MCQ Questions for Class 11 Engineering Physics Kinetic Theory Quiz 1

An ideal gas goes through a reversible cycle

$a\rightarrow b\rightarrow c\rightarrow d$ has the V- T diagram shown below. Process

$d\rightarrow a$ and

$b\rightarrow c$ are adiabatic.
The corresponding P-V diagram for the process is ( all figure are schematic and not drawn to scale).

Explanation

In process

$a \rightarrow b$ and

$c \rightarrow d$ the volume is directly proportional to temperature of gas, hence the process is isobaric, however, the slope of curve is different in each case. using ideal gas equation

$PV=nRT \Rightarrow V=\dfrac{nR}{P} T$

slope

$m=\dfrac{nR}{P}$ for a high pressure the slope of the V-T curve will be less. In given diagram the slope of

$c \rightarrow d$ process is lower than

$a \rightarrow b$ process, hence process

$a \rightarrow b$ is at higher pressure, from which option B and D can be ruled out.

Now initial rate of change of volume w.r.t. temperature is very less from

$b \rightarrow c$ and

$d \rightarrow a$ as shown in V-T diagram.

hence slope of P-V curve should tend to

$\lim_{\Delta V\rightarrow 0}\dfrac{\partial P }{\partial v}=\infty$ as initial change in volume is nearly zero, which can be verified in option A

Hence correct answer is option A.

In the nuclear fusion reaction,

$_{1}^{2}\mathrm{H}+_{1}^{3}\mathrm{H}\rightarrow_2^4{He}$

$+\ {n}$ given that the repulsive potential energy between the two nuclei is

$7.7\times 10^{-14}\ {J}$ , the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann's constant

$\mathrm{k}=1.38\times 10^{-23}\mathrm{J}/\mathrm{K}$ ]

0%
$10^{7}\ {K}$
0%
$10^{5}\ {K}$
0%
$10^{3}\ {K}$
0%
$10^{9}\ {K}$

Explanation

To initiate the reaction, the energy of the gas molecules must be equal to the repulsive potential energy.

We know that the kinetic energy of translation per molecules is given by:

$K.E.=\dfrac32kT$

In order to initiate the fusion reaction,

$\dfrac{3}{2}kT=7.7\times 10^{-14}J$

$\implies T=\dfrac{2\times 7.7\times 10^{-14}}{3\times 1.38\times 10^{-23}}\approx3.7\times 10^{9}K$

An ideal gas is enclosed in a cylinder at pressure of

$2 \,atm$ and temperature,

$300 \,K$ . The mean time between two successive collisions is

$6 \times 10^{-8} \,s$ . If the pressure is doubled and temperature is increased to

$500 \,K$ , the mean time between two successive collisions will be close to:

0%
$4 \times 10^{-8} \,s$
0%
$3 \times 10^{-6} \,s$
0%
$2 \times 10^{-7} \,s$
0%
$0.5 \times 10^{-8} \,s$

Explanation

$t \alpha \dfrac{Volume}{Velocity}$

Volume

$\alpha \dfrac{T}{P}$

$\therefore t \alpha \dfrac{\sqrt{T}}{P}$

$\dfrac{t_1}{6 \times 10^{-8}} = \dfrac{\sqrt{500}}{2P} \times \dfrac{P}{\sqrt{300}}$

$t_1 = 3.8 \times 10^{-8}$

$\approx 4 \times 10^{-8}$

$'N'$ moles of a diatomic gas in a cylinder are at a temperature

$'T'$ . Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas gets converted into monatomic gas. What is the change in the total kinetic energy of the gas ?

0%
$\dfrac{5}{2}nRT$
0%
$\dfrac{1}{2}nRT$
0%
$0$
0%
$\dfrac{3}{2}nRT$

Explanation

$\Rightarrow { KE }_{ initial }=\cfrac { 3 }{ 2 } nRT$

Final moles

$({ n }_{ 2 })=2n$

Since monoatomic gas conversion is there

$\Rightarrow { KE }_{ final }=\cfrac { 3 }{ 2 } \times 2nRT$

$\Rightarrow \Delta KE=\cfrac { 3 }{ 2 } nRT$

Which of the following shows the correct relationship between the pressure

$'P'$ and density

$\rho$ of an ideal gas at constant temperature ?

Explanation

From ideal gas law:

$PV=nRT$

$P=\cfrac{MnRT}{MV}, \quad M:$ Molecular mass number

But

$\rho=Mn/V$

$P=\rho RT/M$

At constant temperature,

$P \propto \rho$

Hence, the correct graph is (d).

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

0%
1 : 4
0%
1 : 2
0%
6 : 9
0%
8 : 9

Explanation

$Sol$ : (D)

$PV =nRT =\dfrac{m }{M }RT$

$\Rightarrow PM =\rho RT$

$\dfrac{\rho _{1}}{\rho _{2}}=\dfrac{P _{1}M _{1}}{P _{2}M _{2}}=(\dfrac{P _{1}}{P _{2}})\times(\dfrac{M _{1}}{M _{2}})=\dfrac{4}{3}\times\dfrac{2}{3}=\dfrac{8}{9}$
Here

$\rho _{1}$ and

$\rho _{2}$ are the densities gases in the vessel containing the mixture.

A gas molecule of mass

$M$ at the surface of the Earth has kinetic energy equivalent to

$0^oC$ . If it were to go up straight without colliding with any other molecules, how high it would rise?
Assume that the height attained is much less than radius of the earth. (

$k_B$ is Boltzmann constant)

0%
$0$
0%
$\dfrac {273 k_B}{2 Mg}$
0%
$\dfrac {546 k_B}{3 Mg}$
0%
$\dfrac {819 k_B}{2 Mg}$

Explanation

Answer is D
Average translational kinetic energy of molecules=

$\displaystyle KE= \frac{3}{2} k_B T= \frac{819}{2} k_B as T= 273K$
Now. KE= PE
Or,

$\displaystyle \frac{819}{2} k_B = Mgh$

$\displaystyle h= \frac{819k_B}{2Mg}$

0%
Statement -1 is True, Statement-2 is True; Statement -2 is a correct explanation for Statement-1.
0%
Statement -1 is True, Statement-2 is True; Statement -2 is NOT a correct explanation for Statement-1.
0%
Statement -1 is True, Statement-2 is False
0%
Statement -1 is False, Statement-2 is True.

Explanation

Average energy of a molecule of an ideal gas at temperature T=

$\dfrac{1}{2}mv^2=\dfrac{3}{2}kT$

Therefore the total energy of the gas =

$\dfrac{3}{2}NkT$

where N is the total number of molecules.

Since Nk=nR,

Average energy =

$\dfrac{3}{2}nRT=\dfrac{3}{2}PV$

The ratio of the specific heats

$\dfrac {C_p}{C_v}=\gamma$ in terms of degree of freedom (n) is given by:

0%
$\left (1+\dfrac {2}{n}\right )$
0%
$\left (1+\dfrac {n}{2}\right )$
0%
$\left (1+\dfrac {1}{n}\right )$
0%
$\left (1+\dfrac {n}{3}\right )$

Explanation

The internal energy for 1 mole of gas is given as

$U=\dfrac{n}{2}RdT=C_vdT$

where n is degrees of freedom

$C_p-C_v= R$

$Cp=(1+\dfrac{n}{2}) R$

and

$\dfrac{C_p}{C_v}=\gamma=\dfrac{(1+\dfrac{n}{2}) R}{\dfrac{n}{2}} =1+\dfrac{2}{n}$

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?

0%
mKT
0%
P/(kT)
0%
Pm/(kT)
0%
P/(kTV)

Explanation

Hint: Use ideal gas equation.

Step 1: Note the given values

Volume =V

Pressure =P

Temperature =T

Mass of each molecule =m

Step 2: Calculate density of gas using ideal gas equation .

From the ideal gas law,

$PV=NkT$

$\implies \dfrac{N}{V}=\dfrac{P}{kT}$

Density of gas=

$\dfrac {\text{Total mass} }{\text volume}$

$\rho= \dfrac{Nm}{V}=\dfrac{Pm}{kT}$

$\textbf{Hence option C correct}$

The amount of heat energy required to raise the temperature of 1 g of Helium at NTP, from

$T_{1}K$ to

$T_{2}K$ is

0%
$\displaystyle \frac{3}{8}N_{a}k_{B}(T_{2}-T_{1})$
0%
$\displaystyle \frac{3}{2}N_{a}k_{B}(T_{2}-T_{1})$
0%
$\displaystyle \frac{3}{4}N_{a}k_{B}(T_{2}-T_{1})$
0%
$\displaystyle \frac{3}{4}N_{a}k_{B}\left ( \frac{T_{2}}{T_{1}} \right )$

Explanation

$\displaystyle Q=\frac{f}{2}nR\Delta T$

$\displaystyle \frac{3}{2}\times \frac{1}{4}\times k_{B}N_{a}\Delta T$

$=\displaystyle \frac{3}{8}N_{a}k_{B}(T_{2}-T_{1})=\frac{3}{8}N_{a}k_{B}(T_{2}-T_{1})$

Degree of freedom for polyatomic gas is

0%
$\ge 4$
0%
$\ge 5$
0%
$\ge 6$
0%
$\ge 7$

Explanation

A diatomic molecule posses

$3$ translational,

$2$ rotational and

$1$ vibrational degree of freedom which gives a total of

$6$ degrees of freedom. A polyatomic gas has more than

$2$ atoms in one molecule of it so it must have more than or equal to

$6$ degrees of freedom.

A vessel of volume

$20\ L$ contains a mixture of hydrogen and helium at temperature of

$27^{\circ}C$ and pressure

$2\ atm$ . The mass of mixture is

$5\ g$ . Assuming the gases to be ideal, the ratio of mass of hydrogen to that of the helium in the given mixture will be

0%
$1 : 2$
0%
$2 : 3$
0%
$2 : 1$
0%
$2 : 5$

Explanation

Let there are

$n_{1}$ moles of hydrogen and

$n_{2}$ moles of helium in the given mixture. As

$Pv = nRT$
Then the pressure of the mixture

$P = \dfrac {n_{1}RT}{V} + \dfrac {n_{2}RT}{V} = (n_{1} + n_{2}) \dfrac {RT}{V}$

$\Rightarrow 2\times 101.3\times 10^{3} = (n_{1} + n_{2}) \times \dfrac {(8.3\times 300)}{20\times 10^{-3}}$
or,

$(n_{1} + n_{2}) = \dfrac {2\times 101.30 \times 10^{3} \times 20\times 10^{-3}}{(8.3)(300)}$
or,

$n_{1} + n_{2} = 1.62 .... (1)$
The mass of the mixture is (in grams)

$n_{1}\times 2 + n_{2} \times 4 = 5$

$\Rightarrow (n_{1} + 2n_{2}) = 2.5 ..... (2)$
Solving the eqns.

$(1)$ and

$(2)$ , we get

$n_{1} = 0.74$ and

$n_{2} = 0.88$
Hence,

$\dfrac {m_{H}}{m_{He}} = \dfrac {0.74\times 2}{0.88\times 4} = \dfrac {1.48}{3.52} = \dfrac {2}{5}$ .

The difference between volume and pressure coefficients of an ideal gas is :

0%
$\dfrac{1}{273}$
0%
$273$
0%
$\dfrac{2}{273}$
0%
Zero

Explanation

Volume coefficient:

$\alpha$
Pressure coefficient:

$\beta$

$\alpha=\beta=\frac{1}{273}$

The gas law

$\left [ \dfrac{PV}{T} \right ]=$ constant is true for

0%
isothermal change only
0%
adiabatic change only
0%
both isothermal & adiabatic
0%
neither isothermal nor adiabatic

Explanation

The given relation is always valid for ideal gas whether the process is adiabatic or isothermal because its the fundamental equation of ideal gas.

$P \times V=n \times R \times T$

The temperature of a gas contained in a closed vessel is increased by 2 K when the pressure is increased by 2%. The initial temperature of the gas is :

0%
200K
0%
100K
0%
200 $^{0}$ C
0%
100 $^{0}$ C

The mass of oxygen gas (in Kilo grams) occupying a volume of 11.2 litre at a temperature 27 $^{0}$ C and a pressure of 76cm of mercury is :

(Molecular weight of oxygen = 32)

0%
0.001456
0%
0.01456
0%
0.1456
0%
1.1456

Explanation

$PV=\dfrac { m }{ M } RT\\ 11.2\times { 10 }^{ -3 }\times 1.01\times { 10 }^{ 5 }=\dfrac { m }{ 32\times { 10 }^{ -3 } } \times 8.314\times 300\\ m=0.01456$

From what minimum height, a block of ice has to be dropped in order that it may melt completely on hitting the ground :

0%
$mgh$
0%
$\dfrac{mgh}{l}$
0%
$\dfrac{l}{g}$
0%
$\dfrac{h}{lg}$

Explanation

Let the mass of the ice be m then

$m \times g \times h = m \times l$ , where l is the latent heat of melting.

$h = \dfrac{l}{g}$

If for a gas

$\dfrac{R}{C_{v}}=0.67$ , then the gas is made up of molecules which are :

0%
Diatomic
0%
Monoatomic
0%
Polyatomic
0%
Mixture of Diatomic & Polyatomic

Explanation

Given

$\dfrac {R}{C_v}=0.67$
or

$\dfrac {R}{f\dfrac{R}{2}}=0.67$
or

$f=\dfrac{2}{0.67}=3$
The degree of freedom is 3. So the given gas is monoatomic.

The P-T graph for the given mass of an ideal gas is shown in figure. Then the volume

0%
increases
0%
decreases
0%
remains constant
0%
data insufficient

A gas in an airtight container is heated from 25 $^{o}$ C to 90 $^{o}$ C. The density of gas will :

0%
increase slightly
0%
increase considerably
0%
remain the same
0%
decrease slightly

The molar gas constant is the same for all gases because at the same temperature and pressure, equal volumes of gases have the same :

0%
number of molecules
0%
average potential energy
0%
ratio of specific heats
0%
density

Explanation

We know,

$R=\frac { PV }{ nT }$
assuming P,V and T to be constant.
Also we know, that equal volume of gases contain equal number of molecules, hence the value of R is also constant.

If the volume of the gas is to be increased by 4 times :

0%
temperature and pressure must be doubled
0%
at constant P, the temperature must be increased by 4 times
0%
at constant T, the pressure must be increased by 4 times
0%
it cannot be increased

Explanation

$PV=RT$
At constant pressure temp has to be increased by 4 times to increase volume by 4 times.

Three closed vessels A, B and C are at the same temperature T and contain gases which obey Maxwell distribution law of velocities. Vessel A contains $O_2$ , B only $N_2$ and C mixture of equal quantities of $O_2$ and $N_2$ . If the average speed of $O_2$ molecules in vessel A is $V_1$ and that of $N_1$ molecules in vessel B is $V_2$ , then the average speed of the $O_2$ molecules in vessel C is

0%
$\frac{{({v_1} + {v_2})}}{2}$
0%
$V_1$
0%
$\sqrt {{v_1}{v_2}}$
0%
None of these

Explanation

Vessel

$A:-O_2$

$B:N_2$

$C:O_2+N_2$
Average speed of one species do not depend on the others

$\therefore$ avj. speed of

$V_2=V_1$

A gas at temperature 27 $^{0}$ C and pressure 30 atmospheres is allowed to expand to one atmospheric pressure. If the volume becomes 10 times its initial volumes, the final temperature becomes :

0%
100 $^{o}$ C
0%
373 K
0%
373 $^{o}$ C
0%
-173 $^{o}$ C

Explanation

$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \\ \dfrac { 30V }{ 300 } =\dfrac { 1\times 10V }{ { T }_{ 2 } } \\ { T }_{ 2 }=100K\\ { T }_{ 2 }=-173^0C$

The Universal gas constant may be expressed as :

a) 8.31 J/mole-K c) 2.00 J/mole-K

b) 8.31 cal/mole-K d) 2.00 cal/mole-K

0%
a,c
0%
a,d
0%
b,c
0%
b,d

The parameter that determine the physical state of gas are :

a) Pressure b) Volume

c) Number of moles d) Temperature

0%
a & b
0%
a,b & c
0%
a,b & d
0%
a,c & d

State whether true or false.

The arrangement of particles in a liquid is less ordered compared to solids. However, there is no order in the arrangement of particles in the gaseous state.

0%
True
0%
False

16 gm of $O_{2}$ gas and x gm of $H_{2}$ gas occupy the same volume at the same temperature and pressure. Then x is :

0%
1/2gm
0%
1gm
0%
8gm
0%
16 gm

Select the correct formula :

(where k=Boltzmann's constant, R= gas constant, n= moles, r = density, M= molecular weight, p= pressure, T= kelvin temperature, V= volume)

a) k=RN $_{av}$

b) $r=\dfrac{nM}{V}$

c) $\dfrac{p}{r}=\dfrac{RT}{M}$

d) R=kN $_{av}$

0%
a,b,c
0%
a,b,d
0%
b,c,d
0%
a,c,d

Explanation

(a)

$k= R/N_{av}$
(b) r is density it is clearly satisfying the relation. nM gives toatal mass and when divided by volume it gives density of the substance
(c)We know that , PV=nRT
PV=wRT/M
PV/w=RT/M
P/r=RT/M
(d)

$k=R/N_{av}$ therefore,

$R=kN_{av}$
hence (b) (c) (d) are correct
Hence option(C)

In the equation PV=constant, the numerical value of constant depends upon

a) temperature b) mass of the gas

c) system of units used d) nature of the gas

0%
a & b
0%
b & c
0%
c & d
0%
all

State whether true or false.

Sponge though compressible, is a solid.

0%
True
0%
False

Two containers of equal volume containing the same gas at pressure $P_{1}$ and $P_{2}$ and absolute temperature $T_{1}$ and $T_{2}$ respectively were connected with narrow capillary tube. The gas reaches a common pressure P and a common temperature T. The ratio P/T is equal to :

0%
$\dfrac{P_{1}}{T_{1}} +\dfrac{P_{2}}{T_{2}}$
0%
$\dfrac{1}{2}\left ( \dfrac{P_{1}}{T_{1}} +\dfrac{P_{2}}{T_{2}}\right )$
0%
$\dfrac{P_{1}T_{2}+P_{2}T_{1}}{T_{1}+T_{2}}$
0%
$\dfrac{P_{1}T_{2}-P_{2}T_{1}}{T_{1}-T_{2}}$

Explanation

Since the mass remains constant before and after opening the tube, moles of the gases remain same.

Thus

$\dfrac{P_1V}{RT_1}+\dfrac{P_2V}{RT_2}=\dfrac{P(2V)}{RT}$

$\implies \dfrac{P}{T}=\dfrac{1}{2}(\dfrac{P_1}{T_1}+\dfrac{P_2}{T_2})$

The value of

$\gamma$ for gas X is 1.66, then x is :

0%
Ne
0%
O $_3$
0%
N $_2$
0%
H $_2$

Explanation

Given that value of gamma is 1.66 i.e

$\dfrac{5}{3}$ which implies that it is a monoatomic gas, and Neon (Ne) is the only monoatomic gas among the given options.

According to the Boltzmann's law of equipartition of energy, the energy per degree of freedom and at a temperature T K is :

0%
(3/2) KT
0%
(2/3) KT
0%
KT
0%
1/2 KT

Explanation

According to the Boltzmann's law of equipartition of energy, the energy per degree of freedom and at a temperature T (in kelvin ) is

$\dfrac{1}{2}$ KT

where K=Boltzmann's constant and

T= temperature (in kelvin )

When the pressure of a gas is changed, then:

0%
The density of the gas also changes.
0%
The ratio of the pressure to the density remains unaffected.
0%
The velocity of the sound remains unaffected.
0%
The value of $\gamma$ changes.

Explanation

When the pressure of a gas is changed, then the ratio of the pressure to the density remains unaffected as

$\displaystyle = \frac{P}{\rho} = \frac{RT}{M}$

A system consists of N particles, which have independent K relations among one another. The number of degrees of freedom of the system is given by :

0%
3 NK
0%
N/3K
0%
3 N/K
0%
3N - K

The value of C

$_v$ for 1 mol of polyatomic gas is (F

$=$ number of degrees of freedom) :

0%
$\displaystyle \frac{fR}{2T}$
0%
$\displaystyle \frac{fR}{2}$
0%
$\displaystyle \frac{fRT}{2}$
0%
$2fRT$

Explanation

The value of C

$_{v}$ for 1 mol of polyatomic gas is

$\dfrac{fR}{2}$

where f is no of degrees of freedom (in a gas for every degree of freedom we have energy distribution as

$\dfrac{kT}{2}$ )

A man is climbing up a spiral type staircase. His degrees of freedom are :

0%
1
0%
2
0%
3
0%
more than 3

The law of equipartition of energy was given by :

0%
Claussius
0%
Maxwell
0%
Boltzmann
0%
Carnot

If the number of molecules in a gas is N then the number of molecules moving in negative X-direction will be :

0%
N/6
0%
N/4
0%
N/3
0%
N

Explanation

Equal number of molecules move in each of the 6(+X,-X,+Y,-Y,+Z,-Z) directions.

Hence, number of molecules moving in -X direction =

$\frac { N }{ 6 }$

Hence, option A is correct

At what temperature will the kinetic energy of gas molecules be double of its value at 27

$^o$ C?

0%
$54^oC$
0%
$108^oC$
0%
$300^oC$
0%
$327^oC$

Explanation

Kinetic energy of gas

$(E) = \dfrac {3}{2} \times k \times T$
Then

$\dfrac {E}{T} = k$

where

$k$ is Boltzmann constant

$\dfrac{{E_1}}{{T}_{1}} = \dfrac{{E}_{2}}{{T}_{2}}$
Given

$E_{1} = E$ ;

$E_{2} = 2\times E$ ;

$T_{1}= 27^{o}C = 300 K$ ;

$T_{2}=?$

Then

$T_{2}=E_{2} \times \dfrac{T_{1}}{E_{1}}=2\times E \times \dfrac{300}{E} = 600 K$

$T_{2} = 327^{o}C$

The internal energy of a monoatomic ideal gas is :

0%
only kinetic
0%
only potential
0%
partly kinetic and partly potential
0%
none

Keeping the number of moles, volume and temperature the same, which of the following are the same for all ideal gases?

0%
rms speed of molecules
0%
Density
0%
Pressure
0%
Average magnitude of momentum

Explanation

We have the relation PV=nRT (ideal gas equation)

by keeping the number of moles, volume and temperature same, pressure will also remains same (

$\because$ R is constant)

Which of the following is/are true on the basis of kinetic theory of matter?

0%
Solids have a definite volume and definite shape
0%
Liquids have a definite volume, but no definite shape
0%
Gases have no definite volume and no definite shape
0%
All of the above

Solids have :

0%
definite mass
0%
definite volume
0%
both A & B
0%
none

The amount of heat required to heat 1 mol of a monoatomic gas from 200

$^o$ C to 250

$^o$ C will be ............. if the heat required to heat the diatomic gas from 200

$^o$ C to 300

$^o$ C is Q.

0%
2Q/3
0%
3Q/5
0%
3Q/10
0%
2Q/5

Explanation

The heat required is given by formulae:

$Q= nc(T_2-T_1)$

For monoatomic:

$Q_1=1\times3/2R\times (250-200)=3/2R\times 50$

For diatomic:

$Q=1\times 5/2R(300-200)=5/2R\times 100$

$Q_1/Q=3\times50/5\times100$

$Q_1/Q=3/10$

$Q_1=3Q/10$

On the basis of kinetic theory of matter :

0%
the solids have a definite volume and definite shape
0%
the liquids have a definite volume but no definite shape
0%
the gases have no definite volume and no definite shape
0%
all of the above

The inter-molecular spaces in a liquid is :

0%
less than a solid
0%
more than a gas
0%
more than a solid
0%
more than a solid and a gas

The correct relation connecting the universal gas constant (R), Avogadro number N

$_A$ and Boltzmann constant (K) is :

0%
$R = NK^2$
0%
$K = NR$
0%
$N = RK$
0%
$R=NK$

Explanation

Units of R, N and K are

$Joule \times mole^{-1} \times K^{-1}$ ,

$mole^{-1}$ and

$Joule \times K^{-1}$
So

$R = N \times K$

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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