MCQ Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 5

If $\rho$ is the density, m is the mass of 1 molecule and K is the Boltzman constant for a gas then the pressure of the gas is:

0%
$P=\dfrac{\rho KT}{m}$
0%
$P=\dfrac{mKT}{\rho }$
0%
$P=\dfrac{\rho mT}{K}$
0%
$P=\rho KmT$

Explanation

$PV=\dfrac { m }{ M } RT\\ P\dfrac { m }{ \rho } =nkT\\ P=\dfrac { \rho nkT }{ m } ,\quad \\ if\quad n=1,\quad P=\dfrac { \rho kT }{ m }$

Its pressure is increased by $0.4$ %. The initial emperature of the gas is

0%
$250^{0}\ C$
0%
$100^{0}\ C$
0%
$-75^{0}\ C$
0%
$-23^{0}\ C$

Explanation

Since it is a closed system the volume remains constant.

Hence,

$\dfrac { { P }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 } }{ { T }_{ 2 } }$

${ P }_{ 2 }=1.004{ P }_{ 1 }\\ \therefore { T }_{ 2 }=1.004{ T }_{ 1 }\\ but\quad { T }_{ 1 }+1={ T }_{ 2 }\\ \therefore { T }_{ 1 }=250K\\ { T }_{ 1 }=-{ 23 }^{ \circ }$

A given amount of a gas is heated till the volume and pressure both increase by $2\%$ each the percentage change in temperature of the gas is equal to nearly

0%
$2\%$
0%
$3\%$
0%
$4\%$
0%
$1\%$

Explanation

$\dfrac { dP }{ P } +\dfrac { dV }{ V } =\dfrac { dT }{ T } \\ Now\quad dP=0.02P,dV=0.02V\\ 0.02+0.02=\dfrac { dT }{ T } \\ \dfrac { dT }{ T } \times 100=4%=percentage\quad change\quad in\quad temperature$

Two sample of Hydrogen and Oxygen of same mass possess same pressure and volume. The ratio of their temperature is

0%
$1:8$
0%
$1:16$
0%
$8:1$
0%
$16:1$

Explanation

${ P }_{ 1 }{ V }_{ 1 }=\dfrac { { m }_{ 1 } }{ { M }_{ 1 } } R{ T }_{ 1 }\\ { P }_{ 2 }{ V }_{ 2 }=\dfrac { { m }_{ 2 } }{ { M }_{ 2 } } R{ T }_{ 2 }\\ \dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { { M }_{ 1 } }{ { M }_{ 2 } } \\ \dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { 1 }{ 16 } \left( { M }_{ 1 }=1gm,{ \ M }_{ 2 }=16gm \right)$

Two gases A&B having same pressure P, volume V and absolute temperature T are mixed. If the mixture has volume and temperature as V & T respectively then the pressure of mixture is

0%
2P
0%
P
0%
P/2
0%
4P

Explanation

Moles of gas A

$=\dfrac{PV}{RT}$

Moles of gas B

$=\dfrac{PV}{RT}$

Moles of the gas mixture

$=\dfrac{P'V}{RT}$

Since the number of moles(mass) remains same before and after mixing,

$\dfrac{PV}{RT}+\dfrac{PV}{RT}=\dfrac{P'V}{RT}$

$\implies P'=2P$

One litre of Helium gas at a pressure of 76 cm - Hg and temperature 27 $^{0}$ C is heated till its pressure and volume are doubled. The final temperature attained by the gas is

0%
900 $^{0}$ C
0%
927 $^{0}$ C
0%
627 $^{0}$ C
0%
327 $^{0}$ C

Explanation

$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \\ \dfrac { PV }{ 300 } =\dfrac { 2P\times 2V }{ { T }_{ 2 } } \\ { T }_{ 2 }=1200K\\ { T }_{ 2 }={ 927 }^{ \circ }C$

The mass of oxygen gas occupy a volume of 11.2 lit at a temperature 27 $^{0}$ C and a pressure of 76 cm of mercury in kilogram is (molecular weight of oxygen =32)

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

Explanation

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

A vessel is filled with an ideal gas at a pressure 27 $^{0}$ C mass of the gas is removed from the vessel & the temp. of the remaining gas is increased to 87 $^{0}$ C . Then the pressure of the gas in the vessel will be

0%
5 atm
0%
6 atm
0%
7 atm
0%
8 atm

Explanation

We have Ideal gas Equation:

$PV=nRT$

First the mass is halved, thus the moles of gas are halved.

$P\propto n$

So,

$P_{1}=\dfrac{P_0}{2}=5atm$

Now the temperature is increased.

$P\propto T$

$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}=\dfrac{360}{300}=\dfrac{6}{5}$

$\implies P_2=6atm$

2 grams of monoatomic gas occupies a volume of 2 litres at a pressure of $8.3\times 10^{5}Pa$ and 127 $^{0}$ C . Find the molecular weight of the gas. (R=8.3 joule/mole/K)

0%
2 gram/mole
0%
16 gram/mole
0%
4 gram/mole
0%
32 gram/mole

Explanation

$PV=\dfrac { m }{ M } RT\\ 2\times { 10 }^{ -3 }\times 8.3\times { 10 }^{ 5 }=\dfrac { 2\times { 10 }^{ 3 } }{ M } \times 8.3\times 400\\ M=\dfrac { 2\times { 10 }^{ 3 }\times 8.3\times 400 }{ 2\times { 10 }^{ -3 }\times 8.3\times { 10 }^{ 5 } } \\ M=4{ gram }/{ mole }$

From a disc of radius

$R$ and mass

$M$ , a circular hole of diameter

$R$ , whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

0%
$\dfrac{9MR^{2}}{32}$
0%
$\dfrac{15MR^{2}}{32}$
0%
$\dfrac{13MR^{2}}{32}$
0%
$\dfrac{11MR^{2}}{32}$

Explanation

$PV=\dfrac { m }{ M } RT\\ PV=\dfrac { 8 }{ 32 } RT\\ PV=\dfrac { RT }{ 4 }$

A reciever has a pressure of 144cm of Hg. After two strokes with an exhaust pump, the pressure is 36cm of Hg. After another two strokes the pressure will be.

0%
9cm of Hg
0%
2.4cm of Hg
0%
6cm of Hg
0%
3cm of Hg

Explanation

${ P }_{ n }=P{ x }^{ n }\\ For\quad two\quad strokes,\\ 36=144{ x }^{ 2 }\\ { x }^{ 2 }=\frac { 1 }{ 4 } \\ \therefore x=\frac { 1 }{ 2 } \\ { { P }_{ n } }^{ \backprime }=144{ x }^{ 4 }\\ { { P }_{ n } }^{ \backprime }=144{ \left( \frac { 1 }{ 2 } \right) }^{ 4 }=9cm\quad of\quad Hg$

Three flasks of identical volume are filled separately by

(a) 1 gram of $H_{2}$

(b) 1 gram of $O_{2}$

They are immersed in a tank of water so that all of them attain same temperature. The pressures $P_{1},P_{2}$ and $P_{3}$ have the on.

0%
$P_{1} > P_{2} > P_{3}$
0%
$P_{1} = P_{2} = P_{3}$
0%
$P_{1} < P_{2} < P_{3}$
0%
$P_{3} > P_{1} > P_{2}$

Explanation

We know,

$PV=\dfrac { m }{ M } RT$
Given the flasks have same volume, and mass m=1gm for all the gases, the temperature attained is also same.
Hence the equation can be reduced to,
PM=Constant, where M is the molecular weight.
since

${ M }_{ 1 }{ <M }_{ 2 }{ <M }_{ 3 }\\ { \therefore P }_{ 1 }{ >P }_{ 2 }{ >P }_{ 3 }$

A closed copper vessel contains water equal to half of its volume when the temperature. Of the vessel is raised to 447 $^{0}$ c the pressure of steam in the vessel is (Treat steam as an ideal gas, R=8310 J/k / mole, density of water =1000 kg/m $^{3}$ molecular weight of water =18 )

0%
33.24 x 10 $^{7}$ pa
0%
16.62 x 10 $^{7}$ pa
0%
10.31 x 10 $^{7}$ pa
0%
8.31x 10 $^{7}$ pa

Explanation

Let V be the volume of the copper vessel.

Then considering steam as an ideal gas,

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

Mass of the water in the vessel=

$m=\rho (\dfrac{V}{2})$

$\implies PV=\dfrac{\rho V}{2M}RT$

$\implies P=\dfrac{\rho RT}{2M}=16.62\times 10^7Pa$

A sample of $O_{2}$ gas and a sample of hydrogen gas both have the same no. of moles, the same volume and same pressure. Assuming them to be perfect gases, the ratio of temperature of oxygen gas to the temperature of hydrogen gas is :

0%
1:1
0%
1:2
0%
2:1
0%
1:4

The density of a gas at N.T.P. is 1.5gm/lit. its density at a pressure of 152cm of Hg and temperature 27 $^{0}$ C

0%
$\frac{273}{100}gm/ltr$
0%
$\frac{150}{273}gm/ltr$
0%
$\frac{1}{273}gm/ltr$
0%
1.5 $gm/lit$

Explanation

Since volume is inversely proportional to density,

$\dfrac { { P }_{ 1 } }{ { { d }_{ 1 }T }_{ 1 } } =\dfrac { { P }_{ 2 } }{ { { d }_{ 2 }T }_{ 2 } } \\ \dfrac { 760 }{ 1.5\times 273 } =\dfrac { 152\times { 10 } }{ { d }_{ 2 }\times 300 } \\ { d }_{ 2 }=\dfrac { 273 }{ 100 }gm/lit$

An electric bulb of $250cc$ was sealed off at a pressure $10^{-3}$ mm of Hg and temperature $$27^{0}$$ C. The number of molecules present in the gas is

0%
$8.02 \times 10^{15}$
0%
$6.023 \times 10^{23}$
0%
$8.021 \times 10^{23}$
0%
$6 \times 10^{22}$

Explanation

$PV=nRT\\ n=\dfrac { PV }{ RT } \\ n=\dfrac { 250\times { 10 }^{ -6 }\times 133.33\times { 10 }^{ -3 } }{ 8.314\times 300 } \\ n=1.336\times { 10 }^{ -8 }\\ N=n\times { N }_{ A }\\ N=1.336\times { 10 }^{ -8 }\times 6.022\times { 10 }^{ 23 }\\ N=8.02\times { 10 }^{ 15 }$

Two identical containers connected by a fine capillary tube contain air at N.T.P. if one of those containers is immersed in pure water, boilling under normal pressure then new pressure is

0%
76 cm of Hg
0%
152 cm of Hg
0%
57 cm of Hg
0%
87.76 cm of Hg

Explanation

Since the number of moles of gas in the system remains constant:

Moles in first container earlier+Moles in second container earlier=Moles in first container later+Moles in second container later

$\implies \dfrac{P_0V}{RT_{0}}+\dfrac{P_0V}{RT_0}=\dfrac{P_1V}{RT_{0}}+\dfrac{P_1V}{RT_1}$

$T_0=273\ K,T_1=373\ K,P_0=76\ cm\ of\ Hg$

$\implies P_1=87.76\ cm\ of\ Hg$

A flask is filled with 13 gm of an ideal gas at 27 $^{0}$ C 52 $^{0}$ C mass of the gas that has to be released to maintain the temperature of the gas in the flask 52 $^{0}$ C and the pressure remaining the same is

0%
2.5 gm
0%
2.0 g
0%
1.5 g
0%
1.0 g

Explanation

Since P, V , M and R are constants,

${ m }_{ 1 }{ T }_{ 1 }={ m }_{ 2 }{ T }_{ 2 }\\ 13\times 300={ m }_{ 2 }\times 325\\ { m }_{ 2 }=12gm$
Hence mass to be removed = 13-12= 1gm

One gram mol of helium at 27

$^{0}$ C is mixed with three gram mols of oxygen at 127

$^{0}$ C at constant pressure. If there is no exchange of heat with the atmosphere then the final temperature will be

0%
375 K
0%
175K
0%
475 K
0%
575K

During an experiment an ideal gas is found to obey an additional gas law VT =constant. The gas is initially at temperature T and pressure P. When it is heated to the temperature 2T, the resulting pressure is

0%
2P
0%
P/2
0%
4P
0%
P/4

Explanation

The gas follows VT=constant. But it also follows PV=nRT.

Thus

$(\dfrac{nRT}{P})T=$ constant

$\implies \dfrac{T^2}{P}=$ constant

Therefore

$\dfrac{T^2}{P}=\dfrac{(2T)^2}{P'}$

$\implies P'=4P$

A cycle tube has volume

$2000 cm^3$ . Initially the tube is filled to 3/4th of its volume by air at pressure of

$105 N/m^{2}$ . It is to be inflated to a pressure of

$6 \times 10^{5} N/m^{2}$ under isothermal conditions. The number of strokes of pump, which gives

$500 cm^3$ air in each stroke, to inflate the tube is

0%
21
0%
12
0%
42
0%
11

Explanation

In this problem, let us conserve total number of moles before and after,

$(6X{ 10 }^{ 5 })(2000)=(105)(1500)+(101325)(500)n$

Where n is total number of strokes of pump.

Solving this equation given

$n\approx21$ .

Equal masses of $N_{2}$ and $O_{2}$ gases are filled in vessel A and B. The volume of vessel B is double of A. The ratio of pressure in vessel A and B will be

0%
16:7
0%
16:14
0%
32:7
0%
32:28

Explanation

The mathematical form of the Ideal Gas Law is:
PV = nRT and n = m/MW
Where:P - pressure
V - volume
n - number of moles
T - temperature
m - mass
MW - Molecular Weight
R - ideal gas constant
So we can write,

$\dfrac{P_a}{ P_b}$ =

$\dfrac{M_b \times V_b}{ M_a \times V_a}$
here,

$V_b = 2 V_a$ and

$M_a$ = atomic wt of

$O_2$ = 8 and

$M_b$ = atomic wt of

$N_2$ = 7
Putting these values we get,

$P_a/ P_b$ = 16/7

If the pressure of a gas contained in a closed vessel increases by x% when heated by 1 $^{0}$ C , its initial temperature is

0%
$\frac{100}{x}$ Kelvin
0%
$\frac{100}{x}$ Celsius
0%
$\frac{x+100}{x}$ Kelvin
0%
$\frac{100-x}{x}$ Celsius

Explanation

Let the initial pressure and temperature be P and T.

From Ideal Gas Law,

$P\propto T$

$\implies \dfrac{(1+\dfrac{x}{100})P}{P}=\dfrac{T+1}{T}$

$\implies T=\dfrac{100}{x}K$

27 $^{0}$ C is 84cm of Hg. If 25% of the gas is now introduced into the same vessel at the same temperature, the final pressure of the gas will be in cm of Hg

0%
105
0%
100
0%
95
0%
90

Explanation

Pressure=

$h\rho g=\dfrac{nRT}{V}$

$\implies h\propto n\propto m$

Therefore

$\dfrac{h_2}{h_1}=\dfrac{1.25m_{0}}{m_0}$

$\implies h_2=1.25h_1=105cm$

A partially inflated balloon contains 500 m $^{3}$ of helium at 27 $^{0}$ C and 1 atm pressure. The volume of the helium at an altitude of 6000 m, where the pressure is 0.5atm and the temperature is -3 $^{0}$ C is

0%
22.4 lit
0%
11.2 lit
0%
900 m $^{3}$
0%
50 m $^{3}$

Explanation

Assuming helium to be an ideal gas, and using relation:

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

$V_2=V_1\dfrac{P_1T_2}{P_2T_1}=900m^3$

A uniform tube with piston in the middle and containing a gas at 0 $^{0}$ C is heated to 100 $^{0}$ C at one side . If the piston moves 5 cm, find the length of the tube containing the gas at 100 $^{0}$ C.

0%
16.75 cm
0%
37.5 cm
0%
28.25 cm
0%
12.5 cm

Explanation

$\dfrac{PV}{T}$ is constant for gas in both sections of the cylinder.

Let the length and area of the tube be

$L$ and

$A$ , respectively.

For section with constant temperature,

$P_{0}V_{0}=P_1V_1$

$\implies P_ (A\dfrac{L}{2})=P_1(A(\dfrac{L}{2}-5))$

Initial temperature

$T_0 = 0^oC = 273 \textrm{K}$

Final temperature

$T_1 = 100^oC = 100+273 = 373 \textrm{K}$

Similarly for section whose temperature increased,

$\dfrac{P_0V_0}{T_0}=\dfrac{P_1V_1}{T_1}$

$\implies \dfrac{P_0(A\dfrac{L}{2})}{273}=\dfrac{P_1(A(\dfrac{L}{2}+5))}{373}$

$\implies L=65cm$

So the length of tube containing gas at

$100^{\circ}$ is

$\dfrac{L}{2}+5=37.5cm$

28gm of $N_2$ gas is contained in a flask at a pressure of 10atm and at a temperature of 57 $^{0}$ C. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to 27 $^{0}$ C. The quantity of N $_{2}$ gas that leaked out is

0%
$\frac{5}{63}gm$
0%
$\frac{63}{5}gm$
0%
$\frac{11}{20}gm$
0%
$\frac{20}{11}gm$

Explanation

Since the volume of the flask remains the same, using Ideal Gas Equation we claim,

$n\propto \dfrac{P}{T}$

$\implies \dfrac{n_2}{n_1}=\dfrac{P_2T_1}{P_1T_2}$

$\implies n_2=\dfrac{11}{20}n_1$

Therefore, fraction of gas leaked out is

$(1-\dfrac{11}{20})=\dfrac{9}{20}$

Thus, amount of gas leaked out=

$\dfrac{9}{20}\times 28gm=\dfrac{63}{5}gm$

In a given species of tobbaco there is 0.1 mg of virus per c.c. The mass of virus is $4 \times 10^{7}$ Kg per kilomol. The number of the molecules of virus present in 1 c.c. will be

0%
$10^{13}$
0%
$10^{11}$
0%
$1.5 \times 10^{12}$
0%
$10^{14}$

Explanation

${ m }_{ v }=\dfrac { M }{ { N }_{ A } } =\dfrac { 4\times { 10 }^{ 7 } }{ 6\times { 10 }^{ 23 }\times { 10 }^{ 3 } } \\ mass\quad of\quad virus\quad per\quad cc=0.1\times { 10 }^{ -6 }kg\\$

Number of molecules in cc

$n=\dfrac { { m }_{ v } }{ { N }_{ A } } =\dfrac { 0.1\times { 10 }^{ -6 }\times 6\times { 10 }^{ 23 }\times { 10 }^{ 3 } }{ 4\times { 10 }^{ 7 } } \\ n=1.5\times { 10 }^{ 12 }$

A vertical cylindrical vessel separated in two parts by a frictionless piston free to move along the length of a vessel. The length of vessel is 90 cm and the piston divides the cylinder in the ratio 2 :Each of the two parts contain 0.1 mole of an ideal gas. The temperature of the gas is $27^{o}$ C in each part. The mass of the piston is

0%
41.5 kg
0%
8.97 kg
0%
9.57 kg
0%
11.57 kg

Explanation

$PV=nRT$

$P = nRT/V$

$PA=nRT/l$

$PA=0.1\times 8.31\times 300/0.6$

$PA=41.5 N$
Therefore option(A) is correct.

$Assertion$ : gases are characterised with two coefficients of expansion

$Reason$ : when heated both volume and pressure increase with the rise in temperature

0%
Both assertion (A) and reason (R) are correct and R gives the correct explanation
0%
Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation
0%
A is true but R is false
0%
A is false but R is true

Explanation

The ideal gas equation

$PV=\eta RT$ tells us that the pressure and volume are directly proportional to the temperature. The coefficients of expansion for pressure is

$\gamma_p$ and that of volume is

$\alpha_v$ .

Follwing operation are carried out on a sample of ideal gas initially at pressure P volume V and kelvin temperature T.

a) At constant volume, the pressure is increased fourfold.

b) At constant pressure, the volume is doubled

c) The volume is doubled and pressure halved.

d) If heated in a vessel open to atmosphere, one-fourth of the gas escapes from the vessel.

Arrange the above operations in the increasing order of final temperature.

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

Explanation

ideal gas equation

$PV = nRT$ eq(1)

where

$P$ = Pressure

$V$ = Volume

$n$ = number of moles

$R$ = universal gas constant

$T$ = Temperature

suppose initial values are

${P}_{1},{V}_{1},{n}_{1},{T}_{1}$ and

final values are

${P}_{2},{V}_{2},{n}_{2},{T}_{2}$

${P}_{1}{V}_{1}={n}_{1}R{T}_{1}$ eq(2)

${P}_{2}{V}_{2}={n}_{2}R{T}_{2}$ eq(3)

(a) At constant volume, the pressure is increased fourfold

${V}_{2} = {V}_{1}$

${P}_{2} = 4 {P}_{1}$

${n}_{2} = {n}_{1}$

$\dfrac{eq(3)}{eq(2)}$

$\dfrac{{T}_{2}}{{T}_{1}} = 4$

${T}_{2} = 4 {T}_{1}$

(b) At constant pressure, the volume is doubled

${V}_{2} = 2 {V}_{1}$

${P}_{2} = {P}_{1}$

${n}_{2} = {n}_{1}$

$\dfrac{eq(3)}{eq(2)}$

$\dfrac{{T}_{2}}{{T}_{1}} = 2$

${T}_{2} = 2 {T}_{1}$

${V}_{2} = 2 {V}_{1}$

${P}_{2} =\dfrac{1}{2} {P}_{1}$

${n}_{2} = {n}_{1}$

$\dfrac{eq(3)}{eq(2)}$

$\dfrac{{T}_{2}}{{T}_{1}} = 1$

${T}_{2} = {T}_{1}$

(d) If heated in a vessel open to atmosphere, onefourth of the gas escapes from the vessel.

for open vessel pressure and volume will be constant.

${V}_{2} = {V}_{1}$

${P}_{2} = {P}_{1}$

one-fourth of gas escaped from vessel, so three-fourth gas left

${n}_{2} = \dfrac{3}{4}{n}_{1}$

$\dfrac{\dfrac{3}{4} {T}_{2}}{{T}_{1}} = 1$

${T}_{2} = \dfrac{4}{3}{T}_{1}$

arranging final temperature in increasing order

(c),(d),(b),(a)

PV = n RT holds good for :

a) Isobaric process b) Isochoric process

c) Isothermal process d) Adiabatic process

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

Assertion: In Joules bulb apparatus, as reservoir is moved up, the mercury level raises into the bulb.

Reason: The pressure on the enclosed gas increases.

0%
Both assertion (A) and reason (R) are correct and R gives the correct explanation
0%
Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation
0%
A is true but R is false
0%
A is false but R is true

Assertion:With increase in temperature, the pressure of given gas increases

Reason:Increase in temperature causes decrease in no. of collision of molecules with walls of container.

0%
Both assertion (A) and reason (R) are correct and R gives the correct explanation
0%
Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation
0%
A is true but R is false
0%
A is false but R is true

Which of the following processes will quadruple the pressure

a) Reduce V to half and double T

b) Reduce V to 1/8th and reduce T to half

c) Double V and half T

d) Increase both V and T to double the values

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

Assertion:PV/T=constant for 1 mole of gas. This constant is same for all gases.

Reason:1 mole of different gases at NTP occupy same volume of 22.4 litres.

0%
Both assertion (A) and reason (R) are correct and R gives the correct explanation
0%
Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation
0%
A is true but R is false
0%
A is false but R is true

Figure shows a cylindrical tube of cross-sectional area A filled with two frictionless pistons. The pistons are connected through wire. The tension in the wire if the temperature rises from T $_{0}$ to 2T $_{0}$ is (Initial pressure is P $_{0}$ , atmospheric pressure)

0%
$P_{0}A$
0%
$\frac{P_{0}A}{2}$
0%
$2P_{0}A$
0%
$\frac{2P_{0}A}{3}$

Explanation

Since the wired would be taut, there would be no change in the volume of enclosed gas, so

$\dfrac{P}{T}=constant$

$\implies \dfrac{P}{2T_0}=\dfrac{P_0}{T_0}$

$\implies P=2P_0$

Thus, an equilibrium to exist, the difference in inside and outside forces must be balanced by the tension in the wire.

Thus,

$T=2P_0A-P_0A=P_0A$

Assertion: PV/T=constant for 1 gram of gas. This constant varies from gas to gas.

Reason:1 gram of different gases at NTP occupy different volumes.

0%
Both assertion (A) and reason (R) are correct and R gives the correct explanation
0%
Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation
0%
A is true but R is false
0%
A is false but R is true

Match List I with List II

List-I List-II

a) 0.00366/ $^{0}$ C e) Avagadros Number

b) 6.023x10 $^{23}$ molecules f) Universal gas constant

c) -273 $^{0}$ C g) Pressure coefficient of gas

d) 8.31 J/K-mole h) Intercept of V-T graph at

constnat pressure

0%
a-g, b-e, c-h, d-f.
0%
a-f, b-g c-e,d-h.
0%
a-g,b-e,c-f, d-h.
0%
a-g,b-f, c-e, d-h.

Explanation

(a)

$0.00366^{o}C$ =

$\frac{1}{273}^{o}C$ = pressure coefficient of gas
a - g
(b)

$6.023 \times {10}^{23}$ is Avagadros Number.
b - e
(c)

$-273^{o}C$ is Intercept of V-T graph at constnat pressure(as shown in image)
c - h
(d)8.31 J/K-mole is Universal gas constant(R).
d - f
Answer is A.

Assertion: Gasses obey Boyle's law at high temperature and low pressure only.

Reason: At low pressure and high temperature, gasses would behave like ideal gases.

0%
Both assertion (A) and reason (R) are correct and R gives the correct explanation
0%
Both assertion (A) and reason (R) are correct but R does give the correct explanation
0%
A is true but R is false
0%
A is false but R is true

The physical factor which distinguishes thermal radiation from visible light is :

0%
wave length
0%
pressure
0%
temperature
0%
amplitude

A wall has two layers

$A$ and

$B$ , each of same thickness and same area of cross-section but made of different materials. The thermal conductivity of

$A$ is three times that of

$B$ . The temperature difference across the wall is

$20$

$^{o}$ C . In thermal equilibrium :

a) the temperature difference across the layer

$A$ is 15

$^{o}$ C
b) the temperature difference across the layer

$B$ is 15

$^{o}$ C
c) the rate of flow of heat across

$A$ is more than across

$B$
d) the rate of flow of heat across

$A$ and

$B$ are same

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

Explanation

Heat transfer from the wall is given as:

${Q}=\dfrac{{KA}({T}_{1}-{T}_{2})}{L}$
Here, area and thickness of both the wall is equal, so for equilibrium condition equate both of them, also

${K}_{1}={3K}_{2}$

${T}_{1}-{T}_{2}={20}^{\circ}C$

${T}_{2}={T}_{1}-{20}^{\circ}C$

${T}_{1}={T}_{2}+{20}^{\circ}C$
Therefor temperature across 1:

$\dfrac { { K }_{ 1}({ T }_{ 1 }-{ T }) }{ { l }_{ 1 } } =\dfrac { { K }_{ 2 }({ T }-{ T }_{ 2 }) }{ { l }_{ 2 } }$
So on substituting the values in above equation we find:

${ { 3K }_{ 2}({ T }_{ 1 }-{ T })} ={ { K }_{ 2 }({ T }-{ T }_{ 1 }+{20}) }$

${T}_{1}-{T}={5}^{\circ}C$
For surface 2:

${ { 3K }_{ 2}({20}+{ T }_{ 2 }-{ T })} ={ { K }_{ 2 }({ T }-{ T }_{ 2}) }$

${T}-{T}_{2}={15}^{\circ}C$

${Q}={K}({T}_{1}-{T}_{2})$
So for surface 1:

${Q} \propto {K}({T}_{1}-{T})={3K}({5})={15}^{\circ}C(K)$
For surface 2:

${Q} \propto {K}({T}-{T}_{2})={K}({15})={15}^{\circ}C(K)$
Same for both as both of them have same length and area.
NOTE:

${T}_{1}=A$ ,

${T}_{2}=B$

The process in which rate of transfer of heat is maximum is :

0%
conduction
0%
convection
0%
radiation
0%
in all these, heat is transferred with the same speed

Explanation

$\textbf{Step 1}$ :

There are three modes of transmission of heat. They are Conduction, convection and radiation.

Conduction: In this mode of transmission. heat dissipates from one place to another by molecular vibration. It is relevant to solid only. Conduction needs a medium to transfer heat.

Convection: Heat is transferred from one place to another by the transfer of molecules. It happens only in liquids or gases. It also needs a medium to transfer heat.

Radiation: It transfers heat in the form of electromagnetic waves. It can heat any form of material. It doesn't need any medium for the transfer of heat.

$\textbf{Step 2}$ :

We know that radiation travels with the speed of light. So, the rate of heat transfer is maximum in radiation in the form of electromagnetic radiation.

Thus option C is correct.

Two rods of same length and cross sections are joined end to end. Their thermal conductivities are in the ratio

$2 : 3$ . If the free end of the first rod is at

$0^{o}$ C and free end of the second rod is at

$100^{o}$ C , the temperature at the junction of the two rods after attaining steady state is:

0%
$33.33^{o}C$
0%
$40^{o}C$
0%
$50^{o}C$
0%
$60^{o}C$

Explanation

From the Fourier's law we know that heat transfer through a plane wall is given as:

${Q}=\dfrac{KA{\Delta}T}{l}$

Here length and area both are same for both the rod.

So,

${Q}={{K}_{1}{\Delta}T}$

So,

${Q}_{1}={{K}_{1}({T}_{1}-{T})}$

${Q}_{2}={{K}_{2}({T}-{T}_{2})}$

On equating both

${Q}_{1}$ and

${Q}_{2}$ we get

${{K}_{1}({T}_{1}-{T})}={{K}_{2}({T}-{T}_{2})}$

$\dfrac{{K}_{1}}{{K}_{2}}=\dfrac{{T}-{T}_{2}}{{T}_{1}-{T}}$

$\dfrac{2}{3}=\dfrac{{T}-{100}}{{0}-{T}}$

${T}={60}^{0}C$

Match List I and List II

List-I List-II

a) P-V graph(T is constant) e) St. line cutting temp axis at - 273 $^{0}$

b) P-T graph(V is constant) f) Rectangular hyperbola

c) V-T graph(pressure P is constant) g) A st.line parallel to axis

d) PV- P garph h) St. line passing trhough origin (T is constant)

0%
a-g,b-e,c-h, d-f.
0%
a-h, b-f c-g, d-e.
0%
a-e, b-g,c-f, d-h.
0%
a-f, b-h, c-e,d-g.

Explanation

(a) P-V graph (T is constant):

$PV = nRT$
at constant temperature

$PV = constant$

$P \alpha \frac{1}{V}$
so it will be Rectangular hyperbola.

(b) P-T Graph (V is constant) :

$PV = nRT$
at constant volume

$P \alpha T$
pressure is proportional to T, so it will be straight line passing through origin.

c) V-T graph(P is constant) :

$PV = nRT$
at constant pressure

$V \alpha T$
volume is proportional to T(Kelvin),so it will be straight line passing through origin, but if temperature is taken in

$^{o}C$ , then it will cut temperature axis at

$-273 ^{o}C$

(d) PV - P Graph (T is constant) :

$PV = nRT$
at constant Temperature

$PV = constant$

$PV$ is constant for constant temperature,so it will be straight line parallel to P axis.

The thermal radiations are similar to :

0%
x-rays
0%
cathode rays
0%
$\alpha$ - rays
0%
$\gamma$ - rays

Explanation

Among the above four options: X-ray can only act as heat rays.

$\alpha$ -rays are nuclear emissions, cathode rays are simply rays made up of electrons.

$\gamma$ -rays are simply photon rays,

$\gamma$ -rays have very high penetrating power so it cannot act as heat ray, So X-ray is the appropriate answer.

If we place our hand below a lighted electric bulb. We feel warmer because of :

0%
convection
0%
radiation
0%
conduction
0%
both A and B

Two metal plates of same area and ratio of thickness

$1 : 2$ and having ratio of thermal conductivities

$1 : 2$ are in contact. If the free sides of the first plate is maintained at

$-20^{o}$ C and the free side of the other plate is maintained at

$+40^{o}$ C ,the temperature of the common surface after attaining steady state is:

0%
$0^{o}C$
0%
$-10^{o}C$
0%
$10^{o}C$
0%
$20^{o}C$

Explanation

From the fourier's law we know that heat transfer through a plane wall is given as:

${Q}=\dfrac{KA{\Delta}T}{l}$

Here, area are same for both the rod,

${Q}=\dfrac{{{K}4{\Delta}T}}{l}$

${Q}_{1}=\dfrac{{K}_{1}({T}_{1}-{T})}{{l}_{1}}$

${Q}_{2}=\dfrac{{K}_{2}({T}-{T}_{2})}{{l}_{2}}$

On equating both

${Q}_{1}$ and

${Q}_{2}$ we get

$\dfrac{{K}_{1}({T}_{1}-{T})}{{l}_{1}}=\dfrac{{K}_{2}({T}-{T}_{2})}{{l}_{2}}$

$({T}_{1}-{T})=({T}-{T}_{2})$

${-20}-{T}={T}-{40}$

${T}={10}^{\circ}C$

A double-pane window used for insulating a room thermally from outside consists of two glass sheets each of area 1m

$^{2}$ and thickness 0.01m separated by 0.05m thick stagnant air space.In the steady state, the room-glass interface and the glass-outdoor interface are at constant temperatures of 270

$^{o}$ C and 0

$^{o}$ C respectively.The thermal conductivity of glass is 0.8 and of air 0.08Wm

$^{-1}$ K

$^{-1}$ .

The rate of flow of heat through the window pane is nearly equal to:

0%
$1000Js^{-1}$
0%
$1224Js^{-1}$
0%
$3000Js^{-1}$
0%
$4000Js^{-1}$

Explanation

Given: A double-pane window used for insulating a room thermally from outside consists of two glass sheets each of area

$1m^2$ and thickness 0.01m separated by 0.05m thick stagnant air space.In the steady state, the room-glass interface and the glass-outdoor interface are at constant temperatures of

$270^oC$ and

$0^oC$ respectively.The thermal conductivity of glass is 0.8 and of air

$0.08Wm^{−1}K^{−1}$

To find the rate of flow of heat through the window pane.

Solution:

As per the given criteria,

Area of the each glass sheet,

$A=1m^2$

Thermal conductivity of air,

$k_a=0.08Wm^{−1}K^{−1}$
Thermal conductivity of glass,

$k_g=0.8Wm^{−1}K^{−1}$

Temperature at room-glass interface,

$T_1=270^\circ C +273=543K$

Temperature at glass-outdoor interface,

$T_4=0^\circ C+273=273K$

Thickness of the glass sheet,

$d_g=0.01m$

Thickness of the air space,

$d_a=0.05m$

By conduction, the equation of heat flow are:

$\dfrac {dQ_1}{dt}=\dfrac {k_gA(T_1-T_2)}{d_g}$

By substituting the values, we get

$\dfrac {dQ_1}{dt}=\dfrac {0.8\times1\times(543-T_2)}{0.01}......(i)$

Similarly at second interface, we get

$\dfrac {dQ_2}{dt}=\dfrac {k_aA(T_2-T_3)}{d_a}$

By substituting the values, we get

$\dfrac {dQ_2}{dt}=\dfrac {0.08\times1\times(T_2-T_3)}{0.05}......(ii)$

Similarly at third interface, we get

$\dfrac {dQ_3}{dt}=\dfrac {k_gA(T_3-T_4)}{d_g}$

By substituting the values, we get

$\dfrac {dQ_3}{dt}=\dfrac {0.8\times1\times(T_3-273)}{0.01}......(iii)$

But in steady state,

$\dfrac {dQ_1}{dt}=\dfrac {dQ_2}{dt}=\dfrac {dQ_3}{dt}$

So equating eqn(i) and (ii), we get

$\dfrac {0.8\times1\times(543-T_2)}{0.01}=\dfrac {0.08\times1\times(T_2-T_3)}{0.05}\\\implies 0.05(0.8\times(543-T_2))=0.01(0.08\times(T_2-T_3))\\\implies 0.04(543-T_2)=0.0008(T_2-T_3)\\\implies (543-T_2)=\dfrac {0.0008}{0.04}(T_2-T_3)\\\implies 543-T_2=0.02T_2-0.02T_3\\\implies 1.02T_2-0.02T_3=543......(iv)$

Now equating eqn(ii) and (iii), we get

$\dfrac {0.08\times1\times(T_2-T_3)}{0.05}=\dfrac {0.8\times1\times(T_3-273)}{0.01}\\\implies 0.01(0.08\times(T_2-T_3))=0.05(0.8\times(T_3-273)\\\implies 0.0008(T_2-T_3)=0.04(T_3-273)\\\implies\dfrac {0.0008}{0.04}(T_2-T_3)= (T_3-273)\\\implies0.02T_2-0.02T_3=T_3-273\\\implies 0.02T_2-1.02T_3=273......(v)$

Multiply, eqn(iv) with

$0.02$ and eqn(v) with

$1.02$ and by subtracting them,we get

$0.0204T_2-0.0004T_3=10.86$

$\underline{-(0.0204T_1-1.0404T_3=278.46)}\\\implies 1.04T_3=-267.6\\\implies T_3=-257.3K$

Substituting the value of

$T_3$ in eqn(v), we get

$0.02T_2-1.02(-257.3)=273\\\implies 0.02T_2=273-262.446\\\implies T_2=527.7K$

So the rate of flow of the heat through the window pane is given by,

$\dfrac {dQ_1}{dt}=\dfrac {0.8\times1\times(543-T_2)}{0.01}$

Substituting the value of

$T_2$ , we get

$\dfrac {dQ_1}{dt}=\dfrac {0.8\times1\times(543-527.7)}{0.01}\\\implies \dfrac {dQ_1}{dt}=1224W$

is the required rate of flow of heat.

Two rods (one semi-circular and other straight) of same material and of same cross sectional area are joined as shown in the figure. The points

$A$ and

$B$ are maintained at different temperatures. The ratio of heat transferred through a cross-section of a semicircular rod to the heat transferred through a cross section of the straight rod in a given time is

0%
$2:$ $\pi$
0%
$1:2$
0%
$\pi$ $:2$
0%
$3:2$

Explanation

We know that for semi-circular rod amount of heat generated is given as:

${ Q }_{ 1 }=\dfrac { KA({ T }_{ 1 }-{ T }_{ 2 })t }{ \pi R }$

For straight rod amount of heat generated is given as:

${ Q }_{ 2 }=\dfrac { KA({ T }_{ 1 }-{ T }_{ 2 })t }{ 2R }$

So taking there ratio, we find

$\dfrac { { Q }_{ 1 } }{ { Q }_{ 2 } } =\dfrac { 2 }{ \pi }$

CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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