NEET Physics Class 11

• Calorimetry And Thermal Expansion Multiple Choice Question And Answers
• Chapter 3 Centre Of Mass Multiple Choice Question And Answers
• Chapter 4 Elasticity And Viscosity Multiple Choice Question And Answers

• Chapter 5 Fluid Mechanics Multiple Choice Questions And Answers
• Chapter 6 Friction Multiple Choice Question And Answers
• Chapter 7 Gravitation Multiple Choice Question Ans Answers
• Chapter 8 Heat Transfer Multiple Choice Question And Answers
• Chapter 9 Kinetic Theory Of Gases And Thermodynamics MCQs
• Chapter 10 Mathematical Tools Multiple Choice Question And Answers

Kinetic Theory Of Gases And Thermodynamics

Kinetic Theory Of Gases:

The kinetic theory of gases is based on the following basic assumptions.

1. A gas consists of a very large number of molecules. These molecules are identical, perfectly elastic, and hard spheres. They are so small that the volume of molecules is negligible compared with the volume of the gas.
2. Molecules do not have any preferred direction of motion, motion is completely random.
3. These molecules travel in straight lines and free motion most of the time. The time of the collision between any two molecules is very small.
4. The collision between molecules and the wall of the container is perfectly elastic. It means kinetic energy is conserved in each collision.
5. The path traveled by a molecule between two collisions is called the free path and the mean of this distance traveled by a molecule is called the mean free path.
6. The motion of molecules is governed by Newton’s law of motion
7. The effect of gravity on the motion of molecules is negligible.

Expression For The Pressure Of A Gas

Let us suppose that a gas is enclosed in a cubical box having length l. Let there be ‘ N ‘ identical molecules, each having mass ‘ m’ since the molecules are of the same mass and perfectly elastic, so their mutual collisions result in the interchange of velocities only.

Only collisions with the walls of the container contribute to the pressure of the gas molecules. Let us focus on a molecule having velocity v₁ and components of velocity v_x1, v_y1, and v_z1 along the x, y, and z-axis as shown in the figure.

The change in momentum of the molecule after one collision with wall BCHE

⇒ \(m v_{x_1}-\left(-m v_{x_1}\right)=2 m v_{x_1}\)

The time taken between the successive impacts on the face BCHE = \(\frac{\text { distance }}{\text { velocity }}\)

⇒ \(\frac{2 \ell}{v_{\mathrm{x}_1}}\)

Time rate of change of momentum due to collision = \(\frac{\text { change in momentum }}{\text { time taken }}\)

⇒ \(\frac{2 m v_{\mathrm{x}_1}}{2 \ell / \mathrm{v}_{\mathrm{x}_1}}=\frac{\mathrm{mv}_{\mathrm{x}_1}^2}{\ell}\)

Hence the net force on the wall BCHE due to the impact of n molecules of the gas is :

⇒ \(\mathrm{F}_{\mathrm{x}}=\frac{\mathrm{mv}_{\mathrm{x}_1}^2}{\ell}+\frac{\mathrm{mv}_{\mathrm{x}_2}^2}{\ell}+\frac{\mathrm{mv}_{\mathrm{x} 3}^2}{\ell}+\ldots \ldots \ldots .+\frac{\mathrm{mv}_{\mathrm{x}_{\mathrm{n}}}^2}{\ell}\)

⇒ \(\frac{m}{\ell}\left(v_{x_1}^2+v_{x_2}^2+v_{x_3}^2+\ldots \ldots \ldots \ldots+v_{x_n}^2\right)\)

⇒ \(\frac{\mathrm{mN}}{\ell}<\mathrm{v}_{\mathrm{x}}^2>\)

where < V²_x > = mean square velocity in x-direction. Since molecules do not favor any particular direction therefore

⇒\(\left\langle\mathrm{v}_{\mathrm{x}}^2\right\rangle=\left\langle\mathrm{v}_{\mathrm{y}}^2\right\rangle=\left\langle\mathrm{v}_{\mathrm{z}}^2\right\rangle\)

But

⇒\(\left\langle\mathrm{v}^2\right\rangle=\left\langle\mathrm{v}_{\mathrm{x}}^2\right\rangle+\left\langle\mathrm{v}_{\mathrm{y}}^2\right\rangle+\left\langle\mathrm{v}_{\mathrm{z}}^2\right\rangle\)

⇒ \(\left\langle\mathrm{v}_{\mathrm{x}}^2\right\rangle\)

⇒ \(\frac{\left\langle\mathrm{v}^2\right\rangle}{3}\) Pressure is equal to force divided by area.

P = \(\frac{\mathrm{F}_{\mathrm{x}}}{\ell^2}=\frac{\mathrm{M}}{3 \ell^3}\left\langle\mathrm{v}^2\right\rangle=\frac{\mathrm{M}}{3 \mathrm{~V}}\left\langle\mathrm{v}^2\right\rangle\).

Pressure is independent of x, y, and z directions.

Where l³ = volume of the container = V

M = total mass of the gas, <c² > = mean square velocity of molecules

⇒ \(P=\frac{1}{3} \rho\left\langle v^2\right\rangle\)

As PV = nRT , then total translational K.E. of gas = \(\frac{1}{2} M<v^2>=\frac{3}{2} P V=\frac{3}{2} n R T\)

Translational kinetic energy of 1 molecule = \(\frac{3}{2} \mathrm{kT}\)(it is independent of nature of gas) kT

⇒ \(\left\langle v^2\right\rangle=\frac{3 P}{\rho}\)

or \(v_{rm s}=\sqrt{\frac{3 P}{\rho}}=\sqrt{\frac{3 R T}{M_{\text {mole }}}}=\sqrt{\frac{3 k T}{m}}\)

Where v_rms is the root mean square velocity of the gas.

Pressure exerted by the gas is P = \(\frac{1}{3} \rho\left\langle v^2\right\rangle=\frac{2}{3} \times \frac{1}{2} \rho<v^2>\)

or \(P=\frac{2}{3} E, E=\frac{3}{2} P\)

Thus, the total translational kinetic energy per unit volume (it is called energy density) of the gas is numerically equal to \(\frac{3}{2} k T\) times the pressure exerted by the gas.

Important Points

1. \(\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\mathrm{T}} \text { and } \mathrm{v}_{\mathrm{rms}} \propto \frac{1}{\sqrt{\mathrm{M}_{\text {mole }}}}\)
2. At absolute zero, the motion of all molecules of the gas stops.
3. At higher temperatures and low pressure or higher temperatures and low density, a real gas behaves as an ideal gas.

Maxwell’s Distribution Law

Distribution Curve – A plot of \(\frac{\mathrm{dN}(\mathrm{v})}{\mathrm{dv}}\)(number of molecules per unit speed interval) against c is known as Maxwell’s distribution curve. The total area under the curve is given by the integral.

⇒ \(\int_0^{\infty} \frac{d N(v)}{d v} d v=\int_0^{\infty} d N(v)=N\)

• Figure shows the distribution curves for two different temperatures. At any temperature, the number of molecules in a given speed interval dv is given by the area under the curve in that interval (shown shaded).
• This number increases, as the speed increases, upto a maximum and then decreases asymptotically toward zero.
• Thus, a maximum number of the molecules have speed lying within a small range centered about the speed corresponding to the peak (A) of the curve. This speed is called the ‘most probable speed’ v_P or v_mp. dN(v)

The distribution curve is asymmetrical about its peak (the most probable speed v_P) because the lowest possible speed is zero, whereas there is no limit to the upper speed a molecule can attain. Therefore, the average speed v is slightly larger than the most probable speed v_P. The root-mean-square speed, v_rms, is still larger

⇒ \(\left(v_{\text {rms }}>\bar{V}>v_p\right)\)

Average (or Mean) Speed:

⇒ \(\bar{v}=\sqrt{\frac{8}{\pi} \frac{k T}{m}}=1.59 \sqrt{\mathrm{kT} / \mathrm{m}}\) (derivation is not in the course)

RMS Speed:

⇒ \(v_{\mathrm{rms}}=\sqrt{\left\langle v^2\right\rangle}=\sqrt{\frac{3 \mathrm{kT}}{\mathrm{m}}}=1.73 \sqrt{\frac{\mathrm{kT}}{\mathrm{m}}}\)

Most Probable Speed:

The most probable speed v_P or v_mp is the speed possessed by the maximum number of molecules and corresponds to the maximum (peak) of the distribution curve. Mathematically, it is obtained by the condition.

⇒ \(\frac{d N(v)}{d v}=0\)[by substitution of formula of dN(v) (which is not in the course)]

Hence the most probable speed is

⇒ \(v_p=\sqrt{\frac{2 k T}{m}}=1.41 \sqrt{\mathrm{kT} / \mathrm{m}}\)

From the above expression, we can see that

⇒\(\mathrm{v}_{\mathrm{rms}}>\overline{\mathrm{v}}>\mathrm{v}_{\mathrm{p}}\)

Degree Of Freedom

A total number of independent coordinates that must be known completely specify the position and configuration of a dynamical system completely is known as “degree of freedom f”. The maximum possible translational degrees of freedom are three i.e.

⇒ \(\left(\frac{1}{2} m V_x^2+\frac{1}{2} m V_y^2+\frac{1}{2} m V_z^2\right)\)

The maximum possible rotational degrees of freedom are three i.e.

⇒ \(\left(\frac{1}{2} I_x \omega_x^2+\frac{1}{2} I_y \omega_y^2+\frac{1}{2} I_z \omega_z^2\right)\)

Vibrational degrees of freedom are two i.e. (Kinetic energy. of vibration and Potential energy of vibration)

Mono Atomic: (all inert gases, He, Ar, etc.) f = 3 (translational)

Diatomic: (gases like H₂, N₂, O₂ etc.) f = 5 (3 translational + 2 rotational)

If temp < 70 K for diatomic molecules, then f = 3

If the temp is between 250 K to 5000 K, then f = 5

If temp > 5000 K f = 7 [ 3 translational.+ 2 rotational + 2 vibrational ]

Maxwell’s Law Of Equpartition Of Energy

Energy associated with each degree of freedom = \(\frac{1}{2} \mathrm{kT}\). If the degree of freedom of a molecule is f, then the total kinetic energy of that molecule U = \(\frac{1}{2} \mathrm{fkT}\)

Internal Energy

The internal energy of a system is the sum of the kinetic and potential energies of the molecules of the system. It is denoted by U. Internal energy (U) of the system is the function of its absolute temperature (T) and its volume (V). i.e. U = f (T, V)

• In the case of an ideal gas, the intermolecular force is zero. Hence its potential energy is also zero.
• In this case, the internal energy is only due to kinetic energy, which depends on the absolute temperature of the gas. i.e. U = f (T).

For an ideal gas internal energy U = \(\frac{f}{2} n R T\)

Question 1. A light container having a diatomic gas enclosed within is moving with velocity v. The Mass of the gas is M and the number of moles is n.

1. What is the kinetic energy of gas w.r.t. center of mass of the system?
2. What is K.E. of gas w.r.t. ground?

Answer:

1. K.E. = \(\frac{5}{2} n R T\)

2. Kinetic energy of gas w.r.t. ground = Kinetic energy of center of mass w.r.t. ground + Kinetic energy of gas w.r.t. center of mass.

K.E. = \(\frac{1}{2} M V^2+\frac{5}{2} n R T\)

Question 2. Two nonconducting containers having volumes V₁ and V₂ contain monoatomic and diatomic gases respectively. They are connected as shown in the figure. Pressure and temperature in the two containers are P₁, T_1, and P₂, T₂ respectively. Initially stop cock is closed, if the stop cock is opened find the final pressure and temperature.

Answer:

⇒ \(n_1=\frac{P_1 V_1}{R T_1}\)

⇒ \(\mathrm{n}_2=\frac{\mathrm{P}_2 \mathrm{~V}_2}{R \mathrm{RT}_2}\)

n = n₁+ n₂(number of moles are conserved)

Finally, the pressure in both parts and the temperature of both gases will become equal.

⇒ \(\frac{P\left(V_1+V_2\right)}{R T}=\frac{P_1 V_1}{R T_1}+\frac{P_2 V_2}{R T_2}\)

From energy conservation

⇒ \(\frac{3}{2} n_1 R T_1+\frac{5}{2} n_2 R T_2\)

⇒ \(\frac{3}{2} n_1 R T+\frac{5}{2} n_2 R T\)

T = \(\frac{\left(3 P_1 V_1+5 P_2 V_2\right) T_1 T_2}{3 P_1 V_1 T_2+5 P_2 V_2 T_1}\)

P = \(\left(\frac{3 P_1 V_1+5 P_2 V_2}{3 P_1 V_1 T_2+5 P_2 V_2 T_1}\right)\left(\frac{P_1 V_1 T_2+P_2 V_2 T_2}{V_1+V_2}\right)\)

Indicator Diagram

A graph representing the variation of pressure or variation of temperature or variation of volume with each other is called or indicator diagram.

1. Every point of the Indicator diagram represents a unique state (P, V, T) of gases.
2. Every curve on the Indicator diagram represents a unique process.

Thermodynamics

Thermodynamics is mainly the study of the exchange of heat energy between bodies and the conversion of the same into mechanical energy and vice versa.

Thermodynamic System

The collection of an extremely large number of atoms or molecules confined within certain boundaries such that it has a certain value of pressure (P), volume (V), and temperature (T) is called a thermodynamic system.

Anything outside the thermodynamic system to which energy or matter is exchanged is called its surroundings. Taking into consideration the interaction between a system and its surroundings thermodynamic system is divided into three classes :

1. Open system: A system is said to be an open system if it can exchange both energy and matter with its surroundings.
2. Closed system: A system is said to be a closed system if it can exchange only energy (not matter with its surroundings).
3. Isolated system: A system is said to be isolated if it can neither exchange nor matter with its surroundings.

Zeroth Law Of Thermodynamics

If two systems (B and C) are separately in thermal equilibrium with a third one (A), then they are in thermal equilibrium with each other.

Equation Of State ( For Ideal Gases)

The relation between the thermodynamic variables (P, V, T) of the system is called an equation of state. The equation of state for an ideal gas of n moles is given by PV = nRT,

Work Done By A Gas

Let P and V be the pressure and volume of the gas. If A is the area of the piston, then the force exerted by a gas on the piston is, F = P × A.

Let the piston move through a small distance dx during the expansion of the gas. Work done for a small displacement dx is dW = F dx = PA dx

Since A dx = dV, the increase in the volume of the gas is dV

⇒ dW = P dV

mg

or \(W=\int d W=\int P d V\)

The area enclosed under the P-V curve gives work done during the process.

Different Types Of Processes

1. Isothermal Process:

T = constant [Boyle’s law applicable] PV = constant

There is an exchange of heat between the system and its surroundings. The system should be compressed or expanded

very slowly so that there is sufficient time for the exchange of heat to keep the temperature constant.

The slope of the P−V curve in the isothermal process:

PV = constant = C

⇒ \(\frac{d P}{d V}=-\frac{P}{V}\)

Work done in the isothermal process:

W = \(n R T \quad \ell n \frac{V_f}{V_i}\)

[If v_f > v_i then W is positive

If v_f < v_i then W is negative]

W = \(\left[2.303 n R T \log _{10} \frac{V_f}{V_i}\right]\)

Internal energy in the isothermal process:

U = f (T) ⇒ ΔU = 0

2. Iso-Choric Process (Isometric Process):

V = constant

⇒ change in volume is zero

⇒ \(\frac{P}{T}\) is constant

⇒ \(\frac{P}{T}\)= const.(Galussac-law)

Work done in the isochoric process:

Since the change in volume is zero therefore dW = p dV = 0

Indicator diagram of the isochoric process:

Change in internal energy in isochoric process: ΔU = \(n \frac{f}{2} R \Delta T\)

Heat given in isochoric process: ΔQ = ΔU = \(n \frac{f}{2} R \Delta T\)

3. Isobaric Process: Pressure remains constant in the isobaric process

∴ P = constant

⇒ \(\frac{\mathrm{V}}{\mathrm{T}}\) = constant

Indicator diagram of the isobaric process:

Work done in the isobaric process:

ΔW = P ΔV = P (V_final – V_initial) = nR (T_final – T_initial)

Change in internal energy in the isobaric process: ΔU = n C_VΔT

Heat given in the isobaric process:

ΔQ = ΔU + ΔW

ΔQ = \(n \frac{f}{2} R \Delta T+P\left[V_f-V_i\right]\)

⇒ \(n \frac{f}{2} R \Delta T+n R \Delta T\)

The above expression gives the idea that to increase the temperature by ΔT in the isobaric process heat required is more than in the isochoric process.

4. Cyclic Process: In the cyclic process initial and final states are the same therefore initial state = final state

Work done = Area enclosed under P-V diagram.

Change in internal Energy ΔU = 0

ΔQ = ΔU + ΔW

∴ ΔQ = ΔW

If the process on the P-V curve is clockwise, then the net work done is (+ve) and vice-versa. The graphs shown below explain when work is positive and when it is negative

Question 1. The cylinder shown in the figure has conducting walls and the temperature of the surroundings is T, the position is initially in equilibrium, and the cylinder contains n moles of a gas. Now the piston is displaced slowly by an external agent to make the volume double the initial. Find work done by an external agent in terms of n, R, T

Answer:

1st Method:

Work done by external agents is positive because Fext and displacement are in the same direction. Since walls are conducting therefore temperature remains constant.

Applying the equilibrium condition when the pressure of the gas is P

⇒ \(P A+F_{\text {ext }}=P_{a t m} A\)

⇒ \(F_{\text {ext }}=P_{a t m} A-P A\)

⇒ \(W_{e x t}=\int_0^d F_{e x t} d x=\int_0^d P_{a t m} A d x-\int_0^d P A d x \)

⇒ \(P_{a t m} A \int_0^d d x-\int_V^{2 V} \frac{n R T}{V} d V=P_{a t m} A d-n R T \ln 2\)

⇒ \(P_{a t m} \cdot V_0^2-n R T \ln 2=n R T(1-\ln 2)\)

2nd Method

Applying the work-energy theorem on the piston

Δk = 0

Wall = Δk

W_gas+ W_atm + W_ext = 0

⇒ \(n R T \ln \frac{V_f}{V_i}-n R T+W_{\text {ext }}=0\)

W_ext = nRT (1 – ln2)

Question 2. Find out the work done in the given graph. Also, draw the corresponding T-V curve and P-T curve.

Answer:

Since in P-V curves area under the cycle is equal to work done therefore work done by the gas is equal to P₀ V₀. Lines A B and CD are isochoric lines, and lines BC and DA are isobaric lines.

∴ The T-V curve and P-T curve are drawn as shown.

Question 3. The t-V curve of the cyclic process is shown below, number of moles of the gas is n to find the total work done during the cycle.

Answer: Since path AB and CD are isochoric therefore work done is zero during path AB and CD. Process BC and DA are isothermal, therefore

⇒ \(W_{B C} =n R 2 T_0 \ln \frac{V_C}{V_B}=2 n R T_0 \ln 2\)

⇒ \(W_{D A} =n R T_0 \ln \frac{V_A}{V_D}=-n R T_0 \ln 2\)

Total work done = \(W_{B C}+W_{D A}=2 n R T_0 \ln 2-n R T_0 \ln 2\)

⇒ \(n R T_0 \ln 2\)

Question 4. The P-T curve of a cyclic process is shown. Find out the work done by the gas in the given process if several moles of the gas are n.

Answer: Since path AB and CD are isochoric therefore work done during AB and CD is zero. Path BC and DA are isobaric.

Hence W_BC = nRΔT = nR(T₃– T₂)

W_DA = nR(T₁– T₄)

Total work done = W_BC + W_DA = nR(T₁+ T₃–T₄– T₂)

Question 5. Consider the cyclic process ABCA on a sample of 2.0 mol of an ideal gas as shown in the figure. The temperatures of the gas at A and B are 300 K and 500 K respectively. A total of 1200 J heat is withdrawn from the sample in the process. Find the work done by the gas in part BC. Take R = 8.3J/mol–K.
Answer:

The change in internal energy during the cyclic process is zero. Hence, the heat supplied to the gas is equal to the work done by it. Hence,

W_AB + W_BC + W_CA = –1200 J. …….(1)

The work done during the process AB is

W_AB = P_A(V_B– V_A)

= nR(T_B– T_A)

= (2.0 mol) (8.3 J/mol–K) (200 K)

= 3320 J

The work done by the gas during the process CA is zero as the volume remains constant. From (1),

3320 J + W_BC = –1200 J

or W_BC = –4520 J.

= –4520 J.

First Law Of Thermodynamics

The first law of thermodynamics is the law of conservation of energy. It states that if a system absorbs heat dQ and as a result the internal energy of the system changes by dU and the system does a work dW, then dQ = dU + dW.

But, do = P dV dQ = dU + P dV

which is the mathematical statement of the first law of thermodynamics.

Heat gained by a system, work done by a system, and an increase in internal energy is taken as positive.

Heat lost by a system, work done on a system, and a decrease in internal energy are taken as negative.

Question 1. 1 gm water at 100ºC is heated to convert into steam at 100ºC at 1 atm. Find out the change in the internal energy of water. It is given that a volume of 1 gm water at 100ºC = 1 cc. volume of 1 gm steam at 100ºC = 1671 cc. Latent heat of vaporization = 540 cal/g. (Mechanical equivalent of heat J = 4.2J/cal.)
Answer:

From the first law of thermodynamic ΔQ = Δu + Δw

ΔQ = mL = 1 × 540 cal. = 540 cal.

ΔW = PΔV = \(\frac{10^5(1671-1) \times 10^{-6}}{4.2}\)

⇒ \(\frac{\left.10^5 \times 1670\right) \times 10^{-6}}{4.2}\) = 40 cal.

Δu = 540 – 40 = 500 cal.

Question 2. Two moles of diatomic gas at 300 K are kept in a non-conducting container enclosed by a piston. Gas is now compressed to increase the temperature from 300 K to 400 K. Find work done by the gas

Answer:

ΔQ = Δu + Δw

Since the container is non-conducting therefore

ΔQ = 0 = Δu + Δw

⇒ ΔW = – Δu = \(-\Delta u=-n \frac{f}{2} R \Delta T\)

⇒ \(-2 \times \frac{5}{2} R(400-300)\)

= –5 × 8.314 × 100 J

= – 5 × 831.4 J

= –4157 J

Question 3. A sample of an ideal gas is taken through the cyclic process of abaca (figure. It absorbs 50 J of heat during part ab, no heat during bc, and rejects 70 J of heat during ca. 40 J of work done on the gas during part bc.

1. Find the internal energy of the gas at b and c if it is 1500 J at a.
2. Calculate the work done by the gas during the part ca.

Answer:

1. In part ab the volume remains constant. Thus, the work done by the gas is zero. The heat absorbed by the gas is 50 J. The increase in internal energy from a to b is

ΔU = ΔQ = 50J.

As the internal energy is 1500 J at a, it will be 1550 J at b. In part bc, the work done by the gas is ΔW = –40J and no heat is given to the system. The increase in internal energy from b to c is

ΔU = –ΔW = 40 J.

As the internal energy is 1550 J at b, it will be 1590 J at c.

2. The change in internal energy from c to a is

ΔU = 1500 J – 1590 J = – 90 J.

The heat given to the system is ΔQ = – 70J.

Using ΔQ = ΔU + ΔW,

ΔW_ca= ΔQ – ΔU

= – 70 J + 90 J

= 20 J.

Question 4. The internal energy of a monatomic ideal gas is 1.5 nRT. One mole of helium is kept in a cylinder of cross-section 8.5 cm². The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of 42 J heat is given to the gas. If the temperature rises through 2ºC, find the distance moved by the piston. Atmospheric pressure = 100 kPa.
Answer:

The change in internal energy of the gas is

ΔU = 1.5 nR (ΔT)

= 1.5 (1 mol) (8.3 J/mol-K) (2K)

= 24.9 J.

The heat is given to the gas = 42 J

The work done by the gas is

ΔW = ΔQ – ΔU

= 42 J – 24.9 J = 17.1 J.

If the distance moved by the piston is x, the work done is

ΔW = (100 kPa) (8.5 cm²) x.

Thus, (10⁵ N/m²) (8.5 × 10^-4 m²) x = 17.1 J

or, x = 0.2m = 20 cm.

Question 5. A sample of ideal gas (f =5) is heated at constant pressure. If an amount of 140 J of heat is supplied to the gas, find

1. The change in internal energy of the gas
2. The work done by the gas.

Answer:

Suppose the sample contains n moles. Also, suppose the volume changes from V₁ to V₂ and the temperature changes from T₁ to T₂.

The heat supplied is

ΔQ = ΔU + PΔV = ΔU + nRΔT = \(\Delta U+\frac{2 \Delta U}{f}\)

1. The change in internal energy is

ΔU = \(n \frac{f}{2} R\left(T_2-T_1\right)=n \frac{f}{2} R\left(T_2-T_1\right)\)

⇒ \(\frac{\mathrm{f}}{2+\mathrm{f}} \Delta Q=\frac{140 \mathrm{~J}}{1.4}\)

= 100 J.

2. The work done by the gas is

ΔW = ΔQ – ΔU

= 140 J – 100 J

= 40 J.

Efficiency Of A Cycle (η):

⇒ \(\eta=\frac{\text { total Mechanical work done by the gas in the whole process }}{\text { Heat absorbed by the gas (only }+ \text { ve) }}\)

⇒ \(=\frac{\text { area under the cycle in } \mathrm{P}-\mathrm{V} \text { curve }}{\text { Heat injected into the system }}\)

⇒ \(\eta=\left(1-\frac{Q_2}{Q_1}\right)\) for Heat Engine,

⇒ \(\eta=\left(1-\frac{T_2}{T_1}\right)\) for Carnot cycle

Question 6. n moles of a diatomic gas have undergone a cyclic process ABC as shown in the figure. The temperature at a is T₀. Find

1. Volume at C?
2. Maximum temperature?
3. Total heat given to gas?
4. Is heat rejected by the gas, if yes how much heat is rejected?
5. Find out the efficiency

Answer:

1. Since triangles OA V₀ and OC V are similar therefore

⇒ \(\frac{2 P_0}{V}=\frac{P_0}{V_0}\)

⇒ \(V=2 V_0\)

2. Since process AB is isochoric hence

⇒ \(\frac{P_A}{T_A}=\frac{P_B}{T_B}\)

⇒ \(T_B=2 T_0\)

Since process BC is isobaric therefore \(\frac{T_B}{V_B}=\frac{T_C}{V_C}\)

⇒ \(T_{\mathrm{C}}=2 \mathrm{~T}_{\mathrm{B}}=4 \mathrm{~T}_0\)

3. Since the process is cyclic therefore

∴ ΔQ = ΔW = area under the cycle = \(\frac{1}{2} P_0 V_0\)

4. Since Δu and ΔW both are negative in process CA

∴ ΔQ is negative in process CA and heat is rejected in process CA

ΔQ_CA = Δw_CA + Δu_CA

⇒ \(-\frac{1}{2}\left[P_0+2 P_0\right] V_0-\frac{5}{2} n R\left(T_c-T_a\right)\)

⇒ \(-\frac{1}{2}\left[P_0+2 P_0\right] V_0-\frac{5}{2} n R\left(\frac{4 P_0 V_0}{n R}-\frac{P_0 V_0}{n R}\right)\)

= –9P₀V₀

= Heat injected.

4. η = efficiency of the cycle = \(=\frac{\text { work done by the gas }}{\text { heat injected }}\)

⇒ \(\eta=\frac{P_0 V_0 / 2}{Q_{\text {injected }}} \times 100\)

ΔQ_inj = ΔQ_AB + ΔQ_BC

⇒ \(\left[\frac{5}{2} n R\left(2 T_0-T_0\right)\right]+\left[\frac{5}{2} n R\left(2 T_0\right)+2 P_0\left(2 V_0-V_0\right)\right]\)

⇒ \(\frac{19}{2} P_0 V_0\)

⇒ \(\eta=\frac{100}{19} \%\)

Specific Heat

The specific heat capacity of a substance is defined as the heat supplied per unit mass of the substance per unit rise in the temperature. If an amount ΔQ of heat is given to a mass m of the substance and its temperature rises by ΔT, the specific heat capacity s is given by the equation

⇒ \(s=\frac{\Delta Q}{m \Delta T}\)

The molar heat capacities of a gas are defined as the heat given per mole of the gas per unit rise in the temperature. The molar heat capacity at constant volume, denoted by C_v, is :

⇒ \(C_v=\left(\frac{\Delta Q}{n \Delta T}\right)_{\text {constant volume }}\)

⇒ \(\frac{f}{2} R\)

and the molar heat capacity at constant pressure, denoted by C_P is,

⇒ \(C_{\mathrm{P}}=\left(\frac{\Delta Q}{\mathrm{n} \Delta T}\right)_{\text {constant volume }}\)

⇒ \(\left(\frac{f}{2}+1\right) R\)

• where n is the amount of the gas in several moles and f is a degree of freedom. Quite often, the term specific heat capacity or specific heat is used for molar heat capacity.
• It is advised that the unit be carefully noted to determine the actual meaning. The unit of specific heat capacity is J/kg-K whereas that of molar heat capacity is J/mol–K.

Molar Heat Capacity Of Ideal Gas In Terms Of R:

1. For a monoatomic gas f = 3

⇒ \(C_v=\frac{3}{2} R\)

⇒ \(C_p=\frac{5}{2} R\)

⇒ \(\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=\frac{5}{3}\)

= 1.67

2. For a diatomic gas f = 5

⇒ \(\mathrm{C}_{\mathrm{V}}=\frac{5}{2} \mathrm{R},\)

⇒ \(\mathrm{C}_{\mathrm{P}}=\frac{7}{2} \mathrm{R}\)

⇒ \(\gamma=\frac{C_P}{C_V}\)

= 1.4

3. For a Triatomic gas f = 6

⇒ \(C_V=3 R, C_P=4 R\)

⇒ \(\gamma=\frac{C_P}{C_V}=\frac{4}{3}\) = 1.33 [Note for CO₂; f = 5, it is linear]

In general, if f is the degree of freedom of a molecule, then,

⇒ \(C_V=\frac{f}{2} R, \quad C_P=\left(\frac{f}{2}+1\right) R\)

⇒ \(\gamma=\frac{C_P}{C_V}=\left[1+\frac{2}{f}\right]\)

Question 1. In a thermodynamic process, the pressure of a certain mass of gas is changed in such a way that 20 Joule heat is released from it and 8 Joule work is done on the gas. If the initial internal energy of the system is 30 joule then the final internal energy will be –
Answer:

dQ = dU + dW ⇒ dQ = U_final– U_initial + dW

U_final = dQ – dW + U_initial or U_final = –20 + 8 + 30 of U_final = 18 Joule

Question 2. A gas is contained in a vessel fitted with a movable piston. The container is placed on a hot stove. A total of 100 cal of heat is given to the gas and the gas does 40 J of work in the expansion resulting from heating. Calculate the increase in internal energy in the process.
Answer:

Heat given to the gas is ΔQ = 100 cal = 418 J.

Work done by the gas is ΔW = 40 J

The increase in internal energy is

ΔU = ΔQ – ΔW

= 418J – 40 J = 378 J

Question 3. A gas is compressed from volume 10 m³ to 4 m³ at a constant pressure of 50N/m². Gas is given 100 J energy by heating then its internal energy.
Answer:

P = 50 N/m²

dV = 10 – 4 = 6 m³

δW = PdV = 6 × 50 = 300 J (Volume is decreasing, δQ = 100 J)

W = – 300 J

δQ = δW + dU

100 + 300 = dU

dU = increased by 400 J

Question 4. Two moles of a diatomic gas at 300 K are enclosed in a cylinder as shown in the figure. The piston is light. Find out the heat given if the gas is slowly heated to 400 K in the following three cases.

1. The piston is free to move
2. If the piston does not move
3. If the piston is heavy and movable.

Answer:

1. Since pressure is constant

∴ ΔQ = nC_P ΔT = 2 × \(\frac{7}{2}\) 2× R × (400 – 300) = 700 R

2. Since volume is constant

∴ ΔW = 0 and ΔQ = Δu (from first law)

ΔQ = Δu = nC_vΔT = 2× \(\frac{5}{2}\) × R × (400 – 300) = 500 R

3. Since pressure is constant

∴ ΔQ = nC_P ΔT = 2 × \(\frac{7}{2}\) × R × (400 – 300) = 700 R

Question 5. The p-V curve of a diatomic gas is shown in the figure. Find the total heat given to the gas in the process AB and BC

Answer:

From the first law of thermodynamics

ΔQ_ABC = Δu_ABC + ΔW_ABC

⇒ \(\Delta W_{A B C}=\Delta W_{A B}+\Delta W_{B C}=0+n R T_B \ln \frac{V_C}{V_B}=n R T_B \ln \frac{2 V_0}{V_0}\)

⇒ \(n R T_B \ln 2=2 P_0 V_0 \ln 2\)

Δu = \(n C_v \Delta T=\frac{5}{2}\left(2 P_0 V_0-P_0 V_0\right)\)

⇒ ΔQ_ABC = \(\frac{5}{2}\) 2P₀V₀+ 2P₀V₀ln 2.

Question 6. Calculate the value of the mechanical equivalent of heat from the following data. The specific heat capacity of air at constant volume = 170 cal/kg-K, γ=C_P/C_v= 1.4, and the density of air at STP is 1.29 kg/m³. Gas consant R = 8.3 J/mol-K.
Answer:

Using pV = nRT, the volume of 1 mole of air at STP is

V = \(\frac{\mathrm{nRT}}{\mathrm{p}}=\frac{(1 \mathrm{~mol}) \times(8.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}) \times(273 \mathrm{~K})}{1.0 \times 10^5 \mathrm{~N} / \mathrm{m}^2}\)

= 0.0224m³

The mass of 1 mole is, therefore,

(1.29 kg/m³) × (0.0224 m³) = 0.029 kg.

The number of moles in 1 kg is \(\frac{1}{0.029}\). The molar heat capacity at constant volume is

⇒ \(C_v=\frac{170 \mathrm{cal}}{(1 / 0.029) \mathrm{mol}-\mathrm{K}}\)= 4.93 cal/mol-K.

Hence, C_P = γC_v= 1.4 × 4.93 cal/mol-K

or, C_P – C_v= 0.4 × 4.93 cal/mol-K

= 1.97 cal/mol-K.

Also, C_P– C_v= R = 8.3 J/mol-K.

Thus, 8.3 J = 1.97 cal.

The mechanical equivalent of heat is

⇒ \(\frac{8.3 \mathrm{~J}}{1.97 \mathrm{cal}}\) =4.2 J/cal.

Average Molar Specific Heat of Metals:

[Dulong and Petit law]

At room temperature average molar specific heat of all metals are same and is nearly equal to 3R ( 6 cal. mol^-1 K^-1).

[Note: Temp. above which the metals have constant C_V is called Debye temp.]

Mayer’s Equation: C_P− C_V= R (for ideal gases only)

Adiabatic Process:

When no heat is supplied or extracted from the system the process is called adiabatic. The process is sudden so there is no time for the exchange of heat. If the walls of a container are thermally insulated no heat can cross the boundary of the system and the process is adibatic.

The equation of the adiabatic process is given by

PV^γ = constant [Poission Law]

Tγ P^1-γ = constant

T V^γ-1 = constant

The slope of P−V−curve in the adiabatic process:

Since PV^γ is a constant

∴ \(\frac{\mathrm{dP}}{\mathrm{dV}}=-\gamma\left(\frac{\mathrm{P}}{\mathrm{V}}\right)\)

The slope of P−T−curve in the adiabatic process: Since T^γ P^1-γ is a constant

∴ \(\frac{d P}{d T}=-\frac{\gamma}{(1-\gamma)} \frac{P}{T}=\frac{(\gamma)}{(\gamma-1)} \frac{P}{T}\)

Slope of T−V−curve:

⇒ \(\frac{d V}{d T}=-\frac{1}{(\gamma-1)} \frac{V}{T}\)

Work Done In Adiabatic Process:

work done by the system is (+ve), if T_i> T_f (hence expansion)

work done on the system is (−ve) if T_i< T_f (hence compression)

Question 7. A quantity of air is kept in a container having walls that are slightly conducting. The initial temperature and volume are 27ºC (equal to the temperature of the surroundings) and 800cm³ respectively. Find the rise in the temperature if the gas is compressed to 200cm³

1. In a short time
2. In a long time. Take γ = 1.4.

Answer:

1. When the gas is compressed in a short time, the process is adiabatic. Thus,

⇒ \(\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1}=\mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}\)

or \(\mathrm{T}_2=\mathrm{T}_1\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}\)

⇒ \((300 \mathrm{~K}) \times\left[\frac{800}{200}\right]^{0.4}\)

= 522K

Rise in temperature = T₂– T₁= 222 K.

2. When the gas is compressed for a long time, the process is isothermal. Thus, the temperature remains equal to the temperature of the surroundings which is 27ºC. The rise in temperature = 0.

Question 8. A monoatomic gas is enclosed in a nonconducting cylinder having a piston that can move freely. Suddenly gas is compressed to 1/8 of its initial volume. Find the final pressure and temperature if the initial pressure and temperature are P₀ and T₀ respectively.

Answer:

Since the process is adiabatic therefore

⇒ \(P_0 V^{\frac{5}{3}}=P_{\text {final }}\left(\frac{V}{8}\right)^{5 / 3}\)

⇒ \(\gamma=\frac{C_P}{C_V}=\frac{5 R}{2} / \frac{3 R}{2}=\frac{5}{3}\)

⇒ \(P_{\text {final }}=32 P_0\)

Since the process is adiabatic therefore

⇒ \(\mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1}\)

⇒ \(\mathrm{T}_0 \mathrm{~V}_0^{2 / 3}=\mathrm{T}_{\text {frall }}\left(\frac{\mathrm{V}_0}{8}\right)^{2 / 3}\)

⇒ \(\mathrm{T}=4 \mathrm{~T}_0\)

Question 9. A cylindrical container having nonconducting walls is partitioned into two equal parts such that the volume of each part is equal to V₀. A movable nonconducting piston is kept between the two parts. The gas on the left is slowly heated so that the gas on the right is compressed upto volume \(\frac{V_0}{8}\). Find pressure and temperature on both sides if the initial pressure and temperature were P₀ and T₀ respectively. Also, find heat given by the heater to the gas. (number of moles in each part is n)

Answer:

Since the process on the right is adiabatic therefore PV^γ = constant

⇒ P₀ V^γ₀ = P_final (V₀/ 8)^γ ⇒ P_final = 32

P₀ T₀ V^γ-1₀ = T_final (V₀/8)^γ-1 ⇒ T_final = 4T₀

Let the volume of the left part be V₁

⇒ \(2 V_0=V_1+\frac{V_0}{8}\)

⇒ \(V_1=\frac{15 V_0}{8}\)

Since several moles on the left parts remain constant therefore for the left part

⇒ \(\frac{\mathrm{PV}}{\mathrm{T}}\) = constant.

The final pressure on both sides will be the same

⇒ \(\frac{P_0 V_0}{T_0}=\frac{P_{\text {final }} V_1}{T_{\text {final }}}\)

⇒ \(\mathrm{T}_{\text {final }}=60 \mathrm{~T}_0\)

⇒ \(\Delta Q=\Delta u+\Delta w\)

⇒ \(\Delta Q=n \frac{5 R}{2}\left(60 T_0-T_0\right)+n \frac{3 R}{2}\left(4 T_0-T_0\right)\)

⇒ \(\Delta Q=\frac{5 n R}{2} \times 59 T_0+\frac{3 n R}{2} \times 3 T_0\)

Free Expansion

If a system, say a gas expands in such a way that no heat enters or leaves the system and also no work is done by or on the system, then the expansion is called the “free expansion”.

ΔQ = 0 , ΔU = 0 and ΔW = 0. Temperature in the free expansion remains constant.

Question 1. A nonconducting cylinder having volume 2V₀ is partitioned by a fixed nonconducting wall in two equal parts. Partition is attached with a valve. The right side of the partition is a vacuum and the left part is filled with a gas having pressure and temperature P₀ and T₀ respectively. If the valve is opened the final pressure and temperature of the two parts are.

Answer: From the first law of thermodynamics ΔQ = Δu + ΔW

Since gas expands freely, therefore, ΔW = 0 since no heat is given to gas ΔQ = 0

⇒ Δu = 0 and temperature remains constant.

T_final = T₀

Since the process is isothermal therefore P₀× V₀= P_final × 2V₀

⇒ Pfinal = P₀/2

Reversible And Irreversible Process

A process is said to be reversible when the various stages of an operation in which it is subjected can be traversed back in the opposite direction in such a way that the substance passes through the same conditions at every step in the reverse process as in the direct process.

A process in which any one of the conditions stated for the reversible process is not fulfilled is called an irreversible process.

Comparison Of Slopes Of Iso-Thermal And Adiabatic Curve

In compression up to the same final volume: | W_adia |> | W_isothermal |

In Expansion up to the same final volume: W_isothermal > W_adia

Limitations Of Ist Law Of Thermodynamics:

The first law of thermodynamics tells us that heat and mechanical work are interconvertible. However, this law fails to explain the following points:

1. It does not tell us about the direction of the transfer of heat.
2. It does not tell us about the conditions under which heat energy is converted into work.
3. It does not tell us whether some process is possible or not.

Mixture Of Non-Reacting Gases:

1. Molecular weight = \(\frac{\mathrm{n}_1 \mathrm{M}_1+\mathrm{n}_2 \mathrm{M}_2}{\mathrm{n}_1+\mathrm{n}_2}\)

M₁ and M₂ are molar masses.

2. Specific heat CV = \(\frac{\mathrm{n}_1 \mathrm{C}_{\mathrm{V}_1}+\mathrm{n}_2 \mathrm{C}_{\mathrm{V}_2}}{\mathrm{n}_1+\mathrm{n}_2}\)

⇒ \(C_p=\frac{n_1 C_{P_1}+n_2 C_{P_2}}{n_1+n_2}\)

3. For mixture, γ = \(\gamma=\frac{\mathrm{C}_{{mix}}}{\mathrm{C}_{\mathrm{v}_{{mix}}}}\)

⇒ \(\frac{n_1 C_{p_1}+n_2 C_{p_2}+\ldots \ldots}{n_1 C_{v_1}+n_2 C_{v_2}+\ldots \ldots \ldots}\)

Question 1. 5 gm air is heated from 4ºC to 6ºC. If the specific heat of air at constant volume is 0.172 cal/gm/ºC, then the increase in the internal energy of air will be –
Answer:

dU = mC_vdT

dU = 5 × 0.172 × 2

dU = 1.72 calorie

Question 2. In the following indicator diagram, the net amount of work done will be –

Answer:

The cyclic process 1 is clockwise and the process 2 is anti clockwise. Therefore W₁ will be positive and W₂ will be negative area 2 > area 1, Hence the network will be negative.

Question 3. Two gram-moles of gas, which are kept at a constant temperature of 0ºC, are compressed from 4 liter to 1 liter. The work done will be
Answer:

W = 2.303 μ RT \(\log _{10} \frac{V_2}{V_1}\)

W = 2.303 × 2 × 8.4 × 273 \(\log _{10} \frac{1}{4}\)

W = 2.303 × 2 × 8.7 × 273 × (log₁₀ – log⁴₁₀)

∵ log⁴₁₀ = 0.6021

∴ W = –6359 Joule

Question 4. Air is filled in a motor car tube at 27ºC temperature and 2-atmosphere pressure. If the tube suddenly bursts then the final temperature will be \(\left[\left(\frac{1}{2}\right)^{2 / 7}=0.82\right]\)
Answer:

⇒ \(\mathrm{T}_2=\mathrm{T}_1\left(\frac{\mathrm{P}_2}{\mathrm{P}_1}\right)^{\frac{\gamma-1}{\gamma}}\)

⇒ \(\mathrm{T}_2=300\left(\frac{1}{2}\right)^{\frac{0.4}{1.4}}\)

⇒ \(300\left(\frac{1}{2}\right)^{2 / 7}\)

= 300×8.2

T₂ = 246K

Question 5. One liter of air at NTP is suddenly compressed to 1 c.c. the final pressure will be.
Answer:

⇒ \(P_2=P_1\left(\frac{V_1}{V_2}\right)^\gamma\)

⇒ \(P_2=10^5\left(10^3\right)^{5 / 3}=10^5 \times 10^5\)

⇒ \(P_2=10^{10} \text { Pascal }\)

Question 6. In the following fig. the work done by the system in the closed path ABCA is Answer: Work done in closed path ABCA

W_ABCA = Area of ΔABC = \(\frac{1}{2}\) AB × BC

W_ABCA = –\(\frac{1}{2}\)(P₂– P₁) (V₂– V₁)

Question 7. According to the fig. if one mole of an ideal gas is in a cyclic process the work done by the gas in the process will be
Answer:

Work done W = area of PV curve

⇒ \(\frac{1}{2}\)[3P₀– P₀][2V₀– V₀]

= P₀V₀

Question 8. In the above question, heat given by the gas is
Answer:

δQ = μC_pdT, μ = 1, dT = T_A– T_C, and for monoatomic ideal gas C_P= 5/2 R

∴ (δQ)_CA = \(\frac{5}{2}\) R [T_A– T_C] = \(\frac{5}{2}\)[P_AV_A– P_CV_C]

But P_A= P₀, V_A= V0, V_C= 2V0, P_C= P₀

and \(\frac{PV}{T}\)

∴ (δQ)C_A = \(\frac{5}{2}\)[P₀V₀– P₀2V₀] = – \(\frac{5}{2}\)P₀V₀

Question 9. In the above question, the heat taken by gas in the path AB will be Answered:

(δQ)_AB = μC_VdT (the process is on constant volume)

C_V= \(\frac{3}{2}\)R, μ = 1

⇒ \((\delta Q)_{A B}=\frac{3}{2} R\left[T_B-T_A\right]\)

⇒ \(\frac{3}{2}\left[3 P_0 V_0-P_0 V_0\right]\)

⇒ \(3 \mathrm{P}_0 \mathrm{~V}_0\)

Question 10. In the above question, the absorbed heat by gas in path BC will be Answered:

If the heat given for the complete process is δQ then

(δQ) = (δQ)_AB + (δQ)_BC + (δQ)_CA

dU = 0 in a cyclic process, thus by the first law of thermodynamics

δQ = δW

∴ (δQ)_AB + (δQ)_BC + (δQ)_AC = δW

(δQ)_BC = δW – (δQ)_AB – (δQ)_BC

⇒ \(P_0 V_0+\frac{5}{2} P_0 V_0-3 P_0 V_0\)

⇒ \(\frac{P_0 V_0}{2}\)

Question 11. For a given cyclic process as shown in fig. the magnitude of absorbed energy for the system is
Answer:

In cyclic process

Q = W (∵ dU = 0)

Q = area of closed loop

Q = 10²π Joule

Second Law Of Thermodynamics

This law gives the direction of heat flow.

According To Classius: It is impossible to make any such machine that can transfer heat from an object with low temperature to an object with high temperature without an external source.

According To Kelvin: It is impossible to obtain work continuously by cooling an object below the temperature of its surroundings.

Statement Of Kelvin-Planck: It is impossible to construct any such machine that works on a cyclic process and absorbs heat from a source, converts all that heat into work, and rejects no heat to sink.

Heat Engine:

The device, used to convert heat energy into mechanical energy, is called a heat engine.

• For the conversion of heat into work with the help of a heat engine, the following conditions have to be met. There should be a body at a higher temperature ‘T₁’ from which heat is extracted. It is called the source.
• The body of the engine contains the working substance. There should be a body at a lower temperature ‘T₂’ to which heat can be rejected. This is called the sink.

Working Of Heat Engine:

Schematic diagram of heat engine

Engine derives an amount ‘Q₁’ of heat from the source.

A part of this heat is converted into work ‘W’.

The remaining heat ‘Q₂’ is rejected to the sink.

Thus Q₁= W + Q₂

or the work done by the engine is given by W = Q₁– Q₂

Efficiency Of Heat Engine:

The efficiency of the heat engine (η) is defined as the fraction of total heat, supplied to the engine which is converted into work.

Mathematically

∴ \(\eta=\frac{W}{Q_1}\)

or \(\eta=\frac{Q_1-Q_2}{Q_1}=1-\frac{Q_2}{Q_1}\)

Carnot Engine And Carnot Cycle

Carnot Engine: The Carnot engine is an ideal heat engine. It consists of the following parts.

Schematic Diagram:

Source: It is a reservoir of heat energy with a conducting top maintained at a constant temperature T₁K. The source is so big that extraction of any amount of heat from it does not change its temperature.

Body Of Heat Engine: It is a barrel having perfectly insulating walls and a conducting bottom. It is fitted with an air-tight piston capable of sliding within the barrel without friction. The barrel contains some quantity of an ideal gas.

Sink: It is a huge body at a lower temperature T₂ having a perfectly conducting top. The size of the sink is so large that any amount of heat rejected to it does not increase its temperature.

Insulating Stand: It is a stand made up of perfectly insulating material such that the barrel when placed over it becomes thoroughly insulated from the surroundings.

Carnot Cycle: As the engine works, the working substance of the engine undergoes a cycle known as the Carnot cycle. The Carnot cycle consists of the following four strokes.

Graphical representation of the Carnot cycle:

First Stroke (Isothermal expansion): In this stroke, the barrel is placed over the source. The piston is gradually pushed back as the gas expands.

The fall of temperature, due to expansion, is compensated by the supply of heat from the source and consequently, temperature remains constant. The conditions of the gas change from A(P₁, V₁) to B(P₂, V₂). If W₁ is the work done during this process, then heat Q₁ derived from the source is given by

⇒ \(\mathrm{Q}_1=\mathrm{W}_1=\text { Area } \mathrm{ABGE}=\mathrm{RT} \log _e\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)\)

Second Stroke (Adiabatic expansion):

The barrel is removed from the source and placed over the insulating stand. The piston is pushed back so that the gas expands adiabatically resulting in a fall of temperature from T₁ to T₂. The conditions of the gas change from B(P₂, V₂) to C(P₃, V₃). If W₂ is the work done in this case then

⇒ \(\mathrm{W}_2=\text { Area } \mathrm{BCHG}=\frac{\mathrm{R}}{\gamma-1}\left(\mathrm{~T}_1-\mathrm{T}_2\right)\)

Third Stroke (isothermal compression): The barrel is placed over the sink. The piston is pushed down there by compressing the gas. The heat generated due to compression flows to the sink maintaining the temperature of the barrel constant.

The state of the gas changes from C(P₃, V₃) to D(P₄, V₄). If W₃ is the work done in this process and Q₂ is the heat rejected to the sink, then

⇒ \(\mathrm{Q}_2=\mathrm{W}_3=\text { Area CDFH }=\mathrm{RT}_2 \log _{\mathrm{e}}\left(\frac{\mathrm{V}_3}{\mathrm{~V}_4}\right)\)

Fourth Stroke (Adiabatic compression): The barrel is placed over the insulating stand. The piston is moved down thereby compressing the gas adiabatically till the temperature of the gas increases from T₂ to T₁.

The state of gas changes from D(P₄, V₄) to A(P₁, V₁). If W₄ is the work done in this process, then

⇒ \(\mathrm{W}_4=\text { Area ADFE }=\frac{\mathrm{R}}{\gamma-1}\left(\mathrm{~T}_1-\mathrm{T}_2\right)\)

Heat Converted Into Work In Carnot Cycle:

During the four strokes, W₁ and W₂ are the work done by the gas, and W₃ and W₄ are the work done on the gas. Therefore the net, work performed by the engine

W = W₁+ W₂– W₃– W₄= Area ABGE + Area BCHG – Area CDFH – Area ADEF = Area ABCD

Thus net work done by the engine during one cycle is equal to the area enclosed by the indicator diagram of the cycle. Analytically

⇒ \(\mathrm{W}=\mathrm{R}\left(\mathrm{T}_1-\mathrm{T}_2\right) \log _e\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)\)

Efficiency Of Carnot Engine:

The efficiency (η) of an engine is defined as the ratio of useful heat (heat converted into work) to the total heat supplied to the engine. Thus.

⇒ \(\eta=\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}\)

or \(\eta=\frac{R\left(T_1-T_2\right) \log _e\left(\frac{V_2}{V_1}\right)}{R T_1 \log _e\left(\frac{V_2}{V_1}\right)}\)

⇒ \(\frac{T_1-T_2}{T_1}\)

or \(\eta=1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1}\)

Some Important Points Regarding Carnot Engine

The efficiency of an engine depends upon the temperatures between which it operates.

η is independent of the nature of the working substance.

η is one only if T₂= 0. Since absolute zero is not attainable, hence even an ideal engine cannot be 100 % efficient.

η is one only if Q₂= 0. But η = 1 is never possible even for an ideal engine. Hence Q₂≠ 0. Thus it is impossible to extract heat from a single body and convert the whole of it into work.

If T₂= T₁, then η = 0

In actual heat engines, there are many losses due to friction, etc. and various processes during each cycle are not quasistatic, so the efficiency of actual engines is much less than that of an ideal engine.

Question 1. A Carnot engine has the same efficiency between

1. 100 K and 500K and
2. Tk and 900 K. The value of T is

Answer:

Efficiency η = \(1-\frac{T_2}{T_1}\)

or η = \(1-\frac{100}{500}\)

⇒ \(1-\frac{T}{900}\)

or \(\frac{100}{500}=\frac{T}{900}\)

∴ T = 180K

Question 2. A Carnot engine takes 10³ kilocalories of heat from a reservoir at 627ºC and exhausts it to a sink at 27ºC. The efficiency of the engine will be.
Answer:

Efficiency of Carnot engine

η = \(1-\frac{T_2}{T_1}\)

⇒ \(1-\frac{300}{900}=\frac{2}{3}\)

or η = 66.6 %

Question 3. In the above problem, the work performed by the engine will be
Answer:

Work performed by the engine

W = \(\eta Q_1=\frac{2}{3} \times 10^6 \times 4.2\)

or W = 2.8 × 10⁶ Joule

Question 4. A Carnot engine has an efficiency of 40% when the sink temperature is 27ºC. The source temperature will be
Answer:

⇒ \(\eta_{\text {efficiency }}=1-\frac{T_2}{T_1}\)

or \(\frac{2}{5}=1-\frac{300}{\mathrm{~T}_1}\)

∴ T₁= 500K

Question 5. A reversible engine takes heat from a reservoir at 527ºC and gives out to the sink at 127ºC. The engine is required to perform useful mechanical work at the rate of 750 watts. The efficiency of the engine is
Answer:

Efficiency η = \(1-\frac{T_2}{T_1}\)

or η = \(1-\frac{400}{800}=\frac{1}{2}\)

η = 50%

Question 6. The efficiency of Carnot’s engine is 50%. The temperature of its sink is 7ºC. To increase its efficiency to 70%. The increase in heat of the source will be
Answer:

Efficiency in first state η = 50% = 1/2

T₂= 273 + 7 = 280 K

Formula η = \(1-\frac{T_2}{T_1}\)

⇒ \(\frac{1}{2}=1-\frac{280}{\mathrm{~T}_1} \Leftrightarrow \frac{280}{\mathrm{~T}_1}=\frac{1}{2}\)

or T₁ = 560ºK (temperature of source)

In the second state (1) \(\frac{70}{100}=1-\frac{280}{\mathrm{~T}_1}\)

∴ \(\mathrm{T}_1=\frac{2800}{3}=933.3 \mathrm{~K}\)

∴ Increase in source temperature = (933.3 – 560) = 373.3 K

Question 7. A Carnot’s engine works at 200ºC and 0ºC and another at 0ºC and –200ºC. The ratio of efficiency of the two is
Answer:

⇒ \(\eta=\frac{\left(T_1-T_2\right)}{T_1}\)

⇒ \(\eta_1=\frac{(473-273)}{473}=\frac{200}{473}\)

and \(\eta_2=\frac{(273-73)}{273}=\frac{200}{273}\)

⇒ \(\frac{\eta_1}{\eta_2}=\frac{273}{473}\)

= 0.577

Question 8. A Carnot engine works as a refrigerator in between 0ºC and 27ºC. How much energy is needed to freeze 10 kg of ice at 0ºC?
Answer:

Heat absorbed by the sink

Q₂= 10 × 10² × 80 = 800 k.cal

Now \(\frac{Q_1}{Q_2}=\frac{T_1}{T_2}, Q_1=Q_2 \cdot \frac{T_1}{T_2}\)

∴ \(Q_1=800 \times \frac{300}{273} \mathrm{k} . \mathrm{cal}\)

= 879 kcal

Question 9. The work efficiency coefficient in the above question
Answer:

Work efficiency coefficient (coefficient of performance)

⇒ \(\beta=\frac{Q_2}{Q_1-Q_2}\)

⇒ \(\frac{800 \times 10^3}{(879-800) \times 10^3}\)

= 10.13

Question 10. A Carnot engine works as a refrigerator in between 250K and 300K. If it acquires 750 calories from a heat source at a low temperature, then the heat is generated at a higher temperature. (in calories) will be.
Solution:

⇒ \(\eta=\frac{Q_2}{Q_1-Q_2}=\frac{T_2}{T_1-T_2}\)

⇒ \(\frac{750}{Q_1-750}=\frac{250}{300-250}\)

Q₁= 900 Calories

Question 11. A vessel of volume 2 x 10^-2 m³ contains a mixture of hydrogen and helium at 47º C temperature and 4.15 x 10⁵ N/m² pressure. The mass of the mixture is 10^-2 kg. Calculate the masses of hydrogen and helium in the given mixture.
Solution:

Let the mass of H₂ is m₁ and He is m₂

∴ m₁+ m₂= 10^-2 kg = 10 × 10^-3 ….(1)

Let P₁, P₂ are partial pressure of H₂ and He

P₁+ P₂= 4.15 × 10⁵ N/m²

for the mixture

⇒ \(\left(P_1+P_2\right) V=\left(\frac{m_1}{n_1}+\frac{m_2}{n_2}\right) R T\)

⇒ \(4.15 \times 10^5 \times 2 \times 10^{-2}\)

⇒ \(\left(\frac{m_1}{2 \times 10^{-3}}+\frac{m_2}{4 \times 10^{-3}}\right) 8.31 \times 320\)

⇒ \(\frac{m_1}{2}+\frac{m_2}{4}=\frac{4.15 \times 2}{8.31 \times 320}=0.00312=3.12 \times 10^{-3}\)

⇒ 2m₁+ m₂= 12.48 × 10^-3 kg …..(2)

Solving (1) and (2)

m₁= 2.48 × 10^-3 kg ≅2.5 × 10^-3 kg

and m = 7.5 × 10^-3 kg.

Question 12. The pressure in a monoatomic gas increases linearly from 4 x 10⁵ N m^-2 to 8 x 10⁵ N m^-2 when its volume increases from 0.2 m³ to 0.5 m³. Calculate the following:

1. Work done by the gas.
2. Increase in the internal energy.

Answer:

1. As here pressure varies linearly with volume, work done by the gas

ΔW = ∫PdV = area under P-V curve

which in the light of Figure 1 becomes:

ΔW = P_I(V_F– V_I) (P_F– P_I) × (V_F– V_I)

i.e., ΔW = \(P_I\left(V_F-V_I\right)+\frac{1}{2}\left(P_F-P_I\right) \times\left(V_F-V_I\right)\)

i.e., ΔW = \(\frac{1}{2}\)(0.5 – 0.2) (8 + 4) × 10⁵

i.e., ΔW = 1.8 × 10⁵ J

2. The change in internal energy of a gas is given by

ΔU = \(\mu C_V \Delta T=\frac{\mu R \Delta T}{(\gamma-1)}\)

⇒ \(\frac{\left(P_F V_F-P_I V_I\right)}{(\gamma-1)}\)

As the gas is monatomic γ = (5/3)

So, ΔU = \(\frac{10^5(8 \times 0.5-4 \times 0.2)}{[(5 / 3)-1]}\)

⇒ \(\frac{3}{2} \times 10^5(4-0.8)\)

i.e., ΔU = 4.8 × 10⁵ J

Refrigerator Or Heat Pump

A refrigerator or heat pump is a heat engine that runs in revenue one direction.
It essentially consists of three parts

1. Source: At higher temperature T₁.
2. Working substance: It is called refrigerant liquid ammonia and freon works as a working substance
3. Sink: At lower temperature T₂.

• The working substance takes heat Q₂ from a sink (contents of the refrigerator) at a lower temperature, has a net amount of work done W in it by an external agent (usually the compressor of the refrigerator), and gives out a larger amount of heat Q₁ to a hot body at temperature T₁(usually atmosphere)
• Thus, it transfers heat from a cold to a hot body at the expense of mechanical energy supplied to it by an external agent. The cold is thus cooled more and more.
• The performance of a refrigerator is expressed using the “coefficient of performance” β which is defined as the ratio of the heat extracted from the cold body to the needed to transfer it to the hot body.

i.e \(\beta=\frac{\text { Heat extracted }}{\text { Work done }}\)

⇒ \(\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}\)

A perfect refrigerator transfers heat from a cold to a hot body without doing work e. W = 0 so that Q₁=Q₂ hence β = ∞

Carnot Refrigerator:

For Carnot refrigerator \(\frac{Q_1}{Q_2}=\frac{T_1}{T_2}\)

⇒ \(\frac{Q_1-Q_2}{Q_2}=\frac{T_1-T_2}{T_2}\)

or \(\frac{Q_2}{Q_1-Q_2}=\frac{T_2}{T_1-T_2}\)

So Coefficient of performance β = \(\frac{T_2}{\mathrm{~T}_1-\mathrm{T}_2}\)

⇒ \(\beta=\frac{T_2}{T_1-T_2}\)

here T₁= temperature of surrounding T₂ = temperature of cold body. It is clear that β = 0 when T₂= 0 i.e. the coefficient of performance will be zero if the cold body is at a temperature equal to absolute zero.

Relation between the coefficient of performance and efficiency of a refrigerator

We know β = \(\frac{Q_2}{Q_1-Q_2}=\frac{Q_2 / Q_1}{1-Q_2-Q_1}\)…………(1)

But the efficiency η = \(\frac{Q_2}{Q_1} \text { or } \frac{Q_2}{Q_1}=1-\eta\)…………(2)

Form (1) and (2) we get, \(\frac{1-\eta}{\eta}\)

Entropy

Entropy is a measure of the disorder of molecular motion of a system. The greater the disorder, the greater the entropy.

The change in entropy i.e

dS = \(=\frac{\text { Heat absorbed by system }}{\text { Absolute temperature }}\)

or \(\mathrm{dS}=\frac{\mathrm{dQ}}{\mathrm{T}}\)

The relation is called the mathematical form of the Second Law of Thermodynamics.

For solids and liquids

1. When heat is given to a substance changes its state at a constant temperature, then changes in entropy

⇒ \(d S=\frac{d Q}{T}= \pm \frac{m L}{T}\)

where the positive sign refers to heat absorption and the negative sign to heat evolution.

2. When heat given to a substance raises its temperature from T₁ to T₂ then changes in entropy

⇒ \(d S=\int \frac{d Q}{T}=\int_{T_1}^{T_2} m c \frac{d T}{T}=m c \log _e\left(\frac{T_2}{T_1}\right)\)

⇒ \(\Delta S=2.303 \mathrm{mc} \log _{10}\left(\frac{T_2}{T_1}\right)\)

For a perfect gas: The perfect gas equation for n moles is PV = nRT

ΔS = \(\int \frac{d Q}{T}=\int \frac{\mu C_V d T+P d V}{T}\) [As dQ = dU+dW]

ΔS = \(\int \frac{\mu C_V d T+\frac{\mu R T}{V} d V}{T}\)

⇒ \(\mu C_V \int_{T_1}^{T_2} \frac{d T}{T}+\mu R \int_{V_1}^{V_2} \frac{d V}{V}\) [ As PV = μ RT]

∴ ΔS = \(\mu C_V \text { long }_e\left(\frac{T_2}{T_1}\right)+\mu \log _e\left(\frac{V_2}{V_1}\right)\)

In terms of T and P, ΔS = \(m C_p{lon}_g\left(\frac{T_2}{T_1}\right)-\mu R \log _e\left(\frac{P_2}{P_1}\right)\)

and in terms of P and V ΔS = \(\mu C_V \log _e\left(\frac{P_2}{P_1}\right)+\mu C_p \log _e\left(\frac{V_2}{V_1}\right)\)

NEET Physics Class 11 Chapter 9 Kinetic Theory Of Gases And Thermodynamics Multiple Choice Question And Answers

Question 1. When an ideal gas is compressed isothermally then its pressure increases because :

1. Its potential energy decreases
2. Its kinetic energy increases and molecules move apart
3. Its number of collisions per unit area with walls of container increases
4. Molecular energy increases

Answer: 3. Its number of collisions per unit area with walls of container increases

Question 2. Which of the following quantities is zero on average for the molecules of an ideal gas in equilibrium?

1. Kinetic energy
2. Momentum
3. Density
4. Speed

Answer: 2. Momentum

Question 3. The average momentum of a molecule in a sample of an ideal gas depends on

1. Temperature
2. Number of moles
3. Volume
4. None of these
5. Answer: 4. None of these

Question 4. The volume of air increases by 5% in its adiabatic expansion. The percentage decrease in its pressure will be –

1. 6%
2. 7%
3. 8%

Answer: 3. 7%

Question 5. The equation for an ideal gas is :

1. PV = RT
2. PV^γ = constant
3. C_p– C_V= R
4. None of these

Answer: 1. PV = RT

Question 6. The temperature and pressure of 2g oxygen are 27° C and 76 cm Hg, then the volume of the gas is:

1. 1.53 litre
2. 2.44 litre
3. 3.08 litre
4. 44.2 litre

Answer: 1. 1.53 liter

Question 7. Significance of a and b in van der Waal’s equation :

1. A and b both show the correction in the volume of gas
2. A and b both show cohesive force between molecules
3. A shows cohesive force while b shows correction in volume
4. A shows correction in volume while b shows cohesive force

Answer: 3. A shows cohesive force while b shows correction in volume

Question 8. In which condition a real gas behaves as an ideal gas?

1. At high pressure
2. At low pressure
3. At low temperature
4. All the above

Answer: 2. At low pressure

Question 9. Which of the following parameters does not characterize the thermodynamic state of matter?

1. Temperature
2. Pressure
3. Work
4. Volume

Answer: 3. Work

Question 10. 1 calorie is the heat required to increase the temperature of 1 gm of water by 1°C from

1. 13.5°C to 14.5°C at 76 mm of Hg
2. 14.5 °C to 15.5°C at 760 mm of Hg
3. 6.5 °C to 7.5°C at 76 mm of Hg
4. 98.5 °C to 99.5°C at 760 mm of Hg

Answer: 2. 14.5 °C to 15.5°C at 760 mm of Hg

Question 11. A real gas behaves like an ideal gas if its

1. Pressure and temperature are both high
2. Pressure and temperature are both low
3. Pressure is high and temperature is low
4. Pressure is low and temperature is high

Answer: 4. Pressure is low and temperature is high

Question 12. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is:

1. 1: 4
2. 1: 2
3. 6: 9
4. 8: 9

Answer: 4. 8: 9

Question 13. The degrees of freedom of a stationary rigid body about its axis will be :

1. One
2. Two
3. Three
4. Four

Answer: 3. Three

Question 14. The temperature of an ideal gas at atmospheric pressure is 300 K and the volume is 1 m³. If temperature and volume become double, then the pressure will be :

1. 10⁵ N/m²
2. 2 × 10⁵ N/m²
3. 0.5 × 10⁵ N/m²
4. 4 × 10⁵ N/m²

Answer: 1. 105 N/m²

Question 15. If 2g of helium is enclosed in a vessel at NTP, how much heat should be added to it to double the pressure? (Specific heat of helium = 3 J/gm K)

1. 1638 J
2. 1019 J
3. 1568 J
4. 836 J

Answer: 1. 1638 J

Question 16. At what temperature volume of an ideal gas at 0ºC become triple?

1. 546ºC
2. 182ºC
3. 819ºC
4. 646ºC

Answer: 1. 546ºC

Question 17. Two balloons are filled, one with pure He gas and the other with air, respectively. If the pressure and temperature of these balloons are the same, then the number of molecules per unit volume is

1. More in the He filled balloon
2. Same in both balloons
3. More in an air-filled balloon
4. In the ratio of 1: 4

Answer: 2. Same in both balloons

Question 18. A diatomic molecule has

1. 1 degree of freedom
2. 3 degrees of freedom
3. 5 degrees of freedom
4. 6 degrees of freedom

Answer: 3. 5 degrees of freedom

Question 19. The equation of state for 5g of oxygen at a pressure P and temperature T, when occupying a volume V, will be :

1. PV = (5/32) RT
2. PV = 5RT
3. PV = (5/2) RT
4. PV = (5/16) RT

Answer: 1. PV = (5/32) RT

Question 20. Two thermally insulated vessels 1 and 2 are filled with air at temperature (T₁, T₂), volume (V₁, V₂), and pressure (P₁, P₂) respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be:

1. \(T_1+T_2\)
2. \(\left(\mathrm{T}_1+\mathrm{T}_2\right) / 2\)
3. \(\frac{T_1 T_2\left(P_1 V_1+P_2 V_2\right)}{P_1 V_1 T_2+P_2 V_2 T_1}\)
4. \(\frac{T_1 T_2\left(P_1 V_1+P_2 V_2\right)}{P_1 V_1 T_1+P_2 V_2 T_2}\)

Answer: 3. \(\frac{T_1 T_2\left(P_1 V_1+P_2 V_2\right)}{P_1 V_1 T_2+P_2 V_2 T_1}\)

Question 21. A gas behaves more closely as an ideal gas at

1. Low pressure and low temperature
2. Low pressure and high temperature
3. High pressure and low temperature
4. High pressure and high temperature

Answer: 2. Low pressure and high temperature

Question 22. The figure shows graphs of pressure vs density for an ideal gas at two temperatures T₁ and T₂.

1. T₁> T₂
2. T₁= T₂
3. T₁< T₂
4. Any of the three is possible

Answer: 1. T₁> T₂

Question 23. Suppose a container is evacuated to leave just one molecule of gas in it. Let ν_a and ν_rms represent the average speed and the rms speed of the gas.

1. ν_a> ν_rms
2. ν_a< ν_rms
3. ν_a= ν_rms
4. νrms is undefined

Answer: 3. ν_a= ν_rms

Question 24. The rms speed of oxygen molecules in a gas is ν. If the temperature is doubled and the O₂ molecule dissociates into oxygen atoms, the rms speed will become

1. ν
2. ν √2
3. 2ν
4. 4ν

Answer: 3. 2ν

Question 25. Consider a mixture of oxygen and hydrogen kept at room temperature. Compared to a hydrogen molecule an oxygen molecule hits the wall

1. With a greater average speed
2. With a smaller average speed
3. With greater average kinetic energy
4. With smaller average kinetic energy.

Answer: 2. With a smaller average speed

Question 26. Consider the quantity MkT / pV of an ideal gas where M is the mass of the gas. It depends on the

1. Temperature of the gas
2. The volume of the gas
3. The pressure of the gas
4. Nature of the gas

Answer: 4. Nature of the gas

Question 27. If the volume of a gas is decreased by 10% during the isothermal process then its pressure will be –

1. Decrease by 10%
2. Increase by 10%
3. Decrease by 11.11%
4. Increase by 11.11%

Answer: 4. Increase by 11.11%

Question 28. The gases carbon-monoxide (CO) and nitrogen at the same temperature have kinetic energies E₁ and E₂ respectively. Then :

1. E₁= E₂
2. E₁> E₂
3. E₁< E₂
4. E₁ and E₂ cannot be compared

Answer: 1. E₁= E₂

Question 29. In equilibrium, the velocity of molecules of a gas depends on its temperature as

1. \(\mathrm{u} \propto \mathrm{T}\)
2. \(\mathrm{u} \propto \frac{1}{\mathrm{~T}}\)
3. \(\mathrm{u} \propto \sqrt{\mathrm{T}}\)
4. \(\mathrm{u} \propto \mathrm{T}^0\)

Answer: 3. \(\mathrm{u} \propto \sqrt{\mathrm{T}}\)

Question 30. A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds \(\left(\frac{v_{\mathrm{rms}}(\text { helium })}{v_{\mathrm{rms}}(\text { argon })}\right)\) is:

1. 0.32
2. 0.45
3. 2.24
4. 3.16

Answer: 4. 3.16

Question 31. The ratio of the average kinetic energy of H₂ and O₂ at a given temperature is :

1. 1: 16
2. 1: 8
3. 1: 4
4. 1: 1

Answer: 4. 1: 1

Question 32. If the temperature of the gas is increased to three times, then its root mean square velocity becomes:

1. 3 times
2. 9 times
3. 12times
4. 3times

Answer: 4. 3times

Question 33. Which of the following statements is incorrect according to assumptions of the kinetic theory of gases?

1. The potential energy of a molecule is zero
2. Molecules move randomly in all directions
3. the kinetic energy of molecules changes when they collide with the wall of a container
4. None of these

Answer: 4. None of these

Question 34. At what temperature is the rms velocity of a hydrogen molecule equal to that of an oxygen molecule at 47º C?

1. 80 K
2. –73 K
3. 3 K
4. 20 K

Answer: 4. 20 K

Question 35. The kinetic energy of one mole gas at 300 K temperature, is E. At 400 K temperature kinetic energy is E’.The value of E’/E is :

1. 1.33
2. \(\sqrt{\left(\frac{4}{3}\right)}\)
3. \(\frac{16}{9}\)
4. 2

Answer: 1. 1.33

Question 36. If the temperature becomes triple, the root mean square velocity of gas molecules will be :

1. \(\text { v } \sqrt{2}\)
2. \(\text { v/ } \sqrt{3}\)
3. \(\sqrt{3} v\)
4. Same

Answer: 3. \(\sqrt{3} v\)

[υ is the root mean square velocity of gas molecules at temperature T]

Question 37. On increasing the temperature of gas contained in a closed vessel by 1°C, the pressure increases by 0.4%. The initial temperature of a gas is:

1. 25 K
2. 250 K
3. 2500° K
4. 250° C

Answer: 2. 250 K

Question 38. When the temperature of a gas is increased then which of the following statements is always true?

1. Work is done on the gas
2. Heat is supplied to a gas
3. The internal energy of the gas is increased
4. The pressure of gas remains unchanged.

Answer: 3. Internal energy of a gas is increased

Question 39. The speed of sound through oxygen at TK is v ms^-1. As the temperature becomes 2T and oxygen gas dissociates into atomic oxygen, the speed of sound :

1. Remains the same
2. Becomes 2v
3. Becomes \(\sqrt{2 v}\)
4. None of these

Answer: 4. None of these

Question 40. When the temperature of an ideal gas is increased from 27ºC to 227ºC, its rms speed is changed from 400 m/s to v_s. The v_s is :

1. 516 m/s
2. 450 m/s
3. 310 m/s
4. 746 m/s

Answer: 1. 516 m/s

Question 41. The root mean square and most probable speed of the molecules in a gas are

1. Same
2. Different
3. Cannot say
4. Depends on the nature of the gas

Answer: 2. Different

Question 42. Hydrogen gas is filled in a container of volume 20 liter. The average translational kinetic energy of all its molecules is 1.5 × 10⁵ J. Pressure of hydrogen in the cylinder is:

1. 2 × 10⁶ N/m²
2. 3 × 10⁶ N/m²
3. 4 × 10⁶ N/m²
4. 5 × 10⁶ N/m²

Answer: 4. 5 × 10⁶ N/m²

Question 43. The root mean square speed of ideal hydrogen gas in a closed chamber at 0°C is 3180 m/s. Its pressure will be (density of hydrogen gas is 8.99 × 10^-2 kg/m³, 1 atm. = 1.01 × 10⁵ N/m²)

1. 1.0 atm.
2. 1.5 atm.
3. 2.0 atm.
4. 3.0 atm.

Answer: 4. 3.0 atm.

Question 44. A flask contains argon and chlorine in the ratio 3: 1 by mass. The temperature of the mixture is 300 K. If the atomic mass of argon = 39.9 u, and the molecular mass of chlorine = 70.9u, then the ratio of average kinetic energy per molecule of argon to chlorine gas is

1. 1: 1
2. 3: 1
3. 1 : 3
4. 39.9: 70.9

Answer: 1. 1: 1

Question 45. At the same temperature and pressure, the densities of two diatomic gases are d₁ and d₂, The ratio of velocities of sound in these gases will be

1. \(\frac{d_1}{d_2}\)
2. \(\sqrt{\frac{d_2}{d_1}}\)
3. \(\sqrt{\frac{d_1}{d_2}}\)
4. \(\frac{d_2^2}{d_1^2}\)

Answer: 2. \(\sqrt{\frac{d_2}{d_1}}\)

Question 46. The ratio of the mean speed of hydrogen molecules to the mean speed of nitrogen molecules in a sample containing a mixture of the two gases.

1. \(\sqrt{14}\)
2. \(\sqrt{7}\)
3. \(\sqrt{28}\)
4. None of these

Answer: 1. \(\sqrt{14}\)

Question 47. Three closed vessels A, B, and C are at the same temperature T, and contain gases that obey the Maxwellian distribution of velocities. Vessel A contains only O₂, B only N_2, and C a mixture of equal quantities of O₂ and N₂. If the average speed of O₂ molecules in vessel A is V₁, that of the N₂ molecules in vessel B is V₂, the average speed of the O₂ molecules in vessel C will be :

1. (V₁+ V₂)/2
2. V₁
3. (V₁V₂)1/2
4. \(\sqrt{3 \mathrm{kT} / \mathrm{M}}\)

Answer: 2. V₁

Question 48. The pressure of an ideal gas is written as P = \(\frac{2 E}{3 V}\). Here E refers to

1. Translational kinetic energy
2. Rotational kinetic energy
3. Vibrational kinetic energy
4. Total kinetic energy.

Answer: 1. Translational kinetic energy

Question 49. Which of the following quantities is the same for all ideal gases at the same temperature?

1. The kinetic energy of 1 mole
2. The kinetic energy of 1 g
3. The number of molecules in 1 mole
4. The number of molecules in 1 g

Answer: 3. The number of molecules in 1 mole

Question 50. Let ΔU₁ and ΔU₂ be the changes in internal energy of the system in the processes A and B then

1. ΔU₁> ΔU₂
2. ΔU₁= ΔU₂
3. ΔU₁< ΔU₂
4. ΔU₁≠ ΔU₂

Answer: 2. ΔU1= ΔU₂

Question 51. The internal energy of a mono-atomic gas is –

1. \(\frac{5 R T}{2}\)
2. \(\frac{3 R T}{2}\)
3. \(\frac{5 R T}{3}\)
4. \(\frac{7 R T}{3}\)

Answer: 2. \(\frac{3 R T}{2}\)

Question 52. The change in internal energy, when a gas is cooled from 927ºC to 27ºC is

1. 100%
2. 200%
3. 75%
4. 400%

Answer: 3. 75%

Question 53. A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes the total internal energy of the system is:

1. 4RT
2. 15RT
3. 9RT
4. 11RT

Answer: 4. 11RT

Question 54. An ideal gas is filled in a closed rigid and thermally insulated container. A coil of 100Ω resistor carrying current 1A for 5 minutes supplies heat to the gas. The change in internal energy of the gas is

1. 10 KJ
2. 20 KJ
3. 30 KJ
4. 0 KJ

Answer: 3. 30 KJ

Question 55. 300 calories of heat is supplied to raise the temperature of 50 gm of air from 20°C to 30°C without any change in its volume. Change in internal energy per gram of air is

1. Zero
2. 0.6 calories
3. 1.2 calories
4. 6.0 calories

Answer: 4. 6.0 calories

Question 56. Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will:

1. Increase
2. Decrease
3. Remain same
4. Decrease for some, while increase for others

Answer: 3. Remain same

Question 57. According to the law of equal distribution of energy, the mean energy of a molecule per degree of freedom is:

1. \(\frac{1}{2} \mathrm{KT}\)
2. KT
3. \(\frac{3}{2} \mathrm{KT}\)
4. \(\frac{5}{2} \mathrm{KT}\)

Answer: 1. \(\frac{1}{2} \mathrm{KT}\)

Question 58. Which of the following statements is correct for any thermodynamic system?

1. The internal energy changes in all processes
2. Internal energy and entropy are state functions
3. The change in entropy can never be zero
4. The work done in an adiabatic process is always zero

Answer: 2. Internal energy and entropy are state functions

Question 59. A system goes from A to B via two processes 1 and 2 as shown in the figure. If ΔU₁ and ΔU₂ are the changes in internal energies in processes 1 and 2 respectively, then:

1. ΔU₁= ΔU₂
2. The relation between ΔU₁ and ΔU₂ cannot be determined
3. ΔU₂> ΔU₁
4. ΔU₂< ΔU₁

Answer: 1. ΔU₁= ΔU₂

Question 60. Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature To, while box B contains one mole of helium at temperature (7/3)To. The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature. (Ignore the heat capacity of boxes). Then, the final temperature of the gases, Tfin terms of T₀ is :

1. \(\mathrm{T}_{\mathrm{f}}=\frac{3}{7} \mathrm{~T}_0\)
2. \(\mathrm{T}_{\mathrm{f}}=\frac{7}{3} \mathrm{~T}_0\)
3. \(T_f=\frac{3}{2} T_0\)
4. \(\mathrm{T}_{\mathrm{f}}=\frac{5}{2} \mathrm{~T}_0\)

Answer: 3. \(T_f=\frac{3}{2} T_0\)

Question 61. An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V₁ and contains ideal gas at pressure p₁ and temperature T₁. The other chamber has volume V₂ and contains ideal gas at pressure p₂ and temperature T₂. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be –

1. \(\frac{T_1 T_2\left(p_1 V_1+p_2 V_2\right)}{p_1 V_1 T_2+p_2 V_2 T_1}\)
2. \(\frac{p_1 V_1 T_1+p_2 V_2 T_2}{p_1 V_1+p_2 V_2}\)
3. \(\frac{p_1 V_1 T_2+p_2 V_2 T_1}{p_1 V_1+p_2 V_2}\)
4. \(\frac{T_1 T_2\left(p_1 V_1+p_2 V_2\right)}{p_1 V_1 T_1+p_2 V_2 T_2}\)

Answer: 1. \(\frac{T_1 T_2\left(p_1 V_1+p_2 V_2\right)}{p_1 V_1 T_2+p_2 V_2 T_1}\)

Question 62. In the following figures to (4), variation of volume by change of pressure is shown. A gas is taken along the path ABCDA. The change in internal energy of the gas will be:

1. Positive in all cases from(1) to (4)
2. Positive in case (1), and but zero in case(4)
3. Negative in cases (1), and but zero in case
4. Zero in all the four cases.

Answer: 4. Zero in all the four cases.

Question 63. An ideal gas changes from state a to state b as shown in Fig. What is the work done by the gas in the process?

1. Zero
2. Positive
3. Negative
4. Infinite

Answer: 1. Zero

Question 64. The process ΔU = 0, for an ideal gas, can be best represented in the form of a graph:

Answer: 2

Question 65. In the following V-T diagram what is the relation between P₁ and P₂:

1. P₂= P₁
2. P₂> P₁
3. P₂< P₁
4. Cannot be predicted

Answer: 3. P₂< P₁

Question 66. In the isothermal expansion of an ideal gas. Select the wrong statement:

1. There is no change in the temperature of the gas
2. There is no change in the internal energy of the gas
3. The work done by the gas is equal to the heat supplied to the gas
4. The work done by the gas is equal to the change in its internal energy

Answer: 4. The work done by the gas is equal to the change in its internal energy

Question 67. In the cyclic process shown on the P – V diagram, the magnitude of the work done is:

1. \(\pi\left(\frac{P_2-P_1}{2}\right)^2\)
2. \(\pi\left(\frac{V_2-V_1}{2}\right)^2\)
3. \(\frac{\pi}{4}\left(P_2-P_1\right)\left(V_2-V_1\right)\)
4. \(\pi\left(P_2 V_2-P_1 V_1\right)\)

Answer: 3. \(\frac{\pi}{4}\left(P_2-P_1\right)\left(V_2-V_1\right)\)

Question 68. A fixed mass of ideal gas undergoes changes in pressure and volume starting at L, as shown in Figure.

Which of the following is correct :

Answer: 2

Question 69. A fixed mass of gas undergoes the cycle of changes represented by PQRSP as shown in Figure. In some of the changes, work is done on the gas, and in others, work is done by the gas. In which pair of the changes work is done on the gas?

1. PQ and RS
2. PQ and QR
3. OR and RS
4. RS and SP.

Answer: 4. RS and SP

Question 70. Consider two processes on a system as shown in Figure. The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let ΔW₁ and ΔW₂ be the work done by the system in the processes A and B respectively.

1. ΔW₁> ΔW₂
2. ΔW₁= ΔW₂
3. ΔW₁< ΔW₂
4. Nothing can be said about the relation between ΔW1 and ΔW₂

Answer: 3. ΔW₁< ΔW₂

Question 71. A mass of an ideal gas undergoes a reversible isothermal compression. Its molecules will then be compared with an initial state, the same

1. Root mean square velocity
2. Mean momentum
3. Mean kinetic energy

1. (1), (2), (3) correct
2. (1), (2) correct
3. (2), (3) correct
4. (1) correct

Answer: 1. (1), (2), (3) correct

Question 72. Find work done by the gas in the process shown in the figure:

1. \(\frac{5}{2} \pi{atm} \mathrm{L}\)
2. \(\frac{5}{2} \mathrm{~atm} \mathrm{~L}\)
3. \(-\frac{3}{2} \pi {atm} \mathrm{L}\)
4. \(-\frac{5}{4} \pi \mathrm{atm} \mathrm{L}\)

Answer: 4. \(-\frac{5}{4} \pi \mathrm{atm} \mathrm{L}\)

Question 73. The change in internal energy of two moles of a gas during adiabatic expansion is found to be –100 joule. The work done during the process is –

1. 100 joule
2. –100 joule
3. Zero
4. 200 joule

Answer: 1. 100 joule

Question 74. The work done in the following figure is –

1. 2 × 10⁵joule
2. 10⁵joule
3. Zero
4. 3 × 10⁵joule

Answer: 2. 105joule

Question 75. The net amount of the work done in the following indicator diagram is –

1. Zero
2. Positive
3. Negative
4. Infinite

Answer: 1. Zero

Question 76. An ideal gas is taken via paths AB, BC, and CA as shown in Fig. The net work done in the whole cycle is-

1. 3P₁V₁
2. –3P₁V₁
3. 6P₁V₁
4. 12P₁V₁

Answer: 2. –3P₁V₁

Question 77. In the indicator diagram shown, the work done along path AB is-

1. Zero
2. 20 joule
3. –20 joule
4. 60 joule

Answer: 2. 20 joule

Question 78. In the above problem work done along path BC is –

1. Zero
2. 40 joule
3. 60 joule
4. None

Answer: 1. Zero

Question 79. In the above problem, the work done along path CA is –

1. 20 joule
2. 30 joule
3. – 30 joule
4. Zero

Answer: 3. –30 joule

Question 80. Starting the same initial conditions, an ideal gas expands from volume V₁ to V₂ in three different ways. The work done by the gas is W₁ if the process is purely isothermal, W₂ if purely isobaric, and W₃ if purely adiabatic. Then:

1. W₂> W₁> W₃
2. W₂> W₃> W₁
3. W₁> W₂> W₃
4. W₁> W₃> W₂

Answer: 1. W₂> W₁> W₃

Question 81. An ideal gas is taken through the cycle A → B → C → A as shown in Fig. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C → A is:

1. – 5 J
2. – 10 J
3. – 15 J
4. – 20 J

Answer: 1. – 5 J

Question 82. The work done by a gas taken through the closed process ABCA, see figure is

1. 6P₀V₀
2. 4P₀V₀
3. P₀V₀
4. Zero

Answer: 1. 6P₀V₀

Question 83. A system is given 400 calories of heat and 1000 joule of work is done by the system, then the change in internal energy of the system will be –

1. 680 joule
2. 680 erg
3. 860 joule
4. – 860 joule

Answer: 1. 680 joule

Question 84. If AB and CD are isothermals and AD and BC are adiabatic then the temperatures of

1. B and C are the same
2. A and C are the same
3. B and D are the same
4. The temperature of A is more than that of D

Answer: 4. Temperature of A is more than that of D

Question 85. An ideal gas initially at a state (P₁, V₁) is allowed to expand isothermally to a state (P₂, V₂). Then the gas is compressed adiabatically to its initial volume V₁. Let the final pressure be P₃ and the work done by the gas during the whole process be W, then

1. P₃> P₁ and W < 0
2. P₃> P₁ and W > 0
3. P₃< P₁ and W > 0
4. P₃< P₁ and W < 0

Answer: 1. P₃> P₁ and W < 0

Question 86. An ideal gas is taken through the process shown in the figure:

1. In process AB, work done by the system is positive
2. In process AB, heat is rejected out of the system.
3. In process AB, internal energy increases
4. In process AB internal energy decreases and in process BC internal energy increases.

Answer: 2. In process AB, heat is rejected out of the system.

Question 87. If heat is supplied to an ideal gas in an isothermal process,

1. The internal energy of the gas will increase
2. The gas will do positive work
3. The gas will do negative work
4. The said process is not possible

Answer: 2. The gas will do positive work

Question 88. A system can be taken from the initial state p₁, V₁ to the final state p₂, V₂ by two different methods. Let ΔQ and ΔW represent the heat given to the system and the work done by the system. Which of the following must be the same in both methods?

1. ΔQ
2. ΔW
3. ΔQ + ΔW
4. ΔQ – ΔW

Answer: 4. ΔQ – ΔW

Question 89. In changing the state of a system from state A to state B adiabatically the work done on the system is 322 joule. If 100 calories of heat are given to the system in bringing it from state B to state A, then the work done on the system in this process will be –

1. 98 joule
2. 38.2 joule
3. 15.9 calorie
4. 15.9 joule

Answer: 1. 98 joule

Question 90. An ideal gas heat engine operates in a Carnot cycle between 227ºC and 127ºC. It absorbs 6 kcal at a higher temperature. The amount of heat (in kcal) converted into work is equal to:

1. 1.6
2. 1.2
3. 4.8
4. 3.5

Answer: 2. 1.2

Question 91. In a closed container of 44.8 liter, the volume of monoatomic gas is filled up. The heat required to raise the temperature by 10°C will be :

1. R
2. 10R
3. 20R
4. 30R

Answer: 4. 30R

Question 92. Two moles of an ideal gas are taken in a cyclic process abcda. During the process, ab and cd temperatures are 500 K and 300 K respectively. Calculate heat absorbed by the system (In 2 = 0.69 and R = 8.3 J/mol-K)

Answer: 2290.3j

Question 93. If Q, E, and W denote respectively the heat added, change in internal energy, and the work done in a closed cycle process, then

1. W = 0
2. Q = W = 0
3. E = 0
4. Q = 0

Answer: 3. E = 0

Question 94. Which of the following is incorrect regarding the first law of thermodynamics?

1. It does not apply to any cycle process
2. It is a restatement of the principle of conservation of energy
3. It introduces the concept of the internal energy
4. It introduces the concept of the entropy

Answer: (1,4)

Question 95. When a system is taken from state I to state f along the path, it is found that Q = 50 cal and W = 20 cal. Along the path ibf Q = 36 cal. W along the path ibf is:

1. 6 cal
2. 16 cal
3. 66 cal
4. 14 cal

Answer: 1. 6 cal

Question 96. When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas is.

1. \(\frac{2}{5}\)
2. \(\frac{3}{5}\)
3. \(\frac{3}{7}\)
4. \(\frac{5}{7}\)

Answer: 4. \(\frac{5}{7}\)

Question 97. Boiling water is changing into steam. Under this condition, the specific heat of water is

1. Zero
2. One
3. Infinite
4. Less than one

Answer: 3. Infinite

Question 98. Supposing the distance between the atoms of a diatomic gas to be constant, its specific heat at constant volume per mole (gram mole) is

1. \(\frac{5}{2} R\)
2. \(\frac{3}{2} R\)
3. R
4. \(\frac{7}{2} R\)

Answer: 1. \(\frac{5}{2} R\)

Question 99. A gas is formed of molecules each molecule possessing f degrees of freedom, then the value of γ = \(\frac{C_p}{C_V}\)is equal to:

1. \(\frac{2}{\mathrm{f}}\)
2. \(1+\frac{2}{f}\)
3. \(1+\frac{f}{2}\)
4. \(\mathrm{f}+\frac{1}{2}\)

Answer: 2. \(1+\frac{2}{f}\)

Question 100. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio C_p/C_V for the gas is:

1. 4/3
2. 2
3. 5/3
4. 3/2

Answer: 4. 3/2

Question 101. Some students find values of C_V and C_P for gas in calorie/gm–mol K. Which pair is most correct?

1. C_V= 3, C_P= 5
2. C_V= 4, C_P= 6
3. C_V= 3, C_P= 2
4. C_V= 3, C_P= 4.2

Answer: 1. C_V= 3, CP= 5

Question 102. The thermal capacity of anybody is:

1. A measure of its capacity to absorb heat
2. A measure of its capacity to provide heat
3. The quantity of heat required to raise its temperature by a unit degree
4. The quantity of heat required to raise the temperature of a unit mass of the body by a unit degree

Answer: 3. The quantity of heat required to raise its temperature by a unit degree

Question 103. 1 mole of a gas with γ = 7/5 is mixed with 1 mole of a gas with γ = 5/3, then the value of γ for the resulting mixture is:

1. 7/5
2. 2/5
3. 24/16
4. 12/7

Answer: 3. 24/16

Question 104. The molar heat capacity at a constant volume of oxygen gas at STP is nearly 2.5 R. As the temperature is increased, it gradually increases and approaches 3.5 R. The most appropriate reason for this behavior is that at high temperatures

1. Oxygen does not behave as an ideal gas
2. Oxygen molecules dissociate in atoms
3. The molecules collide more frequently
4. Molecular vibration gradually becomes effective

Answer: 3. The molecules collide more frequently

Question 105. The amount of heat required to raise the temperature of 100 gm water from 20ºC to 40ºC will be –

1. 100 calorie
2. 2000 calorie
3. 4000 calorie
4. Zero

Answer: 2. 2000 calorie

Question 106. Two moles of ideal helium gas are in a rubber balloon at 30° C. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35°C. The amount of heat required in raising the temperature is nearly (take R = 8.31 J/mol.K)

1. 62J
2. 104 J
3. 124 J
4. 208 J

Answer: 4. 208 J

Question 107. The molar specific heat at constant pressure of an ideal gas is \(\left(\frac{7}{2}\right)\) R. The ratio of specific heat at constant pressure to that at constant volume is:

1. \(\frac{7}{5}\)
2. \(\frac{8}{7}\)
3. \(\frac{5}{7}\)
4. \(\frac{9}{7}\)

Answer: 1. \(\frac{7}{5}\)

Question 108. One mole of ideal monoatomic gas (γ = 5/3) is mixed with one mole of diatomic gas (γ = 7/5). What is γ for the mixture? γ denotes the ratio of specific heat at constant pressure, to that at constant volume.

1. 3/2
2. 23/15
3. 35/23
4. 4/3

Answer: 1. 3/2

Question 109. A gaseous mixture consists of 16 g of helium and 16 g of oxygen. The ratio
Cof the mixture is:

1. 1.59
2. 1.62
3. 1.4
4. 1.54

Answer: 2. 1.62

Question 110. If C_P and CV denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then

1. C_P – C_V = R / 28
2. C_P – C_V = R / 14
3. C_P – C_V = R
4. C_P – C_V = 28R

Answer: 1. C_P – C_V = R/28

Question 111. A gas is contained in a metallic cylinder fitted with a piston. The piston is suddenly moved in to compress the gas and is maintained at this position. As time passes, after this pressure of the gas in the cylinder

1. Increases
2. Decreases
3. Remains constant
4. Increases or decreases depending on the nature of the gas.

Answer: 2. Decreases

Question 112. Two samples A and B are initially kept in the same state. Sample A is expanded through an adiabatic process and sample B through an isothermal process upto the same final volume. The final pressures in A and B are p_A and p_B respectively.

1. p_A> p_B
2. p_A= p_B
3. p_A< p_B
4. The relation between p_A and p_B cannot be deduced.

Answer: 3. p_A< p_B

Question 113. Let T_a and T_b be the final temperature of the samples A and B respectively in the previous question then:

1. T_a< T_b
2. T_a= T_b
3. T_a> T_b
4. The relation between T_a and T_b cannot be deduced.

Answer: 1. T_a< T_b

Question 114. Let ΔW_a and ΔW_b be the work done by the systems A and B respectively in the previous question then:

1. ΔW_a> ΔW_b
2. ΔW_a= ΔW_b
3. ΔW_a< ΔW_b
4. The relation between W_a and W_b cannot be deduced

Answer: 3. ΔW_a< ΔW_b

Question 115. Four curves A, B, C, and D are drawn in Figure. for a given amount of gas. The curves that represent adiabatic and isothermal changes are

1. C and D respectively
2. D and C respectively
3. A and B respectively
4. B and A respectively

Answer: 3. A and B respectively

Question 116. For an ideal gas, the heat capacity at constant pressure is larger than that at constant volume because

1. Positive work is done during the expansion of the gas by the external pressure
2. Positive work is done during expansion by the gas against external pressure
3. Positive work is done during expansion by the gas against intermolecular forces of attraction
4. More collisions occur per unit of time when volume is kept constant.

Answer: 2. Positive work is done during expansion by the gas against external pressure

Question 117. A gas has:

1. One specific heat only
2. Two specific heats only
3. An infinite number of specific heats
4. No specific heat

Answer: 3. Infinite number of specific heats

Question 118. For a solid with a small expansion coefficient,

1. C_p– C_v= R
2. C_p– C_v= R
3. C_p is slightly greater than C_v
4. C_p is slightly less than C_v

Answer: 3. C_p is slightly greater than C_v

Question 119. When an ideal gas undergoes an adiabatic change causing a temperature change ΔT

1. There is no heat gained or lost by the gas
2. The work done by the gas is equal to the change in internal energy
3. The change in internal energy per mole of the gas is Cv ΔT, where Cvis the molar heat capacity at constant volume.

1. (1), (2), (3) correct
2. (1), (2) correct
3. (1), (3) correct
4. (1) correct

Answer: 3. (1), (3) correct

Question 120. The adiabatic bulk modulus of hydrogen gas (γ = 1.4) at NTP is:

1. 1 × 10⁵ N/m²
2. 1 × 10^-5 N/m²
3. 1.4 N/m²
4. 1.4 × 10⁵ N/m²

Answer: 4. 1.4 × 10⁵ N/m²

Question 121. A given quantity of a gas is at pressure P and absolute temperature T. The isothermal bulk modulus of the gas is:

1. \(\frac{2}{3} P\)
2. P
3. \(\frac{3}{2} P\)
4. 2P

Answer: 2. P

Question 122. A and B are two adiabatic curves for two different gases. Then A and B correspond to:

1. Ar and He respectively
2. He and H₂ respectively
3. O₂ and H₂ respectively
4. H₂ and He respectively

Answer: 2. He and H₂ respectively

Question 123. In a cyclic process shown in the figure an ideal gas is adiabatically taken from B and A., the work done on the gas during the process B → A is 30 J, and when the gas is taken from A → B the heat absorbed by the gas is 20 J. The change in internal energy of the gas in the process A → B is:

1. 20 J
2. – 30 J
3. 50 J
4. – 10 J

Answer: 2. –30 J

Question 124. An ideal gas is allowed to expand freely against a vacuum in a rigid insulated container. The gas undergoes:

1. An increase in its internal energy
2. A decrease in its internal energy
3. Neither an increase nor decrease in temperature or internal energy
4. An increase in temperature

Answer: 1. An increase in its internal energy

Question 125. For free expansion of a gas in an adiabatic container which of the following is true?

1. Q = W = 0 and ΔU = 0
2. Q = 0, W > 0 and ΔU = Q
3. W = 0, Q > 0 and ΔU = Q
4. W = 0, Q < 0 and ΔU = 0

Answer: 1. Q = W = 0 and ΔU = 0

Question 126. In an adiabatic process on a gas with γ = 1.4, the pressure is increased by 0.5%. The volume decreases by about

1. 0.36%
2. 0.5%
3. 0.7&
4. 1%

Answer: 1. 0.36%

Question 127. A fixed mass of an ideal gas undergoes the change represented by XYZX below. Which one of the following sets could describe these changes?

Answer: 4

Question 128. Starting with the same initial conditions, an ideal gas expands from volume V₁ to V₂ in three different ways. The work done by the gas is W₁ if the process is isothermal, W₂ if isobaric and W₃ if adiabatic, then :

1. W₂> W₁> W₃
2. W₂> W₃> W₁
3. W₁> W₂> W₃
4. W₁> W₃> W₂

Answer: 1. W₂> W₁> W₃

Question 129. A gas is expanded from volume V₀ to 2V₀ under three different processes. Process 1 is isobaric, process 2 is isothermal and process 3 is adiabatic. Let ΔU₁, ΔU₂, and ΔU₃ be the change in internal energy of the gas in these three processes. Then:

1. ΔU₁> ΔU₂> ΔU₃
2. ΔU₁< ΔU₂< ΔU₃
3. ΔU₂< ΔU₁< ΔU₃
4. ΔU₂< ΔU₃< ΔU₁

Answer: 1. ΔU₁> ΔU₂> ΔU₃

Question 130. The molar heat capacity for the process shown in fig. is

1. C = C_p
2. C = C_v
3. C > C_v
4. C = 0

Answer: 4. C = 0

Question 131. Find the amount of work done to increase the temperature of one mole of ideal gas by 30ºC. if it is expanding under the condition V ∝ T^2/3 (R = 8.31 J/mol – K):

1. 16.62 J
2. 166.2 J
3. 1662 J
4. 1.662 J

Answer: 2. 166.2 J

Question 132. A gas undergoes a process in which its pressure P and volume V are related as VPⁿ = constant. The bulk modulus of the gas in the process is:

1. no
2. P^1/n
3. P/n
4. Pⁿ

Answer: 3. P/n

Question 133. V = \(k\left(\frac{P}{T}\right)^{0.33}\) where k is constant. It is a,

1. Isothermal process
2. Adiabatic process
3. Isochoric process
4. Isobaric process

Answer: 3. Isochoric process

Question 134. For the adiabatic process of an ideal gas the value of \(\frac{d P}{P}\) is equal to –

1. \(-\gamma \frac{d V}{V}\)
2. \(-\gamma \frac{V}{d V}\)
3. \(\frac{d V}{V}\)
4. \(-\gamma^2 \frac{d V}{V}\)

Answer: 1. \(-\gamma \frac{d V}{V}\)

Question 135. The isobaric modulus of elasticity is –

1. ∞
2. Zero
3. 1
4. \(\frac{C_p}{C_v}\)

Answer: 2. Zero

Question 136. Two samples of a gas A and B initially at the same temperature and pressure, are compressed to half their initial volume, A isothermally and B adiabatically. The final pressure in –

1. A and B will be the same
2. A will be more than in B
3. A will be less than B
4. A will be double that in B

Answer: 3. A will be less than in B

Question 137. The isothermal bulk modulus of elasticity of a gas is 1.5 × 10⁵ N/m². Its adiabatic bulk modulus of elasticity will be if γ = 1.4 –

1. 1.5 × 10⁵ N/m²
2. 3 × 10⁵ N/m²
3. 2.1 × 10⁵ N/m²
4. ∞

Answer: 3. 2.1 × 10⁵ N/m²

Question 138. The pressure and volume of a diatomic gas are P and V respectively. It is compressed suddenly to 1/32 of its initial volume then its final pressure will be –

1. 32 P
2. 128 P
3. P/128
4. P/32

Answer: 2. 128 P

Question 139. The work done by gas in an adiabatic process depends on –

1. Change in temperature
2. Change in volume
3. Change in pressure
4. Change is heat

Answer: 1. Change in temperature

Question 140. The volume of a gas is reduced to 1/4 of its initial volume adiabatically at 27ºC. The final temperature of the gas will be if γ = 1.4 –

1. 300 × (4)0.4 K
2. 100 × (4)0.4 K
3. 27 × (4)0.4 K
4. 300 × (1/4)0.4 K

Answer: 1. 300 × (4)0.4 K

Question 141. 1 m³ of gas is compressed suddenly at atmospheric pressure and temperature 27ºC such that its temperature becomes 627ºC. The final pressure of the gas will be (γ = 1.5)-

1. 27 × 10⁶ N/m²
2. 7.2 × 10⁵ N/m²
3. 2.7 × 10⁵ N/m²
4. 27 × 10⁵ N/m²

Answer: 4. 27 × 105 N/m²

Question 142. If 1 kg air (γ = 1.4) is heated adiabatically from 0ºC to 10ºC then the increase in its internal energy will be (C_v= 0.172 cal/gmºC) –

1. 1720 joule
2. 7224 joule
3. 172 calorie
4. 7224 calorie

Answer: 2. 7224 joule

Question 143. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The value of γ for the gas is –

1. \(\frac{5}{3}\)
2. \(\frac{7}{5}\)
3. \(\frac{3}{2}\)
4. \(\frac{11}{9}\)

Answer: 3. \(\frac{3}{2}\)

Question 144. 5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be T₁, the work done in the process is:

1. \(\frac{9}{8} \mathrm{RT}_1\)
2. \(\frac{3}{2} \mathrm{RT}_1\)
3. \(\frac{15}{8} R T_1\)
4. \(\frac{9}{2} R T_1\)

Answer: 1. \(\frac{9}{8} \mathrm{RT}_1\)

Question 145. An ideal gas is expanding such that PT₂ = constant. The coefficient of volume expansion of the gas is

1. \(\frac{1}{\mathrm{~T}}\)
2. \(\frac{2}{\mathrm{~T}}\)
3. \(\frac{3}{\mathrm{~T}}\)
4. \(\frac{4}{\mathrm{~T}}\)

Answer: 3. \(\frac{3}{\mathrm{~T}}\)

Question 146. One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and constant volume is 5/3, the final temperature of the gas will be :

1. (T + 2.4) K
2. (T – 2.4) K
3. (T + 4) K
4. (T – 4) K

Answer: 4. (T – 4) K

Question 147. 2 moles of He gas γ = 5/3 of 20-litre volume at 27ºC subjected to constant pressure is expanded to double the initial volume. Then, it is adiabatically taken to an initial temperature of 27ºC. What will be the work done in the isobaric process? Also find the final pressure, final volume, and work done in the adiabatic process.

1. 7470 J
2. 7074 J
3. 7070 J
4. 7474 J

Answer: 1. 7470 J

Question 148. A Carnot engine works between 600 K and 300 K. In each cycle of operations, the engine draws 1000 joules of energy from the source at 600 K. The efficiency of the engine is –

1. 20%
2. 50%
3. 70%
4. 90%

Answer: 2. 50%

Question 149. In the above problem, the useful work done by the engine is –

1. 100 joule
2. 500 joule
3. 1000 joule
4. 150 joule

Answer: 2. 500 joule

Question 150. In the above problem, the energy rejected by the sink is –

1. 100 joule
2. 500 joule
3. 1000 joule
4. 300 joule

Answer: 2. 500 joule

Question 151. A Carnot engine works between the ice point and the steam point. Its efficiency will be –

1. 26.81 %
2. 53.36 %
3. 71.23 %
4. 85.42 %

Answer: 1. 26.81 %

Question 152. In the above problem, to increase the efficiency of the engine by 20%, its sink temperature will have to be changed by –

1. Increase by 20 K
2. Decrease by 293 K
3. Increase by 20ºC
4. Decrease by 20ºC

Answer: 4. Decrease by 20ºC

Question 153. In the above problem, to increase the efficiency by 20%, its source temperature will have to be changed by –

1. 402.5 K increase
2. 129.5 K decrease
3. 129.5 ºC increase
4. 129.5 ºC decrease

Answer: 3. 129.5 ºC increase

Question 154. A Cannot engine work between 200ºC and 0ºC. Another Carnot engine works between 0ºC and –200ºC. In both cases, the working substance absorbs 4 kilocalories of heat from the source. The efficiency of the first engine will be –

1. \(\frac{100}{473}\)
2. \(\frac{200}{473}\)
3. \(\frac{200}{273}\)
4. \(\frac{273}{373}\)

Answer: 2. \(\frac{200}{473}\)

Question 155. In the above problem, the efficiency of the second engine will be –

1. \(\frac{100}{273}\)
2. \(\frac{173}{273}\)
3. \(\frac{200}{273}\)
4. \(\frac{273}{373}\)

Answer: 3. \(\frac{200}{273}\)

Question 156. In the above problem, the ratio of efficiencies of two engines will be –

1. 0.18
2. 0.38
3. 0.58
4. 0.78

Answer: 3. 0.58

Question 157. In the above problem, the amount of useful work done by the first engine is –

1. 7.1 × 10³ Joule
2. 3.8 × 10⁴ Joule
3. 5.9 × 10⁵ Joule
4. 9.3 × 10⁶ Joule

Answer: 1. 7.1 × 10³ Joule

Question 158. In the above problem, the output work of the second engine is

1. 2.93 × 10³ calorie
2. 12.3 × 10³ calorie
3. 12.3 × 10³ joule
4. 2.93 × 10³ calorie

Answer: 1. 2.93 × 10³ calorie

Question 159. In the above problem, the ratio of outputs of two engines is –

1. 0.577
2. 0.377
3. 0.777
4. 0.177

Answer: 1. 0.577

Question 160. The efficiency of the Carnot engine is 50% and the temperature of the sink is 500 K. If the temperature of the source is kept constant and its efficiency is to be raised to 60%; then the required temperature of the sink will be:

1. 600 K
2. 500 K
3. 400 K
4. 100 K

Answer: 3. 400 K

Question 161. Even the Carnot engine cannot give 100% efficiency because we cannot:

1. Prevent radiation
2. Find ideal sources
3. Reach absolute zero temperature
4. Eliminate friction

Answer: 3. Reach absolute zero temperature

Question 162. “Heat cannot be itself flow from a body at a lower temperature to a body at a higher temperature” is a statement or consequence of :

1. Second law of thermodynamics
2. Conservation of momentum
3. Conservation of mass
4. The first law of thermodynamics

Answer: 1. Second law of thermodynamics

Question 163. A Carnot engine takes 3 × 10⁶ cal of heat from a reservoir at 627ºC and gives it to a sink at 27ºC. The work done by the engine is:

1. 4.2 × 10⁶ J
2. 8.4 × 10⁶ J
3. 16.8 × 10⁶ J
4. Zero

Answer: 2. 8.4 × 106 J

Question 164. In a Carnot engine, the reservoir temperature is 7°C. Its efficiency is 50%. To increase efficiency to 70% by how much temperature of the source is to be raised.

1. 840 K
2. 280 K
3. 560 K
4. 373 K

Answer: 4. 373 K

Question 165. Which statement is incorrect?

1. All reversible cycles have the same efficiency
2. A reversible cycle has more efficiency than an irreversible one
3. Carnot cycle is a reversible one
4. Carnot cycle has the maximum efficiency in all cycles

Answer: 1. All reversible cycles have the same efficiency

Question 166. An ideal gas heat engine operates in cannot cycle between 227ºC and 127ºC. It absorbs 6 × 10⁴ cal of heat at higher temperatures. The amount of heat converted to work is:

1. 2.4 × 10⁴ cal
2. 6 × 10⁴ cal
3. 1.2 × 10⁴ cal
4. 4.8 × 10⁴ cal

Answer: 3. 1.2 × 10⁴ cal

Question 167. A Carnot engine whose sink is at 300 K has an efficiency of 40% By how much should the temperature of the source be increased to increase its efficiency by 50% of the original efficiency:-

1. 275 K
2. 325 K
3. 250 K
4. 380 K

Answer: 3. 250 K

Question 168. An engine has an efficiency of \(\frac{1}{6}\). When the temperature of the sink is reduced by 62ºC, its efficiency is doubled. The temperature of the source is:

1. 124ºC
2. 37ºC
3. 62ºC
4. 99ºC

Answer: 4. 99ºC

Question 169. The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is:

1. \(\frac{1}{2}\)
2. \(\frac{1}{4}\)
3. \(\frac{1}{3}\)
4. \(\frac{2}{3}\)

Answer: 3. \(\frac{1}{3}\)

Question 170. A Carnot engine, having an efficiency of η = 1/10 as a heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at a lower temperature is

1. 99 J
2. 90 J
3. 1 J
4. 100 J

Answer: 2. 90 J

Question 171. The work of 146 kJ is performed to compress one-kilo mole of a gas adiabatically and in this process the temperature of the gas increases by 7C. The gas is (R = 8.3 J mol^-1 K^-1)

1. Diatomic
2. Triatomic
3. A mixture of monoatomic and diatomic
4. Monoatomic

Answer: 1. Diatomic

Question 172. A Carnot working between 300K and 600K has a work output of 800 J per cycle. What is the amount of heat energy supplied to the engine from the source per cycle

1. 1800 J/cycle
2. 1000 J/cycle
3. 2000 J/cycle
4. 1600 J/cycle

Answer: 4. 1600 J/cycle

Question 173. The coefficient of performance of a Carnot refrigerator working between 30° C and 0° C is

1. 10
2. 1
3. 9
4. 0

Answer: 3. 9

Question 174. If the door of a refrigerator is kept open then which of the following is true

1. Room is cooled
2. Room is heated
3. The room is either cooled or heated
4. The room is neither cooled nor heated

Answer: 2. Room is heated

Question 175. An Ideal gas heat engine operated in a Carnot’s cycle between 227° C and 127° C. It absorbs 6 × 10⁴ J at high temperatures. The amount of heat converted into work is

1. 4.8 × 10⁴ J
2. 3.5 × 10⁴ J
3. 1.6 × 10⁴ J
4. 1.2 × 10⁴ J

Answer: 4. 1.2 × 10⁴ J

Question 176. An ideal gas heat engine exhausting heat at 77° C does not have a 30% efficiency. It must take the heat at

1. 127° C
2. 227°C
3. 327° C
4. 673°C

Answer: 2. 227°C

Question 177. The efficiency of the Carnot engine is 100% if

1. T₂= 273 K
2. T₂= 0 K
3. T₁= 273 K
4. T₁= 0 K

Answer: 2. T₂= 0 K

Question 178. The efficiency of Carnot’s engine operating between reservoirs, maintained at temperatures 27°C and 123°C, is

1. 50%
2. 24%
3. 0.75%
4. 0.4%

Answer: 1. 50%

Question 179. A Carnot engine operates between 227°C and 27°C. The efficiency of the engine will be

1. \(\frac{1}{3}\)
2. \(\frac{2}{5}\)
3. \(\frac{3}{4}\)
4. \(\frac{3}{5}\)

Answer: 2. \(\frac{2}{5}\)

Question 180. A Carnot engine has the same efficiency between 800 K to 500 K and x K to 600 K. The value of x is

1. 1000 K
2. 960 K
3. 846K
4. 754 K

Answer: 2. 960 K

Question 181. A scientist says that the efficiency of his heat engine which operates at source temperature 127°C and sink temperature 27°C is 26% then

1. It is impossible
2. It is possible but less probable
3. It is quite probable
4. Data are incomplete

Answer: 1. It is impossible

Question 182. A Carnot’s engine is made to work between 200°C and 0°C first and then between 0°C and –200°C. The ratio of efficiencies of the engine in the two cases is

1. 1.73 :1
2. 1:1.73
3. 1:1
4. 1: 2

Answer: 1. 1.73 :1

Question 183. The efficiency of a Carnot engine is 50% when the temperature of the outlet is 500 K. To increase up to 60% keeping the temperature of intake the same what is the temperature of the outlet

1. 200K
2. 400 K
3. 600K
4. 800 K

Answer: 2. 400 K

Question 184. If an ideal flask containing hot coffee is shaken, the temperature of the coffee will:

1. Decrease
2. Increase
3. Remain same
4. Decrease if temperature is below 4ºC and increase if temperature is equal to or more than 4ºC

Answer: 2. Increase

Question 185. An electric fan is switched on in a closed room. The air in the room is

1. Cooled
2. Heated
3. Maintains its temperature
4. Heated or cooled depending on the atmospheric pressure

Answer: 2. Heated

Question 186. A heat engine employing a Carnot cycle with an efficiency of η = 10% is used as a refrigerating machine, the thermal reservoirs being the same. The refrigerating efficiency ∈ is

1. 12
2. 8
3. \(\frac{1}{10}\)
4. 9

Answer: 4. 9

Question 187. An ideal gas is initially at temperature T and volume V. Its volume is increased by ΔV due to an increase in temperature ΔT, with pressure remaining constant. The quantity \(\delta=\frac{\Delta \mathrm{V}}{\mathrm{V} \Delta \mathrm{T}}\) varies with temperature as:

Answer: 3

Question 188. A monoatomic ideal gas, initially at temperature T₁, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T₂ by releasing the piston suddenly. If L₁ and L₂ are the length of the gas column before and after expansion respectively, Then T₁/T₂ is given by:

1. \(\left(\frac{L_1}{L_2}\right)^{2 / 3}\)
2. \(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\)
3. \(\frac{L_2}{L_1}\)
4. \(\left(\frac{L_2}{L_1}\right)^{2 / 3}\)

Answer: 4. \(\left(\frac{L_2}{L_1}\right)^{2 / 3}\)

Question 189. Which of the following graphs correctly represents the variation of β = –(dV/dP)/V with P for an ideal gas at constant temperature?

Answer: 1

Question 190. An ideal gas undergoes a cyclic process as shown in the given P–T diagram, where the process AC is adiabatic. The process is also represented by :

Answer: 2

Question 191. Statement – 1

The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and volume. because

Statement – 2

The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.

1. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
2. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
3. Statement-1 is True, Statement-2 is False
4. Statement-1 is False, and Statement-2 is True.

Answer: 2. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

Question 192. One mole of a monatomic ideal gas is taken along two cyclic processes E → F → G → E and E → F → H → E as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal, or adiabatic.

Match the paths in List 1 with the magnitudes of the work done in List 2 and select the correct answer using the codes given below the lists.

Codes:

Answer: 1.

Question 193. Two containers of equal volume contain the same gas at pressure p₁ and p₂ and absolute temperature T₁ and T₂ respectively. On joining the vessels the gas reaches a common pressure p and common temperature T. The ratio p/T is equal to

1. \(\frac{p_1}{T_1}+\frac{p_2}{T_2}\)
2. \(\frac{p_1 T_1+p_2 T_2}{\left(T_1+T_2\right)^2}\)
3. \(\frac{p_1 T_2+p_2 T_1}{\left(T_1+T_2\right)^2}\)
4. \(\frac{\mathrm{p}_1}{2 \mathrm{~T}_1}+\frac{\mathrm{p}_2}{2 \mathrm{~T}_2}\)

Answer: 4. \(\frac{\mathrm{p}_1}{2 \mathrm{~T}_1}+\frac{\mathrm{p}_2}{2 \mathrm{~T}_2}\)

Question 194. If a diatomic gas is supplied heat Q in a process, it performs work \(\frac{Q}{4}\). What is the molar heat capacity of the gas in this process?

1. \(\frac{2}{5} R\)
2. \(\frac{5}{2} R\)
3. \(\frac{10}{3} R\)
4. \(\frac{6}{7} R\)

Answer: 3. \(\frac{10}{3} R\)

Question 195. Two samples of air A and B having the same composition and initially at the same temperature T, pressure P, and volume V are taken. A and B are made to undergo the following process :

Case 1: A and B are compressed from volume V to volume V/2. A is compressed isothermally while B is compressed adiabatically. The final pressures are P_AC and P_BC respectively.

Case: 2 A and B are allowed to undergo expansion from volume V to volume 2V. A undergoes while B undergoes adiabatic expansion. The final pressure of A and B are P_AE and P_BE respectively.

1. P_AC = P_BC and P_AE = P_BE
2. P_AC = P_AE and P_BC = P_BE
3. P_AC > P_BC and P_AE < P_BE
4. P_AC < P_BC and P_AE> P_BE

Answer: 4. P_AC < P_BC and P_AE > P_BE

Question 196. 4 moles of an ideal monoatomic gas occupying volume V is adiabatically expanded from temperature 300 K to a volume of \(2 \sqrt{2} \mathrm{~V}\). Find:

1. Final temperature.
2. Change in internal energy (R = 8.3 J/mol K)

Answer:

1. 150K
2. -7500 J

Question 197. The internal energy change in a system that has absorbed 2 kcal of heat and done 500 J of work is

1. 8900 J
2. 6400 J
3. 5400 J
4. 7900 J

Answer: 4. 7900 J

Question 198. If ΔU and ΔW represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true?

1. ΔU = – ΔW, in a adiabatic process
2. ΔU = ΔW, in a isothermal process
3. ΔU = ΔW, in a adiabatic process
4. ΔU = – ΔW, in an isothermal process

Answer: 1. ΔU = – ΔW, in a adiabatic process

Question 199. If Cpand Cvdenote the specific heats (per unit) mass of an ideal gas of molecular weight M, where R is gas constant, then (Mains)]

1. C_p– C_v= R/M²
2. C_p– C_v= R
3. C_p– C_v= R/M
4. C_p– C_v= MR

Answer: 3. C_p– C_v= R/M

Question 200. A monoatomic gas at pressure P₁ and volume V₁ is compressed adiabatically to \(\frac{1}{8} \text { th }\) of its original volume. What is the final pressure of the gas

1. 64P₁
2. P₁
3. 16P₁
4. 32P₁

Answer: 4. 32P₁

Question 201. A mass of diatomic gas (γ = 1.4) at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from 27ºC to 927ºC. The pressure of the gas in a final state is:

1. 28 atm
2. 68.7atm
3. 256 atm
4. 8 atm

Answer: 3. 256 atm

Question 202. A thermodynamic system is taken through the cycle ABCD as shown in the figure. Heat rejected by the gas during the cycle is:

1. 2 PV
2. 4 PV
3. 12PV
4. PV

Answer: 1. 2 PV

Question 203. One mole of an ideal gas goes from an initial state A to final state B via two processes: It first undergoes isothermal expansion from volume V to 3V and then its volume is reduced from 3V to V at constant pressure. The correct P-V diagram representing the two processes is:

Answer: 4

Question 204. An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram:

If Q₁, Q₂, Q₃ indicate the heat absorbed by the gas along the three processes and ΔU₁, ΔU₂, and ΔU₃ indicate the change in internal energy along the three processes respectively, then

1. Q₁> Q₂> Q₃ and ΔU₁= ΔU₂= ΔU₃
2. Q₃> Q₂> Q1and ΔU₁= ΔU₂= ΔU₃
3. Q₁= Q₂= Q₃ and ΔU₁> ΔU₂> ΔU₃
4. Q₃> Q₂> Q₁ and ΔU₁> ΔU₂> ΔU₃

Answer: 1. Q₁> Q₂> Q₃ and ΔU₁= ΔU₂= ΔU₃

Question 205. A gas is taken through the cycle A → B → C → A, as shown. What is the net work done by the gas?

1. 1000 J
2. Zero
3. –2000 J
4. 2000 J

Answer: 1. 1000 J

Question 206. In the given (V–T) diagram, what is the relation between pressure P₁ and P₂?

1. P₂> P₁
2. P₂< P₁
3. Cannot be predicted
4. P₂= P₁

Answer: 2. P₂< P₁

Question 207. The molar-specific heats of an ideal gas at constant pressure and volume are denoted by C_p and C_v, respectively. If γ = \(\frac{C_p}{C_v}\) and R is the universal gas constant, then C_v is equal to:

1. \(\frac{R}{(\gamma-1)}\)
2. \(\frac{(\gamma-1)}{R}\)
3. \(\gamma R\)
4. \(\frac{1+\gamma}{1-\gamma}\)

Answer: 1. \(\frac{R}{(\gamma-1)}\)

Question 208. The amount of heat energy required to raise the temperature of 1g of Helium at NTP, from T₁K to T₂K is:

1. \(\frac{3}{2} \mathrm{~N}_{\mathrm{a}} \mathrm{k}_{\mathrm{B}}\left(\mathrm{T}_2-\mathrm{T}_1\right)\)
2. \(\frac{3}{4} \mathrm{~N}_{\mathrm{a}} \mathrm{k}_{\mathrm{B}}\left(\mathrm{T}_2-\mathrm{T}_1\right)\)
3. \(\frac{3}{4} N_a k_B \frac{T_2}{T_1}\)
4. \(\frac{3}{8} N_a k_B\left(T_2-T_1\right)\)

Answer: 4. \(\frac{3}{8} N_a k_B\left(T_2-T_1\right)\)

Question 209. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of \(\frac{C_p}{C_v}\) for the gas is:

1. 2
2. \(\frac{5}{3}\)
3. \(\frac{3}{2}\)
4. \(\frac{4}{3}\)

Answer: 3. \(\frac{3}{2}\)

Question 210. The mean free path of molecules of a gas (radius ‘r’) is inversely proportional to:

1. r³
2. r²
3. r
4. \(\sqrt{r}\)

Answer: 2. r2

Question 211. One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure,

The change in internal energy of the gas during the transition is:

1. –20 kJ
2. 20 J
3. – 12 kJ
4. 20 kJ

Answer: 1. –20 kJ

Question 212. A Carnot engine, having an efficiency of η=\(\frac{1}{10}\) as a heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at a lower temperature is:

1. 99 J
2. 90 J
3. 1 J
4. 100 J

Answer: 2. 90 J

Question 213. The ratio of the specific heats \(\frac{C_P}{C_v}\)= γin terms of degrees of freedom (n) is given by:

1. \(\left(1+\frac{n}{3}\right)\)
2. \(\left(1+\frac{2}{n}\right)\)
3. \(\left(1+\frac{n}{2}\right)\)
4. \(\left(1+\frac{1}{n}\right)\)

Answer: 2. \(\left(1+\frac{2}{n}\right)\)

Question 214. The figure below shows two paths that may be taken by a gas to go from a state A to a state C.

In process AB, 400J of heat is added to the system, and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process of AC will be:

1. 500 J
2. 460 J
3. 300 J
4. 380 J

Answer: 2. 460 J

Question 215. An ideal gas is compressed to half its initial volume using several processes. Which of the processes results in the maximum work done on the gas?

1. Isobaric
2. Isochoric
3. Isothermal
4. Adiabatic

Answer: 4. Adiabatic

Question 216. The coefficient of performance of a refrigerator is 5 if the temperature inside the freezer is –20°C, and the temperature of the surroundings to which it rejects heat is

1. 41°C
2. 11°C
3. 21°C
4. 31°C

Answer: 4. 31°C

Question 217. Two vessels separately contain two ideal gases A and B at the same temperature the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is:

1. \(\frac{3}{4}\)
2. 2
3. \(\frac{1}{2}\)
4. \(\frac{2}{3}\)

Answer: 1. \(\frac{3}{4}\)

Question 218. A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then:

1. Which of the cases (whether compression through isothermal or through an adiabatic process) requires more work will depend upon the atomicity of the gas
2. Compressing the gas isothermally will require more work to be done
3. Compressing the gas through an adiabatic process will require more work to be done
4. Compressing the gas isothermally or adiabatically will require the same amount of work

Answer: 3. Compressing the gas through adiabatic process will require more work to be done

Question 219. The molecules of a given mass of a gas have r.m.s. velocity of 200 ms^-1 at 27ºC and 1.0×10⁵ Nm^-2 pressure. When the temperature and pressure of the gas are respectively, 127ºC and 0.05×10⁵ Nm², the r.m.s. velocity of velocity of its molecules in ms^-1 is;

1. \(\frac{100}{3}\)
2. \(100 \sqrt{2}\)
3. \(\frac{400}{\sqrt{3}}\)
4. \(\frac{100 \sqrt{2}}{3}\)

Answer: 3. \(\frac{400}{\sqrt{3}}\)

Question 220. One mole of an ideal monatomic gas undergoes a process described by the equation PV³ = constant. The heat capacity of the gas during this process is :

1. R
2. \(\frac{3}{2} R\)
3. \(\frac{5}{2} R\)
4. 2R

Answer: 1. R

Question 221. The temperature inside a refrigerator is t₂ ºC and the room temperature is t₁ ºC. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

1. \(\frac{t_1+t_2}{t_1+273}\)
2. \(\frac{t_1}{t_1-t_2}\)
3. \(\frac{t_1+273}{t_1-t_2}\)
4. \(\frac{t_2+273}{t_1-t_2}\)

Answer: 3. \(\frac{t_1+273}{t_1-t_2}\)

Question 222. A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T. The mass of each molecule of the gas is m. Which of the following gives the density of the gas?

1. mkT
2. P / (kT)
3. Pm / (kT)
4. P / (kTV)

Answer: 3. Pm / (kT)

Question 223. Thermodynamic processes are indicated in the following diagram:

Match the following :

1. 1 → A 2 → C, 3 → D, 4 → B
2. 1 → C, 2 → A, 3 → D, 4 → B
3. 1 → C, 2 → D, 3 → B, 4 → A
4. 1 → D, 2 → B, 3 → A, 4 → C

Answer: 2. 1 → C, 2 → A, 3 → D, 4 → B

Question 224. A gas mixture consists of 2 moles of O₂ and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:

1. 4 RT
2. 15 RT
3. 9 RT
4. 11 RT

Answer: 4. 11 RT

Question 225. A Carnot engine having an efficiency of \(\frac{1}{10}\) as a heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at a lower temperature is:

1. 1 J
2. 90 J
3. 99 J
4. 100 J

Answer: 2. 90 J

Question 226. The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

1. \(\frac{2}{5}\)
2. \(\frac{2}{7}\)
3. \(\frac{1}{3}\)
4. \(\frac{2}{3}\)

Answer: 1. \(\frac{2}{5}\)

Question 227. The efficiency of an ideal heat engine working between the freezing point and boiling point of water is

1. 26.8 %
2. 12.5 %
3. 6.25 %
4. 20 %

Answer: 1. 26.8 %

Question 228. At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere? (Given mass of oxygen molecule (m) = 2.76 × 10^-26 kg Boltzmann’s constant kB =1.38×10^-23 JK^-1

1. 2.508×10⁴ K
2. 1.254×10⁴ K
3. 5.016×10⁴ K
4. 8.360×10⁴ K

Answer: 4. 8.360×10⁴ K

Question 229. In which of the following processes, heat is neither absorbed nor released by a system?

1. Isochoric
2. Isothermal
3. Adiabatic
4. Isobaric

Answer: 3. Adiabatic

Question 230. An increase in the temperature of a gas-filled container would lead to the:

1. Decrease in intermolecular distance
2. Increase in its mass
3. Increase in its kinetic energy
4. Decrease in its pressure

Answer: 3. Increase in its kinetic energy

Question 231. The value of \(\gamma\left(=\frac{C_p}{C_v}\right)\), for hydrogen helium and another ideal diatomic gas X (whose molecules are not rigid but have an additional vibrational mode), are respectively equal to,

1. \(\frac{7}{5}, \frac{5}{3}, \frac{9}{7}\)
2. \(\frac{5}{3}, \frac{7}{5}, \frac{9}{7}\)
3. \(\frac{5}{3}, \frac{7}{5}, \frac{7}{5}\)
4. \(\frac{7}{5}, \frac{5}{3}, \frac{7}{5}\)

Answer: 1. \(\frac{7}{5}, \frac{5}{3}, \frac{9}{7}\)

Question 232. 1 g of water, of volume 1 cm³ at 100ºC, is converted into steam at the same temperature under normal atmospheric pressure ≈1 ×10⁵ Pa. The volume of steam formed equals 1671 cm³. If the specific latent heat of vaporization of water is 2256 J/g, the change in internal energy is:

1. 2423 J
2. 2089 J
3. 167 J
4. 2256 J

Answer: 2. 2089 J

Question 233. The efficiency of a Carnot engine depends upon

1. The temperature of the sink only
2. The temperatures of the source and sink
3. The volume of the cylinder of the engine
4. The temperature of the source only

Answer: 4. The temperature of the source only

Question 234. The P-V diagram for an ideal gas in a piston-cylinder assembly undergoing a thermodynamic process is shown in the figure. The process is

1. Adiabatic
2. Isochoric
3. Isobaric
4. Isothermal

Answer: 4. Isothermal

Question 235. The quantities of heat required to raise the temperature of two solid copper spheres of radii r₁ and r₂ (r₁ = 1.5 r₂) through 1K are in the ratio

1. \(\frac{5}{3}\)
2. \(\frac{27}{8}\)
3. \(\frac{9}{4}\)
4. \(\frac{3}{2}\)

Answer: 3. \(\frac{9}{4}\)

Question 236. Two cylinders A and B of equal capacity are connected via a stop cock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stop cock is suddenly opened. The process is

1. Isobaric
2. Isothermal
3. Adiabatic
4. Isochoric

Answer: 4. Isochoric

Question 237. The mean free path for a gas molecule depends upon the diameter, d of the molecule as

1. \(\ell \propto \frac{1}{\mathrm{~d}^2}\)
2. \(\ell \propto d\)
3. \(\ell \propto \mathrm{d}^2\)
4. \(\ell \propto \frac{1}{d}\)

Answer: 1. \(\ell \propto \frac{1}{\mathrm{~d}^2}\)

Question 238. The average thermal energy for a mono-atomic gas is : (kB is Boltzmann constant and T, absolute temperature)

1. \(\frac{7}{2} k_B T\)
2. \(\frac{1}{2} k_B T\)
3. \(\frac{3}{2} k_B T\)
4. \(\frac{5}{2} k_B T\)

Answer: 3. \(\frac{3}{2} k_B T\)

Question 239. The mean free path for a gas, with molecular diameter d and number density n, can be expressed as

1. \(\frac{1}{\sqrt{2} n^2 \pi^2 d^2}\)
2. \(\frac{1}{\sqrt{2} n \pi d}\)
3. \(\frac{1}{\sqrt{2} n \pi d^2}\)
4. \(\frac{1}{\sqrt{2} n^2 \pi d^2}\)

Answer: 3. \(\frac{1}{\sqrt{2} n \pi d^2}\)

Question 240. Match Column-1 and Column-2 and choose the correct match from the given choices.

1. 1–Q, 2–R, 3–S, 4–P
2. 1–Q, 2–P, 3–S, 4–R
3. 1–R, 2–Q, 3–P, 4–S
4. 1–R, 2–P, 3–S, 4–Q

Answer: 2. 1–Q, 2–P, 3–S, 4–R

Question 241. n moles of a monoatomic gas is carried around the reversible rectangular cycle ABCDA as shown in the diagram. The temperature at A is T₀. The thermodynamic efficiency of the cycle is

1. 15%
2. 50%
3. 20%
4. 25%

Answer: 2. 50%

Question 242. An engine has an efficiency of 1/6. When the temperature of the sink is reduced by 62ºC, its efficiency is doubled. The temperature of the source will be

1. 37ºC
2. 62ºC
3. 99ºC
4. 124ºC

Answer: 3. 99ºC

Question 243. Assertion: The melting point of ice decreases with the increase of pressure. Reason: Ice contracts on melting.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If Assertion is true but the reason is false.
4. If both assertion and reason are false.

Answer: 1. If both assertion and reason are true and reason is the correct explanation of assertion.

Question 244. 1 mole of H₂ gas is contained in a box of volume V = 1.00 m³ at T = 300 K. The gas is heated to a temperature of T = 3000 K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)

1. Same as the pressure initially
2. 2 times the pressure initially
3. 10 times the pressure initially
4. 20 times the pressure initially

Answer: 4. 20 times the pressure initially

Question 245. Assume the gas to be ideal the work done on the gas in taking it from A to B is :

1. 200 R
2. 300 R
3. 400 R
4. 500 R

Answer: 3. 400 R

Question 246. The work done on the gas in taking it from D to A is

1. –414 R
2. + 414 R
3. – 690 R
4. + 690 R

Answer: 2. + 414 R

Question 247. The net work done on the gas in the cycle ABCDA is:

1. Zero
2. 276 R
3. 1076 R
4. 1904 R

Answer: 2. 276 R

Question 248. One kg of a diatomic gas is at a pressure of 8 × 104 N/m². The density of the gas is 4 kg/m³. What is the energy of the gas due to its thermal motion?

1. 5 × 10⁴J
2. 6 × 10⁴J
3. 7 × 10⁴J
4. 3 × 10⁴J

Answer: 1. 5 × 10⁴J

Question 249. A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle, the volume of the gas increases from V to 32 V, the efficiency of the engine is:

1. 0.5
2. 0.75
3. 0.99
4. 0.25

Answer: 2. 0.75

Question 250. A Carnot engine operating between temperatures T₁ and T₂ has effeiciency \(\frac{1}{6}\). When T₂ is lowered by 62 K, its efficiency increases to \(\frac{1}{3}\). Then T₁ and T₂ are, respectively:

1. 372 K and 310 K
2. 372 K and 330 K
3. 330 K and 268 K
4. 310 K and 248 K

Answer: 1. 372 K and 310 K

Question 251. Three perfect gases at absolute temperatures T₁, T_2, and T₃ are mixed. The masses of molecules are m₁, m_2, and m₃ and the number of molecules is n₁,n_2, and n₃ respectively. Assuming no loss of energy, the final temperature of the mixture is:

1. \(\frac{\left(\mathrm{T}_1+\mathrm{T}_2+\mathrm{T}_3\right)}{3}\)
2. \(\frac{\mathrm{n}_1 \mathrm{~T}_1+\mathrm{n}_2 \mathrm{~T}_2+\mathrm{n}_3 \mathrm{~T}_3}{\mathrm{n}_1+\mathrm{n}_2+\mathrm{n}_3}\)
3. \(\frac{n_1 T_1^2+n_2 T_2^2+n_3 T_3^2}{n_1 T_1+n_2 T_2+n_3 T_3}\)
4. \(\frac{n_1^2 T_1^2+n_2^2 T_2^2+n_3^2 T_3^2}{n_1 T_1+n_2 T_2+n_3 T_3}\)

Answer: 2. \(\frac{\mathrm{n}_1 \mathrm{~T}_1+\mathrm{n}_2 \mathrm{~T}_2+\mathrm{n}_3 \mathrm{~T}_3}{\mathrm{n}_1+\mathrm{n}_2+\mathrm{n}_3}\)

Question 252. A thermally insulated vessel contains an ideal gas of molecular mass M and a ratio of specific heat γ. It is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by :

1. \(\frac{(\gamma-1)}{2(\gamma+1) R} M v^2 K\)
2. \(\frac{(\gamma-1)}{2 \gamma R} M^2 K\)
3. \(\frac{\gamma M v^2}{2 R} K\)
4. \(\frac{(\gamma-1)}{2 R} M^2 K\)

Answer: 4. \(\frac{(\gamma-1)}{2 R} M^2 K\)

Question 253. A container with insulating walls is divided into equal parts by a partition fitted with a valve. One part is filled with an ideal gas at a pressure P and temperature T, whereas the other part is completely evacuated. If the valve is suddenly opened, the pressure and temperature of the gas will be:

1. \(\frac{P}{2}, \frac{T}{2}\)
2. P, T
3. \(P, \frac{T}{2}\)
4. \(\frac{P}{2}, T\)

Answer: 4. \(\frac{P}{2}, T\)

Question 254. Helium gas goes through a cycle ABCDA (consisting of two isochoric and isobaric lines) as shown in the figure. The efficiency of this cycle is nearly : (Assume the gas to be close to ideal gas)

1. 15.4%
2. 9.1%
3. 10.5%
4. 12.5%

Answer: 1. 15.4%

Question 255. A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500K. It is desired to have an engine of efficiency of 60%. Then, the intake temperature for the same exhaust (sink) temperature must be:

1. The efficiency of the Carnot engine cannot be made larger than 50%
2. 1200 K
3. 750 K
4. 600 K

Answer: 3. 750 K

Question 256. The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle is:

1. \(\mathrm{P}_0 \mathrm{v}_0\)
2. \(\left(\frac{13}{2}\right) p_0 \mathrm{v}_0\)
3. \(\left(\frac{11}{2}\right) \mathrm{P}_0 \mathrm{v}_0\)
4. \(4 p_0 v_0\)

Answer: 2. \(\left(\frac{13}{2}\right) p_0 \mathrm{v}_0\)

Question 257. One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in the figure. The process BC is adiabatic. The temperatures at A, B, and C are 400K, 800K, and 600 K respectively. Choose the correct statement:

1. The change in internal energy in the whole cyclic process is 250 R.
2. The change in internal energy in the process CA is 700 R
3. The change in internal energy in the process AB is – 350 R
4. The change in internal energy in the process BC is – 500 R

Answer: 4. The change in internal energy in the process BC is – 500 R

Question 258. An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by an additional 46 cm. What will be the length of the air column above the mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)

1. 16 cm
2. 22 cm
3. 38 cm
4. 6 cm

Answer: 1. 16 cm

Question 259. Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume u = \(\frac{U}{V} \propto T^4\) and pressure P = \(\frac{1}{3}\left(\frac{U}{V}\right)\). If the shell now undergoes an adiabatic expansion the relation between T and R is

1. \(\mathrm{T} \propto \mathrm{e}^{-\mathrm{R}}\)
2. \(T \propto e^{-3 R}\)
3. \(T \propto \frac{1}{R}\)
4. \(\mathrm{T} \propto \frac{1}{\mathrm{R}^3}\)

Answer: 3. \(T \propto \frac{1}{R}\)

Question 260. A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways:

1. Sequentially keeping in contact with 2 reservoirs so that each reservoir supplies the same amount of heat.
2. Sequentially keeping in contact with 8 reservoirs, each reservoir supplies the same amount of heat.
3. In both cases, the body is brought from the initial temperature of 100°C to the final temperature of 200°C. Entropy changes of the body in the two cases respectively is

1. ln 2, 4ln2
2. ln 2, ln 2
3. ln 2, 2 ln 2
4. 2 ln 2, 8 ln 2

Answer: 2. ln 2, ln 2

Question 261. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as Vq, where V is the volume of the gas. The value of q is \(\left(\gamma=\frac{C_P}{C_V}\right)\)

1. \(\frac{3 \gamma+5}{6}\)
2. \(\frac{3 \gamma-5}{6}\)
3. \(\frac{\gamma+1}{2}\)
4. \(\frac{\gamma-1}{2}\)

Answer: 3. \(\frac{\gamma+1}{2}\)

Question 262. ‘n’ moles of an ideal gas undergoes a process A▢B as shown in the figure. The maximum temperature of the gas during the process will be:

1. \(\frac{3 P_0 V_0}{2 n R}\)
2. \(\frac{9 P_0 V_0}{2 n R}\)
3. \(\frac{9 P_0 V_0}{n R}\)
4. \(\frac{9 P_0 V_0}{4 n R}\)

Answer: 4. \(\frac{9 P_0 V_0}{4 n R}\)

Question 263. Cp and Cv are specific heats at constant pressure and constant volume respectively. It is observed that

Cp – Cv = a for hydrogen gas

Cp – Cv = b for nitrogen gas

The correct relation between a and b is:

1. a = 28 b
2. \(a=\frac{1}{14} b\)
3. a = b
4. a = 14 b

Answer: 4. a = 14 b

Question 264. The temperature of an open room of volume 30 m³ increased from 17ºC to 27ºC due to the sunshine. The atmospheric pressure in the room remains 1 × 10⁵ Pa. If ni and nf are the number of molecules in the room before and after heating, then nf – ni will be:

1. – 2.5 × 10²⁵
2. – 1.61 × 10²³
3. 1.38 × 10²³
4. 2.5 × 10²⁵

Answer: 1. – 2.5 × 10²⁵

Question 265. Two moles of an ideal monoatomic gas occupy a volume V at 27°C. The gas expands adiabatically to a volume of 2V. Calculate

1. The final temperature of the gas and
2. Change in its internal energy.

1. (1) 189 K (2) –2.7 kJ
2. (1) 195 K (2) 2.7 kJ
3. (1) 189 K (2) 2.7 kJ
4. (1) 195 K (2) –2.7 kj

Answer: 1. (1) 189 K (2) –2.7 kJ

Question 266. A mixture of 2 moles of helium gas (atomic mass = 4u) and 1 mole of argon gas (atomic mass = 40u) is kept at 300 K in a container, the ratio of their rms speeds \(\left[\frac{\mathrm{v}_{\mathrm{rms}}(\text { helium })}{\mathrm{V}_{\mathrm{rms}}(\text { argon })}\right]\), is close to:

1. 0.32
2. 3.16
3. 2.24
4. 0.45

Answer: 2. 3.16

Question 267. A gas can be taken from A to B via two different processes ACB and ADB. When path ACB is used 60J of heat flows into the system and 30J of work is done by the system. If path ADB is used work done by the system is 10J. The heat flow into the system in path ADB is:

1. 80J
2. 100J
3. 20J
4. 40J

Answer: 4. 40J

Question 268. A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature of 27º C. Amount of heat transferred to the gas so that the rms velocity of molecules is doubled, is about: [Take R = 8.3 J/K mole]

1. 0.9 kJ
2. 6 kJ
3. 14 kJ
4. 10 kJ

Answer: 4. 10 kJ

Question 269. Three Carnot engines operate in series between a heat source at temperature T₁ and a heat sink at temperature T₄ (see figure). There are two other reservoirs at temperatures T₂ and T₃ as shown, with T₁ > T₂ > T₃ > T₄. The three engines are equally efficient if:

1. \(\mathrm{T}_2=\left(\mathrm{T}_1 \mathrm{~T}_4^2\right)^{1 / 3} ; \mathrm{T}_3=\left(\mathrm{T}_1^2 \mathrm{~T}_4\right)^{1 / 3}\)
2. \(T_2=\left(T_1 T_4\right)^{1 / 2} ; T_3=\left(\mathrm{T}_1^2 \mathrm{~T}_4\right)^{1 / 3}\)
3. \(T_2=\left(\mathrm{T}_1^3 \mathrm{~T}_4\right)^{1 / 4} ; \mathrm{T}_3=\left(\mathrm{T}_1 \mathrm{~T}_4^3\right)^{1 / 4}\)
4. \(T_2=\left(T_1^2 T_4\right)^{1 / 3} ; T_3=\left(T_1 T_4^2\right)^{1 / 3}\)

Answer: 4. \(T_2=\left(T_1^2 T_4\right)^{1 / 3} ; T_3=\left(T_1 T_4^2\right)^{1 / 3}\)

Question 230. Two kg of a monoatomic gas is at a pressure of 4 × 10⁴ N/m². The density of the gas is 8 kg/m³. What is the order of energy of the gas due to its thermal motion?

1. 105 J
2. 104 J
3. 106 J
4. 103 J

Answer: 4. 103 J

Question 231. Half a mole of an ideal monoatomic gas is heated at a constant pressure of 1 atom from 20°C to 90°C. Work done by the gas is close to: (Gas constant R = 831 J/mol.K)

1. 581 J
2. 146 J
3. 291 J
4. 73 J

Answer: 3. 291 J

Question 232. When 100g of liquid A at 100°C is added to 50 g of liquid B at a temperature of 75°C, the temperature of the mixture becomes 90°C. The temperature of the mixture, if 100g of liquid A at 100°C is added to 50 g of liquid B at 50°C, will be:

1. 80°C
2. 60°C
3. 70°C
4. 85°C

Answer: 1. 80°C

Question 233. A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TVx = constant, then x is:

1. \(\frac{2}{5}\)
2. \(\frac{5}{3}\)
3. \(\frac{3}{5}\)
4. \(\frac{2}{3}\)

Answer: 1. \(\frac{2}{5}\)

Question 234. An ideal gas enclosed in a cylinder at a pressure of 2 atm and temperature, of 300 K. The mean time between two successive collisions is 6 × 10–8 s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to:

1. 3 × 10^-6 s
2. 4 × 10^-8 s
3. 2 × 10^-7 s
4. 05 × 10^-8 s

Answer: 2. 4 × 10^-8 s

Question 235. A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above the piston is l₁, and that below the piston is l₂, such that l₁ > l₂. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass, m, will be given by:

(R is the universal gas constant g is the acceleration due to gravity.)

1. \(\frac{\mathrm{RT}}{\mathrm{g}}\left[\frac{2 \ell_1+\ell_2}{\ell_1 \ell_2}\right]\)
2. \(\frac{\mathrm{RT}}{\mathrm{ng}}\left[\frac{\ell_1-3 \ell_2}{\ell_1 \ell_2}\right]\)
3. \(\frac{\mathrm{nRT}}{\mathrm{g}}\left[\frac{1}{\ell_2}+\frac{1}{\ell_1}\right]\)
4. \(\frac{\mathrm{nRT}}{\mathrm{g}}\left[\frac{\ell_1-\ell_2}{\ell_1 \ell_2}\right]\)

Answer: 4. \(\frac{\mathrm{nRT}}{\mathrm{g}}\left[\frac{\ell_1-\ell_2}{\ell_1 \ell_2}\right]\)

Gravitation Introduction

The motion of celestial bodies such as the sun, the moon, the earth the planets etc. has been a subject of fascination since time immemorial.

• Indian astronomers of ancient times have done brilliant work in this field, the most notable among them being Arya Bhatt the first person to assert that all planets including the Earth revolve around the sun.
• A millennium later the Danish astronomer Tycobrahe (1546-1601) conducted a detailed study of planetary motion which was interpreted by his pupil Johnaase Kepler (1571-1630), ironically after the master himself had passed away.
• Kepler formulated his important findings in three laws of planetary motion

Universal Law Of Gravitation: Newton’s Law

According to this law, “Each particle attracts every other particle. The force of attraction between them is directly proportional to the product of their masses and inversely proportional to the square of the distance between them”.

⇒ \(\mathrm{F} \propto \frac{m_1 m_2}{r^2}\)

or \(\mathrm{F}=G \frac{m_1 m_2}{r^2}\)

where G = 6.67 × 10^-11 Nm² kg^-2 is the universal gravitational constant. This law holds good irrespective of the nature of two objects (size, shape, mass etc.) at all places and at all times. That is why it is known as a universal law of gravitation.

Dimensional Formula Of G:

G = \(\frac{F r^2}{m_1 m_2}\)

⇒ \(\frac{\left[M L T^{-2}\right]\left[L^2\right]}{\left[M^2\right]}\)

= [M^-1 L³ T^-2]

Newton’s Law Of Gravitation In Vector Form:

⇒ \(\overrightarrow{\mathrm{F}}_{12}=\frac{G m_1 m_2}{r^2} \hat{r}_{12}\)

⇒ \(\overrightarrow{\mathrm{F}}_{21}=\frac{G m_1 m_2}{r^2}\)

Where \(\overrightarrow{\mathrm{F}}_{12}\) is the force on mass m₁ exerted by mass m₂ and vice-versa.

⇒ \(\overrightarrow{\mathrm{F}}_{12}=\frac{G m_1 m_2}{r^2} \hat{r}_{12}\)

⇒ \(\overrightarrow{\mathrm{F}}_{21}=\frac{G m_1 m_2}{r^2} \hat{\mathrm{r}}_{21}\)

Now \(\hat{r}_{12}=-\hat{r}_{21}\)

Thus \(\vec{F}_{21}=\frac{-G m_1 m_2}{r^2} \hat{r}_{12}\) Comparing above, we get \(\vec{F}_{12}=-\vec{F}_{21}\)

Important Characteristics Of Gravitational Force

1. Gravitational force between two bodies forms an action and reaction pair i.e. the forces are equal in magnitude but opposite in direction.
2. Gravitational force is a central force i.e. it acts along the line joining the centres of the two interacting bodies.
3. Gravitational force between two bodies is independent of the nature of the medium, in which they lie.
4. The gravitational force between two bodies does not depend upon the presence of other bodies.
5. Gravitational force is negligible in the case of light bodies but becomes appreciable in the case of massive bodies like stars and planets.
6. Gravitational force is long range-force i.e., gravitational force between two bodies is effective even if their separation is very large. For example, the gravitational force between the sun and the earth is of the order of 10²⁷ N although the distance between them is 1.5 × 10⁷ km

Question 1. The centres of two identical spheres are at a distance of 1.0 m apart. If the gravitational force between them is 1.0 N, then find the mass of each sphere. (G = 6.67 × 10^-11 m³ kg^-1 sec^-1)
Answer:

Gravitational force F = \(\frac{G m \cdot m}{r^2}\)

on substituting F = 1.0 N , r = 1.0 m and G = 6.67 × 10^-11 m³ kg^-1 sec^-1

we get m = 1.225 × 10⁵ kg

Principle Of Superposition

The force exerted by a particle or other particle remains unaffected by the presence of other nearby particles in space.

The total force acting on a particle is the vector sum of all the forces acted upon by the individual masses when they are taken alone.

⇒ \(\vec{F}_1=\vec{F}_1+\vec{F}_2+\vec{F}_3+\ldots \ldots\)

Question 2.

Four point masses each of mass ‘m’ are placed on the corner of a square of side ‘a’. Calculate the magnitude of gravitational force experienced by each particle.

Answer:

F_r= resultant force on each particle = 2F cos 45º + F₁

⇒ \(\frac{2 G \cdot m^2}{a^2} \cdot \frac{1}{\sqrt{2}}+\frac{G m^2}{(\sqrt{2} a)^2}=\frac{G \cdot m^2}{2 a^2}(2 \sqrt{2}+1)\)

Gravitational Field

The space surrounding the body within which its gravitational force of attraction is experienced by other bodies is called the gravitational field.

• A gravitational field is very similar to an electric field in electrostatics where charge ‘q’ is replaced by mass ‘m’ and electric constant ‘K’ is replaced by gravitational constant ‘G’.
• The intensity of the gravitational field at a point is defined as the force experienced by a unit mass placed at that point.

⇒ \(\vec{E}=\frac{\vec{F}}{m}\)

The unit of the intensity of the gravitational field is N kg^-1.

Intensity of gravitational field due to point mass:

The force due to mass m on test mass m0 placed at point P is given by :

⇒ \(\mathrm{F}=\frac{\mathrm{GMm}_0}{\mathrm{r}^2}\)

Hence \(E=\frac{F}{m_0}\)

⇒ \(E=\frac{G m}{r^2}\)

In vector form \(\vec{E}=-\frac{G M}{r^2} \hat{r}\)

Dimensional formula of intensity of gravitational field = \(\frac{F}{m}=\frac{\left[M L T^{-2}\right]}{[M]}\)

⇒ \(\left[\begin{array}{ll}M^0 & L T^{-2}\end{array}\right]\)

Question 1. Find the relation between the gravitational field on the surface of two planets A and B of masses m_A, mB and radius R_A and R_B respectively if

1. They have equal mass
2. They have equal (uniform) density

Answer: Let EA and EB be the gravitational field intensities on the surface of planets A and B.

then, \(\mathrm{E}_{\mathrm{A}}=\frac{G m_A}{R_A^2}=\frac{G \frac{4}{3} \pi R_A^3 \rho_A}{R_A^2}\)

⇒ \(\frac{4 G \pi}{3} \rho_A R_A\)

Similarly, \(E_B=\frac{G m_B}{R_{B^2}}=\frac{4 G}{3} \pi \rho_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}\)

For \(m_A=m_B\)

⇒ \(\frac{E_A}{E_B}=\frac{R_B^2}{R_A^2}\)

For \(\rho_{\mathrm{A}}=\rho_{\mathrm{B}}\)

⇒ \(\frac{E_A}{E_B}=\frac{R_A}{R_B}\)

Gravitational Potential

The gravitational potential at a point in the gravitational field of a body is defined as the amount of work done by an external agent in bringing a body of unit mass from infinity to that point, slowly (no change in kinetic energy). Gravitational potential is very similar to electric potential in electrostatics.

Gravitational Potential Due To A Point Mass :

Let the unit mass be displaced through a distance dr towards mass M, then work done is given by

dW = F dr = \(\frac{\mathrm{Gm}}{\mathrm{r}^2} \mathrm{dr}\)

The total work done in displacing the particle from infinity to point P is −

⇒ \(\mathrm{W}=\int d W=\int_{\infty}^r \frac{G M}{r^2} d r=\frac{-G M}{r}\)

Thus gravitational potential, \(V=-\frac{G M}{r}\)

The unit of gravitational potential is J kg^-1. Dimensional Formula of gravitational potential

⇒ \(\frac{\text { work }}{\text { mass }}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{[\mathrm{M}]}\)

= [M°L² T^-2].

Question 1. Find out the potential at P and Q due to the two-point mass system. Find out work done by external agents in bringing unit mass from P to Q. Also find work done by gravitational force.

Answer: V_P1 = potential at P due to mass ‘m’ at ‘l’ = –\(\frac{G m}{\ell}\)

⇒ \(\mathrm{V}_{\mathrm{P} 2}=-\frac{G m}{\ell}\)

∴ \(\mathrm{V}_{\mathrm{p}}=\mathrm{V}_{\mathrm{P} 1}+\mathrm{V}_{\mathrm{P} 2}=-\frac{2 G m}{\ell}\)

⇒ \(V_{Q 1}=-\frac{G M}{\ell / 2}\)

⇒ \(V_{Q 2}=-\frac{G m}{\ell / 2}\)

∴ \(\mathrm{v}_{\mathrm{Q}}=\mathrm{v}_{\mathrm{Q} 1}+\mathrm{v}_{\mathrm{Q} 2}=-\frac{G m}{\ell / 2}-\frac{G m}{\ell / 2}=-\frac{4 G m}{\ell}\)

Force at point Q = 0

Work done by external agent = \(\left(\mathrm{V}_{\mathrm{Q}}-\mathrm{V}_{\mathrm{P}}\right) \times 1=-\frac{2 G M}{\ell}\)

Work done by gravitational force = V_P– V_Q = \(\frac{2 G M}{\ell}\)

Question 2. Find the potential at a point ‘P’ at a distance ‘x’ on the axis away from the centre of a uniform ring of mass M and radius R.
Answer:

The ring can be considered to be made of a large number of point masses (m₁, m₂ ………. etc) Ring

⇒ \(\mathrm{V}_{\mathrm{p}}=-\frac{G m_1}{\sqrt{R^2+x^2}}-\frac{G m_2}{\sqrt{R^2+x^2}}-\ldots \ldots .\)

⇒ \(-\frac{G}{\sqrt{R^2+x^2}}\left(m_1+m_2 \ldots . .\right)=-\frac{G M}{\sqrt{R^2+x^2}}\) where, M = m₁+m₂+m₃+……..

Potential at centre of ring = \(-\frac{G \cdot M}{R}\)

Relation Between Gravitational Field And Potential

The work done by an external agent to move unit mass from one point to another point in the direction of the field E, slowly through an infinitesimal distance dr = Force by external agent × distance moved = – Edr.

Thus dV = – Edr

⇒ E = – \(\frac{d V}{d r}\)

Therefore, the gravitational field at any point is equal to the negative gradient at that point.

Uniform Solid Sphere

Point P inside the shell. r ≤ a, then

V = \(-\frac{\mathrm{Gm}}{2 \mathrm{a}^3}\left(3 \mathrm{a}^2-\mathrm{r}^2\right)\)

and E = \(-\frac{G M r}{a^3}\) ,and at the centre V = \(-\frac{3 G M}{2 a}\) and E = 0

Point P outside the shell. r > a, then V = \(-\frac{G M}{r}\) and E = \(-\frac{G M}{r^2}\)

Uniform Thin Spherical Shell

Point P Inside the shell. r ≤ a , then V = \(-\frac{G M}{a}\) and E = 0

Point P outside shell. r ≥ a, then V = \(-\frac{G M}{r}\) and E = \(-\frac{G M}{r^2}\)

Gravitational Potential Energy

The gravitational potential energy of two mass systems is equal to the work done by an external agent in assembling them, while their initial separation was infinity.

Consider a body of mass m placed at a distance r from another body of mass M. The gravitational force of attraction between them is given by

⇒ \(\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{r}^2}\)

Now, Let the body of mass m is displaced from a point. C to B through a distance ‘dr’ towards the mass M, then work done by internal conservative force (gravitational) is given by,

⇒ \(\mathrm{dW}=\mathrm{F} \mathrm{dr}=\frac{G M m}{r^2} \mathrm{dr}\)

⇒ \(\int d W=\int_{\infty}^r \frac{G M m}{r^2} \mathrm{dr}\)

∴ Gravitational potential energy, \(\mathrm{U}=-\frac{\mathrm{GMm}}{r}\)

Increase In Gravitational Potential Energy:

Suppose a block of mass m on the surface of the earth. We want to lift this block by ‘h’ height. Work required in this process = increase in P.E. = U_f– U_i= m(V_f– V_i)

⇒ \(\mathrm{W}_{\mathrm{ext}}=\Delta \mathrm{U}=(\mathrm{m})\left[-\left(\frac{G M_c}{R_e+h}\right)-\left(-\frac{G M_c}{R_e}\right)\right]\)

W_ext = ΔU = \(\mathrm{GM}_e \mathrm{~m}\left(\frac{1}{R_e}-\frac{1}{R_e+h}\right)=\frac{G M_e m}{R_e}\left(1-\left(1+\frac{h}{R_e}\right)^{-1}\right)\)

(as h << R_e, we can apply the Binomial theorem)

W_ext = ΔU = \(\frac{G M_e m}{R_e}\left(1-\left(1-\frac{h}{R_e}\right)\right)=(\mathrm{m})\left(\frac{G M_e}{R_e^2}\right) \mathrm{h}\)

W_ext = ΔU = mgh

This formula is valid only when h << R_e

Question 1. A body of mass m is placed on the surface of the earth. Find the work required to lift this body by a height

1. h = \(\frac{R_c}{1000}\)
2. h = R_e

Answer: h = \(\frac{R_c}{1000}\) as h << R_e , so

we can apply

W_ext = \(\mathrm{W}_{\mathrm{ext}}=\mathrm{U} \uparrow=\mathrm{mgh}\)

⇒ \(\mathrm{W}_{\mathrm{ext}}=\left(\frac{G M_e}{R_e{ }^2}\right)\left(\frac{R_e}{1000}\right)(\mathrm{m})=\frac{G M_e m}{1000 R_e}\)

h = R_e, in this case, h is not much less than R_e, so we cannot apply ΔU = mgh so we cannot apply ΔU = mgh

W_ext = U↑ = U_f– U_i= m(V_f– V_i)

⇒ \(\mathrm{W}_{\mathrm{ext}}=\mathrm{m}\left[\left(-\frac{G M_e}{R_e+R_e}\right)-\left(-\frac{G M_e}{R_e}\right)\right]\)

⇒ \(\mathrm{W}_{\mathrm{ost}}=-\frac{G M_e m}{2 R_e}\)

Acceleration Due To Gravity

It is the acceleration, a freely falling body near the earth’s surface acquires due to the earth’s gravitational pull.

• The property by virtue of which a body experiences or exerts a gravitational pull on another body is called gravitational mass mG, and the property by virtue of which a body opposes any change in its state of rest or uniform motion is called its inertial mass
• m₁ thus if \(\overrightarrow{\mathrm{E}}\) is the gravitational field intensity due to the earth at a point P, and g is acceleration due to gravity at the same point, then \(m_1 \vec{g}=m_G \vec{E}\)
• Now the value of inertial and gravitational mass happens to be exactly the same to a great degree of accuracy for all bodies. Hence,\(\vec{g}=\vec{E}\)

The gravitational field intensity on the surface of the earth is therefore numerically equal to the acceleration due to gravity (g), there. Thus we get,

⇒ \(g=\frac{G M_e}{R_e^2}\)

where, M_e= Mass of earth

R_e = Radius of earth

Note: Here the distribution of mass in the earth is taken to be spherical and symmetrical so that its entire mass can be assumed to be concentrated at its centre for the purpose of calculation of g.

Variation Of Acceleration Due To Gravity

Effect of Altitude

Acceleration due to gravity on the surface of the earth is given by,

g = \(\frac{G M_e}{R_e^2}\)

Now, consider the body at a height ‘h’ above the surface of the earth, then the acceleration due to gravity at height ‘h’ given by

⇒ \(\mathrm{g}_{\mathrm{h}}=\frac{G M_e}{\left(R_e+h\right)^2}=\mathrm{g}\left(1+\frac{h}{R_e}\right)^{-2} \simeq \mathrm{g}\left(1-\frac{2 h}{R_e}\right)\)

when h << R.

The decrease in the value of ‘g’ with height h = g – g_h= \(\frac{2 g h}{R_e}\)

The percentage decrease in the value of \(‘ \mathrm{~g}^{\prime}=\frac{g-g_h}{g} \times 100=\frac{2 h}{R_e} \times 100 \%\)

Effect Of Depth

The gravitational pull on the surface is equal to its weight i.e. mg = \(\frac{G M_e m}{R_e^2}\)

∴ \(\mathrm{mg}=\frac{G \times \frac{4}{3} \pi R_e^3 \rho m}{R_e^2}\)

or \(\mathrm{g}=\frac{4}{3} \pi \mathrm{GR}_{\mathrm{e}} \rho\) ………(1)

When the body is taken to a depth d, the mass of the sphere of radius (R_e – d) will only be effective for the gravitational pull and the outward shall will have no resultant effect on the mass. If the acceleration due to gravity on the surface of the solid sphere is gd, then

⇒ \(g_d=\frac{4}{3} \pi G\left(R_e-d\right) \rho\) ……………(2)

By dividing equation (2) by equation (1)

⇒ \(\mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{d}{R_e}\right)\)

Important Points

At the center of the earth, d = R_e, so g_centre = \(g\left(1-\frac{R_e}{R_e}\right)\) = 0.

Thus, the weight (mg) of the body at the centre of the earth is zero.

Percentage decrease in the value of ‘g’ with the depth

⇒ \(\left(\frac{g-g_d}{g}\right) \times 100=\frac{d}{R_e} \times 100\)

Effect Of The Surface Of Earth:

The equatorial radius is about 21 km longer than its polar radius.

We know, g = \(\frac{G M_e}{R_e^2}\) Hence g_pole > g_equator.

The weight of the body increases as the body is taken from the equator to the pole.

Effect of rotation of the Earth: The earth rotates around its axis with angular velocity ω. Consider a particle of mass m at latitude θ. The angular velocity of the particle is also ω.

⇒ \(\mathrm{g}^{\prime}=\mathrm{g}\left[1-\frac{R_e \omega^2}{g} \cos ^2 \theta\right]\)

At pole θ = 90° ⇒ g_pole = g,

At equator θ = 0 ⇒ g_equator = g \(\left[1-\frac{R_e \omega^2}{g}\right]\)

Hence g _pole > g _equator

If the body is taken from the pole to the equator, then g′ = g\(\left(1-\frac{R_e \omega^2}{g}\right)\)

Hence % change in weight = \(\frac{m g-m g\left(1-\frac{R_e \omega^2}{g}\right)}{m g} \times 100\)

⇒ \(\frac{m R_e \omega^2}{m g} \times 100\)

⇒ \(\frac{R_e \omega^2}{g} \times 100\)

Escape Speed

The minimum speed required to send a body out of the gravity field of a planet (send it to r → ∞)

Escape Speed At Earth’s Surface:

Suppose a particle of mass m is on the earth’s surface

We project it with a velocity V from the earth’s surface so that it just reaches r → ∞ (at r → ∞, its velocity becomes zero)

Applying energy conservation between the initial position (when the particle was at the earth’s surface) and finding positions (when the particle just reaches r → ∞)

K_i+ U_i= K_f+ U_f

⇒ \(\frac{1}{2} m v^2+m_0\left(-\frac{G M_e}{R}\right)\)

⇒ \(0+\mathrm{m}_0\left(-\frac{G M_e}{(r \rightarrow \infty)}\right)\)

⇒ \(\mathrm{v}=\sqrt{\frac{2 G M_0}{R}}\)

Escape speed from earth is surface \(V_e=\sqrt{\frac{2 G M_e}{R}}\)

If we put the values of G, M_e, and R we get

V_e = 11.2 km/s.

Escape Speed Depends On :

1. Mass (Me) and size (R) of the planet
2. Position from where the particle is projected.

Escape Speed Does Not Depend On :

1. Mass of the body which is projected (m0)
2. The angle of projection.

If a body is thrown from the Earth’s surface with escape speed, it goes out of the earth’s gravitational field and never returns to the Earth’s surface. But it starts revolving around the sun.

Kepler’s Law For Planetary Motion

Suppose a planet is revolving around the sun, or a satellite is revolving around the earth, then the planetary motion can be studied with the help of Kepler’s three laws.

Kepler’s Law Of Orbit

Each planet moves around the sun in a circular path or elliptical path with the sun at its focus. (In fact circular path is a subset of an elliptical path)

Law Of Areal Velocity:

To understand this law, let us understand the angular momentum conservation for the planet.

• If a planet moves in an elliptical orbit, the gravitation force acting on it always passes through the centre of the sun. So torque of this gravitation force about the centre of the sun will be zero.
• Hence we can say that the angular momentum of the planet about the centre of the sun will remain conserved (constant) τ about the sun = 0

⇒ \(\frac{d J}{d t}\) = 0

J_planet/ sun = constant

⇒ mvr sinθ = constant

Now we can easily study Kapler’s law of areal velocity.

If a planet moves around the sun, the radius vector \((\vec{r})\) also rotates are sweeps area as shown in the figure. Now let’s find a rate of area swept by the radius vector \((\vec{r})\).

Suppose a planet is revolving around the sun and at any instant its velocity is v, and the angle between radius vector \((\vec{r})\) and velocity \((\vec{v})\). In dt time, it moves by a distance vdt, during this dt time, the area swept by the radius vector will be OAB which can be assumed to be a triangle

dA = 1/2 (Base) (Perpendicular height)

dA = 1/2 (r) (vdtsinθ)

so rate of area swept \(\frac{d A}{d t}=\frac{1}{2} \mathrm{vr} \sin \theta\)

we can write \(\frac{d A}{d t}=\frac{1}{2} \frac{m v r \sin \theta}{m}\)

where mvr sinθ = angular momentum of the planet about the sun, which remains conserved (constant)

⇒ \(\frac{d A}{d t}=\frac{L_{\text {planet } / \text { sun }}}{2 m}\)= constant

so the Rate of area swept by the radius vector is constant

Kepler’s Law Of Time Period: Suppose a planet is revolving around the sun in a circular orbit

then \(\frac{m_0 v^2}{r}=\frac{G M_s m_0}{r^2}\)

v = \(\sqrt{\frac{G M_s}{r}}\)

Time period of the revolution is

⇒ \(\frac{2 \pi r}{v}=2 \pi \mathrm{r} \sqrt{\frac{r}{G M_s}}\)

⇒ \(\mathrm{T}^2=\left(\frac{4 \pi^2}{G M_s}\right) \mathrm{r}^3\)

For all the planets of a sun, T² ∝ r³

Circular Motion Of A Satellite Around A Planet

Suppose a satellite of mass m0 is at a distance r from a planet. If the satellite does not revolve, then due to the gravitational attraction, it may collide with the planet.

To avoid a collision, the satellite revolves around the planet, in a circular motion of a satellite.

⇒ \(\frac{G M_e m_0}{r^2}=\frac{m_0 v^2}{r}\) ….

⇒ v = \(\sqrt{\frac{G M_e}{r}}\) this velocity is called orbital velocity (v0)

⇒ \(\mathrm{v}_0=\sqrt{\frac{G M_e}{r}}\)

Total Energy Of The Satellite Moving In Circular Orbit:

KE = \(\frac{1}{2} m_0 v^2\) and from equation …..(1)

⇒ \(\frac{m_0 v^2}{r}=\frac{G M_e m_0}{r^2}\)

⇒ \(\mathrm{m}_0 \mathrm{v}^2=\frac{G M_e m_0}{r}\)

⇒ \(\mathrm{KE}=\frac{1}{2} m_0 v^2=\frac{G M_e m_0}{2 r}\)

Potential energy

⇒ \(\mathrm{U}=-\frac{G M_e m_0}{r}\)

Total energy = KE + PE = \(\left(\frac{G M_e m_0}{2 r}\right)+\left(\frac{-G M_e m_0}{r}\right)\)

⇒ \(\mathrm{TE}=-\frac{G M_e m_0}{2 r}\)

The total energy is –ve. It shows that the satellite is still bounded by the planet.

Geo – Stationery Satelite

We know that the earth rotates about its axis with angular velocity ω_earth and time period T_earth = 24 hours.

Suppose a satellite is set in an orbit which is in the plane of the equator, whose ω is equal to ω_earth, (or its T is equal to T_earth = 24 hours) and whose direction is also the same as that of earth. Then as seen from Earth, it will appear to be stationary. This type of satellite is called geostationary satellite. For a geo-stationery satellite,

w_satelite = w_earth

⇒ T_satelite = T_earth= 24 hr.

So time period of a geo-stationery satellite must be 24 hours. To achieve T = 24 hours, the orbital radius geo-stationery satellite :

⇒ \(\mathrm{T}^2=\left(\frac{4 \pi^2}{G M_e}\right) \mathrm{r}^3\)

Putting the values, we get an orbital radius of geo stationary satellite r = 6.6 Re(here Re = radius of the earth) height from the surface h = 5.6 Re.

Path Of A Satellite According To Different Speed Of Projection

Suppose a satellite is at a distance r from the centre of the earth. If we give different velocities (v) to the satellite, its path will be different

If v < v₀ \(\left(\text { or } v<\sqrt{\frac{G M_e}{r}}\right)\) then the satellite will move in an elliptical path and strike the earth’s surface.

But if the size of the earth were small, the satellite would complete the elliptical orbit, and the centre of the earth would be at its farther focus.

If v = v₀ \(\left(\text { or } \quad v=\sqrt{\frac{G M_e}{r}}\right)\), then the satellite will revolve in a circular orbit.

If v₀ > V > v₀ \(\left(\text { or } \sqrt{\frac{2 G M_e}{r}}>v>\sqrt{\frac{G M_e}{r}}\right)\), then the satellite will revolve in an elliptical orbital, and the centre of the earth will be at its nearer focus.

If v = v_e \(\left(\begin{array}{ll}\text { or } and v=\sqrt{\frac{2 G M_e}{r}}
\end{array}\right)\), then the satellite will just escape with a parabolic path.

Question 1. Suppose a planet is revolving around the sun in an elliptical path given by \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). Find a period of revolution. The angular momentum of the planet about the sun is L.

Answer: Rate of area swept = \(\frac{d A}{d t}=\frac{L}{2 m}\) constant

⇒ \(\mathrm{dA}=\frac{L}{2 m} d t\)

⇒ \(\int_{A=0}^{A=\pi a b} d A=\int_{t=0}^{t=T} \frac{L}{2 m} d t\)

⇒ \(\pi \mathrm{ab}=\frac{L}{2 m} \mathrm{~T}\)

⇒ \(\mathrm{T}=\frac{2 \pi m a b}{L}\)

Question 2. The Earth and Jupiter are two planets of the sun. The orbital radius of the earth is 10⁷ m and that of Jupiter is 4 × 10⁷ m. If the time period of the revolution of earth is T = 365 days, find the time period of revolution of the Jupiter.

Answer:

For both the planets T² ∝ r³

⇒ \(\left(\frac{T_{\text {jupitar }}}{T_{\text {earth }}}\right)^2=\left(\frac{T_{\text {jupiter }}}{r_{\text {earhh }}}\right)^3\)

⇒ \(\left(\frac{T_{\text {juppicr }}}{365 \text { days }}\right)^2=\left(\frac{4 \times 10^7}{10^7}\right)^3\)

T_jupiter = 8 × 365 days

Question 3.

Suppose earth has radius R and mass M. A point mass m₀ is at a distance r from the centre. Find the gravitational potential energy of the mass due to the earth.
Answer:

U_g= (m₀) (V_earth)

⇒ \(\mathrm{U}_{\mathrm{g}}=\left(\mathrm{m}_0\right)\left(-\frac{G M_e}{r}\right)=\left(-\frac{G M_e m_0}{r}\right)\)

Question 4.

Suppose the earth has mass and radius R. A small groove in made and point mass m0is placed at the centre of the sphere. With what minimum velocity should we project the particle so that it may escape out of the gravity field (reaches to r → ∞)
Answer:

Suppose the particle is projected with speed v, and to send it to infinity, its velocity should be zero at r → ∞. Applying energy conservation between its initial position (centre) and final position (r → ∞) K_i+ U_i= k_f+ U_f

⇒ \(\frac{1}{2} m_0 v^2+\left(m_0\right)\left(v_{\text {earth }}\right)\)

⇒ \(\frac{1}{2} \mathrm{~m}_0 \mathrm{v}^2+\left(\mathrm{m}_0\right)\left(-\frac{3 G M_e}{2 R}\right)\)

⇒ 0+m₀(0)

⇒ \(\mathrm{v}=\left(\sqrt{\frac{3 G M_e}{R}}\right)\)

Summary

Newton’s Law Of Gravitation:

Gravitational attraction force between two point masses

⇒ \(\mathrm{F}_{\mathrm{g}}=\frac{G m_1 m_2}{r^2}\) and its direction will be attractive.

Gravitational force on (1) due to (2) in vector form

⇒ \(\vec{F}_{12}=\frac{G m_1 m_2}{r^2}\)

Gravitational Field: Gravitational force acting on unit mass.

⇒ \(\mathbf{g}=\frac{F}{m}\)

Gravitational Potential:

⇒ \(\mathrm{v}_{\mathrm{g}}=\frac{U}{m}\)

Gravitation potential energy of unit mass

⇒ \(\mathrm{g}=-\frac{d V_g}{d r}\)

and \(\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=-\int_A^B \vec{g} \cdot d \vec{r}\)

For point mass: GM

⇒ \(\mathrm{g}=\frac{G M}{r^2}, \mathrm{~V}=-\frac{G M}{r}\)

For circular ring g = \(\frac{G M x}{\left(R^2+x^2\right)^{3 / 2}}\)

⇒ \(\mathrm{v}=-\frac{G M}{\sqrt{R^2+x^2}}\)

For thin circular disc

⇒ \(\mathrm{g}=\frac{2 G M}{R^2}\left(1-\frac{1}{\sqrt{1+\left(\frac{R}{x}\right)^2}}\right)\)

⇒ \(\mathrm{v}=\frac{-2 G M}{R^2}\left(\sqrt{R^2+x^2}-x\right)\)

Uniform thin spherical shell: 

⇒ \(\mathrm{g}_{\text {out }}=\frac{G M}{r^2}\)

⇒ \(\mathrm{g}_{\text {surface }}=\frac{G M}{R^2}\)

g_in = 0

Potential:

⇒ \(\mathrm{v}_{\text {out }}=-\frac{G M}{r}\)

⇒ \(\mathrm{V}_{\text {surtace }}=-\frac{G M}{R}\)

⇒ \(\mathrm{v}_{\mathrm{in}}=-\frac{G M}{R}\)

Uniform solid sphere: (Most Important)

⇒ \(\mathrm{g}_{\text {out }}=\frac{G M}{r^2}\)

⇒ \(\mathrm{g}_{\text {surface }}=\frac{G M}{R^2}\)

⇒ \(\mathrm{g}_{\mathrm{in}}=\frac{G M}{R^3} r\)

⇒ \(\mathrm{g}_{\text {centre }}=0\)

Potential: \(\mathrm{V}_{\text {out }}=-\frac{G M}{r}\)

⇒ \(\mathrm{V}_{\mathrm{in}}=-\frac{G M}{2 R^3}\left(3 R^2-r^2\right)\)

⇒ \(\mathrm{V}_{\text {surtace }}=-\frac{G M}{R}\)

⇒ \(\mathrm{V}_{\text {centre }}=-\frac{3}{2} \frac{G M}{R}\)

Self Energy:

Surface = \(\mathrm{U}_{\text {self }}=-\frac{1}{2} \frac{G M^2}{R}0\)

Gravitational Self energy of a Uniform Sphere = U_self= \(-\frac{3}{5} \frac{G M^2}{R}\)

Escape speed from earth’s surface

⇒ \(\mathrm{V}_{\mathrm{e}}=\sqrt{\frac{2 G M_e}{R}}\)

= 11.2km/sec.

If a satellite is moving around the Earth in a circular orbit, then its orbital speed is

⇒ \(\mathrm{V}_0=\sqrt{\frac{G M_e}{r}}\)

where r is the distance of the satellite from the centre of the earth.

PE . of the satellite = – \(\frac{G M_e m}{r}\)

KE of the satellite = \(\frac{1}{2} m v_0^2=\frac{G M_e m}{2 r}\)

TE of the satellite = \(-\frac{G M_e m}{2 r}\)

Time Period of Geo-stationary satellite = 24 hours

Kapler’s laws: 

1. Law of Orbit: If a planet is revolving around a sun, its path is either elliptical (or circular)
2. Law of Area :

View (1) If a planet is revolving around a sun, the angular momentum of the planet about the sun remains conserved

View (2) The radius vector from the sum to the planet sweeps the area at a constant rate

Areal velocity = \(\frac{d A}{d t}=\frac{L}{2 m}\) = constant

(3) For all the planets of a sun T² ∝ R³

⇒ \(\mathrm{T}^2=\left(\frac{4 \pi^2}{G M_s}\right) \mathrm{R}^3\)

Factors Affecting Acceleration Due to Gravity

1. Effect Of Altitude: \(\mathrm{g}_{\mathrm{n}}=\frac{G M_e}{\left(R_e+h\right)^2}\)

\(\mathrm{g}\left(1+\frac{h}{R_e}\right)^{-2} \simeq \mathrm{g}\left(1-\frac{2 h}{R_e}\right)\) when h << R.

2. Effect Of Depth: \(\mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{d}{R_e}\right)\)

3. Effect Of The Surface Of Earth

The equatorial radius is about 21 km longer than its polar radius.

We know, g = \(\frac{G M_e}{R_e^2}\)

Hence g_pole > g_equator.

4. Effect Of Rotation Of The Earth

Consider a particle of mass m at latitude θ. g′ = g – ω²R_e cos²θ

At pole θ = 90°

⇒ g_pole = g , At equator θ = 0

⇒ g_equator = g – ω²Re.

Hence g_pole > g_equator

NEET Physics Class 11 Chapter 7 Gravitation Multiple Choice Question Ans Answers

Question 1. Weight of an object is :

1. Normal reaction between ground and the object
2. Gravitational force exerted by earth on the object.
3. Dependent on frame of reference.
4. Net force on the object

Answer: 2. Gravitational force exerted by earth on the object

Question 2. The weight of a body at the centre of the earth is –

1. Zero
2. Infinite
3. Same as on the surface of earth
4. None of the above

Answer: 1. Zero

Question 3. If the distance between two masses is doubled, the gravitational attraction between them.

1. Is doubled
2. Becomes four times
3. Is reduced to half
4. Is reduced to a quarter

Answer: 4. Is reduced to a quarter

Question 4. The gravitational force between two stones of mass 1 kg each separated by a distance of 1 metre in vacuum is –

1. Zero
2. 6.675 × 10^-5 Newton
3. 6.675 × 10^-11 Newton
4. 6.675 × 10^-8 Newton

Answer: 3. 6.675 × 10^-11 newton

Question 5. Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is –

1. \(\mathrm{v}=\frac{1}{2 R} \sqrt{\frac{1}{G m}}\)
2. \(\mathrm{v}=\sqrt{\frac{G m}{2 R}}\)
3. \(\mathrm{v}=\frac{1}{2} \sqrt{\frac{G m}{R}}\)
4. \(\mathrm{v}=\sqrt{\frac{4 G m}{R}}\)

Answer: 3. \(\mathrm{v}=\frac{1}{2} \sqrt{\frac{G m}{R}}\)

Question 6. Reason of weightlessness in a satellite is –

1. Zero gravity
2. Centre of mass
3. Zero reaction force by satellite surface
4. None

Answer: 3. Zero reaction force by satellite surface

Question 7. The gravitational force Fgbetween two objects does not depend on –

1. Sum of the masses
2. Product of the masses
3. Gravitational constant
4. Distance between the masses

Answer: 1. Sum of the masses

Question 8. A mass M splits into two parts m and (M – m), which are then separated by a certain distance. What ratio (m/M) maximies the gravitational force between the parts?

1. \(\frac{2}{3}\)
2. \(\frac{3}{4}\)
3. \(\frac{1}{2}\)
4. \(\frac{1}{3}\)

Answer: 3. \(\frac{1}{2}\)

Question 9. On a planet (whose size is the same as that of Earth and mass 4 times of the Earth) the energy needed to lift a 2kg mass vertically upwards through a 2m distance on the planet is (g = 10m/sec² on surface of earth)

1. 16 J
2. 32 J
3. 160 J
4. 320 J

Answer: 3. 160 J

Question 10. The dimensions of universal gravitational constant are :

1. [M^-1L³T^-2]
2. [ML²T^-1]
3. [M^-2L³T^-2]
4. [M^-2L²T^-1]

Answer: 1. [M^-1L³T^-2]

Question 11. If the change in the value of ‘g’ at a height h above the surface of the earth is the same as at a depth x below it, then (both x and h being much smaller than the radius of the earth) –

1. x = h
2. x = 2h
3. x = \(\frac{h}{2}\)
4. x = h²

Answer: 2. x = 2h

Question 12. The moon’s radius is 1/4 that of the Earth and its mass is 1/80 time that of the earth. If g represents the acceleration due to gravity on the surface of the earth, that on the surface of the moon is

1. g/4
2. g/5
3. g/6
4. g/8

Answer: 2. g/5

Question 13. Assuming the earth to be a homogeneous sphere of radius R, its density in terms of G (constant of gravitation) and g (acceleration due to gravity on the surface of the earth) is

1. 3g/(4πRG)
2. 4πg/(3RG)
3. 4πRg/(3G)
4. 4πRG/(3g)

Answer: 1. 3g/(4πRG)

Question 14. An object is placed at a distance of R/2 from the centre of earth. Knowing mass is distributed uniformly, acceleration of that object due to gravity at that point is: (g = acceleration due to gravity on the surface of earth and R is the radius of earth)

1. g
2. 2 g
3. g/2
4. None of these

Answer: 3. g/2

Question 15. Altitude at which acceleration due to gravity decreases by 0.1% approximately : (Radius of earth = 6400 km)

1. 3.2 km
2. 6.4 km
3. 2.4 km
4. 1.6 km

Answer: 1. 3.2 km

Question 16. An iron ball and a wooden ball of the same radius are released from a height ‘h’ in vacuum. The time taken by both of them to reach the ground is –

1. Unequal
2. Exactly equal
3. Roughly equal
4. Zero

Answer: 2. Exactly equal

Question 17. The correct answer to above question is based on –

1. Acceleration due to gravity in vacuum is same irrespective of size and mass of the body
2. Acceleration due to gravity in v
3. acuum depends on the mass of the body
4. There is no acceleration due to gravity in vacuum
5. In vacuum there is resistance offered to the motion of the body and this resistance depends on the mass of the body

Answer: 1. Acceleration due to gravity in vacuum is same irrespective of size and mass of the body

Question 18. When a body is taken from the equator to the poles, its appearent weight –

1. Remains constant
2. Increases
3. Decreases
4. Increases at N-pole and decreases at S-pole

Answer: 2. Increases

Question 19. A body of mass m is taken to the bottom of a deep mine. Then –

1. Its mass increases
2. Its mass decreases
3. Its weight increases
4. Its weight decreases

Answer: 4. Its weight decreases

Question 20. As we go from the equator to the poles, the value of g

1. Remains the same
2. Decreases
3. Increases
4. Decreases upto a latitude of 45º

Answer: 3. Increases

Question 21. Force of gravity is least at

1. The equator
2. The poles
3. A point in between equator and any pole
4. None of these

Answer: 1. The equator

Question 22. Spot the wrong statement :

1. The acceleration due to gravity ‘g’ decreases if –
2. We go down from the surface of the earth towards its centre
3. We go up from the surface of the earth
4. We go from the equator towards the poles on the surface of the earth
5. The rotational velocity of the earth is increased

Answer: 3. We go from the equator towards the poles on the surface of the earth

Question 23. Choose the correct statement from the following : Weightlessness of an astronaut moving in a satellite is a situation of –

1. Zero g
2. No gravity
3. Zero mass
4. Free fall

Answer: 4. Free fall

Question 24. If the earth suddenly shrinks (without changing mass) to half of its present radius, the acceleration due to gravity will be –

1. g/2
2. 4g
3. g/4
4. 2g

Answer: 2. 4g

Question 25. The moon’s radius is 1/4 that of the earth and its mass is 1/80 times that of the earth. If g represents the acceleration due to gravity on the surface of the earth, that on the surface of the moon is –

1. g/4
2. g/5
3. g/6
4. g/8

Answer: 2. g/5

Question 26. The radius of the earth is around 6000 km. The weight of a body at a height of 6000 km from the earth’s surface becomes –

1. Half
2. One-fourth
3. One third
4. No change

Answer: 2. One-fourth

Question 27. At what height from the ground will the value of ‘g’ be the same as that in a 10 km deep mine below the surface of the earth –

1. 20 km
2. 10 km
3. 15 km
4. 5 km

Answer: 4. 5 km

Question 28. At what distance from the centre of the earth, the value of acceleration due to gravity g will be half that on the surface (R = Radius of earth)

1. 2R
2. R
3. 1.414 R
4. 0.414 R

Answer: 3. 1.414 R

Question 29. What will be the acceleration due to gravity at height h if h >> R. Where R is the radius of the earth and g is the acceleration due to gravity on the surface of the earth.

1. \(\frac{g}{\left(1+\frac{h}{R}\right)^2}\)
2. \(g\left(1-\frac{2 h}{R}\right)\)
3. \(\frac{g}{\left(1-\frac{h}{R}\right)^2}\)
4. \(g\left(1-\frac{h}{R}\right)\)

Answer: 1. \(\frac{g}{\left(1+\frac{h}{R}\right)^2}\)

Question 30. If the density of the earth is doubled keeping its radius constant then acceleration due to gravity will be (g = 9.8 m/s²)

1. 19.6 m/s²
2. 9.8 m/s²
3. 4.9 m/s²
4. 2.45 m/s²

Answer: 1. 19.6 m/s²

Question 31. The acceleration due to gravity at the pole and equator can be related as –

1. gp< ge
2. gp= ge= g
3. gp= ge< g
4. gp> ge

Answer: 4. gp> ge

Question 32. The depth at which the effective value of acceleration due to gravity is \(\frac{g}{4}\) is

1. R
2. \(\frac{3 R}{4}\)
3. \(\frac{R}{2}\)
4. \(\frac{R}{4}\)

Answer: 2. \(\frac{R}{2}\)

Question 33. Two bodies of mass 100 kg and 104 kg are lying one meter apart. At what distance from a 100 kg body will the intensity of the gravitational field be zero

1. \(\frac{1}{9} m\)
2. \(\frac{1}{10} m\)
3. \(\frac{1}{11} m\)
4. \(\frac{10}{11} \mathrm{~m}\)

Answer: 3. \(\frac{1}{11} m\)

Question 34. Figure shows a hemispherical shell having uniform mass density. The direction of gravitational field intensity at point P will be along:

1. a
2. b
3. c
4. d

Answer: 3. c

Question 35. Two bodies of mass 10² kg and 10³ kg are lying 1m apart. The gravitational potential at the mid-point of the line joining them is

1. 0
2. –1.47 Joule/kg
3. 1.47 Joule/kg
4. –1.47 × 10^-9 joule/kg

Answer: 4. –1.47 × 10^-9 joule/kg

Question 36. A simple pendulum has a period T₁ when on the earth’s surface, and T₂ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of T₂/T₁ is:

1. 1
2. \(\sqrt{2}\)
3. 4
4. 2

Answer: 4. 2

Question 37. Near earth time period of a satellite is 4 h. Find its time period at a distance 4R from the centre of earth:

1. 32 h
2. \(\left(\frac{1}{8^3 \sqrt{2}}\right) h\)
3. \(8^3 \sqrt{2} h\)
4. 16 h

Answer: 1. 32 h

Question 38. The radius of the orbit of a planet is two times that of the Earth. The time period of a planet is:

1. 4.2 T
2. 2.8 T
3. 5.6 T
4. 8.4 T

Answer: 2. 2.8 T

Question 39. In the case of earth:

1. The field is zero, both at the centre and infinity
2. The potential is zero, both at the centre and infinity
3. The potential is the same, both at centre and infinity but not zero
4. The potential is maximum at the centre

Answer: 1. Field is zero, both at the centre and infinity

Question 40. What would be the angular speed of the earth, so that bodies lying on the equator may appear weightless? (g = 10m/s² and radius of earth = 6400 km)

1. 1.25 × 10^-3 rad/sec
2. 1.25 × 10^-2 rad/sec
3. 1.25 × 10^-4 rad/sec
4. 1.25 × 10^-1 rad/sec

Answer: 1. 1.25 × 10^-3 rad/sec

Question 41. The speed with which the earth has to rotate on its axis so that a person on the equator would weigh (3/5)th as much as present will be (Take the equatorial radius as 6400 km.)

1. 3.28 × 10^-4 rad/sec
2. 7.826 × 10^-4 rad/sec
3. 3.28 × 10^-3 rad/sec
4. 7.28 × 10^-3 rad/sec

Answer: 2. 7.826 × 10^-4 rad/sec

Question 42. A body of mass m is lifted up from the surface of the earth to a height three times the radius of the earth. The change in potential energy of the body is (g = gravity field at the surface of the earth)

1. mgR
2. \(\frac{3}{4} \mathrm{mgR}\)
3. \(\frac{1}{3} \mathrm{mgR}\)
4. \(\frac{2}{3} \mathrm{mgR}\)

Answer: 2. \(\frac{3}{4} \mathrm{mgR}\)

Question 43. The change in potential energy when a body of mass m is raised to a height n R from the earth’s surface is (R = Radius of earth)

1. mgR
2. nmgR
3. \(\mathrm{mgR} \frac{n^2}{n^2+1}\)
4. \(\mathrm{mgR} \frac{n}{n+1}\)

Answer: 4. \(\mathrm{mgR} \frac{n}{n+1}\)

Question 44. If the mass of the earth is M, the radius is R and the gravitational constant is G, then work done to take 1 kg mass from the earth’s surface to infinity will be –

1. \(\sqrt{\frac{G M}{2 R}}\)
2. \(\frac{G M}{R}\)
3. \(\sqrt{\frac{2 G M}{R}}\)
4. \(\frac{G M}{2 R}\)

Answer: 2. \(\frac{G M}{R}\)

Question 45. A rocket is launched with a velocity of 10 km/s. If the radius of the earth is R, then the maximum height attained by it will be

1. 2R
2. 3R
3. 4R
4. 5R

Answer: 3. 3R

Question 46. What is the intensity of the gravitational field at the centre of a spherical shell –

1. Gm/r²
2. g
3. Zero
4. None of these

Answer: 3. Zero

Question 47. The escape velocity of a body of 1 kg mass on a planet is 100 m/sec. The gravitational Potential energy of the body on the Planet is –

1. – 5000 J
2. – 1000 J
3. – 2400 J
4. 5000 J

Answer: 1. – 5000 J

Question 48. The kinetic energy needed to project a body of mass m from the earth’s surface (radius R) to infinity is –

1. mgR/2
2. 2 mgR
3. mgR
4. mgR/4

Answer: 3. mgR

Question 49. The escape velocity of a sphere of mass m from Earth having mass M and radius R is given by –

1. \(\sqrt{\frac{2 G M}{R}}\)
2. \(2 \sqrt{\frac{G M}{R}}\)
3. \(\sqrt{\frac{2 G M m}{R}}\)
4. \(\sqrt{\frac{G M}{R}}\)

Answer: 1. \(\sqrt{\frac{2 G M}{R}}\)

Question 50. If g is the acceleration due to gravity at the earth’s surface and r is the radius of the earth, the escape velocity for the body to escape out of the earth’s gravitational field is –

1. gr
2. \(\sqrt{2 g r}\)
3. g/r
4. r/g

Answer: 2. \(\sqrt{2 g r}\)

Question 51. For the moon to cease to remain the earth’s satellite, its orbital velocity has to increase by a factor of –

1. 2
2. \(\sqrt{2}\)
3. \(1 / \sqrt{2}\)
4. \(\sqrt{3}\)

Answer: 2. \(\sqrt{2}\)

Question 52. Escape velocity on a planet is ve. If the radius of the planet remains the same and the mass becomes 4 times, the escape velocity becomes –

1. 4v_e
2. 2v_e
3. v_e
4. v_e

Answer: 2. 2v_e

Question 53. How many times is the escape velocity (Ve), of orbital velocity (V0) for a satellite revolving near Earth –

1. \(\sqrt{2}\) times
2. 2 times
3. 3 times
4. 4 times

Answer: 1. \(\sqrt{2}\) times

Question 54. If the radius of a planet is R and its density is, the escape velocity from its surface will be –

1. \(v_e \propto R\)
2. \(\mathrm{v}_{\mathrm{e}} \propto \mathrm{R} \sqrt{p}\)
3. \(\mathrm{v}_{\mathrm{e}} \propto \frac{\sqrt{\rho}}{R}\)
4. \(\mathrm{v}_{\mathrm{e}} \propto \frac{1}{\sqrt{\rho R}}\)

Answer: 2. \(\mathrm{v}_{\mathrm{e}} \propto \mathrm{R} \sqrt{p}\)

Question 55. If V, R and g denote respectively the escape velocity from the surface of the earth radius of the earth, and acceleration due to gravity, then the correct equation is –

1. \(v=\sqrt{g R}\)
2. \(V=\sqrt{\frac{4}{3} g R^3}\)
3. \(\mathrm{V}=\mathrm{R} \sqrt{g}\)
4. \(V=\sqrt{2 g R}\)

Answer: 4. \(V=\sqrt{2 g R}\)

Question 56. If the radius of a planet is four times that of Earth and the value of g is the same for both, the escape velocity on the planet will be –

1. 11.2 km/s
2. 5.6 km/s
3. 22.4 km/s
4. None

Answer: 3. 22.4 km/s

Question 57. If the radius and acceleration due to gravity both are doubled, the escape velocity of the earth will become.

1. 11.2 km/s
2. 22.4 km/s
3. 5.6 km/s
4. 44.8 km/s

Answer: 2. 22.4 km/s

Question 58. If g is the acceleration due to gravity on the earth’s surface, the gain in P.E. of an object of mass m raised from the surface of the earth to a height of the radius R of the earth is

1. mgR
2. 2mgR
3. 12mgR
4. 14mgR

Answer: 3. 12mgR

Question 59. A missile is launched with a velocity less than the escape velocity. The sum of kinetic energy and potential energy will be

1. Positive
2. Negative
3. Negative or positive, uncertain
4. Zero

Answer: 2. Negative

Question 60. If v_e is escape velocity and v₀ is the orbital velocity of a satellite for orbit close to the earth’s surface, then these are related by :

1. \(\mathrm{v}_0=\sqrt{2} v_e\)
2. \(v_0=v_e\)
3. \(v_e=\sqrt{2 v_0}\)
4. \(v_e=\sqrt{2} v_0\)

Answer: 4. \(v_e=\sqrt{2} v_0\)

Question 61. An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy E0. Its potential energy is :

1. − Eº
2. 1.5 Eº
3. 2 Eº
4. Eº

Answer: 3. 2 Eº

Question 62. The mass and radius of the earth and moon are M₁, R₁ and M₂, R₂ respectively. Their centres are d distance apart. With what velocity should a particle of mass m be projected from the midpoint of its centres so that it may escape out to infinity?

1. \(\sqrt{\frac{G\left(M_1+M_2\right)}{d}}\)
2. \(\sqrt{\frac{2 G\left(M_1+M_2\right)}{d}}\)
3. \(\sqrt{\frac{4 G\left(M_1+M_2\right)}{d}}\)
4. \(\sqrt{\frac{G M_1 M_2}{d}}\)

Answer: 3. \(\sqrt{\frac{4 G\left(M_1+M_2\right)}{d}}\)

Question 63. A satellite has to revolve around the earth in a circular orbit of radius 8 × 10³ km. The velocity of projection of the satellite in this orbit will be

1. 16 km/sec
2. 8 km/sec
3. 3 km/sec
4. 7.08 km/sec

Answer: 4. 7.08 km/sec

Question 64. The ratio of the radius of the earth to that of the moon is 10. The ratio of g an earth to the moon is 6. The ratio of the escape velocity from the Earth’s surface to that from the moon is approximately

1. 10
2. 8
3. 4
4. 2

Answer: 2. 8

Question 65. Acceleration due to gravity on a planet is 10 times the value on the Earth. Escape velocity for the planet and the earth are V_p and V_e respectively Assuming that the radii of the planet and the earth are the same, then

1. \(V_{\mathrm{P}}=10 \mathrm{~V}_{\mathrm{e}}\)
2. \(V_P=\sqrt{10} V_e\)
3. \(V_P=\frac{V_e}{\sqrt{10}}\)
4. \(V_P=\frac{V_e}{10}\)

Answer: 3. \(V_P=\frac{V_e}{\sqrt{10}}\)

Question 66. A space shuttle is launched in a circular orbit near the Earth’s surface. The additional velocity given to the space shuttle to get free from the influence of gravitational force will be

1. 1.52 km/s
2. 2.75 km/s
3. 3.28 km/s
4. 5.18 km/s

Answer: 3. 3.28 km/s

Question 67. A body of mass m is situated at a distance 4R_e above the earth’s surface, where R_e is the radius of the earth. How much minimum energy be given to the body so that it may escape

1. mgR_e
2. 2mgR_e
3. \(\frac{m g R_e}{5}\)
4. \(\frac{m g R_e}{16}\)

Answer: 3. \(\frac{m g R_e}{5}\)

Question 68. The potential energy of a body of mass 3kg on the surface of a planet is 54 joule. The escape velocity will be

1. 18m/s
2. 162 m/s
3. 36 m/s
4. 6 m/s

Answer: 4. 6 m/s

Question 69. The escape velocity from a planet is v₀. The escape velocity from a planet having twice the radius but the same density will be

1. 0.5 v₀
2. v₀
3. 2v₀
4. 4v₀

Answer: 3. 2v₀

Question 70. If the kinetic energy of a satellite orbiting around the earth is doubled then

1. The satellite will escape into space.
2. The satellite will fall down on the earth
3. The radius of its orbit will be doubled
4. The radius of its orbit will become half

Answer: 1. The satellite will escape into the space.

Question 71. The escape velocity from the earth does not depend upon

1. Mass of earth
2. Mass of the body
3. Radius of earth
4. Acceleration due to gravity

Answer: 2. Mass of the body

Question 72. There is no atmosphere on the moon because

1. It is near the earth
2. It is orbiting around the earth
3. There was no gas at all
4. The escape velocity of gas molecules is less than their root-mean-square velocity

Answer: 4. The escape velocity of gas molecules is less than their root-mean-square velocity

Question 73. The escape velocity is

1. 2gR
2. gR
3. \(\sqrt{g R}\)
4. \(\sqrt{2g R}\)

Answer: 4. \(\sqrt{2g R}\)

Question 74. A particle of mass m is taken through the gravitational field produced by a source S, from A to B, along the three paths as shown in the figure. If the work done along the paths 1, 2 and 3 is W₁, W₂ and W₃ respectively, then

1. W₁= W₂= W₃
2. W₂> W₃= W₂
3. W₃= W₂> W₁
4. W₁> W₂> W₃

Answer: 1. W₁= W₂= W₃

Question 75. The escape velocity of a particle of mass m varies as :

1. m²
2. m
3. m⁰
4. m^-1

Answer: 3. m⁰

Question 76. Acceleration due to gravity at the earth’s surface is g ms^-2. Find the effective value of gravity at a height of 32 km from sea level : (R_e= 6400 km) (R_e= 6400 km)

1. 0.5 g ms^-2
2. 0.99 g ms^-2
3. 1.01 g ms^-2
4. 0.90 g ms^-2

Answer: 2. 0.99 g ms^-2

Question 77. The radius of orbit of the satellite of earth is R. Its kinetic energy is proportional to :

1. \(\frac{1}{R}\)
2. \(\frac{1}{\sqrt{R}}\)
3. R
4. \(\frac{1}{R^{3 / 2}}\)

Answer: 1. \(\frac{1}{R}\)

Question 78. A cosmonaut is orbiting earth in a spacecraft at an altitude h = 630 km with a speed of 8 km/s. If the radius of the earth is 6400 km, the acceleration of the cosmonaut is

1. 9.10 m/s²
2. 9.80 m/s²
3. 10.0 m/s²
4. 9.88 m/s²

Answer: 1. 9.10 m/s²

Question 79. A very very large number of particles of the same mass m are kept at horizontal distances of 1m, 2m, 4m, 8m and so on from (0,0) point. The total gravitational potential at this point is (addition of G.P. of infinite terms = \(\frac{a}{1-r}\) where a = first term, r = common ratio) :

1. – 8G m
2. – 3G m
3. – 4G m
4. – 2G m

Answer: 4. – 2G m

Question 80. A body starts from rest at a point, distance R₀ from the centre of the earth of mass M, radius R. The velocity acquired by the body when it reaches the surface of the earth will be

1. \(\mathrm{GM}\left(\frac{1}{R}-\frac{1}{R_0}\right)\)
2. \(2 \mathrm{GM}\left(\frac{1}{R}-\frac{1}{R_0}\right)\)
3. \(\sqrt{2 G M\left(\frac{1}{R}-\frac{1}{R_0}\right)}\)
4. \(2 \mathrm{GM} \sqrt{\left(\frac{1}{R}-\frac{1}{R_0}\right)}\)

Answer: 3. \(\sqrt{2 G M\left(\frac{1}{R}-\frac{1}{R_0}\right)}\)

Question 81. The relation between the escape velocity from the earth and the velocity of a satellite orbiting near the earth’s surface is

1. v_e = 3v
2. v_e= v
3. v_e= 2v
4. v_e= v/2

Answer: 2. v_e= v

Question 82. A body attains a height equal to the radius of the earth. The velocity of the body with which it was projected is :

1. \(\sqrt{\frac{G M}{R}}\)
2. \(\sqrt{\frac{2 G M}{R}}\)
3. \(\sqrt{\frac{5}{4} \frac{G M}{R}}\)
4. \(\sqrt{\frac{3 G M}{R}}\)

Answer: 1. \(\sqrt{\frac{G M}{R}}\)

Question 83. A satellite of mass m is circulating around the earth with constant angular velocity. If the radius of the orbit is R₀ and the mass of the earth is M, the angular momentum about the centre of the earth is

1. \(M \sqrt{G M R_0}\)
2. \(M \sqrt{G m R_0}\)
3. \(M \sqrt{\frac{G M}{R_0}}\)
4. \(M \sqrt{\frac{G M}{R_0}}\)

Answer: 1. \(M \sqrt{G M R_0}\)

Question 84. Which of the following quantities is conserved for a satellite revolving around the earth in a particular orbit?

1. Angular velocity
2. Force
3. Angular momentum
4. Velocity

Answer: 3. Angular momentum

Question 85. The distance of Neptune and Saturn from sun are nearly 10¹³ and 10¹² meters respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio –

1. \(\sqrt{10}\)
2. 100
3. \(10 \sqrt{10}\)
4. \(1 / \sqrt{10}\)

Answer: 3. \(10 \sqrt{10}\)

Question 86. The period of a satellite in a circular orbit of radius R is T, and the period of another satellite in a circular orbit of radius 4R is –

1. 4T
2. T/4
3. 8T
4. T/8

Answer: 3. 8T

Question 87. If a body describes a circular motion under an inverse square field, the time taken to complete one revolution T is related to the radius of the circular orbit as –

1. T ∝ r
2. T ∝ r²
3. T² ∝ r³
4. T ∝ r⁴

Answer: 3. T² ∝ r³

Question 88. The escape velocity of a sphere of mass m from Earth having mass M and radius R is given by –

1. \(\sqrt{\frac{2 G M}{R}}\)
2. \(2 \sqrt{\frac{G M}{R}}\)
3. \(\sqrt{\frac{2 G M m}{R}}\)
4. \(\sqrt{\frac{G M}{R}}\)

Answer: 1. \(\sqrt{\frac{2 G M}{R}}\)

Question 89. The escape velocity from the earth is about 11 km/second. The escape velocity from a planet having twice the radius and the same mean density as the Earth is –

1. 22km/sec
2. 11 km/sec
3. 5.5 km/sec
4. 15.5 km/sec

Answer: 1. 22km/sec

Question 90. A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of the earth in the new orbit is 2 times that of the earlier orbit. The time period in the second orbit is –

1. 4.8 hours
2. \(48 \sqrt{2}\) hours
3. 24 hrs
4. Infinite

Answer: 2. \(48 \sqrt{2}\) hours

Question 91. Two satellites A and B go around planet P in circular orbits having radial 4R and R respectively. If the speed of the satellite A is 3V, the speed of the satellite B will be

1. 12 V
2. 6 V
3. 4/3 V
4. 3/2 V

Answer: 2. 6 V

Question 92. The escape velocity for a rocket from earth is 11.2 km/sec. Its value on a planet where the acceleration due to gravity is double that on the earth and the diameter of the planet is twice that of the earth will be in km/sec-

1. 11.2
2. 5.6
3. 22.4
4. 53.6

Answer: 3. 22.4

Question 93. A satellite revolves around the earth in an elliptical orbit. Its speed

1. Is the same at all points in the orbit
2. Is greatest when it is closest to the earth
3. Is greatest when it is farthest from the earth
4. Goes on increasing or decreasing continuously depending upon the mass of the satellite

Answer: 2. Is greatest when it is closest to the earth

Question 94. Time period of revolution of a satellite around a planet of radius R is T. The Period of revolution around another planet, whose radius is 3R but has the same density is –

1. \(\frac{T}{3 \sqrt{3}}\)
2. 3T
3. 9T
4. \(3 \sqrt{3} T\)

Answer: 1. \(\frac{T}{3 \sqrt{3}}\)

Question 95. If Veand V₀ represent the escape velocity and orbital velocity of a satellite corresponding to a circular orbit of radius R, then –

1. \(V_e=V_0\)
2. \(\sqrt{2} V_0=V_e\)
3. \(V_e=\frac{1}{\sqrt{2}} V_o\)
4. None of these

Answer: 2. \(\sqrt{2} V_0=V_e\)

Question 96. A spherical planet far out in space has a mass of M₀ and a diameter D₀. A particle will experience acceleration due to gravity which is equal to

1. GM₀/D_0²
2. 2mGM₀/D_0²
3. 4GM₀/D_0²
4. GmM₀/D_0²

Answer: 3.4GM₀/D_0²

Question 97. A satellite can be in a geostationary orbit around a planet at a distance r from the centre of the planet. If the angular velocity of the planet about its axis doubles, a satellite can now be in a geostationary orbit around the planet if its distance from the centre of the planet is

1. \(\frac{r}{2}\)
2. \(\frac{r}{2 \sqrt{2}}\)
3. \(\frac{r}{(4)^{1 / 3}}\)
4. \(\frac{r}{(2)^{1 / 3}}\)

Answer: 3. \(\frac{r}{(4)^{1 / 3}}\)

Question 98. Consider a satellite going around the earth in an orbit. Which of the following statements is wrong –

1. It is a freely falling body
2. It suffers no acceleration
3. It is moving at a constant speed
4. Its angular momentum remains constant

Answer: 2. It suffers no acceleration

Question 99. The period of a satellite in a circular orbit around a planet is independent of –

1. The mass of the planet
2. The radius of the planet
3. The mass of the satellite
4. All the three parameters (1), (2) and (3)

Answer: 3. The mass of the satellite

Question 100. A small satellite is revolving near the earth’s surface. Its orbital velocity will be nearly.

1. 8 km/sec
2. 11.2 km/sec
3. 4 km/sec
4. 6 km/sec

Answer: 1. 8 km/sec

Question 101. A satellite revolves around the earth in an elliptical orbit. Its speed.

1. Is the same at all points in the orbit
2. Is greatest when it is closest to the earth
3. Is greatest when it is farthest from the earth
4. Goes on increasing or decreasing continuously depending upon the mass of the satellite

Answer: 2. Is greatest when it is closest to the earth

Question 102. If the height of a satellite from the earth is negligible in comparison to the radius of the earth R, the orbital velocity of the satellite is –

1. gR
2. gR/2
3. \(\sqrt{g / R}\)
4. \(\sqrt{g R}\)

Answer: 4. \(\sqrt{g R}\)

Question 103. Orbital velocity of an artificial satellite does not depend upon –

1. Mass of the earth
2. Mass of the satellite
3. Radius of the earth
4. Acceleration due to gravity

Answer: 2. Mass of the satellite

Question 104. The time period of a geostationary satellite is –

1. 24 hours
2. 12 hours
3. 365 days
4. One month

Answer: 1. 24 hours

Question 105. Which one of the following statements regarding artificial satellites of the earth is incorrect –

1. The orbital velocity depends on the mass of the satellite
2. A minimum velocity of 8 km/sec is required by a satellite to orbit quite close to the earth
3. The period of revolution is large if the radius of its orbit is large
4. The height of a geostationary satellite is about 36000 km from earth

Answer: 1. The orbital velocity depends on the mass of the satellite

Question 106. Two identical satellites are at R and 7R away from the earth’s surface, the wrong statement is (R = Radius of the earth)

1. The ratio of total energy will be 4
2. The ratio of kinetic energies will be 4
3. The ratio of potential energies will be 4
4. The ratio of total energy will be 4 but the ratio of potential and kinetic energies will be 2

Answer: 4. Ratio of total energy will be 4 but the ratio of potential and kinetic energies will be 2

Question 107. For a satellite escape velocity is 11 km/s. If the satellite is launched at an angle of 60º with the vertical, then the escape velocity will be –

1. 11 km/s
2. \(11 \sqrt{3} \mathrm{~km} / \mathrm{s}\)
3. \(\frac{11}{\sqrt{3}} \mathrm{~km} / \mathrm{s}\)
4. 33 km/s

Answer: 1. 11 km/s

Question 108. The distance of a geo-stationary satellite from the centre of the earth (Radius R = 6400 km) is nearest to –

1. 5 R
2. 7 R
3. 10 R
4. 18 R

Answer: 2. 7 R

Question 109. Periodic time of a satellite revolving above Earth’s surface at a height equal to R, radius of Earth, is [g is acceleration due to gravity at Earth’s surface]

1. \(2 \pi \sqrt{\frac{2 R}{g}}\)
2. \(4 \sqrt{2} \pi \sqrt{\frac{R}{g}}\)
3. \(2 \pi \sqrt{\frac{R}{g}}\)
4. \(8 \pi \sqrt{\frac{R}{g}}\)

Answer: 2. \(4 \sqrt{2} \pi \sqrt{\frac{R}{g}}\)

Question 110. Given the radius of Earth ‘R’ and length of a day ‘T’ the height of a geostationary satellite is [G-Gravitational constant. M-Mass of Earth]

1. +R
2. – R
3. – R
4. None

Answer: 3. – R

Question 111. The distance of a geostationary satellite from the surface of the earth radius (Re= 6400 km) in terms of Re is –

1. 13.76 R_e
2. 10.76 R_e
3. 5.56 R_e
4. 2.56 R_e

Answer: 3. 5.56 R_e

Question 112. The orbital velocity of a planet revolving close to the earth’s surface is –

1. \(\sqrt{2 g R}\)
2. \(\sqrt{g R}\)
3. \(\sqrt{\frac{2 g}{R}}\)
4. \(\sqrt{\frac{g}{R}}\)

Answer: 2. \(\sqrt{g R}\)

Question 113. A satellite moves around the earth in a circular orbit of radius r with speed v. If the mass of the satellite is M, its total energy is –

1. \(-\frac{1}{2} \mathrm{Mv}^2\)
2. \(\frac{1}{2} M v^2\)
3. \(\frac{3}{2} M v^2\)
4. Mv²

Answer: 1. \(-\frac{1}{2} \mathrm{Mv}^2\)

Question 114. If a satellite is shifted towards the earth. The time period of the satellite will be –

1. Increase
2. Decrease
3. Unchanged
4. Nothing can be said

Answer: 2. Decrease

Question 115. Two satellites A and B go around a planet in circular orbits having radii 4R and R, respectively. If the speed of satellite A is 3v, then the speed of satellite B is –

1. \(\frac{3 v}{2}\)
2. \(\frac{4 v}{2}\)
3. 6v
4. 12v

Answer: 3. 6v

Question 116. If gravity field due to a point mass follows \(\mathrm{g} \propto \frac{1}{r^3}\) instead of \(\frac{1}{r^2}\), then the relation between time period of a satellite near earth’s surface and radius of its orbit r will be –

1. T² ∝ r³
2. T ∝ r²
3. T² ∝ r
4. T ∝ r

Answer: 2. T ∝ r²

Question 117. A satellite appears to be at rest when seen from the equator. Its height from the earth’s surface is nearly

1. 35800km
2. 358000 km
3. 6400km
4. Such a satellite cannot exist

Answer: 1. 35800km

Question 118. A body is dropped by a satellite in its geostationary orbit

1. It will burn on entering the atmosphere
2. It will remain in the same place with respect to the earth
3. It will reach the earth in 24 hours
4. It will perform uncertain motion

Answer: 2. It will remain in the same place with respect to the earth

Question 119. A satellite of Earth can move only in those orbits whose plane coincides with

1. The plane of a great circle of the earth
2. The plane passing through the poles of the earth
3. The plane of a circle at any latitude on earth
4. None of these

Answer: 1. The plane of a great circle of the earth

Question 120. A satellite launching station should be

1. Near the equatorial region
2. Near the polar region
3. On the polar axis
4. All locations are equally good

Answer: 1. Near the equatorial region

Question 121. The minimum number of satellites needed to be placed at the surface of the earth for worldwide communication between any two locations is –

1. 6
2. 4
3. 3
4. 5

Answer: 3. 3

Question 122. A geostationary satellite orbits around the earth in a circular orbit with a radius of 36000 km. Then, the time period of a spy satellite orbiting a few hundred kilometres above the earth’s surface (R_Earth = 6400 km) will approximately be :

1. 1/2 hr
2. 1 hr
3. 2 hr
4. 4 hr

Answer: 3. 2 hr

Question 123. A satellite is moving with a constant speed ‘V’ in a circular orbit about the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is

1. \(\frac{1}{2} m V^2\)
2. mV²
3. \(\frac{3}{2} m V^2\)
4. 2mV²

Answer: 2. mV²

Question 124. Two satellites of earth, S₁ and S₂ are moving in the same orbit. The mass of S₁ is four times the mass of S₂. Which one of the following statements is true:

1. The time period of S₁ is four times that of S₂
2. The potential energies of the earth and satellite in the two cases are equal
3. S₁ and S₂ are moving at the same speed
4. The kinetic energies of the two satellites are equal

Answer: 3. S₁ and S₂ are moving at the same speed

Question 125. The orbital speed of a satellite revolving near the earth is :

1. \(\sqrt{2 g R}\)
2. \(\sqrt{g R}\)
3. \(\sqrt{g / R}\)
4. \(\sqrt{2 g / R}\)

Answer: 2. \(\sqrt{g R}\)

Question 126. If the radius of the earth is decreased by 1% and mass remains constant, then the acceleration due to gravity

1. Decrease by 2%
2. Decrease by 1%
3. Increase by 1%
4. Increase by 2%

Answer: 4. Increase by 2%

Question 127. The escape velocity for a rocket is 11.2 km/s. If it is taken to a planet where the radius and acceleration due to gravity are double that of earth, then the escape velocity will be :

1. 5.6 m/s
2. 11.2 m/s
3. 22.4 km/s
4. 44.2 m/s

Answer: 3. 22.4 km/s

Question 128. Suppose the radius of the moon’s orbit around the earth is doubled. Its period around the earth will become:

1. 1/2 times
2. \(\sqrt{2}\) times
3. 22/3 times
4. 23/2 times

Answer: 4. 23/2 times

Question 129. In the case of an orbiting satellite if the radius of the orbit is decreased :

1. Its Kinetic Energy decreases
2. Its Potential Energy increase
3. Its Mechanical Energy decreases
4. Its speed decreases

Answer: 3. Its Mechanical Energy decreases

Question 130. A satellite of the earth is revolving in a circular orbit with a uniform speed v. If the gravitational force suddenly disappears, the satellite will

1. Continue to move with velocity v along the original orbit
2. Move with a velocity v, tangentially to the original orbit
3. Fall down with increasing velocity
4. Ultimately come to rest somewhere in the original orbit

Answer: 2. Move with a velocity v, tangentially to the original orbit

Question 131. The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period becomes

1. 10 hour
2. 80 hour
3. 40 hour
4. 20 hour

Answer: 3. 40 hour

Question 132. The escape velocity for a body projected vertically upwards from the surface of the earth is 11 km/s. If the body is projected at an angle of 45º with the vertical, the escape velocity will be :

1. \(11 \sqrt{2} \mathrm{~km} / \mathrm{s}\)
2. 22 km/s
3. 11 km/s
4. \(11 / \sqrt{2} \mathrm{~m} / \mathrm{s}\) m/s

Answer: 3. 11 km/s

Question 133. A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is :

1. gx
2. \(\frac{g R}{R-x}\)
3. \(\frac{g R^2}{R+x}\)
4. \(\left(\frac{g R^2}{R+x}\right)^{1 / 2}\)

Answer: 4. \(\frac{g R}{R-x}\)

Question 134. The time period of an earth satellite in a circular orbit is independent of :

1. The mass of the satellite
2. The radius of its orbit
3. Both the mass and radius of the orbit
4. Neither the mass of the satellite nor the radius of its orbit

Answer: 1. The mass of the satellite

Question 135. If g is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth, is :

1. 2mgR
2. \(\frac{1}{2} m g R\)
3. \(\frac{1}{4} m g R\)
4. mgR

Answer: 2. \(\frac{1}{2} m g R\)

Question 136. The change in the value of ‘g’ at a height ‘h’ above the surface of the earth is the same as at a depth ‘d’ below the surface of the earth. When both ‘d’ and ‘h’ are much smaller than the radius of the earth, then, which one of the following is correct?

1. \(\mathrm{d}=\frac{h}{2}\)
2. \(\mathrm{d}=\frac{3 h}{2}\)
3. d = 2h
4. d = h

Answer: 3. d = 2h

Question 137. A particle of mass 10 kg is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between them, to take the particle far away from the sphere (you may take G = 6.67 × 10^-11 Nm²/kg²);

1. 13.34 × 10^-10 J
2. 3.33 × 10^-10 J
3. 6.67 × 10^-9 J
4. 6.67 × 10^-7 J

Answer: 4. 6.67 × 10^-7 J

Question 138. If g_E and g_m are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan’s oil drop experiment could be performed on the two surfaces, one will find the ratio

1. 1
2. 0
3. g_E/g_M
4. g_M/g_E

Answer: 1. 1

Question 139. A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is 11 km s^-1, the escape velocity from the surface of the planet would be

1. 11 km s^-1
2. 110 km s^-1
3. 0.11 km s^-1
4. 1.1 km s^-1

Answer: 2. 110 km s^-1

Question 140. The distance of neptune and saturn from the sun is nearly 10¹³ and 10¹² meters respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio –

1. \(\sqrt{10}\)
2. 100
3. 10
4. 1/10

Answer: 3. 10

Question 141. The period of a satellite in a circular orbit of eradius R is T, and the period of another satellite in a circular orbit of radius 4R is.

1. 4T
2. T/4
3. 8T
4. T/8

Answer: 3. 8T

Question 142. Two planets move around the sun. The periodic times and the mean radii of the orbits are T₁, T₂ and r₁, r₂ respectively. The ratio T₁/ T₂ is equal to –

1. (r₁/ r₂)^1/2
2. r₁/ r₂
3. (r₁/ r₂)²
4. (r₁/ r₂)^3/2

Answer: 4. (r₁/ r₂)^3/2

Question 143. The rotation period of an earth satellite close to the surface of the earth is 83 minutes. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be –

1. 83 minute
2. 83 × \(\sqrt{8}\) minutes
3. 664 minutes
4. 249 minutes

Answer: 3. 664 minutes

Question 144. A satellite of mass m is circulating around the earth with constant angular velocity. If the radius of the orbit is R₀ and the mass of the earth is M, the angular momentum about the centre of the earth is –

1. \(\mathrm{m} \sqrt{G M R_0}\)
2. \(\mathrm{M} \sqrt{G M R_0}\)
3. \(\mathrm{m} \sqrt{\frac{G M}{R_0}}\)
4. \(\mathrm{M} \sqrt{\frac{G M}{R_0}}\)

Answer: 1. \(\mathrm{m} \sqrt{G M R_0}\)

Question 145. A planet revolves around the sun whose mean distance is 1.588 times the mean distance between the earth and the sun. The revolution time of the planet will be –

1. 1.25 years
2. 1.59 years
3. 0.89 years
4. 2 years

Answer: 4. 2 years

Question 146. If the mass of a satellite is doubled and the time period remains constant the ratio of orbit in the two cases will be –

1. 1: 2
2. 1: 1
3. 1 : 3
4. None of these

Answer: 2. 1:1

Question 147. The earth revolves around the sun in one year. If the distance between them becomes double, the new period of revolution will be –

1. 1/2 years
2. \(2 \sqrt{2}\) years
3. 4 years
4. 8 years

Answer: 2. \(2 \sqrt{2}\) years

Question 148. A body revolved around the sun 27 times faster than the earth what is the ratio of their radii

1. 1/3
2. 1/9
3. 1/27
4. 1/4

Answer: 2. 1/9

Question 149. The orbital angular momentum of a satellite revolving at a distance r from the centre is L. If the distance is increased to 16r, then the new angular momentum will be –

1. 16 L
2. 64 L
3. \(\frac{L}{4}\)
4. 4 L

Answer: 4. 4 L

Question 150. The ratio of the distance of two planets from the sun is 1.38. The ratio of their period of revolution around the sun is –

1. 1.38
2. 1.38^3/2
3. 1.38^1/2
4. 1.38³

Answer: 2. 1.38^3/2

Question 151. Kepler’s second law (law of areas) is nothing but a statement of –

1. Work energy theorem
2. Conservation of linear momentum
3. Conservation of angular momentum
4. Conservation of energy

Answer: 3. Conservation of angular momentum

Question 152. In an elliptical orbit under gravitational force, in general.

1. Tangential velocity is constant
2. Angular velocity is constant
3. Radial velocity is constant
4. Areal velocity is constant

Answer: 4. Areal velocity is constant

Question 153. What does not change in the field of central force?

1. Potential energy
2. Kinetic energy
3. Linear momentum
4. Angular momentum

Answer: 4. Angular momentum

Question 154. A planet is moving in an elliptic orbit. If T, V, E, L stand respectively for their kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the centre of force, which of the following statements is correct

1. T is conserved
2. V is always positive
3. E is always negative
4. L is conserved but the direction of the vector changes continuously

Answer: 3. E is always negative

Question 155. Three identical stars of mass M are located at the vertices of an equilateral triangle with side L. The speed at which they will move if they all revolve under the influence of one another’s gravitational force in a circular orbit circumscribing the triangle while still preserving the equilateral triangle :

1. \(\sqrt{\frac{2 G M}{L}}\)
2. \(\sqrt{\frac{G M}{L}}\)
3. \(2 \sqrt{\frac{G M}{L}}\)
4. Not possible at all

Answer: 2. \(\sqrt{\frac{G M}{L}}\)

Question 156. Periodic-time of a satellite revolving very near to the surface of the earth is – (ρ is the density of the earth)

1. Proportional to \(\frac{1}{\rho}\)
2. Proportional to \(\frac{1}{\sqrt{\rho}}\)
3. Proportional ρ
4. Does not depend on ρ.

Answer: 2. Proportional to \(\frac{1}{\sqrt{\rho}}\)

Question 157. A satellite is moving around the earth. In order to make it move to infinity, its velocity must be increased by

1. 20%
2. It is impossible to do so
3. 82.8%
4. 41.4%

Answer: 4. 41.4%

Question 158. If the radius of the earth is to decrease by 4% and its density remains the same, then its escape velocity will

1. Remain same
2. Increase by 4%
3. Decrease by 4%
4. Increase by 2%

Answer: 3. Decrease by 4%

Question 159. An earth satellite is moved from one stable circular orbit to another higher stable circular orbit. Which one of the following quantities increases for the satellite as a result of this change

1. Gravitational force
2. Gravitational potential energy
3. Centripetal acceleration
4. Linear orbital speed

Answer: 2. Gravitational potential energy

Question 160. The relay satellite transmits the television programme continuously from one part of the world to another because its

1. A period is greater than the period of rotation of the earth
2. Period is less than the period of rotation of the earth about its axis
3. Period has no relation with the period of the earth about its axis
4. Period is equal to the period of rotation of the earth about its axis

Answer: 4. Period is equal to the period of rotation of the earth about its axis

Question 161. If the universal constant of gravitation were decreasing uniformly with time, then a satellite in orbit would still maintain its

1. Radius
2. Tangential speed
3. Angular momentum
4. Period of revolution

Answer: 3. Angular momentum

Question 162. A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth :

1. The acceleration of S is always directed towards the centre of the earth
2. The angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant
3. The total mechanical energy of S varies periodically with time
4. The linear momentum of S remains constant in magnitude.

Answer: 1. The acceleration of S is always directed towards the centre of the earth

Question 163. The moon revolves around the Earth 13 times in one year. If the ratio of sun-earth distance to earth-moon distance is 392, then the ratio of masses of sun and earth will be

1. 3.56 × 10⁵
2. 3.56 × 10⁶
3. 3.56 × 10⁷
4. 3.56 × 10⁸

Answer: 1. 3.56 × 10⁵

Question 164. The earth revolves around the sun in an elliptical orbit. If \(\frac{O A}{O B}\)= x, the ratio of speeds of earth at B and A will be

1. x
2. \(\sqrt{x}\)
3. x2
4. \(x \sqrt{x}\)

Answer: 1. x

Question 165. The time period of a satellite of earth is 5 h. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become

1. 10 h
2. 80 h
3. 40 h
4. 20 h

Answer: 3. 40 h

Question 166. If two spheres of the same masses and radius are brought in contact, then the force of attraction between them will be proportional to (for a given density ρ) :

1. r²
2. r³
3. r⁶
4. r⁴

Answer: 4. r⁴

Question 167. Assume the earth to be a sphere of radius R. If g is the acceleration due to gravity at any point on the earth’s surface, the mass of the earth is :

1. \(\frac{g R}{G}\)
2. \(\frac{g^2 R^2}{G}\)
3. \(\frac{g R^2}{G}\)
4. \(\frac{g^2 R}{G}\)

Answer: 3. \(\frac{g R^2}{G}\)

Question 168. Energy required to transfer a 400 kg satellite in a circular orbit of radius 2 R to a circular orbit of radius 4 R, where R is the radius of the earth. [Given g = 9.8 ms^-2, R = 6.4 × 10⁶ m]

1. 1.65 × 10⁹ J
2. 3.13 × 10⁹ J
3. 6.26 × 10⁹ H
4. 4.80 × 10⁹ J

Answer: 2. 3.13 × 10⁹ J

Question 169. Suppose the gravitational force varies inversely as the 4th power of the distance. If a satellite describes a circular orbit of radius R under the influence of this force, then the time period T of the orbit is proportional to

1. R^3/2
2. R^5/2
3. R²
4. R^7/2

Answer: 2. R^5/2

Question 170. A double star system consists of two stars A and B which have time periods T_A and T_B. Radius R_A and R_B and mass M_A and M_B. Choose the correct option.

1. If T_A> T_B then R_A> R_B
2. If T_A> T_B then M_A> M_B
3. \(\left(\frac{T_A}{T_B}\right)^2=\left(\frac{R_A}{R_B}\right)^3\)
4. T_A = T_B

Answer: 4. T_A = T_B

Question 171. Mass M is uniformly distributed only on the curved surface of a thin hemispherical shell. A, B and C are three points on the circular base of a hemisphere, such that A is the centre. Let the gravitational potential at points A, B and C be V_A, V_B, V_C respectively. Then

1. V_A> V_B>V_C
2. V_C> V_B>V_A
3. V_B>V_A and V_B> V_C
4. V_A= V_B=V_C

Answer: 4. V_A= V_B=V_C

Question 172. The figure shows the elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If t1is the time for the planet to move from C to D and t₂ is the time to move from A to B, then:

1. t₁> t₂
2. t₁= 4t₂
3. t₁= 2t₂
4. t₁= t₂

Answer: 3. t₁= 2t₂

Question 173. A particle of mass M is situated at the centre of a spherical shell of the same mass and radius a. The gravitational potential at a point situated at \(\frac{a}{2}\) distance from the centre, will be

1. \(-\frac{3 G M}{a}\)
2. \(-\frac{2 G M}{a}\)
3. \(-\frac{G M}{a}\)
4. \(-\frac{4 G M}{a}\)

Answer: 1. \(-\frac{3 G M}{a}\)

Question 174. The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M, to transfer it from a circular orbit of radius R₁ to another of radius R₂(R₂> R₁) is

1. \(G m M\left(\frac{1}{R_1^2}-\frac{1}{R_2^2}\right)\)
2. \({GmM}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
3. \(-\frac{G M}{a}\)
4. \(-\frac{4 G M}{a}\)

Answer: 4. \(-\frac{4 G M}{a}\)

Question 175. The radii of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If the speed of satellite A is 3V, then the speed of satellite B will be

1. \(\frac{3 V}{4}\)
2. 6V
3. 12 V
4. \(\frac{3 V}{2}\)

Answer: 2. 6V

Question 176. The dependence of acceleration due to gravity g on the distance r from the centre of the earth, assumed to be a sphere of radius R of uniform density is as shown in the figures below. The correct figure is.

Answer: 4

Question 177. A planet moving along an elliptical orbit is closest to the sun at a distance r₁ and farthest away at a distance of r₂. If v₁ and v₂ are the linear velocities at these points respectively, then the ratio \(\frac{v_1}{v_2}\) is:

1. (r₁/r₂)²
2. r₂/r₁
3. (r₂/r₁)²
4. r₁/r₂

Answer: 2. r₂/r₁

Question 178. A body projected vertically from the Earth reaches a height equal to the earth’s radius before returning to the earth. The power exerted by the gravitational force is greatest :

1. At the highest position of the body
2. At the instant just before the body hits the earth
3. It remains constant all through
4. At the instant just after the body is projected

Answer: 2. At the instant just before the body hits the earth

Question 179. A particle of mass m is thrown upwards from the surface of the earth, with a velocity u. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth, is :

1. \(\sqrt{\frac{2 G M}{R}}\)
2. \(\sqrt{\frac{2 G M}{R^2}}\)
3. \(\sqrt{2 g R^2}\)
4. \(\sqrt{\frac{2 G M}{R^2}}\)

Answer: 1. \(\sqrt{\frac{2 G M}{R}}\)

Question 180. A particle of mass M is situated at the centre of a spherical shell of mass and radius a. The magnitude of the gravitational potential at a point situated at a/2 distance from the centre will be:

1. \(\frac{2 G M}{a}\)
2. \(\frac{3 G M}{a}\)
3. \(\frac{4 G M}{a}\)
4. \(\frac{G M}{a}\)

Answer: 2. \(\frac{3 G M}{a}\)

Question 181. Which one of the following plots represents the variation of gravitational field on a particle with distance r due to a thin spherical shell of radius R ? (r is measured from the centre of the spherical shell)

Answer: 2

Question 182. The height at which the weight of a body becomes 1/16th, its weight on the surface of the earth (radius R), is :

1. 5R
2. 15R
3. 3R
4. 4R

Answer: 3. 3R

Question 183. A spherical planet has a mass MP and diameter DP. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity, equal to :

1. 4GMP/DP²
2. GMPm/DP²
3. GMP/DP²
4. 4GMPm/DP²

Answer: 1. 4GMP/DP²

Question 184. A geostationary satellite is orbiting the earth at a height of 5R above the surface of the earth, R is the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is :

1. 5
2. 10
3. \(6 \sqrt{2}\)
4. \(\frac{6}{\sqrt{2}}\)

Answer: 3. \(6 \sqrt{2}\)

Question 185. A body of mass ‘m’ is taken from the earth’s surface to a height equal to twice the radius (R) of the earth. The change in potential energy of the body will be:

1. \(\frac{2}{3} m g R\)
2. 3mgR
3. \(\frac{1}{3} m g R\)
4. mg2R

Answer: 1. \(\frac{2}{3} m g R\)

Question 186. Infinite number of bodies, each of mass 2 kg are situated on the x-axis at distances 1m, 2m, 4m, 8m, …….. respectively, from the origin. The resulting gravitational potential due to this system at the origin will be :

1. \(-\frac{8}{3} G\)
2. \(-\frac{4}{3} G\)
3. –4G
4. – G

Answer: 3. –4G

Question 187. A block hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass = 5.98×10²⁴ kg) have to be compressed to be a black hole?

1. 10^-9 m
2. 10^-6 m
3. 10^-2 m
4. 100 m

Answer: 3. 10^-2 m

Question 188. Dependence of intensity of gravitational field (E) of the earth with distance (r) from the centre of the earth is correctly represented by:

Answer: 1

Question 189. Kepler’s third law states that the square of the period of revolution (T) of a planet around the sun, is proportional to the third power of average distance r between the sun and planet i.e. T² = Kr³ here K is constant. If the masses of the sun and planet are M and m respectively then as per Newton’s law of gravitation force of attraction between them is \(\mathrm{F}=\frac{G M m}{r^2}\) r, here G is gravitational constant. The relation between G and K is described as:

1. GMK = 4π²
2. K = G
3. \(\mathrm{K}=\frac{1}{G}\)
4. GK = 4π²

Answer: 1. GMK = 4π²

Question 190. A remote-sensing satellite of the earth revolves in a circular orbit at a height of 0.25 × 10⁶ m above the surface of the earth. If the earth’s radius is 6.38 × 10⁶ m and g = 9.8 ms^-2, then the orbital speed of the satellite is :

1. 8.56 km s^-1
2. 9.13 km s^-1
3. 6.67 km s^-1
4. 7.76 km s^-1

Answer: 4. 7.76 km s^-1

Question 191. At what height from the surface of the earth are the gravitational potential and the value of g –5.4 × 107 J kg^-2 and 6.0 ms^-2 respectively? Take the radius of the earth as 6400 km.

1. 2000 km
2. 2600 km
3. 1600 km
4. 1400 km

Answer: 2. 2600 km

Question 192. A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Then,

1. The total mechanical energy of S varies periodically with time.
2. The linear momentum of S remains constant in magnitude.
3. The acceleration of S is always directed towards the centre of the earth.
4. The angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant.

Answer: 3. The acceleration of S is always directed towards the centre of the earth.

Question 193. The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice that of earth is:

1. \(1: \sqrt{2}\)
2. 1: 2
3. \(1: 2 \sqrt{2}\)
4. 1: 4

Answer: 3. \(1: 2 \sqrt{2}\)

Question 194. Starting from the centre of the earth having radius r, the variation of g (acceleration due to gravity) is shown by

Answer: 3

Question 195. A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g0, the value of acceleration due to gravity at the earth’s surface, is

1. \(-\frac{2 m g_0 R^2}{R+h}\)
2. \(\frac{\mathrm{mg}_0 \mathrm{R}^2}{2(\mathrm{R}+\mathrm{h})}\)
3. \(-\frac{m g_0 R^2}{2(R+h)}\)
4. \(\frac{\mathrm{Rmg}_0 \mathrm{R}^2}{\mathrm{R}+\mathrm{h}}\)

Answer: 3. \(-\frac{m g_0 R^2}{2(R+h)}\)

Question 196. A physical quantity of the dimensions of length that can be formed out of c, G and \(\frac{e^2}{4 \pi \epsilon_0}\) is the velocity of light, G is a universal constant of gravitation and e is charge] :

1. \(\frac{1}{c^2}\left[G \frac{e^2}{4 \pi \epsilon_0}\right]^{1 / 2}\)
2. \(c^2\left[G \frac{e^2}{4 \pi \epsilon_0}\right]^{1 / 2}\)
3. \(\frac{1}{c^2}\left[\frac{e^2}{G 4 \pi \epsilon_0}\right]^{1 / 2}\)
4. \(\frac{1}{c} G \frac{e^2}{4 \pi \epsilon_0}\)

Answer: 1. \(\frac{1}{c^2}\left[G \frac{e^2}{4 \pi \epsilon_0}\right]^{1 / 2}\)

Question 197. Suppose the charge of a proton and an electron differ slightly. One of them is – e, and the other is (e + Δe). If the net of electrostatic force and the gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of [Given the mass of hydrogen mh = 1.67 × 10^-27 kg]

1. 10^-20 C
2. 10^-23 C
3. 10^-37 C
4. 10^-47 C

Answer: 3. 10^-37 C

Question 198. Two astronauts are floating in gravitational-free space after having lost contact with their spaceship. The two will :

1. Keep floating at the same distance between them
2. Move towards each other
3. Move away from each other
4. Will become stationary

Answer: 2. Move towards each other

Question 199. The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB and KC, respectively. AC is the major axis and SB is perpendicular to AB at the position of the Sun S as shown in the figure. Then

1. K_A < K_B < K_C
2. K_B > K_A > K_C
3. K_B < K_A < K_C
4. K_A > K_B > K_C

Answer: 4. K_A > K_B > K_C

Question 200. If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?

1. Raindrops will fall faster.
2. ‘g’ on the Earth will not change
3. Time period of a simple pendulum on the Earth would decrease.
4. Walking on the ground would become more difficult.

Answer: 2. ‘g’ on the Earth will not change

Question 201. The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is :

1. \(\frac{3}{2} \mathrm{mgR}\)
2. mgR
3. 2 mgR
4. \(\frac{1}{2} \mathrm{mgR}\)

Answer: 4. \(\frac{1}{2} \mathrm{mgR}\)

Question 202. The time period of a geostationary satellite is 24 h, at a height of 6R_E (R_E is the radius of the earth) from the surface of the earth. The time period of another satellite whose height is 2.5 R_E from the surface will be :

1. \(6 \sqrt{2} h\)
2. \(12 \sqrt{2} h\)
3. \(\frac{24}{2.5} h\)
4. \(\frac{12}{2.5} h\)

Answer: 1. \(6 \sqrt{2} h\)

Question 203. Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final – initial) of an object of mass m, when taken to a height h from the surface of the earth (of radius R), is given by :

1. \(-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}\)
2. \(\frac{\mathrm{GMmh}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\)
3. mgh
4. \(\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}\)

Answer: 2. \(\frac{\mathrm{GMmh}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\)

Question 204. A body weight of 72N on the surface of the earth. What is the gravitational force on it at a height equal to half the radius of the earth?

1. 24N
2. 48 N
3. 32 N
4. 30 N

Answer: 3. 32 N

Question 205. The escape velocity from the Earth’s surface is υ. The escape velocity from the surface of another planet having a radius, four times that of Earth and the same mass density is:

1. 2υ
2. 3υ
3. 4υ
4. υ

Answer: 3. 4υ

Question 206. A particle of mass ‘m’ is projected with a velocity u = kV_e (k < 1) from the surface of the earth. (V_e = escape velocity) The maximum height above the surface reached by the particle is

1. \(\mathrm{R}\left(\frac{\mathrm{k}}{1+\mathrm{k}}\right)^2\)
2. \(\frac{\mathrm{R}^2 \mathrm{k}}{1+\mathrm{k}}\)
3. \(\frac{\mathrm{Rk}^2}{1-\mathrm{k}^2}\)
4. \(R\left(\frac{k}{1-k}\right)^2\)

Answer: 3. \(\frac{\mathrm{Rk}^2}{1-\mathrm{k}^2}\)

Question 207. The height at which the acceleration due to gravity becomes \(\) (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is

1. \(\frac{R}{\sqrt{2}}\)
2. \(\frac{R}{2}\)
3. \(\sqrt{2} R\)
4. 2R

Answer: 4. 2R

Question 208. Two bodies of masses m and 4 m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is:

1. Zero
2. \(-\frac{4 G m}{r}\)
3. \(-\frac{6 G m}{r}\)
4. \(-\frac{9 G m}{r}\)

Answer: 4. \(-\frac{9 G m}{r}\)

Question 209. Two particles of equal mass ‘m’ go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to its centre of mass is:

1. \(\sqrt{\frac{G m}{4 R}}\)
2. \(\sqrt{\frac{G m}{3 R}}\)
3. \(\frac{G m M}{2 R}\)
4. \(\frac{G m M}{3 R}\)

Answer: 1. \(\sqrt{\frac{G m}{4 R}}\)

Question 210. What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

1. \(\frac{5 G m M}{6 R}\)
2. \(\frac{2 G m M}{3 R}\)
3. \(\frac{G m M}{2 R}\)
4. \(\frac{G m M}{3 R}\)

Answer: 1. \(\frac{5 G m M}{6 R}\)

Question 211. Four particles, each of mass M and equidistant from each other move along a circle of radius R under the action of their mutual gravitational attraction. the speed of each particle is:

1. \(\sqrt{\frac{G M}{R}}\)
2. \(\sqrt{2 \sqrt{2} \frac{G M}{R}}\)
3. \(\sqrt{\frac{G M}{R}(1+2 \sqrt{2})}\)
4. \(\frac{1}{2} \sqrt{\frac{G M}{R}(1+2 \sqrt{2})}\)

Answer: 4. \(\frac{1}{2} \sqrt{\frac{G M}{R}(1+2 \sqrt{2})}\)

Question 212. From a solid sphere of mass M and radius R, a spherical portion of radius R/2 is removed, as shown in the figure. Taking gravitational potential V = 0 at r = ∞, the potential at the centre of the cavity thus formed is :(G = gravitational constant)

1. \(\frac{-G M}{2 R}\)
2. \(\frac{-G M}{R}\)
3. \(\frac{-2 G M}{3 R}\)
4. \(\frac{-2 G M}{R}\)

Answer: 2. \(\frac{-G M}{R}\)

Question 213. If the angular momentum of a planet of mass m, moving around the sun in a circular orbit is L, about the centre of the Sun, its areal velocity is:

1. \(\frac{2 L}{m}\)
2. \(\frac{4 \mathrm{~L}}{\mathrm{~m}}\)
3. \(\frac{L}{2 m}\)
4. \(\frac{\mathrm{L}}{\mathrm{m}}\)

Answer: 3. \(\frac{L}{2 m}\)

Question 214. The energy required to take a satellite to a height ‘h’ above Earth’s surface (radius or Earth = 6.4 × 10³ km) is E₁ and the kinetic energy required for the satellite to be in a circular orbit at this height is E₂. The value of h for which E₁ and E₂ are equal is :

1. 1.28 × 10⁴ km
2. 6.4 × 10³ km
3. 3.2 × 10³ km
4. 1.6 × 10³ km

Answer: 3. 3.2 × 10³ km

Question 215. A satellite is moving with a constant speed v in a circular orbit around the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is :

1. \(\frac{3}{2} m v^2\)
2. \(\frac{1}{2} m v^2\)
3. 2mv²
4. mv²

Answer: 4. mv²

Question 216. Two stars of masses 3 × 10³¹ kg each, and at distance 2 × 10¹¹ m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star’s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that a meteorite should have at O is (Take Gravitational constant G = 6.67 × 10^-11 Nm2 kg^-2) What is the order of energy of the gas due to its thermal motion?

1. 3.8 × 10⁴ m/s
2. 1.4 × 10⁵ m/s
3. 2.8 × 10⁵ m/s
4. 2.4 × 10⁴ m/s

Answer: 3. 2.8 × 10⁵ m/s

Question 217. A satellite is revolving in a circular orbit at a height h from the earth’s surface, such that h<<R where R is the radius of the earth. Assuming that the effect of the earth’s atmosphere can be neglected. The minimum increase in the speed required so that the satellite could escape from the gravitational field of the earth is :

1. \(\sqrt{g R}(\sqrt{2}-1)\)
2. \(\sqrt{\frac{g R}{2}}\)
3. \(\sqrt{2 \mathrm{gR}}\)
4. \(\sqrt{g R}\)

Answer: 1. \(\sqrt{g R}(\sqrt{2}-1)\)

Question 218. The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be: 

1. \(\frac{3}{2} s\)
2. \(\frac{2}{\sqrt{3}} \mathrm{~s}\)
3. \(\frac{\sqrt{3}}{2} s\)
4. \(2 \sqrt{3} \mathrm{~s}\)

Answer: 4. \(2 \sqrt{3} \mathrm{~s}\)

Question 219. A straight rod of length L extends from x = a to x = L + a. The gravitational force it exerts on a point mass ‘m’ at x = 0, if the mass per unit length of the rod is A + Bx², is given by

1. \(G m\left[A\left(\frac{1}{a+L}-\frac{1}{a}\right)-B L\right]\)
2. \({Gm}\left[A\left(\frac{1}{a+L}-\frac{1}{a}\right)+B L\right]\)
3. \({Gm}\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)+B L\right]\)
4. \(G m\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)-B L\right]\)

Answer: 3. \({Gm}\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)+B L\right]\)

Question 220. A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision. the subsequent motion of the combined body will be :

1. In the same circular orbit of radius R
2. Such that it escapes to infinity
3. In an elliptical orbit
4. In a circular orbit of a different radius

Answer: 3. In an elliptical orbit

Question 221. Two satellites, A and B, have masses of m and 2m respectively. A is in a circular orbit of radius R, and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, \(\frac{T_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\) is:

1. 1
2. \(\sqrt{\frac{1}{2}}\)
3. 2
4. \(\frac{1}{2}\)

Answer: 1. 1

Friction Contact Force

When two bodies are kept in contact, electromagnetic forces act between the charged particles (molecules) at the surfaces of the bodies.

Thus, each body exerts a contact force on the other. The magnitudes of the contact forces acting on the two bodies are equal but their directions are opposite and therefore the contact forces obey Newton’s third law.

• The direction of the contact force acting on a particular body is not necessarily perpendicular to the contact surface.
• We can resolve this contact force into two components, one perpendicular to the contact surface and the other parallel to it.
• In the figure, the perpendicular component to the contact surface is called the normal contact force or normal force (generally written as N) and the parallel component is called friction (generally written as f). Therefore if R is the contact force then

⇒ \(\mathrm{R}=\sqrt{f^2+N^2}\)

Reasons For Friction

1. Ιnter-locking of extended parts of one object into the extended parts of the other object.
2. Bonding between the molecules of the two surfaces or objects in contact.

Friction Force Is Of Two Types.

1. Kinetic
2. Static

1. Kinetic Friction Force

Kinetic friction exists between two contact surfaces only when there is relative motion between the two contact surfaces. It stops acting when relative motion between two surfaces ceases.

Direction Of Kinetic Friction On An Object

• Ιt is opposite to the relative velocity of the object considered with respect to the other object in contact.
• Note that its direction is not opposite to the force applied it is opposite to the relative motion of the body considered which is in contact with the other surface.

Magnitude Of Kinetic Friction

The magnitude of the kinetic friction is proportional to the normal force acting between the two bodies. We can write

f_k= μ_k N

where N is the normal force. The proportionality constant μ_k is called the coefficient of kinetic friction and its value depends on the nature of the two surfaces in contact.

Question 1. Find the direction of the kinetic friction force

1. On the block, exerted by the ground.
2. On the ground, exerted by the block.

Answer:

where f₁ and f₂ are the friction forces on the block and ground respectively.

Question 2. In the above example, the correct relation between the magnitude of f₁ and f₂ is

1. f₁> f₂
2. f₂> f₁
3. f₁= f₂
4. It is not possible to decide due to insufficient data.

Answer: By Newton‘s third law the above friction forces are action-reaction pairs, equal but opposite to each other in direction. Hence (3).

Also, note that the direction of kinetic friction has nothing to do with applied force F.

Question 3. All surfaces as shown in the figure are rough. Draw the friction force on A and B

Answer:

Kinetic friction acts in such a way so as to reduce relative motion.

Question 4. Find out the distance travelled by the blocks shown in the figure before it stops.

Answer:

N – 10 g = 0

N = 100 N

f_k= µ_kN

f_k= 0.5 × 100 = 50 N

f_k= ma

50 = 10 a

⇒ a = 5

∴ v² = u² + 2as

0² = 10² + 2 (–5) (S)

∴ S = 10 m

Question 5. Find out the distance travelled by the block on the incline before it stops. The initial velocity of the block is 10 m/s and the coefficient of friction between the block and incline is μ = 0.5.

Answer:

N = mg cos37°

∴ mg sin 37° + µN = ma

a = 10 m/s² down the incline

Now v² = u² + 2as

0 = 10² + 2(–10) S

∴ S = 5 m

Question 6. Find the time taken in the above example by the block before it stops.

Answer:

a = g sin 37° + µg cos 37°

∴ a = 10 m/s² down the incline

∴ S = \(u t+\frac{1}{2} at^2\)

5 = \(\frac{1}{2} \times 10 \times \mathrm{t}^2\)

∴ t = 1sec.

Question 7. A block is given a velocity of 10 m/s and a force of 100 N in addition to friction force is also acting on the block. Find the retardation of the block?

Answer:

As there is relative motion

∴ kinetic friction will act to reduce this relative motion.

f_k = µN = 0.1 × 10 × 10 = 10 N

100 + 10 = 10a

a = 11 m/s²

Static Friction: Ιt exists between the two surfaces when there is a tendency of relative motion but no relative motion occurs along the two contact surfaces.

• For Question consider a bed inside a room; when we gently push the bed with a finger, the bed does not move.
• This means that the bed has a tendency to move in the direction of the applied force but does not move as there exists static friction force acting in the opposite direction of the applied force.

Question 8. What is the value of static friction force on the block?

Answer:

In the horizontal direction acceleration is zero.

Therefore Σ F = 0.

∴ ƒ = 0

Direction Of Static Friction Force: The static friction force on an object is opposite to its impending motion relative to the surface.

The following steps should be followed in determining the direction of static friction force on an object.

1. Draw the free body diagram with respect to the other object on which it is kept.
2. Include pseudo force also if the contact surface is accelerating.
3. Decide the resultant force and the component parallel to the surface of this resultant force.
4. The direction of static friction is opposite to the above component of the resultant force.

Note: Here once again the static friction is involved when there is no relative motion between two surfaces.

Question 9. In the following figure, an object of mass M is kept on a rough table as seen from above. Forces are applied to it as shown. Find the direction of static friction if the object does not move.

Answer:

In the above problem, we first draw the free-body diagram to find the resultant force.

As the object does not move this is not a case of limiting friction. The direction of static friction is opposite to the direction of the resultant force F_R as shown in the figure by f_s. Its magnitude is equal to 25 N.

Magnitude Of Kinetic And Static Friction

Kinetic Friction:

The magnitude of the kinetic friction is proportional to the normal force acting between the two bodies. We can write

f_k= μ_k N

where N is the normal force. The proportionality constant μ_k is called the coefficient of kinetic friction and its value depends on the nature of the two surfaces in contact.

If the surfaces are smooth μ_k will be small, if the surfaces are rough μ_k will be large. It also depends on the materials of the two bodies in contact.

Static Friction:

The magnitude of static friction is equal and opposite to the external force exerted, till the object at which force is exerted is at rest. This means it is a variable and self-adjusting force. However, it has a maximum value called limiting friction.

f_max = μ_sN

The actual force of static friction may be smaller than μ_sN and its value depends on other forces acting on the body. The magnitude of frictional force is equal to that required to keep the body at relative rest.

0, f_s, f_smax

Here μsand μk are proportionality constants. μ_s is called the coefficient of static friction and μ_k is called the coefficient of kinetic friction. They are dimensionless quantities independent of shape and area of contact. It is a property of the two contact surfaces.

Note: μ_s> μ_k for a given pair of surfaces. If not mentioned then μ_s= μ_k can be taken. The value of μ can be from 0 to ∞.

The following table gives a rough estimate of the values of the coefficient of static friction between certain pairs of materials.

The actual value depends on the degree of smoothness and other environmental factors. For example, wood may be prepared at various degrees of smoothness and the friction coefficient will vary.

table

Rolling Friction: When a body (say wheel) rolls on a surface the resistance offered by the surface is called rolling friction.

• Rolling friction forces arise as, for example, a rubber tyre rolls on pavement, primarily because the tyre deforms as the wheel rolls. The sliding of molecules against each other within the rubber causes energy to be lost.
• The velocity of the point of contact with respect to the surface remains zero.
• The rolling friction is negligible in comparison to static or kinetic friction which may be present simultaneously i.e., µ_R< µ_k< µ_S

Angle Of Friction

The angle of friction is the angle which the resultant of limiting friction F_S and normal reaction N makes with the normal reaction. It is represented by λ, Thus from the figure.

⇒ \(\tan \lambda=\frac{F_S}{N}\) (∵ F_s= µ N) ortan

or \(\frac{F_S}{N}\) = θtan λ = µ

For smooth surfaces, λ = 0 (zero)

Angle Of Repose (θ)

If a body is placed on an inclined plane and if its angle of inclination is gradually increased, then at some angle of inclination θ the body will just begin to slide down this angle is called the angle of repose (θ).

F_S = mg sinθ and N = mg cosθ

So, or µ = tanθ

Relation between angle of friction (λ) and angle of repose (θ)

We know that tan λ = µ and µ = tan θ

hence tan λ = tan θ or θ = λ

Thus, angle of repose = angle of friction

Question 1. Find the acceleration of the block. Initially, the block is at rest.

Answer: Zero

Question 2. Find out the acceleration of the block. Initially, the block is at rest.

Answer:

N + 24 – 100 = 0 for vertical direction

∴ N = 76 N

Now 0 ≤ f_s ≤ µ_s N

0 ≤ f_s ≤ 76 × 0.5

0 ≤ f_s≤ 38 N

∴ 32 < 38 Hence f = 32

∴ acceleration of the block is zero.

Question 3. Find out the acceleration of the block for different ranges of F.

Answer:

0 ≤ f ≤ µ_sN

⇒ 0 ≤ f ≤ µ_smg

a = 0 if F ≤ µ_smg

⇒ a = \(\frac{F-\mu M g}{M}\) if F > µMg M

Question 4. Find out the acceleration of the block. Initially, the block is at rest.

Answer:

0 ≤ f_s ≤ µ_sN

⇒ 0 ≤ f_s≤ 50

Now 51 > 50

∴ The block will move but if the block starts moving then kinetic friction is involved.

f_k= µ_k N = 0.3 × 100 = 30 N

∴ 51 – 30 = 10 a

∴ a = 2.1 m/s²

Question 5. Find out the minimum force that must be applied on the block vertically downwards so that the block doesn’t move.

Answer:

100-f_s = 0

∴ f_s= 100 ……..

F + 10 g = N ⇒ N = 100 + F ……….. (2)

Now 0 ≤ f_s≤ μN

100 ≤ 0.5 N

100 ≤ 0.5 [100 + F]

200 ≤ 100 + F

F ≥ 100 N

∴ Minimum F = 100 N

Question 6. A particle of mass 5 kg is moving on a rough fixed inclined plane with a constant velocity of 5 m/s as shown in the figure. Find the friction force acting on a body by plane.

Answer:

f_k= μ_kN = μ_k mg cos 37° = mg sin30° = 5 (10)\(\left(\frac{1}{2}\right)\)