac{m_{\mathrm{NH}_{4} \mathrm{NO}_{3}}}{M_{\mathrm{NH}_{4} \mathrm{NO}_{3}}}\)
= \(\frac{132}{80}\)
= 1.65 mol
For 1 mol NH₄NO_3(s) Δn = 3 mol
∴ For 1.65 mol NH₄NO_3(s) Δn = 3 × 1.65 = 4.95 mol
W = -ΔnRT = -4.95 × 8.314 × 373
= – 15350 J
= – 15.35 kJ
Ans. Work is done by the system.
Work done = – 15.35 kJ

Question ix.
Calculate standard enthalpy of reaction,
Fe₂O_3(s) + 3CO_(g) → 2 Fe_(s) + 3CO_2(g),
from the following data.
∆_fH⁰(Fe₂O₃) = -824 kJ/mol,
∆_fH⁰(CO) = -110 kJ/mol,
∆_fH⁰(CO₂) = -393 kJ/mol
Answer:
Given : ∆_fH⁰Fe₂O₃ = -824 kJ/mol^-1;
∆_fH⁰(CO) = – 110 kJ mol^-1
∆_fH⁰(CO₂) = – 393 kJ/mol^-1; ∆_fH⁰ = ?
Required equation,
Fe₂O_3(s) + 3CO_(g) → 2Fe_(s) + 3CO_2(g)
∆H₁ = ? – (I)
Given equations :

= -(-824) -3 (-110) + 3(-393)
= 824 + 330 – 1179
∆_fH⁰ = -25 kJ
Ans. ∆_fH⁰ = -25 kJ

Question x.
For a certain reaction ∆H⁰ =219 kJ and ∆S⁰ = -21 J/K. Determine whether the reaction is spontaneous or nonspontaneous.
Answer:
Given : ∆H⁰ = 219 kJ; ∆S⁰ = -21 J/K, ∆G⁰ = ?
For standard state, T = 298 K
∆G⁰ = ∆H⁰ – T∆S⁰
= 219 – 298 × (-21) × 10^-3
= 219 + 6.258
= 225.3 kJ
Since ∆S < 0 and ∆G⁰ > 0, the reaction is non-spontaneous.

Question xi.
Determine whether the following reaction is spontaneous under standard state conditions.
2 H₂O_(l) + O_2(g) → 2H₂O_2(l)
if ∆H⁰ = 196 kJ, ∆S⁰ = -126 J/K
Does it have a cross-over temperature?
Answer:
Given : 2H₂O_(l) + O_2(g) → 2H₂O_2(l)
∆H⁰ = +196 kJ
∆S⁰ = -126 JK-1 =0.126 kj K-1
T= 298 K
∆G⁰ = ?
Cross over temperature = T = ?
∆G⁰ = ∆H⁰ – T∆S⁰
= 196 – 298 (-0.126)
= 196 + 37.55
= + 233.55 kJ
∵ ∆G⁰ > 0, the reaction is non-spontaneous.
∆H⁰ > 0, ∆S⁰ < 0,
Since at all temperatures, ∆G⁰ > 0, there is no cross over temperature.
Ans. The reaction is non-spontaneous.
There is no cross-over temperature for the reaction.

Question xii.
Calculate ∆U at 298 K for the reaction,
C₂H_4(g) + HCl_(g) → C₂H₅Cl_(g), ∆H = -72.3 kJ
How much PV work is done?
Answer:
Given : C₂H_4(g) + HCl_(g) → C₂H₅Cl_(g)
T = 298 K; ∆H = -72.3 kJ; PV = ?;
∆U = ?
∆n = Σn_{2gaseous products} – Σn_{1gaseous reactants}
= 1 – (1 + 1)= -1 mol
For PV work :
W = -∆nRT
= – (- 1) × 8.314 × 298
= 2478 J = 2.478 kJ
∆H = ∆U + ∆nRT
∴ ∆U = ∆H – ∆nRT
= – 72.3 – (-2.478)
= – 69.82 kJ
Ans. PV work = 2.478 kJ
∆U = -69.82 kJ

Question xiii.
Calculate the work done during synthesis of NH₃ in which volume changes from 8.0 dm3 to 4.0 dm³ at a constant external pressure of 43 bar. In what direction the work energy flows?
Answer:
Given : V₁ = 8.0 dm³; V₂ = 4.0 dm³; P_ex = 43 bar
W = ? What direction work energy flows ?
W = -P_ex(V₂ – V₁)
= -43 (4 – 8)
= 172 dm³ bar
= 172 × 100 J
= 17200 J
= 17.2 kJ
In this compression process, the work is done on the system and work energy flows into the system.

Question xiv.
Calculate the amount of work done in the
(a) oxidation of 1 mole HCl_(g) at 200 °C according to reaction.
4HCl_(g) + O_2(g) → 2 Cl_2(g) + 2 H₂O_(g)
(b) decomposition of one mole of NO at 300 °C for the reaction
2 NO_(g) → N_2(g) + O₂
Answer:
Given :
(a) 4HCl_(g) + O_2(g) → 2Cl_2(g) + 2H₂O_(g)
n_HCl = 1 mol; T = 273 + 200 = 473 K, W = ?
For 4 mol HCl ∆n = (2 + 2) – (4 + 1) = – 1 mol
∴ For 1 mol HCl ∆n = –\(\frac {1}{4}\) = -0.25 mol
W = -∆nRT = – (-0.25) × 8.314 × 473 = 983.11
(b) ∆n = (1 + 1) – 2 = 0 mol
W = -∆nRT = -(0) × 8.314 × 473 = 0
Ans. (a) W = 983.1 J
(b) W = 0.0 J

Question xv.
When 6.0 g of O₂ reacts with CIF as per
2CIF_(g) + O_2(g) → Cl₂O_(g) + OF_2(g)
The enthalpy change is 38.55 kJ. What is standard enthalpy of the reaction ?
Answer:
Given : The given reaction is for 1 mol O₂ or 32 g O₂.
∵ For 6.0 g O₂
∆ H⁰ = 38.55 kJ
∴ For 32 g O₂
∆ H⁰ = \(\frac{32 \times 38.55}{6}\)
= 205.6 kJ
Ans. ∆H⁰ = 205.6 kJ

Question xvi.
Calculate the standard enthalpy of formation of CH₃OH_(l) from the following data:
i. CH₃OH_(l) + \(\frac {3}{2}\) O_2(g) → CO_2(g) + 2H₂O_(l), ∆H⁰ = -726 kJ mol^-1
ii. C (Graphite) + O_2(g) → CO_2(g), ∆_cH⁰ = -393 kJ mol^-1
iii. H_2(g) + \(\frac {1}{2}\) O_2(g) → H₂O_(l), ∆_fH⁰ = -286 kJ mol^-1
Answer:


∴ ∆H⁰
= –\(\Delta H_{2}^{0}\) + \(\Delta H_{3}^{0}\) + 2∆\(\Delta H_{4}^{0}\)
= – (- 726) + (- 393) + 2(- 286)
= 726 – 393 – 572
= – 239 kJ mol^-1
Ans. Standard enthalpy of formation = ∆_fH⁰= -239 kJ mol^-1

Question xvii.
Calculate ∆H0 for the following reaction at 298 K
H₂B₄O_7(s) + H₂O_(l) → 4HBO_2(aq)
i. 2H₃BO_3(aq) → B₂O_3(s) + 3H₂O_(l), ∆H⁰ = 14.4 kJ mol^-1
ii. H₃BO_3(aq) → HBO_2(aq) + H₂O_(l), ∆H⁰ = -0.02 kJ mol^-1
iii. H₂B₄O_7(s) → 2B₂O_3(s) + H₂O_(l), ∆H⁰ = 17.3 kJ mol^-1
Answer:
Given equations :
i. 2H₃BO_3(aq) → B₂O_3(s) + 3H₂O_(l), ……….(i)
∆H⁰ = 14.4 kJ mol^-1
ii. H₃BO_3(aq) → HBO_2(aq) + H₂O_(l) ……….(ii)
∆H⁰ = -0.02 kJ mol^-1
iii. H₂B₄O_7(s) → 2B₂O_3(s) + H₂O_(l), ……….(iii)
∆H⁰ = 17.3 kJ mol^-1
Required equation :
(iv) H₂B₄O_7(s) + H₂O_(l) → 4HBO_2(aq) ……. (iv)
\(\Delta H_{4}^{0}=?\)
To obtain eq. (iv) add 4 times equation (ii) and eq.
(iii) and subtract 2 times equation (i).
∴ eq. (iv) = 4 eq. (ii) + eq. (iii) – 2eq. (i)
∴ \(\Delta H_{4}^{0}=4 \Delta H_{2}^{0}+\Delta H_{3}^{0}-2 \Delta H_{1}^{0}\)
= 4(-0.02) + 17.3 – 2(14.4)
= -0.08 + 17.3 – 28.8
= -11.58 kJ
∴ Enthalpy change for the reaction
= ∆_rH⁰ = -11.58 kJ
Ans. ∆_rH⁰ for the given reaction = -11.58 kJ

Question xviii.
Calculate the total heat required (a) to melt 180 g of ice at 0 °C, (b) heat it to 100 °C and then (c) vapourise it at that temperature. Given ∆_fusH_(ice) = 6.01 kJ mol^-1 at 0 °C, ∆_vapH_(H2O) = 40.7 kJ mol^-1 at 100 °C specific heat of water is 4.18 J g^-1 K^-1.
Answer:
Given : Mass of ice = m = 180 g
T₁ = 273 + 0 °C = 273 K
T₂ = 273 + 100 °C = 373 K
∆_fusH_(ice) = ∆_fusH_(H2O)(s) = 6.01 kJ mol^-1
∆_vapH_H2O(l) = 40.7 kJ mol^-1
Specific heat of water = C = 4.18 J g^-1 K^-1
For converting 180 g ice into vapour, ∆ H_Total = ?
Number of moles of H₂O = \(\frac {180}{18}\) = 10 mol
The total process can be represented as,

(i) ∆H₁ = ∆_fusH = 10 mol × 6.01 kJ mol^-1
= 60.1 kJ
(ii) When the temperature of water is raised from 0 °C to 100 °C (i.e., 273 K to 373 K), then
∆ H₂ = m × C × ∆T
= m × C × (T2 – T1)
= 180 g × 4.18 Jg^-1K^-1 × (373 – 273) × 10^-3 kJ = 75.24 kJ
∆ H₃ = ∆_vapH = 10 mol × 40.7 kJ mol^-1 = 407 kJ
Hence total enthalpy change,
∆ H_Total = ∆H₁ + ∆H₂ + ∆H₃
= 60.1 + 75.24 + 407
= 542.34 kJ
Ans. Total heat required = 542.34 kJ

Question xix.
The enthalpy change for the reaction,
C₂H_4(g) + H_2(g) → C₂H_6(g)
is -620 J when 100 ml of ethylene and 100 mL of H₂ react at 1 bar pressure. Calculate the pressure volume type of work and ∆U for the reaction.
Answer:
Given :
\(\begin{aligned}
&\mathrm{C}_{2} \mathrm{H}_{4(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6(\mathrm{~g})} \\
&100 \mathrm{~mL} \quad 100 \mathrm{ml} \quad 100 \mathrm{ml}
\end{aligned}\)
∆H = – 620 J; V_C2H4 = 100 mL; V_H2 = 100 mL
P_ex= 1 bar; W=?; ∆U = ?
∆V = 100 – (100 + 100) = -100 mL = -0.1 dm³
W = -P_ex(V₂ – V₁)
= -P_ex × ∆V
= -1 × (-0.1)
= 0.1 dm³ bar
= 0.1 × 100 J
= +10 J
∆H = ∆U + P∆V
∴ ∆U = ∆H – P∆V = -620 – (+10) = -610 J
Ans. W = +10 J; ∆U = -610 J

Question xx.
Calculate the work done and comment on whether work is done on or by the system for the decomposition of 2 moles of NH₄NO₃ at 100 °C
NH₄NO_3(s) → N₂O_(g) + 2H₂O_(g)
Answer:
Given : NH₄NO_3(s) → N₂O_(g) + 2H₂O_(g)
n_NH4NO3 = 2 mol; T = 273 + 100 = 373 K
W = ? Comment on work = ?
∆n_reaction = (1 + 2) – 0 = 3 mol
∵ For 1 mol of NH₄NO₃ ∆n_reaction = 3 mol
∴ For 2 mol of NH₄NO₃ ∆n_reaction = 6 mol
Due to 6 moles of gaseous products from 2 mol NH₄NO₃, there is work of expansion, hence work is done by the system.
W = -∆nRT
= – 6 × 8.314 × 373 = -18606 J
= -18.606 kJ
Ans. Work is done by the system.
W= -18.606 kJ

12th Chemistry Digest Chapter 4 Chemical Thermodynamics Intext Questions and Answers

(Textbook page No. 73)

Question 1.
Under what conditions ∆H = ∆U ?
Answer:
(a) ∆H = ∆U + P∆V
when ∆V = 0, ∆H = ∆U
(b) ∆H = ∆U + ∆nRT
when ∆n = 0, ∆H = ∆U

Try this… (Textbook page No. 71)

Question 1.
25 kJ of work is done on the system and it releases 10 kJ of heat. What is ∆U?
Answer:
W = 25 kJ; Q= -10 kJ
∆U = Q + W = -10 + 25
∆U = + 15 kJ

Try this… (Textbook page No. 75)

Question 1.
For KCl, ∆_LH = 699 kJ/mol^-1 and ∆_hydH = -681.8 kJ/mol^-1. What will be its enthalpy of solution?
Answer:
Enthalpy of solution :
∆_solnH = ∆_LH + ∆_hydH
= 699 + (-681.8)
∆_solnH = +17.2 kJ mol^-1

Try this… (Textbook page No. 76)

Question 1.
Given the thermochemical equation,
C₂H_2(g) + \(\frac {5}{2}\) O_2(g) → 2CO_2(g)+ H₂O_(l), ∆_rH⁰ = -1300 kJ
Write thermochemical equations when
i. Coefficients of substances are multiplied by 2.
ii. equation is reversed.
Answer:
(i) 2C₂H_2(g) + 5O_2(g) → 4CO_2(g) + 2H₂O_(l)
∆_rH⁰ = -2 × 1300 kJ
= – 2600 kJ
(ii) 2CO_2(g) + H₂O_(l) → C₂H_2(g) + \(\frac {5}{2}\)O_2(g)
∆_rH⁰ = +1300 KJ

Try this… (Textbook page No. 78)

Question 1.
(i) Write thermochemical equation for complete oxidation of one mole of H_2(g). Standard enthalpy change of the reaction is -286 kJ.
(ii) Is the value -286 kJ, enthalpy of formation or enthalpy of combustion or both? Explain.
Answer:
(i) H_2(g) + \(\frac {1}{2}\)O_2(g) → H₂O_(l) ∆_cH⁰ = -286 KJ mol^-1
(ii) The value -286 kJ is the standard enthalpy of formation of H₂O_(l) or standard enthalpy of combustion of H_2(g).

Question 2.
Write equation for bond enthalpy of Cl-Cl bond in Cl₂ molecule ∆_rH⁰ for dissociation of Cl₂ molecule is 242.7 kJ.
Answer:
Equation for bond enthalpy :
Cl_2(g) → 2Cl_(g) ∆_rH⁰ = 242.7 kJ mol^-1
∴ Bond enthalpy of Cl₂ = 242.7 kJ mol^-1

Try this… (Textbook page No. 82)

Question 1.
State whether ∆S is positive, negative or zero for the following reactions.
i. 2H_2(g) + O_2(g) → 2H₂O_(l)
ii. CaCO_3(s) → CaO_(s) + CO_2(g)
Answer:
(i) 2H_2(g) + O_2(g) → 2H₂O_(l)
Since the system is converted from gaseous state to a liquid state, the disorder is decreased, hence ∆S < O (negative).

(ii) CaCO_3(s) → CaO_(s) + CO_2(g)
Since molecules of solid CaCO₃ break giving gaseous CO₂, disorder is increased hence ∆S > O (positive).

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 3 Inheritance and Variation Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 3 Inheritance and Variation

1. Multiple Choice Questions

Question 1.
Phenotypic ratio of incomplete dominance in Mirabilis jalapa.
(a) 2 : 1 : 1
(b) 1 : 2 : 1
(c) 3 : 1
(d) 2 : 2
Answer:
(b) 1 : 2 : 1

Question 2.
In dihybrid cross, F₂ generation offspring show four different phenotypes while the genotypes are ……………….
(a) six
(b) nine
(c) eight
(d) sixteen
Answer:
(b) nine

Question 3.
A cross between an individual with unknown genotype for a trait with recessive plant for that trait is ……………….
(a) back cross
(b) reciprocal cross
(C) monohybrid cross
(d) test cross
Answer:
(d) test cross

Question 4.
When phenotypic and genotypic ratios are the same, then it is an example of ……………….
(a) incomplete dominance
(b) complete dominance
(c) multiple alleles
(d) cytoplasmic inheritance
Answer:
(a) incomplete dominance

Question 5.
If the centromere is situated near the end of the chromosome, the chromosome is called ……………….
(a) Metacentric
(b) Acrocentric
(c) Sub-Metacentric
(d) Telocentric
Answer:
(d) Telocentric

Question 6.
Chromosomal theory of inheritance was proposed by ……………….
(a) Sutton and Boveri
(b) Watson and Crick
(c) Miller and Urey
(d) Oparin and Halden
Answer:
(a) Sutton and Boveri

Question 7.
If the genes are located in a chromosome as p-q-r-s-t, which of the following gene pairs will have least probability of being inherited together ?
(a) p and q
(b) r and s
(c) s and t
(d) p and s
Answer:
(d) p and s

Question 8.
Find the mismatched pair:
(a) Down’s syndrome = 44 + XY
(b) Turner’s syndrome = 44 + XO
(c) Klinefelter’s syndrome = 44 + XXY
(d) Super female = 44 + XXX
Answer:
(a) Down’s syndrome = 44 + XY

Question 9.
A colourblind man marries a woman, who is homozygous for normal colour vision, the probability of their son being colour blind is ……………….
(a) 0%
(b) 25%
(c) 50%
(d) 100%
Answer:
(a) 0%

2. Very Short Answer Questions

Question 1.
Explain the statements
a. Test cross is back cross but back cross is not necessarily a test cross.
b. Law of dominance is not universal.
Answer:
a. (1) Test cross is the cross between F₁ hybrid and its homozygous recessive parent.
(2) Back cross is the cross of offspring with any one of the parents, either dominant or recessive.
(3) Therefore, test cross can be a back cross – but back cross cannot be a test cross.

b. (1) There are many traits in many organisms which show dominance. For example, widow’s peak in human beings is a dominant trait. Yellow seed colour and round seed shape are dominant traits in pea plant.
(2) However, there are characters which are either co-dominant, such as genes for human blood group A and B or incompletely dominant as in flower colour of Mirabilis jalapa.
(3) Therefore the law of dominance is not universally applicable.

Question 2.
Define the following terms:
a. Dihybrid cross
b. Homozygous
c. Heterozygous
d. Test cross
Answer:
a. A cross between parents differing in two heritable traits is called dihybrid cross.
b. An individual possessing identical alleles for a particular trait is called homozygous or pure for that trait. E.g. TT for tallness and tt for dwarfness.
c. An individual possessing contrasting allele for a particular trait is called heterozygous. E.g. Tt showing tallness.
d. The cross of F₁ progeny with homozygous recessive parent is called a test cross.

Question 3.
What are allosomes?
Answer:
Allosomes are the chromosomes which decide the sex of an organism.

Question 4.
What is crossing over?
Answer:
Crossing over is the process of forming new recombinations by interchanging and exchanging non-sister chromatid arms of the homologous chromosomes.

Question 5.
Give one example of autosomal recessive disorder.
Answer:
Thalassemia is an example of autosomal recessive disorder.

Question 6.
What are X-linked genes?
Answer:
Genes located on the non-homologous region of X chromosome are called X-linked genes.

Question 7.
What are holandric traits?
Answer:
Genes located on the non-homologous region of Y chromosome are called Y-linked genes. The traits due to such genes are called holandric traits which are seen only in male sex.

Question 8.
Give an example of chromosomal disorder caused due to non-disjunction of autosomes.
Answer:
Down’s syndrome is an example of chromosomal disorder caused due to non-disjunction of autosomes.

Question 9.
Give one example of complete sex linkage.
Answer:
Sex linkage can be complete X linkage and complete Y linkage. X linkage is haemophilia and Y linkage is hypertrichosis.

3. Short Answer Questions

Question 1.
Enlist seven traits of pea plant selected / studied by Mendel.
Answer:
Seven traits in pea selected by Mendel:

1. Tall habit versus dwarf habit (Height of the plant).
2. Purple flowers versus white flowers. (Colour of flowers)
3. Yellow seeds versus green seeds. (Colour of seeds)
4. Round seeds versus wrinkled seeds. (Shape of seeds)
5. Green pods versus yellow pods. (Colour of pods)
6. Inflated pods versus constricted pods. (Shape of pods)
7. Axial flower versus terminal flower. (Position of a flower)

Question 2.
Why law of segregation is also called the law of purity of gametes?
Answer:
(1) Mendel’s law of segregation is also called Law of purity of gametes because, during formation of gametes, the alleles separate/ segregate from each other and only one allele enters a gamete.

(2) The separation of one allele does not affect other. Since single allele enters a gamete means gametes will be pure for a trait.
E.g. The contrasting characters such as tall (T) and dwarf (t) present in F₁ hybrid (Tt) segregate during the formation of gametes.

(3) Owing to this, two types of gametes i.e. T and t are formed which are pure for the characters which they carry.
(4) Thus for example:

Question 3.
Pleiotropy.
Answer:

1. When a single gene controls two or more different traits, it is called a pleiotropic gene and the phenomenon is known as pleiotropy or pleiotropism.
2. The pleiotropic ratio is always 1 : 2 instead . of normal 3 : 1.
3. Sickle-cell anaemia is caused by the gene HbS. The healthy or normal gene which is dominant is HbA. The heterozygotes or carriers i.e., HbA/Hbs show anaemia as there is deficiency of haemoglobin due to sickling of RBCs. Abnormally low concentration of oxygen can cause sickling of RBCs.
4. The homozygotes possessing the recessive gene HbS die because of fatal anaemia because the gene for sickle-cell anaemia is lethal in homozygous condition and causes sickle-cell trait in heterozygous carrier.
5. Thus a single gene produces two different expressions.
6. When two carriers are married they will produce normal carriers and Sickle-cell anaemic children in the ratio of 1 : 2 : 1. Out of these three children sickle-cell anaemic child will die leaving the ratio 1 : 2 instead of 3 : 1.

Question 4.
What are the reasons of Mendel’s success?
Answer:
Reasons for Mendel’s success:

1. Mendel planned his experiments carefully and these experiments consisted of large sample.
2. He always recorded the results of number of plants of each type and their ratios.
3. The contrasting characters that he chose were easily recognizable.
4. The seven pairs of contrasting characters that he selected were under control of a single factor each. They were present on separate chromosomes and were transmitted from one generation to the next.
5. Mendel studied and introduced concept of dominance and recessiveness.

Question 5.
“Father is responsible for determination of sex of child and not the mother”. Justify.
Answer:

1. Human made is heterogame tic, i.e. he produces two different types of sperms. One is bearing X chromosome along with 22 autosomes and the other is Y bearing sperm with 22 autosomes.
2. Mother, on the other hand, is homogametic, producing all similar types of ova, i.e 22 + X chromosomal combination.
3. If 22+X bearing sperm fertilise an egg, female child is formed while if Y bearing sperm fertilizes an egg, male child is formed.
4. Thus the sex of the child is dependent upon type of sperm that father gives, therefore, it is said that father is responsible for determination of sex of a child and not the mother.

Question 6.
What is linkage? How many linkage groups do occur in human being and maize?
Answer:

1. Linkage is defined as the tendency of the genes to be inherited together because they are present in the same chromosome. Linkage group is group of genes situated on a chromosome.
2. Humans have 23 linkage groups because they have 23 pairs of chromosomes.
3. Maize plant has 10 linkage groups because they have 10 pairs of chromosomes.

Question 7.
PKU.
Answer:

1. PKU means phenylketonuria which is an autosomal recessive inborn error.
2. In this disorder the metabolism of phenylalanine does not occur due to deficiency of phenylalanine hydroxylase (PAH) enzyme.
3. This enzyme is necessary to metabolize the amino acid phenylalanine to the amino acid tyrosine.
4. When PAH activity is reduced, phenylalanine accumulates in blood and cerebrospinal fluid and is converted into phenylpyruvate or phenyl-ketone which is a toxic compound. This may cause mental retardation. Excess phenylalanine is excreted in urine, hence this disease is called phenylketonuria.
5. PKU is caused by mutations in the PAH gene on chromosome no. 12.
6. Untreated PKU causes abnormal phenotype which includes growth failure, poor skin pigmentation, microcephaly, seizures, global developmental delay and severe intellectual impairment. However, at birth if an infant is checked for PKU, the further abnormalities can be avoided.

Question 8.
Compare X-chromosome and Y-chromosome.
Answer:

X-chromosome	Y-chromosome
1. X-chromosome is straight, rod like and longer 1. than Y chromosome. It is metacentric.	1. Y-chromosome is shorter chromosome which is acrocentric.
2. X-chromosome has large amount of euchromatin and small amount of heterochromatin.	2. Y-chromosome has small amount of euchromatin and large amount of heterochromatin.
3. X-chromosome has large amount of DNA, hence it is genetically active due to more genes.	3. Y-chromosome has less amount of DNA, hence it is genetically less active or inert due to lesser genes.
4. Non-homologous region of X-chromosome is longer and contains more genes.	4. Non-homologous region of Y-chromosome is shorter and contains lesser genes.
5. Contains X-linked genes on non-homologous region.	5. Contains Y-linked genes on non-homologous region.
6. X-chromosome is present in men as well as women.	6. Y-chromosome is present only in men.

Question 9.
Explain the chromosomal theory of inheritance.
Answer:
Chromosomal theory of inheritance was put forth by Sutton and Boveri after studying paraillel behaviour of genes and chromosomes during meiotic division. This theory states following points:

1. Chromosomal theory identifies chromosomes as the carrier of genetic material.
2. All the hereditary characters are transmitted by gametes. Nucleus of gametes, i.e. sperms and ova of the parents contain chromosomes which transmit the heredity to offspring.
3. Chromosomes are found in pairs in somatic or diploid cells.
4. During gamete formation, homologous chromosomes pair and segregate independently at meiosis. The diploid condition is converted into haploid condition. Thus each gamete contains only one chromosome of a pair.
5. During fertilization, the union of sperm and egg restores the diploid number of chromosomes.

Question 10.
Observe the given pedigree chart and answer the following questions

(a) Identify whether the trait is sex-linked or autosomal.
(b) Give an example of a trait in human beings which shows such a pattern of inheritance.
Answer:
Pedigree given above shows:

1. First Generation : Carrier woman marrying a sufferer man. Their three children are in following birth order.
2. Second generation : First son is normal, second daughter is carrier and third daughter is sufferer.
3. Third generation : The sufferer daughter marries a normal man. Her children are normal daughter and sufferer son.

(a) The above pedigree show sex-linked (X-linked) trait. Since criss-cross inheritance is seen in the trait, it must be sex-linked inheritance.
(b) Such trait and its inheritance can be seen in colour blindness.

4. Match the Columns

rewrite the matching pairs.

Column I	Column II
(1) 21 trisomy	(a) Turner’s syndrome
(2) X-monosomy	(b) Klinefelter’s syndrome
(3) Holandric traits	(c) Down’s syndrome
(4) Feminized male	(d) Hypertrichosis

Answer:

Column I	Column II
(1) 21 trisomy	(c) Down’s syndrome
(2) X-monosomy	(a) Turner’s syndrome
(3) Holandric traits	(d) Hypertrichosis
(4) Feminized male	(b) Klinefelter’s syndrome

5. Long Answer Questions

Question 1.
What is dihybrid cross? Explain with suitable example and checker board method.
Answer:
1. A cross which involves two pairs of alleles is called a dihybrid cross. A phenotypic ratio of 9 : 3 : 3 : 1 obtained in the F₂ generation of a dihybrid cross is called a dihybrid ratio.

(2) Thus for example, when we cross a true breeding pea plant bearing round and yellow seeds with a true breeding pea plant bearing wrinkled and green seeds we get pea plants bearing round and yellow seeds in the F₁ generation.

(3) When F₁ plants are selfed, we get a ratio of 9 : 3 : 3 : 1 in the F₂ generation, where 9 plants bear yellow round seeds, 3 plants bear yellow wrinkled seeds, 3 plants bear green round seeds and 1 plant bears green wrinkled seeds.

(4) Parents (P₁) : RRYY × rryy
Gametes of P₁ RY and ry
F₁ generation : RrYy(Yellow round)
On selfing F₁ : RrYy × RrYy
Gametes of F₁ : RY, Ry, rY, ry

P₂ generation:

Round Yellow : 9 Round green : 3 Wrinkled yellow : 3 Wrinkled green : 1
Phenotypic ratio : 9 : 3 : 3 : 1
Genotypic ratio : 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1

Question 2.
Explain with suitable example an independent assortment.
Answer:
(1) The law of independent assortment states that when hybrid possessing two or more pairs of contrasting characters bearing alleles form gametes, the alleles in each pair segregate independently of the other pair. Therefore, the inheritance of one pair of characters is independent of that of the other pair of characters.
(2) For example, when we cross a pea plant which is tall and having purple flowers with dwarf plant having white flowers we obtain all tall plants with purple flowers in F₁ generation. When F₁ generation are selfed, 9 : 3 : 3 : 1 ratio was obtained in F₂ generation with 9 tall and purple flower, 3 tall with white flowers, 3 dwarf with purple flowers and 1 which was dwarf and white. Tallness and purple colour are dominant traits while dwarfness and white colour are recessive traits.

(i) Homozygous tall purple – TTPP
(ii) Homozygous dwarf white – ttpp

Tall purple = 9. Tall white = 3
Dwarf purple = 3, Dwarf white = 1,
Phenotypic ratio = 9 : 3 : 3 : 1
Results : The offspring of F₁ generation will be in the proportion of 9 : 3 : 3 : 1, where 9 are tall purple, 3 are tall white, 3 are dwarf purple and 1 is dwarf white.

Question 3.
Define test cross and explain its significance.
Answer:
1. Definition of test cross : A cross between F₁ offspring and its homozygous recessive parent is called a test cross.
2. Significance of test cross:

• Test cross can be used to find out the genotype of any plant which shows dominant characters.
• Whether the plant is homozygous or heterozygous can be understood by performing test cross.
• Test cross is used to introduce useful recessive traits in the hybrids of self- pollinated plants.
• Test cross is quicker method to improve the variety of crop plants and thus it is useful for breeders and geneticists.
• Test cross can be used for verifying the laws of inheritance.

Question 4.
What is parthenogenesis? Explain the haplodiploid method of sex determination in honey bee.
Answer:
I. Parthenogenesis is a natural form of asexual reproduction in which growth and development of embryos occur without fertilization by sperm. In some insects like honey bees, parthenogenesis means development of an embryo from an unfertilized egg cell.

II. In honey bee:

1. Sex determination is by haplodiploid system.
2. Sex is determined by the number of sets of chromosomes received by an individual.
3. The egg which is fertilized by sperm, becomes diploid and develops into female.
4. The egg which is not fertilized develops by parthenogenesis and develops into a male.
5. The queen and worker bee therefore contain 32 chromosomes. The drone, i.e. male bears 16 chromosomes.
6. The sperms are produced by mitosis while eggs are produced by meiosis.

Question 5.
In the answer for inheritance of X-linked. genes, Madhav had shown carrier male. His answer was marked incorrect. Madhav was wondering why his marks were cut. Explain the reason.
Answer:
Males can never be carriers. They have single X and other Y chromosome. In X linked inheritance, the genes are present on the non-homologous region of X chromosome. Males do not have other X and hence if the genes are present on his X chromosome, they will not be suppressed in them. The Y chromosome does not have dominant gene to hide this expression as there is no homolorous region too. But in case of females, there are double X chromosomes and hence if X-linked gene is recessive, the other X can hide the expression of such X-linked gene.

Thus she becomes a carrier without showing any physical characters. She is physically normal and does not suffer from such X-linked recessive disorder. Thus, Madhav will get his answer wrong due to incorrect concept.

Question 6.
With the help of neat labelled diagram, describe the structure of chromosome.
Answer:

(1) A chromosome is best visible during metaphase, when it is highly condensed.

(2) Chromosome shows two identical halves, called sister chromatids. Chromatids are held together at centromere which is also called primary constriction.

(3) Primary constriction has disc shaped plate called kinetochore. This plate is useful for attachment of spindle fibres at the time of cell division.

(4) Additional narrow areas called secondary constrictions are seen in some chromosomes which are known as nucleolar organizers. They help in the formation of nucleolus. At secondary constriction (i) there is nucleolar organising region. Secondary constriction (ii) shows attachment of satellite body or SAT body.

(5) Each chromatid is made up of sub¬chromatids called chromonemata. Each chromonema consists of a long, unbranched, slender, highly coiled DNA thread. This double stranded DNA molecule extends throughout the length of the chromosome.

(6) The ends of the chromatid arms are called telomeres.

Question 7.
What is criss-cross inheritance? Explain with suitable example.
Answer:
Criss-cross inheritance is the type of inheritance in which the genes are passed on from father to daughter and then to her son, i.e. from male to female and from female to male (grandson). In other words, it is also said that the transmission is from the grandfather to his grandson through his daughter.

I. Inheritance of Colour blindness show criss-cross pattern.
(1) Colour blindness is a sex-linked disorder in which the person concerned cannot distinguish between red and green colours.

(2) It is recessively X-linked disorder, which is expressed in males. It is rarely seen in females.

(3) The genes for normal vision are dominant whereas those for colour blindness are recessive.

(4)

• Gene for normal vision : X^C
• Gene for colour blindness : X^c
• Normal female : X^CX^C
• Normal male : X^CY
• Colour blind female : X^cX^c
• Carrier female : X^CX^c
• Colour blind male : X^c Y

II. Crosses showing the inheritance of colour blindness:
(i) A cross between normal female and colour-blind male.

(ii) A cross of carrier female with normal male.

(1) Normal female with Colour blind male. Such cross produces 50% carrier daughters and 50% normal sons.

(2) Carrier female with normal male. Such a cross produces 25% normal daughters, 25% normal sons, 25% carrier daughters and 25% colour blind sons.

(3) Colour blind father transmits the disorder to his grandson through his carrier daughter. The inheritance of characters from the father to his grandson through his daughter is called criss-cross inheritance.

Question 8.
Describe the different types of chromosomes.
Answer:
I. Chromosomes are classified into the following four types according to the position of the centromere in them:
(1) Metacentric : In metacentric chromosome, the centromere is situated in the middle of the chromosome. The two arms of the chromosome are nearly equal. It appears ‘V’-shaped during anaphase.

(2) Sub-metacentric : In sub-metacentric chromosome, the centromere is situated some distance away from the middle. Due to this, one arm of the chromosome is shorter than the other. It appears T-shaped during anaphase.

(3) Acrocentric : In acrocentric chromosome, the centromere is situated near the end of the chromosome. One arm of the acrocentric chromosome is very short while the other is long making it appear like ‘J’-shaped during anaphase.

(4) Telocentric : In telocentric chromosome, the centromere is situated at the tip of the chromosome. Telocentric chromosome has only one arm thus it appears rod-shaped.

II. Based on the functions, chromosomes are divided into autosomes and allosomes. Autosomes are somatic chromosomes which decide the body characters. Allosomes are sex chromosomes which decide the sex of the individual.

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 1 Introduction to Partnership and Partnership Final Accounts Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 1 Introduction to Partnership and Partnership Final Accounts

I. Objective Questions:

A. Select the most appropriate alternatives from the following and rewrite the sentences:

Question 1.
When there is no partnership agreement between partners, the division of profits takes place in ____________ ratio.
(a) equal
(b) capital ratio
(c) initial contribution
(d) experience and tenure of partners
Answer:
(a) equal

Question 2.
To find out Net Profit or Net Loss of the business ____________ Account is prepared.
(a) Trading
(b) Capital
(c) Current
(d) Profit & Loss
Answer:
(d) Profit & Loss

Question 3.
A ____________ is an Intangible Asset.
(a) Goodwill
(b) Stock
(c) Cash
(d) Furniture
Answer:
(a) Goodwill

Question 4.
In the absence of an agreement, interest on a loan advanced by the partner to the firm is allowed at the rate of ____________
(a) 5%
(b) 6%
(c) 10%
(d) 9%
Answer:
(b) 6%

Question 5.
Liability of partners in a partnership business is ____________
(a) limited
(b) unlimited
(c) limited and unlimited
(d) none of the above
Answer:
(b) unlimited

Question 6.
The Indian Partnership Act is in force since ____________
(a) 1932
(b) 1881
(c) 1956
(d) 1984
Answer:
(a) 1932

Question 7.
Maximum number of Partners in a firm are ____________ according to Companies Act, 2013.
(a) 10
(b) 25
(c) 20
(d) 50
Answer:
(d) 50

B. Write the word/phrase/term, which can substitute each of the following statements.

Question 1.
Persons who form the partnership firm.
Answer:
Partners

Question 2.
Amount of cash or goods withdrawn by partners from the business from time to time.
Answer:
Drawings

Question 3.
An association of two or more persons according to Indian Partnership Act 1932.
Answer:
Partnership firm

Question 4.
Act under which partnership firms are regulated.
Answer:
Indian Partnership Act

Question 5.
Process of entering the name of the partnership firm in the register of the Registrar.
Answer:
Registration

Question 6.
Partnership agreement in written form.
Answer:
Partnership Deed

Question 7.
Under this method capital, balances of partners remain constant.
Answer:
Fixed Capital Method

Question 8.
Proportion in which partners share profit.
Answer:
Profit-Sharing Ratio

Question 9.
Such a capital method in which only Capital Account is maintained for each partner.
Answer:
Fluctuating Capital Method

Question 10.
The account to which all adjustments are made when capital is fixed.
Answer:
Current Account

Question 11.
Expenses that are paid before they are due.
Answer:
Prepaid expenses

Question 12.
The accounts are prepared at the end of each accounting year.
Answer:
Final Accounts

Question 13.
An asset that can be converted into cash easily.
Answer:
Current Assets or Liquid Assets

Question 14.
Order in which fixed assets are recorded first in the Balance Sheet.
Answer:
Order of liquidation

Question 15.
The account in which selling expenses of the business are recorded.
Answer:
Profit and Loss Account

Question 16.
Debit balance of Trading Account.
Answer:
Gross loss

Question 17.
The credit balance of Profit and Loss Account.
Answer:
Net profit

C. State whether the following statements are True or False with reasons:

Question 1.
A partnership firm is a Non-Trading concern.
Answer:
This statement is False.
The main aim of a partnership firm Is to earn maximum profit. The partnership is a trading concern. It undertakes either manufacturing or distributive activities with the sole aim of earning profit and distribute that profit among the partners in a specific ratio. It is never formed for charitable purposes.

Question 2.
A profit and Loss Account is a Real Account.
Answer:
This statement is False.
Account of expenses, losses, gains, and incomes is called a Nominal account. The profit and Loss Account contains all indirect expenses and indirect incomes of the firm. Therefore, a Profit and Loss Account is a Nominal Account and not a real account.

Question 3.
Carriage inward is carriage on purchase.
Answer:
This statement is True.
Total transport expenses incurred on bringing the goods from market to the place of business is called the carriage. When goods are purchased, the carriage is supposed to be borne by the firm. It is known as carriage inward. It means carriage paid on purchase.

Question 4.
Adjustments are recorded in Partners Current Account in Fixed Capital Method.
Answer:
This statement is True.
In Fixed Capital Method, as the name suggests capital balances (opening and closing) are generally remain fixed. Under this method, adjustments are not to be recorded in Capital Account. All adjustments are recorded in a separate account called Partners’ Current Accounts.

Question 5.
Prepaid expenses are treated as liabilities.
Answer:
This statement is False.
Prepaid expenses are expenses that are paid before they are due. Therefore, they are considered an asset of the business organization.

Question 6.
If the partnership deed is silent, partners share profits and losses in proportion to their capital.
Answer:
This statement is False.
As per the provisions made under the Indian Partnership Act 1932, when a partnership deed is silent about profit and loss sharing ratio, partners are supposed to share profits and losses in equal proportion, and not in their capital ratio.

Question 7.
Balance Sheet is an Account.
Answer:
This statement is False.
A financial statement showing all assets and liabilities is called a Balance sheet. It is not an account. It is a position statement that shows various assets owned by the firm and various liabilities owned by it. On the left-hand side, all liabilities are listed and on the right-hand side, all assets are recorded.

Question 8.
Wages paid for the installation of machinery is a Revenue expenditure.
Answer:
This statement is False.
Wages paid for the installation of machinery is a capital expenditure and therefore it is added to the cost of machinery. It is generally, paid once in a life of an asset. It is a long-term and capital expenditure.

Question 9.
Income received in advance is a liability.
Answer:
This statement is True.
When Income in respect to next year, it received in the current year, it is known as income received in advance. So, in next year firm will not be able to receive that amount and therefore it is considered as a liability for the current year.

Question 10.
R.D.D. is created on Creditors.
Answer:
This statement is Raise.
R.D.D. stands for Reserve for Doubtful Debts. It is created on the value of debtors. Such provision is made against profit and loss accounts. In the future, if the loss is incurred on account of bad debts, such an amount is used to run the business.

Question 11.
Depreciation is not calculated on Current Assets.
Answer:
This statement is True.
Current Assets mean liquid assets having no fixed tenure therefore depreciation cannot be calculated on it. Depreciation is calculated and charged on fixed assets for their use, wear and tear, etc.

Question 12.
Goodwill is an intangible asset.
Answer:
This statement is True.
Goodwill is a reputation of business computed in terms of money. Reputation can be experienced but can’t be seen or felt. Therefore, Goodwill is an intangible asset.

Question 13.
Indirect expenses are debited to Trading Account.
Answer:
This statement is Raise.
Indirect expenses mean expenses that are not directly related to the production of goods and services. Therefore, indirect expenses cannot be debited to Trading Account. All indirect expenses are debited to the Profit and Loss Account.

Question 14.
A bank loan is a current liability.
Answer:
This statement is Raise.
A loan usually taken for the period of more than 1 year say 5 years from the bank is called Bank Loan. It is a long term loan. It is not repaid within 1 year but paid in installments over a number of years. It might be paid in lumpsum at the expiry of the term.

Question 15.
Net profit is the debit balance of Profit and Loss Account.
Answer:
This statement is Raise.
In a Profit and Loss Account, when the credit side total i.e. a total of incomes is more than the debit side total, i.e. expenses it is known as a credit balance. When incomes exceed expenses there is profit. Therefore credit balance of the Profit and Loss Account indicates net profit.

D. Find an odd one.

Question 1.
Wages, Salary, Royalty, Import Duty
Answer:
Salary

Question 2.
Postage, Stationery, Advertising, Purchases
Answer:
Purchases

Question 3.
Capital, Bills Receivable, Reserve fund, Bank overdraft
Answer:
Bills Receivable

Question 4.
Building, Machinery, Furniture, Bills Payable
Answer:
Bill Payable

Question 5.
Discount received, Dividend received, Interest received, Depreciation
Answer:
Depreciation

E. Complete the sentences.

Question 1.
Partners share profits & losses in ____________ ratio in the absence of partnership deed.
Answer:
equal

Question 2.
Registration of partnership is ____________ in India.
Answer:
optional

Question 3.
Partnership business must be ____________
Answer:
lawful

Question 4.
Liabilities of partners in partnership firm is ____________
Answer:
unlimited

Question 5.
The balance of the Drawings Account of a partner is transferred to his ____________ account under the Fixed Capital Method.
Answer:
Current

Question 6.
The interest on capital of a partner is debited to ____________ account.
Answer:
Profit and Loss

Question 7.
Partners are ____________ liable for the debts of the firm.
Answer:
joint & several

Question 8.
Partnership Deed is an ____________ of Partnership.
Answer:
Article

Question 9.
The withdrawal by the partner for personal use from the firm is ____________ to his account.
Answer:
debited

Question 10.
Commission payable to partner is ____________ to the firm.
Answer:
liability/outstanding expense

Question 11.
When partners adopt Fixed Capital Method then they have to operate ____________ Account.
Answer:
Partner’s Current

Question 12.
If the partners Current Account shows ____________ balance it is shown to the Liability side of the Balance Sheet.
Answer:
credit

Question 13.
The expenses paid for trading purpose are known as ____________ expenses.
Answer:
trade

Question 14.
Cash receipts which are recurring in nature are called as ____________ Receipts.
Answer:
Revenue

Question 15.
Return outward are deducted from ____________
Answer:
purchase

Question 16.
Expenses which are paid before due date are called as ____________
Answer:
Prepaid Expenses

Question 17.
Assets which are held in the business for a long period are called ____________
Answer:
Fixed Assets

Question 18.
Trading Account is prepared on the basis of ____________ expenses.
Answer:
direct

Question 19.
When commission is allowed to any partner, it is ____________ of the business.
Answer:
expenditure

Question 20.
When goods are distributed as free samples, it is treated as ____________ of the business.
Answer:
advertisement expense

F. Answer in one sentence only:

Question 1.
What is Fluctuating Capital?
Answer:
When capital balances of the partners go on changing every year due to transactions of partners with the firm, it is known as Fluctuating Capital.

Question 2.
Why is Partnership Deed necessary?
Answer:
Partnership Deed is necessary to prevent disputes or misunderstandings among the partners in the future.

Question 3.
If the Partnership Deed is silent, in which ratio, the partners will share the profit or loss?
Answer:
If the Partnership Deed is silent, partners will share profits and losses in equal ratio.

Question 4.
What is the Fixed Capital Method?
Answer:
Fixed Capital Method is one in which capital balances of the partners remain the same at the end of every financial year unless any amount of additional capital is introduced or part of the capital is withdrawn by the partner from the business.

Question 5.
How many partners are required to form a partnership firm?
Answer:
Minimum two persons are required to form a partnership firm.

Question 6.
What is Partnership Deed?
Answer:
A partnership deed is a written agreement duly stamped and signed document containing the terms and conditions of the partnership.

Question 7.
What are the objectives of the Partnership Firm?
Answer:
To earn a maximum profit is the main objective of the partnership firm.

Question 8.
What rate of interest is allowed on a partner’s loan in the absence of an agreement?
Answer:
6 % is the rate of interest to be allowed on a partner’s loan in the absence of an agreement.

Question 9.
What is the minimum number of partners in a partnership firm according to the Indian Partnership Act 1932?
Answer:
Minimum two persons are required a number of partners in a partnership firm according to Indian Partnership Act 1932.

Question 10.
What is the liability of a partner?
Answer:
The liability of a partner (except minor partner) is unlimited.

Question 11.
In the absence of Partnership Deed, what is the rate of interest on a loan advanced by the partner to the firm is allowed?
Answer:
In the absence of Partnership Deed, 6% is the rate of interest on a loan advanced by the partner to the firm.

Question 12.
What do you mean by pre-received income?
Answer:
Income that is received by the partnership firm before it is due is called pre-received income.

Question 13.
What is the effect of the adjustment of provision for discount on debtors in the final accounts of partnership?
Answer:
The effects of the adjustment of provision for discount on debtors in the final accounts of partnership are as follows:
Debit Profit and Loss A/c and deduct the amount of provision for discount on debtors from the number of debtors.

Question 14.
When are the Partners Current Account is opened?
Answer:
When Fixed Capital Method is adopted by the firm, Partners’s Current Account is opened.

Question 15.
As per which principle of accounting, closing stock is valued at cost price or at market price whichever is less?
Answer:
As per the Conservatism principle of accounting, the closing stock is valued at cost price or at market price whichever is less.

Question 16.
What is the provision of the Indian Partnership Act with regard to Interest on Capital?
Answer:
As per the provision of the Indian Partnership Act, Interest in Capital is not to be allowed.

Question 17.
Why is the Balance Sheet prepared?
Answer:
The Balance Sheet is prepared to know the financial position of the business in the form of its assets and liabilities on a particular date.

Question 18.
Why wages paid for the installation of machinery are not shown in Trading Account?
Answer:
Wages paid for the installation of machinery is a capital expenditure and it is not to be recorded in Trading Account.

Question 19.
What do you mean by indirect incomes?
Answer:
All incomes other than direct incomes are called indirect incomes.
[e.g. Interest received on investments, Incomes like discount, commission, dividend, rent, etc. received].

Question 20.
Why partners capital is treated as a long-term liability of business?
Answer:
Partner’s Capital is not refunded during the existence of the partnership firm unless the partner is retired or expired.

G. Do you agree/disagree with the following statements:

Question 1.
When Partnership Deed is silent, partners share profits of the firm according to capital ratio.
Answer:
Disagree

Question 2.
The current Account always shows a debit balance.
Answer:
Disagree

Question 3.
It is compulsory to have a partnership agreement in writing.
Answer:
Disagree

Question 4.
Partnership Firm is a trading concern.
Answer:
Agree

Question 5.
Interest in the capital is an expenditure for the partnership firm.
Answer:
Agree

Question 6.
A partnership is an association of two or more persons.
Answer:
Agree

Question 7.
Partners are entitled to get a Salary or Commission.
Answer:
Disagree

Question 8.
The balance of the Capital Account remains constant under Fixed Capital Method.
Answer:
Agree

Question 9.
The Indian Partnership Act came into existence in the year 1945.
Answer:
Disagree

Question 10.
Profit and Loss Account reflects the true financial position.
Answer:
Disagree

Question 11.
The amount borrowed by a partner from his business will be debited to the Current Account.
Answer:
Agree

Question 12.
Sold but undispatched goods must be part of the valuation of closing stock.
Answer:
Disagree

Question 13.
Carriage inward is a selling and distribution overhead.
Answer:
Disagree

Question 14.
Gross profit is an operating profit.
Answer:
Disagree

Question 15.
All financial expenditures are debited to the Profit and Loss Account.
Answer:
Agree

Question 16.
Free distribution of goods is debited to Trading Account.
Answer:
Disagree

H. Calculate the following:

Question 1.
Undervaluation of closing stock by 10%, closing stock was ₹ 30,000. Find out the value of the closing stock.
Solution:
Undervaluation of closing stock by 10 %
Revised value = \(\frac{\text { Book value }}{100-\% \text { of undervaluation }} \times 100\)
= \(\frac{30,000}{100-10} \times 100\)
= ₹ 33,333.
∴ Value of closing stock = ₹ 33,333.

Question 2.
Calculate 12.5% P.A. depreciation on Furniture:
(a) on ₹ 220,000 for 1 year
(b) on ₹ 10,000 for 6 months
Solution:
Depreciation = Amount of asset × Period × %
(a) Depreciation on furniture = 220,000 × 1 × \(\frac{12.5}{100}\) = ₹ 27,500
∴ Deprecation on furniture for 1 year = ₹ 27,500

(b) Depreciation on furniture = 10,000 × \(\frac{6}{12} \times \frac{12.5}{100}\) = ₹ 625
∴ Depreciation on furniture for 6 months = ₹ 625

Question 3.
The insurance premium is paid for the year ending on 1st September 2019 amounted to ₹ 1500. Calculate prepaid insurance assuming that the year-end is 31st March 2019.
Solution:
From 31st March to 1st September, 5 months period prepaid insurance amount we have to find.
An insurance premium paid for the 12 months = ₹ 1500
∴ for 5 months period it is 1500 × \(\frac{5}{12}\) = ₹ 625
Thus, prepaid insurance premium amount = ₹ 625.

Question 4.
Find out Gross Profit/Gross Loss: Purchases ₹ 30,000, Sales ₹ 15,000, Carriage inward ₹ 2400, Opening stock ₹ 10,000, Purchase return ₹ 1000, Closing stock ₹ 36,000.
Solution:

Question 5.
A borrowed loan from Bank of Maharashtra ₹ 2,00,000 on 1st October 2019 @15 % p.a. Calculate interest on a bank loan for the year 2019 – 20 assuming that the financial year ends on 31st March, every year.
Solution:
From 1st October to 31st March, 6 months period interest on loan is to be calculated.
Interest (I) = \(\frac{\text { PRN }}{100}\)
∴ Interest on loan = 2,00,000 × \(\frac{15}{100} \times \frac{6}{12}\) = ₹ 15,000
∴ Interest on loan on ₹ 2,00,000 for 6 months = ₹ 15,000

Practical Problems

Question 1.
Amitbhai and Narendrabhai are in Partnership Sharing Profits and Losses equally. From the following Trial Balance and Adjustments given below, you are required to prepare the Trading and Profit and Loss Account for the year ended 31st March 2019 and Balance Sheet as of that date.

Trial Balance as of 31st March 2019

Adjustments:
1. Stock on hand on 31st March 2019 was valued at ₹ 43,000.
2. Uninsured goods worth ₹ 8,000 were stolen.
3. Create R.D.D. at 2 % on sundry debtors.
4. Mr. Patil, our customer becomes insolvent and could not pay his debts of ₹ 500.
5. Outstanding Expenses – Rent ₹ 800 and salaries ₹ 300.
6. Depreciate Factory Building by ₹ 2,500 and Furniture by ₹ 1,800.
Solution:
In the books of Amitbhai and Narendrabhai

Balance Sheet as of 31st March 2019


Notes:
1. Import duty, Motive power, and Depreciation on Factory building are recorded in the Trading A/c.

2. 10% govt, the bond is an investment. It was purchased on 1 – 10 – 2018.
∴ Interest is calculated for six months.
Interest on Govt. Bond = \(\frac{40,000}{1} \times \frac{6}{12} \times \frac{10}{100}\) = ₹ 2,000

3. Adv. exp. paid for 2 years from 01 – 01 – 2019. Upto 31 – 3 – 2019, 3 months adv. exp. is written off to Profit and Loss A/c. It is calculated as below:
= 10,000 × \(\frac{3}{24}\) = ₹ 1,250
∴ Prepaid adv. exp. = 10,000 – 1,250 = ₹ 8,750

Question 2.
From the following Trial Balance of M/s Mitesh and Mangesh, you are required to prepare Trading and Profit and Loss Account for the year ended 31st March 2019 and Balance Sheet as of that date.

Trial Balance as of 31st March 2019

Adjustments:
1. Mitesh and Mangesh are sharing profit and losses in the ratio 3 : 1.
2. Partners are entitled to get commission @ 1% each on gross profit.
3. The closing stock is valued at ₹ 23,700.
4. Outstanding Expenses – Audit fees ₹ 400; Carriage ₹ 600.
5. Building is valued at ₹ 46,500.
6. Furniture is depreciated by 5%.
7. Provide interest on partner’s capital at 2.5% p.a.
8. Goods of ₹ 900 were taken by Mangesh for his personal use.
9. Write off ₹ 1,000 as Bad debts and maintain R.D.D. at 3 % on Sundry Debtors.
Solution:
In the books of M/s Mitesh and Mangesh



Balance Sheet as of 31st March 2019


Working Notes:
1. In this problem, Current Account balances are given. So, the total amount of fixed capital is directly shown in the Liabilities side of the Balance Sheet. Effects of adjustments related to commission to partners, interest on capital, goods are withdrawn by Mangesh are given in the Current Account. Closing balances of the Current Account are shown separately on the Liability side of the Balance Sheet.

2. Building is valued at ₹ 46,500 whereas the opening balance of Building given is ₹ 48,500. Therefore, a difference of the amount of ₹ 2,000 (48,500 – 46,500) is nothing but Depreciation charged on Building.

3. Return Inward ⇒ Sales Return
Return Outward ⇒ Purchase Return

4. Commission payable to partners:
Mitesh = 1% on Gross Profit = \(\frac{1}{100} \times \frac{99,000}{1}\) = ₹ 990/-
Mangesh = 1% on Gross Profit = \(\frac{1}{100}\) × 99,000 = ₹ 990/-

Question 3.
From the following Trial Balance and Adjustments given below of Reena and Aarti, you are required to prepare Trading and Profit and Loss Account for the year ended 31st March 2019 and Balance Sheet as of that date.
Trial Balance as of 31st March 2019

Adjustments:
1. Closing stock valued at ₹ 22,000.
2. Write off ₹ 900 for bad & doubtful debts and create a provision for reserve for doubtful debts ₹ 1,000.
3. Create a provision for discount on debtors @ 3 % and on creditors @ 5%.
4. Outstanding Expenses – Wages ₹ 700 and Salaries ₹ 800.
5. Insurance is paid for 15 months, w.e.f. 1st April 2018.
6. Depreciate land and building @ 5%.
7. Reena & Aarti are sharing Profits & Losses in their Capital Ratio.
Solution:
In the books of Reena and Aarti

Balance Sheet as of 31st March 2019

Working Notes:
1. Insurance premium ₹ 1,500 is paid for 15 months, i.e. prepaid insurance premium for 3 months = ₹ 300.

2. Reserve for Discount on Debtors = 3% on (Debtors – New Bad debts – New Reserve)
= \(\frac{3}{100}\) × (40,000 – 900 – 1,000)
= \(\frac{3}{100}\) × (40,000 – 1,900)
= \(\frac{3}{100}\) × 38,100
= ₹ 1,143

3. Reserve for Discount on Creditors = 5% on (Value of Creditors)
= \(\frac{5}{100}\) × 25,700
= ₹ 1,285

4. Profit and Loss ratio = Capital ratio = 50,000 : 30,000 = 5 : 3

Question 4.
From the following Trial Balance of M/s Meera and Madhav. Prepare Trading and Profit and Loss Account for the year ended 31st March 2019 and Balance Sheet as on that date.
Trial Balance as of 31st March, 2019

Adjustments:
1. Closing stock is valued at ₹ 32,000.
2. Provide provision for doubtful debts ₹ 2,000.
3. Create a reserve for a discount on debtors @ 3%.
4. Value of leasehold premises on 31st March 2019 ₹ 1,00,000.
5. Outstanding Expenses: Printing & stationery ₹ 500.
Solution:
In the books of M/s Meera and Madhav

Balance Sheet as of 31st March 2019

Working Notes:
1. Advertisement expenses written off to Profit and Loss account during the year 2018-19 for six months i.e. from 1/10/18 to 31/03/19.
Advertisement expenses W/off = (Advertisement bill paid) × \(\frac{1}{3} \times \frac{6}{12}\)
= 4,800 × \(\frac{1}{3} \times \frac{6}{12}\)
= ₹ 800.
Prepaid advertisement = 4,800 – 800 = ₹ 4,000.

2. Reserve for Discount on Debtors = 3% (Balance in debtors)
= \(\frac{3}{100}\) × (80,500 – 2,000)
= \(\frac{3}{100}\) × 78,500
= ₹ 2,355.

3. Difference between the opening balance (₹ 1,10,000) and the closing balance (₹ 1,00,000) for leasehold premises is to be considered as written off on leasehold premises.

Question 5.
Sucheta & Gayatri are partners sharing Profit and Losses in the ratio 3 : 2. From the following Trial Balance and additional information, you are required to prepare Trading and Profit and Loss Account for the year ended 31st March 2019 and Balance Sheet as of that date.
Trial Balance as of 31st March 2019

Adjustments:
1. Stock on 31st March 2019 was valued at ₹ 19,700.
2. Goods costing ₹ 3,000 distributed as a free sample.
3. Motive power includes ₹ 500 paid for deposit of Power Meter.
4. Depreciate building @ 5 %.
5. Write off ₹ 2000 for bad debts and maintain R.D.D. at 3% on debtors.
6. Bills receivable included dishonored of Bill of ₹ 4,000.
Solution:
In the books of Sucheta and Gayatri


Balance Sheet as of 31st March, 2019

Working Notes:
1. Rate of interest on the partner’s loan is not mentioned, therefore interest on the loan is calculated at 6% p.a.
∴ Interest on Sucheta’s Loan = 6,150 × 1 × \(\frac{6}{100}\) = ₹ 369
2. Add dishonored bill amount to debtors amount and then calculate B.D. and R.D.D.
3. Subtract dishonored bill amount from bills receivable amount.

Question 6.
Archana and Prerana are partners, sharing Profits and Losses in the ratio 2 : 1 with the help of the following Trial Balance and Adjustments given below. You are required to prepare a Trading and Profit and Loss Account for the year ended 31st March 2019 and a Balance Sheet as of that date.
Trial Balance as of 31st March, 2019

Adjustments:
1. Stock on 31st March 2019 is valued at Cost Price ₹ 12,000 and Market Price ₹ 17,000.
2. Our customer Mr. Shekhar failed to pay his dues of ₹ 800.
3. 1/8th of Patents are to be written off.
4. A part of Furniture ₹ 5,000 is purchased on 1st Oct. 2018.
5. Depreciation on Land & Building 10% and on Furniture 5%.
6. Outstanding Expenses Wages ₹ 300 and Electricity Charges ₹ 200.
7. Allow Interest on Capital 3%.
Solution:
In the books of Archana and Prema


Balance Sheet as of 31st March 2019

Working Notes:
1. Stationery stock is an asset.
2. Depreciation of furniture:

∴ Total Depreciation = 400 + 125 = ₹ 525
3. \(\frac{1}{8}\) patents to be written off = 2,000 × \(\frac{1}{8}\) = ₹ 250.
4. As no other expenses are given, Trade Expense is recorded in Profit and Loss Account.

Question 7.
Satish and Pramod are partners. Prepare Trading Account and Profit and Loss Account for the year 31st March 2019. You have to find out Gross Profit and Net Profit only.
Trial Balance as of 31st March 2019

Adjustments:
1. The Closing stock is valued at ₹ 15,400.
2. Outstanding wages ₹ 500.
3. Create provision for Bad debts ₹ 800 and maintain R.D.D. 3 % on Sundry Debtors.
4. Goods of ₹ 1,800 distributed as a free sample.
5. Goods of ₹ 2,000 were sold and delivered on 31st March 2019 but no entry is passed in the Books of Account.
Solution:
In the books of Satish and Pramod

Working Notes :
1. Here only gross profit and net profit is to find out. Therefore, the Balance Sheet is not prepared.

2. Interest on a 10% bank loan is calculated for 9 months (From 1/7/2018 to 31/3/2019)
I = \(\frac{\mathrm{PRN}}{100}\) = 8,000 × \(\frac{10}{100} \times \frac{9}{12}\) = ₹ 600

3. Goods distributed as free samples is an advertisement expense for business.

4. Sundry Debtors = 40,000
Add: Unrecorded Sales = 2,000
Less: Provision for Bad Debts = 800
Total = 41,200
Less: R.D.D. (New) (3% of 41,200 = 1,236) = 39,964

Question 8.
Nana and Nani are partners in a Partnership Firm sharing Profits and Losses equally. You are required to give effects of Adjustments in Profit & Loss A/c and Balance Sheet with the help of the following information.
Trial Balance as of 31st March 2019

Adjustments:
1. Gross profit amounted to ₹ 34,500.
2. Insurance paid for 15 months w.e.f. 1-4-2018.
3. Depreciate Land and Building at 10 % p.a. and Furniture at 5% p.a.
4. Write off ₹ 1,000 for Bad debts and maintain R.D.D. at 5 % on Sundry debtors.
5. Closing stock is valued at ₹ 34,500.
Solution:
In the books of Nana and Nani

Balance Sheet as of 31st March 2019

Working Notes:
1. Here, the Profit and Loss Account and Balance Sheet are to be prepared. Therefore, Trading Account is not prepared. Gross profit (given) is recorded on the Credit side of the Profit and Loss Account.

2. Land and Building

Total Depreciation = ₹ 4,500

3. Interest on 10% bank loan is calculated for 6 months. (From 1/10/2018 to 31/3/2019)
I = \(\frac{\text { PRN }}{100}\)
= 30,000 × \(\frac{10}{100} \times \frac{6}{12}\)
= ₹ 1,500

4. Prepaid insurance = \(\frac{3}{15}\) × (Insurance Amount)
= \(\frac{3}{15}\) × 15,000
= ₹ 3,000

5. RDD = 5% on (Debtors – New Bad debts)
= \(\frac{5}{100}\) × (26,000 – 1,000)
= \(\frac{5}{100}\) × 25,000
= ₹ 1,250

Question 9.
Sun and Moon are partners in a Partnership Firm sharing Profits and Losses equally. You are required to give effects of Adjustments with the help of the following information:
Trial Balance as of 31st March 2019

Adjustments:
1. Partners are entitled to get a salary of ₹ 6,000 p.a. in addition to their profit & loss sharing.
2. Depreciation on Land & Building, Furniture and Machinery @ 10%, 5% and 3% respectively.
3. Interest in Capital 5% p.a.
4. Closing stock ₹ 60,743.
5. Wages included ₹ 1,000 as advance is given to workers.
6. Interest due but not paid ₹ 800.
7. Total net profit amounted to ₹ 38,113.
You are required to prepare the Balance Sheet and Partners Current A/c only.
Solution:
In the books of Sun and Moon

Balance Sheet as of 31st March 2019


Working Notes:
1. Depreciation on machinery is calculated for 9 months. (i.e. from 1/7/18 to 31/3/19)
Depreciation = 40,000 × \(\frac{3}{100} \times \frac{9}{12}\) = ₹ 900

2. Interest on 8% debentures, calculated for 6 months. (i.e. from 1/10/18 to 31/3/19)
I = \(\frac{\text { PRN }}{100}\)
= 8,000 × \(\frac{8}{100} \times \frac{6}{12}\)
= ₹ 320

3. Advance given to workers (by firm) ₹ 1,000 is an asset for the firm, so, it is shown on the Assets side.

4. Interest due but not paid is a liability for the firm, so, it is shown on the Liabilities side.

Question 10.
Kshipra and Manisha are partners sharing Profit and Losses in their Capital ratio. You are required to prepare Trading Account and Profit and Loss Account for the year ended 31st March 2019 and a Balance Sheet as of that date.
Trial Balance as of 31st March 2019

Adjustments:
1. Stock on 31st March 2019 was at ₹ 37,000.
2. Sales include the sale of machinery of ₹ 2,000, which is sold on 1st April 2018.
3. Depreciation on fixed assets @ 5%
4. Each partner is entitled to get a commission at 1% of Gross profit and interest on Capital 5% p.a.
5. Outstanding Expenses wages ₹ 200 & Salaries ₹ 500.
6. Create provision for Doubtful debts @ 3% on Sundry debtors.
Solution:
In the books of Kshipra and Manisha


Balance Sheet as of 31st March 2019

Working Notes:
1. Depreciation on fixed assets means depreciation on Furniture, Plant & Machinery, and Building.

2. Sales includes the sale of Machinery of ₹ 2,000 is subtracted from sales and from Plant & Machinery.
On balance amount of Plant & Machinery ₹ 58,000, calculate 5 % depreciation i.e. 60,000 – 2,000 = ₹ 58,000 × 5% = ₹ 2,900

3. Here on gross profit calculate 1% commission for partners and record it to Profit and Loss A/c and in Current A/cs. Commission payable to each partner = \(\frac{1}{100}\) × Gross profit
= \(\frac{1}{100}\) × 81,700
= ₹ 817.

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 13 Amines

1. Choose the most correct option.

Question i.
The hybridisation of nitrogen in primary amine is ………………………. .
a. sp
b. sp²
c. sp³
d. sp³d
Answer:
c. sp³

Question ii.
Isobutylamine is an example of ………………………. .
a. 2° amine
b. 3° amine
c. 1° amine
d. quaternary ammonium salt.
Answer:
a. 2° amine

Question iii.
Which one of the following compounds has the highest boiling point?
a. n-Butylamine
b. sec-Butylamine
c. isobutylamine
d. tert-Butylamine
Answer:
a. n-Butylamine

Question iv.
Which of the following has the highest basic strength?
a. Trimethylamine
b. Methylamine
c. Ammonia
d. Dimethylamine
Answer:
d. Dimethylamine

Question v.
Which type of amine does produce N₂ when treated with HNO₂?
a. Primary amine
b. Secondary amine
c. Tertiary amine
d. Both primary and secondary amines
Answer:
a. Primary amine

Question vi.
Carbylamine test is given by
a. Primary amine
b. Secondary amine
c. Tertiary amine
d. Both secondary and tertiary amines
Answer:
a. Primary amine

Question vii.
Which one of the following compounds does not react with acetyl chloride?
a. CH₃-CH₂-NH₂
b. (CH₃-CH₂)₂NH
c. (CH₃-CH₂)₃N
d. C₆H₅-NH₂
Answer:
Answer:
c. (CH₃ – CH₂)₃N

Question viii.
Which of the following compounds will dissolve in aqueous NaOH after undergoing reaction with Hinsberg reagent?
a. Ethylamine
b. Triethylamine
c. Trimethylamine
d. Diethylamine
Answer:
a. Ethyl amine

Question ix.
Identify ‘B’ in the following reactions

Answer:
d. CH₃-CH₂-OH

Question x.
Which one of the following compounds contains azo linkage?
a. Hydrazine
b. p-Hydroxyazobenzene
c. N-Nitrosodiethylamine
d. Ethylenediamine
Answer:
b. p-Hydroxyazobenzene

2. Answer in one sentence.

Question i.
Write reaction of p-toluenesulfonyl chloride with diethylamine.
Answer:

Question ii.
How many moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine?
Answer:
2 moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine.

Question iii.
Which amide does produce ethanamine by Hofmann bromamide degradation reaction?
Answer:
Propanamide (CH₃ – CH₂ – CONH₂) produces ethanamine by Hofmann bromamide degradation reaction.

Question iv.
Write the order of basicity of aliphatic alkylamine in gaseous phase.
Answer:
The order of basicity of aliphatic alkyl amines in the gaseous follows the order : tertiary amine > secondary amine > primary amine > NH₃.

Question v.
Why are primary aliphatic amines stronger bases than ammonia?
Answer:
The alkyl group tends to increase the electron density on the nitrogen atom. As a result, amines can donate the lone f pair of electrons on nitrogen more easily than ammonia. Hence, aliphatic amines are stronger bases than ammonia.

Question vi.
Predict the product of the following reaction. Nitrobenzene Sn/Conc. HCl?
Answer:
The product is aniline/

Question vii.
Write the IUPAC name of benzylamine.
Answer:
The IUPAC name is Phenylmethanamine.

Question viii.
Arrange the following amines in an increasing order of boiling points. n-propylamine, ethylmethyl amine, trimethylamine.
Answer:
Amines in an increasing order of boiling points : trimethyl amine, ethyl methyl amine, n-propyl amine

Question ix.
Write the balanced chemical equations for the action of dil H₂SO₄ on diethylamine.
Answer:

Question x.
Arrange the following amines in the increasing order of their _pK_b values. Aniline, Cyclohexylamine, 4-Nitroaniline
Answer:
Cyclohexyl amine (_pK_A 3.34), aniline (_pK_A 9.13) 4-nitroaniline (_pK_A 12.99)

3. Answer the following

Question i.
Identify A and B in the following reactions.

Answer:

Question ii.
Explain the basic nature of amines with suitable example.
Answer:
The basic strength of amines is expressed in terms of K_b or _pK_b value. According to Lowry-Bron-sted theory the basic nature of amines is explained by the following equilibrium equation.

In this equilibrium amine accepts H⁺, hence an amine is a Lowry-Bronsted base.

According to Lewis theory, the species which donates a pair of electrons is called a base.

The nitrogen atom in amiqes has a lone pair of electrons, which can be donated to suitable acceptor like proton H⁺.

The aqueous solutions of amines are basic in nature due to release of free OH^– ions in solutions. Hence amines are Lewis bases. There exists an equilibrium in their aqueous solutions as follows :

R – NH₂ + H₂O ⇌ RNH₃ + OH^–

Since OH^– is a stronger base, equilibrium shifts towards left-hand side giving less concentration of OH^–.

Here, K_b value is smaller and _pK_b value is larger.

Hence amines are weak bases.

Question iii.
What is diazotisation? Write diazotisation reaction of aniline.
Answer:
Aryl amines react with nitrous acid in cold condition (273 – 278 K) forms arene diazonium salts. The conversion of primary aromatic amine into diazonium salts is called diazotisation.

Diazotisation of aniline :

Question iv.
Write reaction to convert acetic acid into methylamine.
Answer:

Question v.
Write a short note on coupling reactions.
Answer:

Reactions involving retention of diazo group : (Coupling reactions) :

Question vi.
Explain Gabriel phthalimide synthesis.
Answer:
Phthalimide is reacted with alcoholic KOH to form potassium phthalimide. Further potassium phthalimide is treated with an ethyl iodide. The product N-ethylphthalimide is hydrolysed with aq NaOH to form ethyl amine. This reaction is known Gabriel phthalimide synthesis.

Question vii.
Explain carbylamine reaction with suitable examples.
Answer:
Aliphatic or aromatic primary amines on heating with chloroform and alcoholic potassium hydroxide solution form carbyl amines or isocyanides with extremely unpleasant smell. This reaction is a test for primary amines.

Secondary and tertiary amines do not give this test.

Question viii.
Write reaction to convert
(i) methanamine into ethanamine
(ii) Aniline into p-bromoaniline.
Answer:
(1) Methanamine into ethanamine

(2) Aniline into p-bromo aniline

Question ix.
Complete the following reactions :
a. C₆H₅N₂ Cl + C₂H₅OH →
b. C₆H₅NH₂ + Br₂(aq) → ?
Answer:
(a)

(b)

Question x.
Explain Ammonolysis of alkyl halides.
Answer:
When an alkyl halide is heated with alcoholic ammonia in a sealed tube under pressure at 373 K, a mixture of primary, secondary, tertiary amines and a quaternary ammonium salt is obtained. In this reaction, breaking of C – X bond by ammonia is called ammonolysis of alkyl halides. The reaction is also known as alkylation. For example, when methyl bromide is heated with alcoholic ammonia at 373 K, it gives a mixture of methylamine (a primary amine), dimethylamine (a secondary amine), trimethyl amine (a tertiary amine) and tetramethylam- monium bromide (a quaternary ammonium salt).

The order of reactivity of alkyl halides with ammonia is R – I > R – Br > R – Cl.

Question xi.
Write reaction to convert ethylamine into methylamine.
Answer:

4. Answer the following.

Question i.
Write the IUPAC names of the following amines :

Answer:

Question ii.
What are amines? How are they classified?
Answer:
Amines are classified on the basis of the number of hydrogen atoms of ammonia that are replaced by alkyl group. Amines are classified as primary (1°), secondary (2°) and tertiary (3°).

(1) Primary amines (1° amines) : The amines in which only one hydrogen atom of ammonia is replaced by an alkyl group or aryl group are called primary (1°) amines.

Examples :
(i) CH₃ – NH₂ methylamine
(ii) CH₃ – CH₂ – NH₂ ethylamine

(2) Secondary amines (2° amines) : The amines in which two hydrogen atoms of ammonia are replaced by two, same or different alkyl or aryl groups are called secondary (2°) amines.

Examples :
(i) C₂H₅ – NH – CH₃ ethylmethylamine
(ii) CH₃ – NH – CH₃ dimethylamine

(3) Tertiary amines (3° amines) : The amines in which all the three hydrogen atoms of ammonia are replaced by three same or different alkyl or aryl groups are called tertiary (3°) amines.

Examples :

Secondary and tertiary amines are further classified as (1) Simple or symmetrical amines (2) Mixed or unsymmetrical amines.

(i) Simple or symmetrical amines : In simple amines same alkyl groups are attached to the nitrogen e.g.

(ii) Mixed or unsymmetrical amines : In mixed amines different alkyl groups are attached to the nitrogen.

Question iii.
Write IUPAC names of the following amines.

Answer:

Question iv.
Write reactions to prepare ethanamine from
a. Acetonitrile
b. Nitroethane
c. Propionamide
Answer:
a. Ethanamine from acetonitrile :

b. Ethanamine from nitroethane :

c. Ethanamine from Propionamide :

Question v.
What is the action of acetic anhydride on ethylamine, diethylamine and triethylamine?
Answer:
Acetylation of amines : The reaction in which the H atom attached to nitrogen in amine is replaced by acetyl group is called acetylation of amines.

(1) Ethylamine on reaction with acetic anhydride forms monoacetyl derivative, N-acetylethylamine.

(2) Diethylamine (a secondary amine) on reaction with acetic anhydride forms a monoacetyl derivative, N-acetyldiethyl amine (or N,N-diethyl acetamide).

(3) Triethylamine does not react with acetic anhydride as it does not have any H atom attached nitrogen atom of amin e

Question vii.
Distinguish between ethylamine, diethylamine and triethylamine by using Hinsberg’s reagent?
Answer:
This reaction is useful for the distinction of primary, secondary and tertiary amines.

(i) Primary amine (like ethyl amine) is treated with Hinsberg’s reagent (benzene sulphonyl chloride) forms N-alkyl benzene sulphonamide which dissolve in aqueous KOH solution to form a clear solution of potassium salt and upon acidification gives insoluble N-alkyl benzene sulphonamide.

(ii) Secondary amine like diethyl amine is treated with benzene sulphonyl chloride forms N,N-diethyl benzene which sulphonyl amide remains insoluble in aqueous KOH and does not dissolve in acid.

(iii) Tertiary amine like triethyl amine does not react with benzene sulphonyl chloride and remains insoluble in KOH, however it dissolves in dil. HCl to give a clear solution due to formation of ammonium salt.

Question viii.
Write reactions to bring about the following conversions :
a. Aniline into p-nitroaniline
b. Aniline into sulphanilic acid?
Answer:
(1) Aniline into p-nitroaniline

(2) Aniline into sulphanilic acid.

Activity :

• Prepare a chart of azodyes, colours and its application.
• Prepare a list of names and structures of N-containing ingredients of diet.

12th Chemistry Digest Chapter 13 Amines Intext Questions and Answers

Use your brain power! (Textbook Page No 282)

Question 1.
Classify the following amines as simple/mixed; 1°, 2°, 3° and aliphatic or aromatic. (C₂H₅)₂NH, (CH₃)₃N, C₂H₅ – NH – CH₃,

Answer:

(A) Common Names : Rules

1. According to common naming system, the amines are named as alkylamines.
2. The common name of a primary amine is obtained by writing the name of the alkyl group followed by the word ‘amine’.
   Example : CH₃ – NH₂ : methyl-amine
3. The simple {symmetrical) secondary and tertiary amines are written by adding prefix ‘di- (forpresence of two alkyl groups) and ‘tri’- (for presence of three alkyl groups) respectively to the name of alkyl groups.
   Examples: (i) CH₃ – NH – CH₃ dimethylamine, (ii) (C₂H₅)₃ N triethylamine
4. The mixed (or unsymmetrical) secondary and tertiary amines are given names by writing the names of alkyl groups in alphabetical order, followed by the word ‘amine’.
   Example : CH₃ – CH₂ – NH – CH₃ ethyhnethylamine

(B) IUPAC names : Rules

1. According to IUPAC system of nomenclature of amines, aliphatic amines are named as alkanarnines.
2. The name of the amine is obtained by replacing the suffix ‘e’ from parent alkane’s name by ‘amine’.
3. The position of the amino group is indicated by the lowest possible locant.
   Example :
   
4. In case of secondary and tertiary amines, the largest alkyl group is considered to be the parent alkane and other alkyl groups are written as N-substituents.
   Example : ClH₅NH – CH₃ N – Methylethanamine
5. A complete name of amine is written as one word.

Try this….. (Textbook Page No 283)

Question 1.
Draw possible structures of all the isomers of C₄H₁₁N. Write their common as well as IUPAC names.
Answer:

Use your brain power! (Textbook Page No 283)

Question 1.
Write chemical equations for

(i) reaction of alc. NH with C₂H₅I.
Answer:

(ii) Amonolysis of benzyl chloride followed by the reaction with 2 moles of CH3I.
Answer:

(2) Ammonolysis of alkyl halides is not suitable method to prepare primary amines.
Answer:
In the laboratory, ammonolysis of alkyl halides is not a suitable method to prepare primary amines as it gives a mixture of primary, secondary, tertiary amines and quaternary ammonium salts. (Refer to the reaction in answer to Question 16). The separation of primary amine becomes difficult.

Problem 13.1 : (Textbook Page No 285)

Question 1.
Write reaction to convert methyl bromide into ethyl amine? Also, comment on the number of carbon atoms in the starting compound and the product.
Solution :
Methyl bromide can be converted into ethyl amine in two stage reaction sequence as shown below.

The starting compound methyl bromide contains one carbon atom while the product ethylamine contains two carbon atoms. A reaction in which number of carbons increases involves a step up reaction. The overall conversion of methyl bromide into ethyl amine is a step up conversion.

Use your brain power! (Textbook Page No 285)

Identify ‘A’ and ‘B’ in the following conversions.

Answer:

Use your brain power! (Textbook Page No 286)

Question 1.
Write the chemical equations for the following conversions :
(1) Methyl chloride to ethylamine.
(2) Benzamide to aniline.
(3) 1, 4-Dichlorobutane to hexane-1, 6-diamine.
(4) Benzamide to benzylamine.
Answer:
(1) Methyl chloride to ethylamine

(2) Benzamide to aniline

(3) 1, 4-Dichlorobutane to hexane-1, 6-diamine

(4) Benzamide to benzylamine

Use your brain power! (Textbook Page No 287)

Question 1.
Arrange the following :
(1) In decreasing order of the boiling point C₂H₅ – OH, C₂H₅ – NH₂, (CH₃)₂ NH
(2) In increasing order of solubility in water: C₂H₅ – NH₂, C₃H₇ – NH₂, C₆H₅ – NH₂
Answer:
(1) Decreasing order of the boiling point : C₂H₅ — OH, C₂H₅ — NH₂, (CH₃)₂ NH
(2) Increasing order of solubility in water : C₆H₅NH₂, C₃H₇ — NH₂, C₂H₅ — NH₂

Use your brain power! (Textbook Page No 288)

Question 1.
Refer to pK_b values and answer which compound from the following pairs is the stronger base?
(1) CH₃ – NH₂ and (CH₃)₂ NH
(2) (C₂H₅)₂ NH and (C₂H₅)₃ N
(3) NH₃ and (CH₃)₂ CH – NH₂
Answer:
(1) CH₃ -NH₂ and (CH₃)₂ NH
(CH₃)₂ NH is a stronger base

(2) (C₂H₅)₂ NH and (C₂H₅)₃ N
(C₂H₅)₂ NH is a stronger base

(3) NH₃ and (CH₃)₂ CHNH₂
(CH₃)₂ CHNH₂ is a stronger base

Use your brain power! (Textbook Page No 290)

Question 1.
Arrange the following amines in decreasing order of their basic strength :
NH₃, CH₃ – NH₂, (CH₃)₂ NH, C₆H₅NH₂
Answer:
Decreasing order of basic strength :
(CH₃)₂NH, CH₃ -NH₂, NH₃, C₆H₅NH₂

Use your brain power! (Textbook Page No 291)

Question 1.

Answer:

Use your brain power! (Textbook Page No 291)

Question 1.
Complete the following reaction :

Answer:

Use your brain power! (Textbook Page No 292)

Answer:

Use your brain power! (Textbook Page No 292)

Question 1.
Write the carbylamine reaction by using aniline as starting material.
Answer:

Can you tell? (Textbook Page No 292)

(1) What is the formula of nitrous acid ?
(2) Can nitrous acid be stored in bottle ?
Answer:
(1) Formula of nitrous acid : H – O – N = O
(2) Nitrous acid cannot be stored in bottle.

Use your brain power! (Textbook Page No 294)

Question 1.
How will you distinguish between methyl amine, dimethylamine and trimethylamine by Hinsberg’s test?
Answer:
(1) Methyl amine (primary amine) reacts with benzene sulphonyl chloride to form N-methylbenzene sulphona- mide

(2) Dimethyl amine reacts with benzene sulphonyl chloride to give N, N – dimethylbenzene sulphonamide.

(3) Trimethyl amine does not react with benzene sulphonyl chloride and remains insoluble in KOH

Problem 13.1 : (Textbook Page No 295)

Question 1.
Write the scheme for preparation of p-bromoaniline from aniline. Justify your answer.
Solution :
NH₂ – group in aniline is highly ring activating and o – /p – directing due to involvement of the lone pair of electrons on ‘N’ in resonance with the ring. As a result, on reaction with Br₂ it gives 2,4,6-tribromoaniline. To get a monobromo product, it is necessary to decrease the ring activating effect of – NH₂ group. This is done by acetylation of aniline. The lone pair of ‘N’ in acetanilide is also involved in resonance in the acetyl group. To that extent, ring activation decreases.

Hence, acetanilide on bromination gives a monobromo product p-bromoacetanilide. After monobromination the original – NH₂ group is regenerated. The protection of – NH₂ group in the form of acetyl group is removed by acid catalyzed hydrolysis to get p-bromoaniline, as shown in the following scheme.

Use your brain power! (Textbook Page No 296)

Question 1.
(1) Can aniline react with a Lewis acid?
(2) Why aniline does not undergo Frledel – Craft’s reaction using aluminium chloride?
Answer:
(1) Aniline reacts with a Lewis acid, forms salt.
(2) Aniline does not undergo Friedcl-Crafr’s reaction (alkylation and acetylation) due to salt formation with aluminium chloride (Lewis acid), which is used as catalyst. Due to this, nitrogen of anime acquires + ve charge and hence acts as strong deactivating effect on the ring and makes it difficult for electrophilic attack.

Can you tell? (Textbook Page No 294)

(1) Do tertiary amines have ‘H’ bonded to ‘N?
(2) Why do tertiary amines not react with benzene sulfonyl chloride?
Answer:
(1) Tertiary amines do not have ‘H’ bonded to ‘N’.
(2) Tertiary amine does not undergo reaction with benzene sulphonyl chloride as it does not have any H atom attached to nitrogen atom of amine.

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 13 Amines Textbook Exercise Questions and Answers.

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 2 Accounts of ‘Not for Profit’ Concerns Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 2 Accounts of ‘Not for Profit’ Concerns

1. Objective Type Questions:

A. Select the most appropriate alternatives from the following and rewrite the sentences:

Question 1.
Not for Profit concern renders ___________ services to public at large.
(a) commercial
(b) social
(c) individual
(d) group
Answer:
(b) social

Question 2.
Donation for Scholarship Fund is ___________
(a) Capital Receipt
(b) Revenue Receipt
(c) Capital Expenditure
(d) Revenue Expenditure
Answer:
(a) Capital Receipt

Question 3.
Income and Expenditure Account is a ___________ Account.
(a) Capital
(b) Real
(c) Personal
(d) Nominal
Answer:
(d) Nominal

Question 4.
Outstanding subscription at the end of the Accounting Year represents ___________
(a) Liability
(b) an Expenditure
(c) an Asset
(d) Capital Fund
Answer:
(c) an Asset

Question 5.
Subscription received in advance during the accounting year is ___________
(a) an Income
(b) an Expense
(c) an Asset
(d) a Liability
Answer:
(d) a Liability

Question 6.
Excess of Income over Expenditure is termed as ___________
(a) Deficit
(b) Profit
(c) Surplus
(d) Loss
Answer:
(c) Surplus

Question 7.
Not for Profit concern prepares ___________ Account instead of Profit and Loss Account to know the result.
(a) Trading
(b) Income and Expenditure
(c) Cash
(d) Receipt and Payments
Answer:
(b) Income and Expenditure

Question 8.
The closing balance of Receipts and Payments Account usually represent ___________
(a) Closing stock
(b) Cash and Bank Balance
(c) Surplus
(d) Deficit
Answer:
(b) Cash and Bank Balance

Question 9.
Not for Profit organization is also called ___________ organization.
(a) service
(b) trading
(c) profit-making
(d) commercial
Answer:
(a) service

Question 10.
Expenditure on Purchase of Building is a ___________ Expenditure.
(a) Capital
(b) Revenue
(c) General
(d) Recurring
Answer:
(a) Capital

B. Write the Word/Phrase/Term, which can substitute each of the following Statements.

Question 1.
The Form of Organization providing services to the society only.
Answer:
Not for Profit concern

Question 2.
An account which is prepared by Not for Profit concern instead of Profit and Loss Account.
Answer:
Income and Expenditure Account

Question 3.
Donations are received for a specific purpose.
Answer:
Specific donation/Capital Receipt

Question 4.
The Receipts are not recurring in nature.
Answer:
Capital Receipt

Question 5.
An Account that records only revenue items in case of Not a for-profit concern.
Answer:
Income and Expenditure Account

Question 6.
Accounts which records only cash transactions in case of Not for Profit concerns.
Answer:
Receipts and Payments Account

Question 7.
The income is earned during the year but not received during the year.
Answer:
Outstanding income

Question 8.
The credit balance of Income and Expenditure Account.
Answer:
Surplus

Question 9.
To excess of total assets over total liabilities of a Not for Profit concern.
Answer:
Capital Fund

Question 10.
All such receipts are non-recurring in nature and not forming a part of a regular flow of income.
Answer:
Capital Receipts

C. State whether the following statements are True or False with reasons.

Question 1.
Not for Profit concerns do not have a profit motive.
Answer:
This statement is True.
Not for profit concerns, the main aim is to give services to its members or to the society at large. They do not carry any Trading activity or Manufacturing activity so there is no question of having a profit motive for ‘Not for Profit’ concerns.

Question 2.
Charitable Institutions prepare Profit and Loss Accounts at the end of every financial year.
Answer:
This statement is False.
Charitable Institutions, Not for Profit concerns, do not undertake any trading activities and hence instead of Profit and Loss Account prepare Income-Expenditure Account to record all revenue expenses/losses and revenue incomes/gains of the current year.

Question 3.
There is no difference between Receipts and Payments Account and Income and Expenditure Account.
Answer:
This statement is False.
In the Receipts and Payments Account, all receipts and payments transactions in cash or through the bank are recorded irrespective of the current year, previous year, or next year while in Income-Expenditure Account, only the current year’s incomes and expenses (Revenue) are recorded.

Question 4.
Income and Expenditure Account represents either surplus or deficit.
Answer:
This statement is True.
In the Income and Expenditure Account, all revenue incomes and expenses are recorded and at the end of the specified period, the difference is found out which is known as ‘Surplus’ (Revenue incomes are more than Revenue expenses) or ‘Deficit’ (Revenue expenses are more than Revenue incomes).

Question 5.
Receipts and Payments Accounts do not have any opening balance.
Answer:
This statement is False.
Receipts and Payments Account is just like a cash book of trading concern and opening balance (Cash or Bank or Cash and Bank) must be there to start recording of transactions.

Question 6.
Not for Profit concerns do not prepare a Balance Sheet.
Answer:
This statement is False.
To know the financial position of the organization, at the end of the particular period, Not for Profit concerns prepare Balance Sheet.

Question 7.
Purchases of Sports Equipments is a Capital Expenditure.
Answer:
This statement is True.
Generally, the life span of sports equipment is more than one year, so the purchase of sports equipment is considered a capital expenditure.

Question 8.
Income and Expenditure Account is a Real Account.
Answer:
This statement is False.
In the Income and Expenditure Account, all the revenue incomes and revenue expenses are recorded and therefore it is a Nominal Account and not a Real Account.

Question 9.
The Receipts and Payments Account contains only the transactions relating to the current year.
Answer:
This statement is False.
In the Receipts and Payments Account, transactions of not only the current year but of the previous year or of the next year are also recorded.

Question 10.
Excess of Assets over liabilities is called Capital Fund.
Answer:
This statement is True.
For ‘Not for Profit’ concerns in the Balance Sheet, when a total of Assets is more than the total of Liabilities, the difference of amount is considered as ‘Capital Fund’.

D. Fill in the blanks.

Question 1.
Not for Profit organization is never engaged in ___________ activities.
Answer:
trading

Question 2.
Not for Profit organization is called ___________ organization.
Answer:
service

Question 3.
Receipts and Payments Account falls under the category of ___________ Account.
Answer:
Real

Question 4.
In Receipts and Payment Account the summary of ___________ transactions are recorded.
Answer:
cash

Question 5.
Income and Expenditure Account is similar to the ___________ Account of Trading concern.
Answer:
Profit and Loss

Question 6.
Credit side of Receipts and Payments Account shows cash ___________
Answer:
payments

Question 7.
Income and Expenditure Account is a ___________ Account.
Answer:
Nominal

Question 8.
Mumbai University prepares ___________ Account instead of a Profit and Loss account.
Answer:
Income and Expenditure

Question 9.
Subscription received from the members is considered as ___________ receipts.
Answer:
revenue

Question 10.
The transactions recorded in the Income and Expenditure Account related only to the ___________ year.
Answer:
current

E. Answer in one sentence only.

Question 1.
What do you mean by ‘Not for Profit’ Concern?
Answer:
A concern or organization which is formed and established to serve its members and society or the general public by undertaking various activities without any profit motive is called a ‘Not for Profit’ concern.

Question 2.
Which organizations prepare Income and Expenditure Account?
Answer:
‘Not for profit’ concern prepares Income and Expenditure Account.

Question 3.
What is Receipts and Payments Account?
Answer:
An account that is prepared by a ‘Not for Profit Concern’ to record a summary of all types of cash receipts and cash payments inclusive of bank transactions is called Receipts and Payments Account.

Question 4.
Why Income and Expenditure Account is prepared?
Answer:
Income and Expenditure Account is prepared to ascertain, whether the concern has sufficient income to meet its expenses, or not.

Question 5.
What is Capital Fund?
Answer:
Excess of Assets over Liabilities is known as Capital Fund which also consists of contributions, subscription, entrance fees, surplus income, etc.

Question 6.
What is a Subscription?
Answer:
Subscription is the periodical payment made by the members to the ‘Not for Profit’ concern for maintaining their membership.

Question 7.
What is ’Legacy’?
Answer:
Any asset, property, or amount of cash which ‘Not for Profit’ concern receives as per the provisions made in the will of the donor after his death is called Legacy.

Question 8.
What is Surplus?
Answer:
Excess of income over expenditure shown by Income and Expenditure Account represents Surplus for the financial year.

Question 9.
What do you mean by Non-recurring Expenses?
Answer:
Non-recurring expenses are the expenses that are made for the acquisition of fixed assets that gives benefits for a long period.

Question 10.
To which account ‘Surplus’ or ‘Deficit’ is transferred?
Answer:
‘Surplus’ or ‘Deficit’ is transferred to the Balance Sheet by adding it or subtracting it from Capital Fund.

F. I. Complete the Table:

Question 1.

Sr. No.	Income (₹)	Expenditure (₹)	Surplus/Deficit (₹)
1	10,000	?	5,000 (Deficit)
2	8,000	?	4,000 (Surplus)
3	?	15,000	8,000 (Surplus)
4	7,500	9,000	?
5	15,000	11,300	?

Solution:

Sr.No.	Income (₹)	Expenditure (₹)	Surplus/Deficit (₹)
1	10,000	15,000	5,000 (Deficit)
2	8,000	4,000	4,000 (Surplus)
3	23,000	15,000	8,000 (Surplus)
4	7,500	9,000	1,500 (Deficit)
5	15,000	11,300	3,700 (Surplus)

II. Salaries paid during the year:

Question 1.

Sr.No.	Total (₹)	Prepaid/Outstanding	₹	Expenditure for the year
1	1,100	Prepaid	100	?
2	2,700	Prepaid	?	2,000
3	8,250	Prepaid	?	6,650
4	1,200	Outstanding	200	?
5	?	Outstanding	600	5,100
6	1,800	Outstanding	?	2,200

Solution:

Sr. No.	Total (₹)	Prepaid/Outstanding	₹	Expenditure for the year
1	1,100	Prepaid	100	1,000
2	2,700	Prepaid	700	2,000
3	8,250	Prepaid	1,600	6,650
4	1,200	Outstanding	200	1,400
5	4,500	Outstanding	600	5,100
6	1,800	Outstanding	400	2,200

III. Rent received during the year:

Question 1.

Sr.No.	Total Received (₹)	Rent Received in Advance/Accrued	₹	Income for the year (₹)
1	1,300	Received in Advance	200	?
2	?	Received in Advance	400	1,400
3	2,650	Received in Advance	?	2,000
4	?	Accrued	290	3,190
5	1,700	Accrued	?	2,150
6	2,600	Accrued	500	?

Solution:

Sr.No.	Total Received (₹)	Rent received in Advance/Accrued	₹	Income for the year (₹)
1	1,300	Received in Advance	200	1,100
2	1,800	Received in Advance	400	1,400
3	2,650	Received in Advance	650	2,000
4	2,900	Accrued	290	3,190
5	1,700	Accrued	450	2,150
6	2,600	Accrued	500	3,100

G. Calculate the following:

Question 1.
10 % p.a. Depreciation on Furniture ₹ 50,000 (for three months)
Solutions:
Depreciation = Cost of Asset × Rate × Period
= 50,000 × \(\frac{10}{100}\) × \(\frac{3}{12}\)
= ₹ 1250 Depr. for 3 months
Thus, Depreciation on furniture @ 10 % on ₹ 50,000 for 3 months = ₹ 1250.

Question 2.
12 % p.a. Interest on Bank loan ₹ 80,000 for 1 year.
Answer:
I = \(\frac{\mathrm{PRN}}{100}\)
= 80,000 × \(\frac{12}{100}\) × 1
= ₹ 9600.
Thus, interest on a Bank loan ₹ 80,000 for 1 year = ₹ 9600.

Question 3.
Opening stock of stationery ₹ 5,000, purchases of stationery ₹ 7000, outstanding stationery bill ₹ 12,000, closing stock ₹ 1000. What is the amount of stationery consumed?
Answer:

Question 4.
Salary ₹ 10,000, outstanding salary ₹ 5,000. Calculate the salary to be debited to the Income and Expenditure Account.
Answer:

Question 5.
Library Books ₹ ……………? Less 10% Depreciation ₹ 5,000 = ₹ 45,000.
Answer:
Library books ₹ 50,000. Less 10% Depreciation ₹ 5,000 = ? 45,000

H. Find odd one:

Question 1.
Trading Account, Profit and Loss Account, Receipts and Payments Account, Balance Sheet.
Answer:
Receipts and Payments Account

Question 2.
Machinery, Furniture, Computers, Salaries.
Answer:
Salaries

Question 3.
Subscription, Stationery, Interest Received, Locker Rent received.
Answer:
Stationery

Question 4.
Reliance Industries, Venna Vidya Mandir, Laxmi Hospital, Manoj Sports club.
Answer:
Reliance Industries

Question 5.
Surplus, Deficit, Net Profit, Capital fund.
Answer:
Net Profit

Practical Problems

Question 1.
Calculation of stationery consumed during the year

Adjustments:
1. ₹ 1,000 outstanding for the stationery bill.
2. Stock of Stationery as of 31 – 03 – 2019 was valued at ₹ 1,800.
With the above information, calculate the amount of Stationery consumed during the year and show its presence in final Accounts of a concern.
Solution:
In the books of ___________________

Balance Sheet as of 31 – 03 – 2019

Question 2.
Presentation of Subscription only

Adjustments:
The outstanding subscription for 2018-19 is ₹ 32,000.
With the above information, present the item Subscription in Income and Expenditure Account for the year ended 31 -03-2019 and Balance Sheet as on the date.
Solution:
In the books of ___________________

Balance Sheet as of 31st March 2019

Working Note:
Outstanding subs, given in the balance sheet as on 01-04-2018 are ₹ 26,000 (for 2016-17) and ₹ 35,000 (for 2017-18). Against that, as shown in Receipt – Payment A/c ₹ 23,000 and ₹ 30,000 are received respectively. Means ₹ 3,000 and ₹ 5,000 are still outstanding which are known in the current year balance sheet.

Question 3.

Adjustments:
Subscription outstanding for the year 2017-18 is ₹ 6,000.
During the previous year subscription received in advance for 2017-18 is ₹ 2,000.
The outstanding subscription of 2016-17 is ₹ 2,500.
With the help of the above information present the item Subscription in Income and Expenditure Account for the year ending 31-03-2018 and Balance Sheet as on that date.
Solution:
In the books of ___________________

Balance Sheet as of 31st March 2018

Question 4.
Preparation of Income and Expenditure Account only
Following is the Receipts and Payments Account of “Satara Sports Club” Satara.
Prepare Income and Expenditure Account for the year ending 31-03-2019.

Adjustments:
1. Outstanding subscription for the current year is ₹ 4,500.
2. Outstanding rent for the current year amounted to ₹ 1,000.
3. Entrance fees are to be treated as Revenue Income.
4. Stock of sports material as of 01-04-2018 ₹ 6,000 and on 31 – 03 – 2019 ₹ 14,000.
Solution:
In the books of ‘Satara Sports Club’ Satara

Working Notes:
1. Entrance fees are to be treated as Revenue income. Therefore entire amount is recorded on the income side.
2. Since the selling price of Furniture ₹ 4,500 is lower than its cost price of ₹ 8,000, there is a Loss in the sale of furniture.
It is calculated as follows:
Loss on sale of furniture = Book value (cost) – Selling price
= 8,000 – 4,500
= ₹ 3,500
It is debited to Income and Expenditure A/c.

Question 5.
“Bhartiya Kala Kendra”, Solapur gives you the following information for the year ended on 31-03-2018. Prepare Income and Expenditure Account for the year ending 31-03-2018.

Additional Information:
1. Legacies are to be capitalized.
2. Outstanding salary ₹ 200.
3. 50% of Entrance fees are to be capitalized.
Solution:
In the books of Bhartiya Kala Kendra, Solapur

Question 6.
Accounts of a Charitable Hospital
From the following particulars relating to “Radha-Krishna Charitable Hospital”, Pune.
Prepare Income and Expenditure Account for the year ending 31-03-2020 and Balance Sheet as of that date.

Additional Information:

Provide Depreciation on Equipments ₹ 1,900 and on Building ₹ 1,500.
Solution:
In the Books of Radha-Krishna Charitable Hospital, Pune

Balance Sheet as of 31st March 2020

Working Notes:
1. To find medicines consumed, here in the opening stock, purchases are added and the closing stock of medicine is subtracted.
2. For equipment, in opening balance, add equipment purchased during the year and subtract depreciation to get the closing balance of equipment.
3. Interest ₹ 10,000 is received on 10% investments means there is no outstanding interest.

Question 7.
From the following transactions of Receipts and Payments Account of “Pavan-putra Hanuman Vyayamshala”, Parbhani and the adjustments are given, you are required to prepare Income and Expenditure Account and Balance Sheet as of 31st March 2019.

Adjustments:
1. There are 500 members paying an annual subscription of ₹ 50 each.
2. Outstanding salary was ₹ 1,200.
3. The Assets on 01-04-2018 were as follows:
Building ₹ 50,000, Furniture ₹ 15,000
4. Provide depreciation on Building and Furniture at 5% and 10% respectively.
5. 50% Entrance fees are to be capitalized.
6. Interest on investment at 5% p.a. has accrued for 6 months.
7. Capital fund ₹ 70,000 on 01-04-2018.
Solution:
In the Books of Pavan-putra Hanuman Vyayamshala, Parbhani

Balance Sheet as of 31st March 2019

Working Notes:
1. Interest on investment is receivable for 6 months:
I = \(\frac{\text { PRN }}{100}\)
= 15,000 × \(\frac{5}{100} \times \frac{6}{12}\)
= ₹ 375 (outstanding interest on investment)
2. 50 % of entrance fees (i.e. \(\frac{6200}{2}\) = ₹ 310o) is to be capitalised means add it to capital fund.
3. Total subscription of current year = 500 members × ₹ 50 = ₹ 25,000
But actual subscription received = ₹ 18,000
means difference (25,000 – 18,000) of ₹ 7,000 is outstanding subscription.

Question 8.
Newly Started Art Circle
“Jeevan Jyoti Art Circle” a newly established concern has presented the following information:

Adjustments:
1. Subscription outstanding for the year was ₹ 5,000.
2. Depreciate Furniture @ 10% p.a.
3. Full amount of admission fees and 50% donations are to be capitalized.
You are required to prepare an Income and Expenditure Account for the year ending 31-03-2018 and a Balance Sheet as of that date.
Solution:

Balance Sheet as of 31st March 2018

Working Notes:
The full amount of admission fees and 50% of donations are added to the surplus amount to get capital funds. (Opening balance of the capital fund is not given.)

Question 9.
Given below is the Receipts and Payments Account of “Vithai Mahila Mandat”, Pandharpur for the year ending 31-03-2018. Prepare an Income and Expenditure Account for the year ended 31-03-2018 and Balance Sheet as of that date.

Adjustments:
1. Capital fund on 01-04-2017 was ₹ 90,000.
2. Outstanding subscription ₹ 4,000.
3. Entrance fees are to be capitalized.
4. Rent paid includes ₹ 800 paid for April 2018.
5. They have the following Assets and Liabilities as of 01-04-2017:
Furniture ₹ 9,000, Building ₹ 70,000, and Outstanding Expenses ₹ 12,000.
Solution:
In the books of Vithai Mahila Mandal, Pandharpur

Balance Sheet as of 31st March 2018

Working Notes:
1. Outstanding subscription of ₹ 4,000 is first added to subscription received on the credit side of Income and Expenditure A/c and then it is shown on the Assets side of Balance Sheet.
2. Entire amount of entrance fees ₹ 3,500 is added to the capital fund.
3. Prepaid ₹ 800 is first deducted from rent paid on the debit side of Income & Expenditure A/c and then shown on the Assets side of the Balance Sheet.
4. Outstanding expenses ₹ 12,000 is directly shown on the Liabilities side of the Balance Sheet.

Question 10.
From the following Receipts and Payments Account “K.B.P. Engineering College”, Nashik for the year ending on 31 – 03 – 2019 and additional information, prepare Income and Expenditure Account for the year ending 31-03-2019 and Balance Sheet as on that date.

Additional Information:

1. 50% of donations are for the Building funds and the balance is to be treated as Revenue income.
2. Outstanding subscription ₹ 5,300.
3. Life membership fees are to capitalize.
Solution:
In the books of K.B.P. Engineering College, Nashik

Balance Sheet as of 31st March 2019

Working Notes:
1. Life membership fees are to be capitalized, which means add the entire amount to the capital fund.

2. 50% of Donations of ₹ 7,00,000 i.e., ₹ 3,50,000 is to be added to the Building Fund, and the remaining amount of donation i.e., ₹ 3,50,000 is credited to Income and Expenditure A/c.

3. The depreciation on Fixed assets is calculated by using the following formula:
Depreciation = Opening balance + Purchases – Closing value
∴ Depreciation on Books = 6,00,000 + 61,000 – 6,00,000
= 6,61,000 – 6,00,000
= ₹ 61,000
∴ Depreciation on Furniture = 3,19,000 + 51,000 – 3,00,000
= 3,70,000 – 3,00,000
= ₹ 70,000

4. Fixed deposit: Opening balance given = ₹ 9,10,000
Fixed deposit (31-03-2019) = ₹ 8,50,000
(Newly purchased)
∴ Total fixed deposits = ₹ 17,50,000

5. Admission Fees ₹ 60,000 are recorded on the credit side of Income and Expenditure A/c because it is taken as Revenue income.

Question 11.
Account of a School
From the following Balance Sheet and Receipts and Payments Account of “New English School”, Barshi, prepare Income and Expenditure Account for the year ending 31 – 03 – 2020 and a Balance Sheet as on that date.


Additional Information:
1. Outstanding salary of ₹ 9,000.
2. Outstanding tuition fees ₹ 15,000.
3. Depreciate library books by ₹ 9,000 and Furniture by ₹ 10,000.
Solution:
In the books of New English School, Barshi

Balance Sheet as of 31st March 2020

Working Notes:
1. Donation (Capital) is added to the Capital fund.
2. Government Grant ₹ 1,20,000 is recorded on the credit side of Income & Expenditure A/c because it is the revenue income of the organization.
3. Outstanding tuition fees ₹ 15,000 and outstanding salary ₹ 9,000 are added to the respective head of Account and then they are shown separately on the Assets side and Liabilities side of the Balance Sheet respectively.

Question 12.
Account of a Library
Following is the Receipts Payments Account of “Dhananjay Library”, Mumbai for the year ending 31-03-2020.

You are required to prepare an Income and Expenditure Account for the year ended 31-03-2020 and a Balance Sheet as of that day.
The following information is also made available to you.
1. On 31 -03-2019, the library had the following Assets also; Books ₹ 50,000, Furniture ₹ 6,500, and Machinery of ₹ 30,000.
2. Subscription received in advance amounted to ₹ 500.
3. Outstanding salaries ₹ 1300 and Rent ₹ 950.
4. 50% of the admission fees should be capitalized.
5. Furniture to be depreciated at 10% p.a.
6. Library books were purchased on 1st April 2019 charge depreciation at 10% p.a.
7. The Investments were purchased on 01-04-2019 and they carry interest at 20% p.a.
Solution:
In the books of Dhananjay Library, Mumbai

Balance Sheet as of 31st March 2020

Working Notes:
1. Opening Balance Sheet is prepared to find out opening capital fund:
Balance Sheet as of 31st March 2019

2. Outstanding expenses of the previous year, paid in the current year so no entry for outstanding expenses (2018-19)

3. Interest on Investment @ 20% on ₹ 6,000 = ₹ 1,200
Interest on Investment received = ₹ 900
Outstanding interest on investment = ₹ 300

4. Depreciation on library books at 10% p.a.
On opening balance of ₹ 50,000 (for whole year) = ₹ 5,000
On purchases on 01-04-2019 (for whole year) = ₹ 500
Total Depreciation = ₹ 5,500

5. Subscriptions received in advance ₹ 500 is deducted from subscription received on the credit side of Income and Expenditure A/c and then subscription received in advance is shown separately on the Liabilities side of Balance Sheet.

6. Outstanding salaries and outstanding rent are added to the respective head of Account on the debit side of the Income & Expenditure Account and both the outstanding items are recorded on the Liabilities side of the Balance Sheet.

Question 13.
Outstanding Expenses and Prepaid Expenses
From the following information supplied to you, prepare Income and Expenditure Account for the year ending 31-03-2020 and Balance Sheet as on that date for “Morya Sports Club”, Thane.
Balance Sheet as on 01-04-2019


Adjustments:
1. Subscription received includes ₹ 3,000 for 2018-19 and outstanding subscription for 2019-20 was ₹ 14,000.
2. On 31-03-2020, the Prepaid insurance premium was ₹ 2,500.
3. Depreciate Furniture by ₹ 3,000.
4. Locker rent outstanding for 2019-20 is ₹ 400.
Answer:
In the books of ‘Morya Sports Club’ Thane

Balance Sheet as of 31st March 2020

Working Notes:
1. Outstanding subscription (2018-19) ₹ 8,000 given in b/s against that ₹ 3,000 received in 2019-20.
Means still receivable subscription = ₹ 5,000 (8,000 – 3,000).

2. Prepaid insurance premium (2018-19) ₹ 2,000 is for the current year. Therefore, in the current year’s insurance premium, ₹ 2,000 is to be added and then subtract the current year’s prepaid insurance premium.

3. Outstanding salary of (2018-19), ₹ 4,000 of the previous year is to be subtracted from current year’s salary.

4. Total amount of donations and entrance fees are to be capitalized so add the entire amount of both the items to Capital fund.

5. Outstanding locker’s rent (2018-19) ₹ 500 is given in the Balance Sheet. It is to be subtracted from the current year’s locker’s rent and then adds the current year’s outstanding locker rent.

Question 14.
Charitable Hospital
Following information has been provided by “Vivekanand Charitable Hospital”, Latur. You are required to prepare an Income and Expenditure Account for the year ending 31-03-2019 and the Balance Sheet as of that date.


Adjustments:
1. On 31-03-2019 stock of drugs was valued at ₹ 22,000.
2. Depreciation on Building at 5% p.a. and on Ambulance ₹ 30,000.
3. Life membership fees are to be capitalized.
Answer:
In the books of Vivekanand Charitable Hospital, Latur

Balance Sheet as of 31st March 2019

Working Notes:
1. Purchase of drugs ₹ 2,00,000 includes ₹ 40,000 of 2017-18 and in the Balance Sheet of 2017-18, the outstanding bill of drugs is ₹ 50,000 given. So, ₹ 10,000 is still outstanding.

2. Consumption of drugs:
Opening stock (2017-18) = ₹ 42,000
Add: Purchase of drugs = ₹ 1,60,000
Total = 1,60,000 + 42,000 = ₹ 2,02,000
Less: Closing stock of drugs = ₹ 22,000
Consumption of drugs = 1,80,000