MCQ Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 6

Three rods of same material and having same crosssection have been joined as shown in fig.Each rod is of same length. The left and right rods are kept at 0

$^{o}$ C and 90

$^{o}$ C respectively.The temperature of the junction of the three rods will be

0%
$45^{o}C$
0%
$60^{o}C$
0%
$30^{o}C$
0%
$20^{o}C$

Explanation

Let the temperature of the junction will be

${X}$
So assume that total incoming temperature at the junction will be equal to total outgoing temperature.
So

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

$({0}-{X})=({X}-{90})+({X-90})$

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

Four identical rods of same material are joined at ends so as to form a square. If the temperature difference at the ends of a diagonal is 100

$^{o}$ C, then the temperature difference across the ends of another diagonal will be

0%
$0^{o}C$
0%
$25^{o}C$
0%
$50^{o}C$
0%
$100^{o}C$

Explanation

Four identical rods A, B, C, D which having AC & BD as a diagonal point.

Let the temperature at

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

Therefore, temperature at

${C}={t+100}^{0}C$

Heat is conducted from A to C via B or D. B & D lies midway along with the passing of heat conduction through ABC and ADC respectively.

Hence, temperature of

${B}=\dfrac{t+(t+100)}{2}={t+50}^{0}C$

similarly,temperature of

${D}=\dfrac{t+(t+100)}{2}={t+50}^{0}C$

So, the difference of temperature between diagonal

${BD}={0}$

Choose the correct statements from the following

0%
All bodies emit thermal radiations at all temperatures
0%
Thermal radiations are electromagnetic waves
0%
Thermal radiations are not reflected from a mirror
0%
Thermal radiations travel in free space with a velocity of $3\times 10$ $^{8}$ ms $^{-1}$ .

Explanation

Every body emit thermal radiation at any temperature

$T>0K$ . These thermal radiations are electromagnetic in nature which can travel in vacuum and whose wavelength decreases if the temperature of body is increased.

As these radiations are electromagnetic, they get reflected form the mirror, just like visible light.

So option A, B and D are correct.

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

$^{2}$ and thickness

$0.01$ m separated by

$0.05$ m thick stagnant air space.In the steady state, the room-glass interface and the glass-outdoor interface are at constant temperatures of 27

$^{o}$ C and 0

$^{o}$ C respectively.The thermal conductivity of glass is

$0.8$ and of air 0.08Wm

$^{-1}$ K

$^{-1}$

The temperature of the outer glass-air interface is

0%
$26.5^{o}C$
0%
$25.5^{o}C$
0%
$24.5^{o}C$
0%
$23.5^{o}C$

Explanation

${T}_{2}$ = temperature of glass-1 and air interface

${T}_{3}$ = temperature of air and glass-2 interface

The equation of heat flow are:

$\displaystyle \frac { { Q }_{ 1 } }{ t } =\frac { { K }_{ 1 }A({ T }_{ 1 }-{ T }_{ 2 }) }{ { d }_{ 1 } }=\frac { { 0.8 }{\times}{1}({ 27}-{ T }_{ 2 }) }{ 0.01 }$

$\displaystyle \frac { { Q }_{ 2 } }{ t } =\frac { { K }_{ 2 }A({ T }_{ 2 }-{ T }_{ 3 }) }{ { d }_{ 2} }=\frac { { 0.08 }{\times}{1}({T}_{2}-{ T }_{ 3 }) }{ 0.05 }$

$\displaystyle \frac { { Q }_{ 3 } }{ t } =\frac { { K }_{ 3 }A({ T }_{ 3 }-{ T }_{ 4 }) }{ { d }_{ 3 } }=\frac { { 0.8 }{\times}{1}({T}_{3}) }{ 0.01 }$

In steady state:

$\displaystyle \frac { { q }_{ 1 } }{ t }=\frac { { q }_{ 2 } }{ t }=\frac { { q }_{ 3 } }{ t }$

On equating

$({1}) ({2})$ and

${1} {3}$ ,we find the following:

${1350}={51{T}_{2}}-{T}_{3}$ ----------(a)

${27}-{T}_{2}={T}_{3}$ ----------(b)

The temperature of outer glass interface

${T}_{2}={26.48}^{0}C$

Three cylindrical rods

$A, B, C$ of equal lengths and equal diameters are joined in series as shown in the figure. Their thermal conductivities are

$2k, k, 0.5k$ respectively. In the steady state, the free ends of rods

$A$ and

$C$ are at 100

$^{o}$ C and 0

$^{o}$ C respectively. Neglect loss of heat from the curved surfaces of rods.

The temperature of the junction between rods

$A$ and

$B$ is :

0%
$55.7^{o}C$
0%
$65.7^{o}C$
0%
$75.7^{o}C$
0%
$85.7^{o}C$

Explanation

Let temperature between

${A}$ and

${B}={T}_{1}$ and temperature between

${B}$ and

${C}={T}_{2}$

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

$\dfrac { { K }_{ A }A({ 100}-{ T }_{ 1 }) }{ l } =\dfrac { { K }_{ B }A({ T }_{ 1 }-{ T }_{ 2 }) }{ l } =\dfrac { { K }_{ C }A({ T }_{ 2 }-{ 0 }) }{ l }$
Given that:

${K}_{A}={2K}$ ,

${K}_{B}={K}$ ,

${K}_{C}={0.5K}$

${200}-{2}{{T}_{1}}={T}_{1}-{T}_{2}={0.5}{{T}_{2}}$

We find:

${T}_{1}=85.71 ^0C$

${T}_{2}=57.14 ^0C$

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 constanttemperatures of 27

$^{o}$ C and 0

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

$^{-1}$ K

$^{-1}$

The temperatures of the inner glass-air interface is

0%
$2.5^{o}C$
0%
$2.0^{o}C$
0%
$1.5^{o}C$
0%
$0.5^{o}C$

Explanation

${T}_{2}$ = temperature of glass-1 and air interface

${T}_{3}$ = temperature of air and glass-2 interface

The equation of heat flow are:

$\displaystyle \frac { { Q }_{ 1 } }{ t } =\frac { { K }_{ 1 }A({ T }_{ 1 }-{ T }_{ 2 }) }{ { d }_{ 1 } }=\frac { { 0.8 }{\times}{1}({ 27}-{ T }_{ 2 }) }{ 0.01 }$

$\displaystyle \frac { { Q }_{ 2 } }{ t } =\frac { { K }_{ 2 }A({ T }_{ 2 }-{ T }_{ 3 }) }{ { d }_{ 2} }=\frac { { 0.08 }{\times}{1}({T}_{2}-{ T }_{ 3 }) }{ 0.05 }$

$\displaystyle \frac { { Q }_{ 3 } }{ t } =\frac { { K }_{ 3 }A({ T }_{ 3 }-{ T }_{ 4 }) }{ { d }_{ 3 } }=\frac { { 0.8 }{\times}{1}({T}_{3}) }{ 0.01 }$

In steady state:

$\displaystyle \frac { { q }_{ 1 } }{ t }=\frac { { q }_{ 2 } }{ t }=\frac { { q }_{ 3 } }{ t }$

On equating

$({1}) ({2})$ and

${1} {3}$ , we find the following :

${1350}={51{T}_{2}}-{T}_{3}$ ----------(a)

${27}-{T}_{2}={T}_{3}$ ----------(b)

The temperature of outer glass interface,

${T}_{2}={26.48}^{0}C$
So, from equation (b), we find temperature of inner glass interface

${T}_{3}={27}-{{T}_{2}}={27}-{26.48}={0.52}^{0}C$

Three rods A,B and C have the same dimensions.Their conductivities are

$K_{A}, K_{B}$ and

$K_{C}$ respectively. A and B are placed end to end, with their free ends kept at certain temperature difference. C is placed separately with its ends kept at same temperature difference. The two arrangements conduct heat at the same rate

$K_{c}$ must be equal to

0%
$K_{A}+K_{B}$
0%
$\dfrac{K_{A}+K_{B}}{K_{A}K_{B}}$
0%
$\dfrac{1}{2}(K_{A}+K_{B})$
0%
$\dfrac{2K_{A}K_{B}}{K_{A}+K_{B}}$

Explanation

We know that

$\dfrac { Q }{ t } =\dfrac { KA(\Delta T) }{ L }$

so all rods having same dimensions,therefor for rod

${A} and {B}$

equivalent thermal conductivity is given as

${ K }_{ equivalent }=\dfrac { { K }_{ A }+{ K }_{ B } }{ { K }_{ A }{ K }_{ B } }$

$\dfrac { Q }{ t } =\dfrac { 2{ k }_{ a }{ k }_{ b } }{ ({ k }_{ a }+{ k }_{ b }) } \times \dfrac { A({ T }_{ A }-{ T }_{ B }) }{ 2L }$

now for rod

${C}$ we say

$\dfrac { Q }{ t } =\dfrac { {K}_{c}A({ T }_{ A }-{ T }_{ B }) }{ L }$ ,here it is mention that both are at same temperature difference.

so equating both of them, we find

${ K }_{ C }=\dfrac { 2{ K}_{ a }{ K }_{ b } }{ ({ K }_{ a }+{ K }_{ b }) }$

Three rods made of same material and having the same cross section and lengths have been joined as shown in the figure. The temperature at the junction of the rods will be :

0%
$45^{o}C$
0%
$60^{o}C$
0%
$30^{o}C$
0%
$20^{o}C$

Explanation

Let the temperature of the junction be

${\theta}$
So assume that total incoming heat at the junction will be equal to total outgoing heat.
So,

$({T}_{1}-{\theta})=({\theta}-{T}_{2})+({\theta}-{T}_{3})$

$({0}-{\theta})=({\theta}-{90})+({\theta-90})$

${\theta}={60}^{0}C$

$\dfrac{KA(\theta - 0)}{\ell} = \dfrac{KA(90-\theta)}{\ell} = \dfrac{KA(90-\theta)}{\ell}$

An electric kettle takes 4 A current at 220 V.
How much time will it take to boil 1 kg of water
from temperature

$20^{0}C ?$ The temperature of
boiling water is

$100^{0}C ?$

0%
12.6 min
0%
4.2min
0%
6.3 min
0%
8.4 min

Explanation

Heat required to boil water

$Q = mc \Delta T$

$= 1 \times 1 \times (100 20) = 80kcal$
Heat given by supply

$H = V.i.t = 220 \times 4 \times t$

$H = Q$

$\Rightarrow 220 \times 4 \times t = 80,000$

$t=\frac{80,000}{220 \times 4}$

$=\frac{1000}{11}sec$

$=\frac{1000}{11 \times 60}=1.5 min$

A vessel contains air and saturated vapor. The pressure of air is

$\mathrm{p}_{2}$ and

$\mathrm{p}_{1}$ is the S.V. P. On compressing the mixture to one-fourth of its original volume, what is the increase in pressure of the mixture?

0%
$2\mathrm{p}_{1}$
0%
$2\mathrm{p}_{2}$
0%
$3\mathrm{p}_{1}$
0%
$3\mathrm{p}_{2}$

Explanation

Initially total pressure

$=(P_{1}+P_{2})$
When compressed to

$\dfrac{1}{4}$ of its volume, the pressure of air becomes

$4P_{2}.$
Pressure of saturated vapour remains the same.
Total pressure after compression

$=4P_{2}+P_{1}$
Difference in pressure

$=(4P_{2}+P_{1})-(P_{1}+P_{2})$

$=3P_{2}$

A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat

$Q$ in time

$t$ . The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time

$t$ ?

0%
$\dfrac{Q}{4}$
0%
$\dfrac{Q}{16}$
0%
$2Q$
0%
$\dfrac{Q}{2}$

Three perfect gases at absolute temperatures

$\mathrm{T}_{1},\ \mathrm{T}_{2}$ and

$\mathrm{T}_{3}$ are mixed. The masses of molecules are

$\mathrm{m}_{1},\ \mathrm{m}_{2}$ and

$\mathrm{m}_{3}$ and the number of molecules are

$\mathrm{n}_{1},\ \mathrm{n}_{2}$ and

$\mathrm{n}_{3}$ respectively. Assuming no loss of energy, the final temperature of the mixture is:

0%
$\displaystyle \frac{(\mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3})}{3}$
0%
$\displaystyle \frac{\mathrm{n}_{1}\mathrm{T}_{\mathrm{l}}+\mathrm{n}_{2}\mathrm{T}_{2}+\mathrm{n}_{3}\mathrm{T}_{3}}{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}}$
0%
$\displaystyle \frac{\mathrm{n}_{1}\mathrm{T}_{1}^{2}+\mathrm{n}_{2}\mathrm{T}_{2}^{2}+\mathrm{n}_{3}\mathrm{T}_{3}^{3}}{\mathrm{n}_{1}\mathrm{T}_{1}+\mathrm{n}_{2}\mathrm{T}_{2}+\mathrm{n}_{3}\mathrm{T}_{3}}$
0%
$\displaystyle \frac{\mathrm{n}_{1}^{2}\mathrm{T}_{1}^{2}+\mathrm{n}_{2}^{2}\mathrm{T}_{2}^{2}+\mathrm{n}_{3}\mathrm{T}_{3}^{3}}{\mathrm{n}_{1}\mathrm{T}_{1}+\mathrm{n}_{2}\mathrm{T}_{2}+\mathrm{n}_{3}\mathrm{T}_{3}}$

Explanation

Number of moles of first gas

$=\displaystyle \frac{\mathrm{n}_{1}}{\mathrm{N}_{\mathrm{A}}}$
Number of moles of second gas

$=\displaystyle \frac{\mathrm{n}_{2}}{\mathrm{N}_{\mathrm{A}}}$

Number of moles of third gas

$=\displaystyle \frac{\mathrm{n}_{3}}{\mathrm{N}_{\mathrm{A}}}$

If no loss of energy then

$\mathrm{P}_{1}\mathrm{V}_{1}+\mathrm{P}_{2}\mathrm{V}_{2}+\mathrm{P}_{3}\mathrm{V}_{3}=$

$PV$

$\displaystyle \frac{\mathrm{n}_{1}}{\mathrm{N}_{\mathrm{A}}}\mathrm{R}\mathrm{T}_{1}+\frac{\mathrm{n}_{2}}{\mathrm{N}_{\mathrm{A}}}\mathrm{R}\mathrm{T}_{2}+\frac{\mathrm{n}_{3}}{\mathrm{N}_{\mathrm{A}}}\mathrm{R}\mathrm{T}_{3}$

$\displaystyle =\frac{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}}{\mathrm{N}_{\mathrm{A}}}\mathrm{R}\mathrm{T}_{mix}$

$\implies \displaystyle \mathrm{T}_{mix}=\frac{\mathrm{n}_{1}\mathrm{T}_{1}+\mathrm{n}_{2}\mathrm{T}_{2}+\mathrm{n}_{3}\mathrm{T}_{3}}{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}}$ .

Statement-1
Radiation involves transfer of heat by electromagnetic waves.
Statement-2
Electromagnetic waves do not required any material medium for propagation.

0%
Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1.
0%
Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Staement-1.
0%
Statement-1 is true, Statement-2 is false.
0%
Statement-1 is false, Statement-2 is true.

Consider two rods of same length and different specific heats (S

$_{1}$ , S

$_{2}$ ), conductivity (K

$_{1}$ , K

$_{2}$ ) and area of cross-sections (A

$_{1}$ , A

$_{2}$ ) and both having temperature T

$_{1}$ and T

$_{2}$ at their ends. If rate of loss of heat due to conduction is equal, then:

0%
$K_{1}A_{1}=K_{2}A_{2}$
0%
$\frac{K_{1}A_{1}}{S_{1}}=\frac{K_{2}A_{2}}{S_{2}}$
0%
$K_{2}A_{1}=K_{1}A_{2}$
0%
$\frac{K_{2}A_{1}}{S_{2}}=\frac{K_{1}A_{2}}{S_{1}}$

Explanation

Given:

For rod 1

Cross-sectional area

$=A_1$

Specific Heat

$=S_1$

Temperaturedifference between the ends of the rod

$=T_1-T_2$

Thermal Conductivity

$= K_1$

For rod 2:

Cross-sectional area

$=A_2$

Specific Heat

$=S_2$

Temperature difference at ends of the rod

$=T_1-T_2$

Thermal Conductivity

$= K_2$

By Fourier's law of heat conduction:

For 1st rod

$\dfrac{dQ_1}{dt}=-K_1\dfrac{A_1dT}{L}$

For 2nd rod:

$\dfrac{dQ_2}{dt}=-K_2\dfrac{A_2dT}{L}$

The rate of heat transfer is same;

Thus,

$-K_1\dfrac{A_1dT}{L}=-K_2\dfrac{A_2dT}{L}$

$K_1A_1=K_2A_2$

A cylinder of radius r and thermal conductivity

$K_1$ is surrounded by a cylindrical shell of inner radius r and outer radius 2 r, whose thermal conductivity is

$K_2$ . There is no loss of heat across cylindrical surfaces, when the ends of the combined system are maintained at temps.

$T_1$ and

$T_2$ . The effective thermal conductivity of the system, in the steady state is

0%
$\cfrac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}$
0%
$K_1+K_2$
0%
$\cfrac{{{K_1} + 3{K_2}}}{4}$
0%
$\cfrac{{3{K_1} + 3{K_2}}}{4}$

If rc and rs respectively represents cuticular and stomatal resistance, the total resistance (R) could be expressed as

0%
$R=rc+rs$
0%
$R=r-rs$
0%
$\frac{1}{R}=\frac{1}{rc}+\frac{1}{rs}$
0%
$\frac{r}{R}=\frac{1}{rc}-\frac{1}{rc}$

$100$

$\mathrm{g}$ of water is heated from

$30^{\mathrm{o}}\mathrm{C}$ to

$50^{\mathrm{o}}\mathrm{C}$ . Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is

$4184$

$\mathrm{J}/\mathrm{K}\mathrm{g}/\mathrm{K}$ )

0%
$4.2$ $\mathrm{k}\mathrm{J}$
0%
$8.4$ $\mathrm{k}\mathrm{J}$
0%
$84$ $\mathrm{k}\mathrm{J}$
0%
$2.1$ $\mathrm{k}\mathrm{J}$

Explanation

$\Delta Q= \Delta U +\Delta W$

Since

$\Delta W =0$ ,

$\Delta Q= \Delta U$

$\Delta U=ms \Delta T= 0.1\times 4.184 \times 20=8.368 kJ$

There are two lead spheres, the ratio c being

$1 : 2$ . If both are at the same tempe then ratio of heat contents is

0%
$1 : 1$
0%
$1 : 2$
0%
$1 : 4$
0%
$1 : 8$

Explanation

We know that the heat content is proportional to the mass and specific heat capacity of the substance as well as temperature. As both the spheres are of the same material and are at the same temperature, heat content will be dependent on mass only.

$m = \rho \times v$

where $\rho$ is the density of the sphere and $v$ the volume.

And $v$ = $\dfrac {4 \pi r^3}{3}$

The given radius ratio is $1:2$ .

Hence heat contents ratio is in the ratio $1:8$

Two plates of same area are placed in contact. Their thickness as well as thermal conductivities are in the ratio 2 :The outer surface of one plate is maintained at

$10^\circ C$ and that of other at

$0^\circ C$ . What is the temp.of the common surface?

0%
$0^\circ C$
0%
$2.5^\circ C$
0%
$5^\circ C$
0%
$6.5^\circ C$

Explanation

Let $A$ be the area of the both the plates.

Given ,

$\dfrac{k_1}{k_2}$ = $\dfrac {2}{3}$ ------- $(i)$

where $k_1$ and $k_2$ are the thermal conductivities of plate 1 and 2 respectively.

And

$\dfrac{l_1}{l_2}$ = $\dfrac {2}{3}$ ------- $(ii)$

where $l_1$ and $l_2$ are the thermal conductivities of plate 1 and 2 respectively.

Let the temperature of the common surface be $T$

Equating the rate of heat transfers through both the plates

$\dfrac{k_1 A (10-T)}{L_1}$ = $\dfrac{k_2 A (T-0)}{L_2}$

Using $(i)$ and $(ii)$

We have ,

$10-T = T-0$

Hence , $T = 5^0C$

Producers gas is a mixture of

0%
Carbon monoxide and nitrogen gas
0%
Carbon monoxide and hydrogen gas
0%
Carbon monoxide and water vapour
0%
Carbon monoxide and nitrous oxide

An ideal gas can be expanded from an initial state to a certain volume through two different processes (i)

$PV^2 = constant\:and\:(ii)P = KV^2$ where K is a positive constant. Based on the given situation, choose the correct statements

0%
Final temperature in (ii) will be greater than in (i)
0%
Final temperature in (ii) will be less than in (i)
0%
Total heat given to the gas in case (ii) is greater than in (i)
0%
Total heat given to the gas in case (ii) is less than in (i)

Explanation

Since, the question says that this is an ideal gas,
the molecules will always follow the ideal gas equation irrespective of the law that is used for the transformation.
Hence, we use P =

$\dfrac{nRT}{V}$
This gives us a relation between T and V that can show us the variation of T as the gas is expanded.
TV = C {for the first case}
T = C

${V}^{3}$ {for the second case}
Now, in both the cases,
V is increasing as expansion is occuring.
In first case T will decrease, whereas in the second case T will increase.

In a room there are four objects a wooden dish, steel dish, glass dish and a copper dish. If a fire is lit in the room so that it burns at 300 degrees centigrade, and is equidistant from all the four dishes, then after a long time the dishes can be listed in the increasing order of temperature. Which is the correct order of temperature of dishes?

0%
wooden dish > steel dish > glass dish > copper dish
0%
steel dish > copper dish > glass dish > wooden dish
0%
copper dish > steel dish > glass dish > wooden dish
0%
none of the above

At pressure P and absolute temperature T a mass M of an ideal gas fills a closed container of volume V. An additional mass 2M of the same gas is added into the container and the volume is then reduced to

$\dfrac{v}{3}$ and the temperature to

$\dfrac{T}{3}.$ The pressure of the gas will now be :

0%
$\dfrac{P}{3}$
0%
$P$
0%
$3 P$
0%
$9 P$

Explanation

$M_{0}$ is molecular mass of the gas then for initial condition PV

$=\dfrac{M}{M_{0}}.RT ........(1)$
After 2M mass has been added

${P}'.\dfrac{V}{3}=\dfrac{3M}{M_{0}}.R.\frac{T}{3} .........(2)$
By dividing (2) and (1)

${P}'=3P$

The rate of heat conduction is proportional to the cross-sectional area and temperature
gradient (temperature difference per unit length). On a typical day during the World Cup
tournament in South Africa, the air in a room is heated to

$25^{o}C$ while the air outside is

$-2^{o}C$ . The area of the window of the room is

$2 m^{2}$ and it is made of crown glass with
thickness 2 mm and thermal conductivity

$1.0 WK^{-1}m^{-1}$ . What is the heat power loss
through the window?

0%
1.2 kW
0%
2.7 kW
0%
27 kW
0%
50 kW

Choose the correct statements from the following:

0%
Good reflectors are good emitters of thermal radiations.
0%
Burns caused by water at $100^{\circ}C$ are more severe than those caused by steam at $100^{\circ}C$ .
0%
All bodies emit thermal radiations at all temperatures greater than 0 K.
0%
It is impossible to construct a heat engine of 100% efficiency.

Explanation

(A) Reflectivity (

$e$ ) +Emissivity (

$\epsilon$ ) =1

$\therefore$ a good reflector is a poor emitter.
(B) Steam will contain more heat energy due to addition of

$540^oC$ of latent heat of vaporization though both are at same temp of

$100^oC$ . Hence, the burn will be severe in case of steam compared to water.
(C)All bodies emit thermal radiations at all temperatures greater than

$0 K$ .
(D)Efficiency is equal to

$1-\dfrac{T_2}{T_1}$ where

$T_2$ is the temp of sink and

$T_1$ is the temp of source. since

$T_2$ cannot be zero, efficiency cannot be

$100\%$

A diatomic gas is filled inside a conducting cylinder. Now we push the piston slowly to make volume of gas half of initial. Pick correct statements

0%
Pressure of gas Increases because there is more average change in linear momentum of molecule in each collision
0%
Pressure force on side wall of container increased
0%
Pressure force on piston is increased
0%
More molecules collide with piston per unit time

A rigid container has a hole in its wall. When the container is evacuated, its weight is 100 gm. When someair is filled in it at 27C, its weight becomes 200 gm. Now the temperature of air inside is increased by

$\Delta$ T, the weight becomes 150 gm.

$\Delta$ T should be :

0%
$27^{\circ}$
0%
$\dfrac{27^{\circ}} {4}$ C
0%
$300^{\circ}$
0%
$327^{\circ}$

Explanation

Initially mass of air is 200 - 100 = 100gm. finally mass of air is 150 - 100 = 50 gm. As there is a hole in the wall, pressure inside the container will remain constant =

$P_{0}$

$PV = nRT \Rightarrow T \propto \dfrac{1} {n}$
as number of moles of gas is halved, the temperature should be doubled (in K)

$T_{i} = 300K$ So,

$T_{f} = 600 K \Rightarrow \Delta T = 300 K = 300^{\circ}C$

A cuboid

$ABCDEFGH$ is anisotropic with

$\alpha _{x}=1 \times 10^{-5} /^\circ C,\:\alpha _{y}=2 \times 10^{-5} /^\circ C,\:\alpha _{z}=3 \times 10^{-5} /^\circ C$ . Coefficient of superficial expansion of faces can be

0%
$\beta _{ABCD}=5 \times 10^{-5}/ ^\circ C$
0%
$\beta _{BCGH}=4 \times 10^{-5}/ ^\circ C$
0%
$\beta _{CDEH}=3 \times 10^{-5}/ ^\circ C$
0%
$\beta _{EFGH}=2 \times 10^{-5}/ ^\circ C$

Explanation

Coefficient of superficial expansion of face ABCD=

$\alpha_x+\alpha_z=4\times 10^{-5}/^{\circ}C$

Coefficient of superficial expansion of face BCGH=

$\alpha_y+\alpha_z=5\times 10^{-5}/^{\circ}C$

Coefficient of superficial expansion of face CDEH=

$\alpha_x+\alpha_y=3\times 10^{-5}/^{\circ}C$

Coefficient of superficial expansion of face EFGH=

$\alpha_x+\alpha_z=4\times 10^{-5}/^{\circ}C$

A vessel of volume 4 litres contains a mixture of 8 g of O

$_2$ , 14 g of N

$_2$ and 22g of CO

$_2$ at 27

$^o$ C. The pressure exerted by the mixture is

0%
$5 \times 10^6 N/m^2$
0%
$6 \times 10^3 N/m^2$
0%
10 atmosphere
0%
$7.79 \times 10^5 N/m^2$

Explanation

Moles of

$O_2=\dfrac{8}{32}=\dfrac{1}{4}$

Moles of

$N_2=\dfrac{14}{28}=\dfrac{1}{2}$

Moles of

$CO_2=\dfrac{22}{44}=\dfrac{1}{2}$

Thus total moles in the mixture=

$\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{5}{4}$

Using Ideal Gas Equation,

$PV=nRT$

$\implies P=\dfrac{nRT}{V}=\dfrac{\dfrac{5}{4}\times 8.3\times 300}{4\times 10^{-3}}N/m^2\approx 7.79\times 10^{5}N/m^2$

Sunlight contains ultraviolet radiations which makes it feel hot.

0%
True
0%
False

A vessel contains

$1$ mole of O

$_2$ (molar mass

$32 gm$ ) at a temperature

$T$ . The pressure is

$P$ . An identical vessel containing

$1$ mole of He (molar mass

$4 gm$ ) at a temperature

$2T$ has a pressure :

0%
P
0%
$\displaystyle \frac{P}{8}$
0%
2P
0%
8P

Explanation

Using

$PV=nRT$

$\therefore P_{He} = 2 P$ as temperature of He is doubled.

The temperature of the sun, if pressure is

$1.4 \times 10^9$ atm, density is

$1.4 gcm^{-3}$ and average molecular weight is 2, will be

$[Given\ R = 8.4 J mol^{-1} K^{-1}]$

0%
$1.2 \times 10^7 K$
0%
$2.4 \times 10^7 K$
0%
$0.4 \times 10^7 K$
0%
$0.2 \times 10^7 K$

Explanation

$PV = nRT n = \displaystyle \frac{m}{M}$ and

$\rho = \displaystyle \frac{m}{V}$
or

$\displaystyle T = \frac{PV}{nR} = \frac{PM}{\rho R}$

$= \displaystyle \frac{1.4 \times 10^9 \times 1.01 \times 10^5 \times 2 \times 10^{-3}}{1.4 \times 10^3 \times 8.4}$

$= 2.4 \times 10^7 K$

What is new gauge pressure?

0%
1.82 atm
0%
1.70 atm
0%
1.92 atm
0%
none

Explanation

$\displaystyle \frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$
or

$\displaystyle P_2 = \frac{2.72 \times 150 \times 318}{278 \times 159}$

$= 2.94 atm$
Gauge pressure

$= 2.94 - 1.02$

$= 1.92 atm$

A barometer tube 90 cm long contains some air above mercury. The reading is 74.8 cm when true atmospheric pressure is 76 cm and temperature is 30

$^o$ C. If the reading is observed to be 75.4 cm on a day when temperature is 10

$^o$ C, then find the true pressure.

0%
76.07 cm
0%
75.6 cm
0%
76.57 cm
0%
77.123 cm

Explanation

Let A be the area of the cross-section

$V_1 = (90 - 74.8) A = 15. 2 A cm^3$

$P_1 = 76 - 74.8 = 1.2$ cm of Hg

$P_2 = (P - 75.4)$ cm of Hg

$V_2 = (90 - 75.4) A = 14.6 A cm^3$

$\displaystyle \frac{P_1V_1}{T_1} = \frac{P_2 V_2}{T_2} = \frac{1.2 \times 15.2 \times 283}{303 \times 14.6} = 75.4 + 1.17$

$= 76.57$ cm of Hg

Two containers of equal volume contain the same gas at pressures

$P_1$ and

$P_2$ and absolute temperatures

$T_1$ and

$T_2$ , respectively. On joining the vessels, the gas reaches a common pressure

$P$ and common temperature

$T$ . The ratio

$\displaystyle\frac{P}{T}$ is equal to

0%
$\displaystyle\frac{P_1}{T_1}+\frac{P_2}{T_2}$
0%
$\displaystyle\frac{P_1T_1+P_2T_2}{{(T_1+T_2)}^2}$
0%
$\displaystyle\frac{P_1T_2+P_2T_1}{{(T_1+T_2)}^2}$
0%
$\displaystyle\frac{P_1}{2T_1}+\frac{P_2}{2T_2}$

Explanation

$P_1V=n_1RT_1$

$P_2V=n_2RT_2$

where V is the volume of each vessel.

When the vessels are joined,

$P(2V)=(n_1 + n_2)RT$

$\therefore \dfrac{P}{T}= \dfrac{1}{2}\dfrac{(n_1 + n_2)R}{V}=\dfrac{1}{2}(\dfrac{P_1}{T_1} + \dfrac{P_2}{T_2})=\dfrac{P_1}{2T_1} + \dfrac{P_2}{2T_2}$

One mole of an ideal gas undergoes a process

$P = \displaystyle \frac{P_o}{1 + \left ( \displaystyle\frac{V}{V_o} \right )^2}$ , where

$P_o$ and

$V_o$ are constants. Find the temperature of the gas when

$V = V_o$ .

0%
$\displaystyle \frac{P_o V_o}{R}$
0%
$\displaystyle \frac{2P_o V_o}{3R}$
0%
$\displaystyle \frac{2 P_o V_o}{R}$
0%
$\displaystyle \frac{P_o V_o}{2R}$

Explanation

Since the gas is ideal, therefore

$P_o V_o = RT$

Using the relation given when

$V = V_o$

$P = \displaystyle \frac{P_o}{2}$

Thus we get

$\displaystyle \frac{P_o}{2} (V_o) = RT$

$T = \displaystyle \frac{P_oV_o}{2R}$ .

The African bombardier beetle stenaptinus insignis can emit a jet of defensive spray from the movable tip of its abdomen. The beetle's body has reservoirs of two different chemicals. When the beetle is disturbed, these chemicals are combined in a reaction chamber producing a compounds that is warmed from

$20^o$ C to

$100^o$ C by the heat of reaction. The high pressure produced allows the compound to the sprayed out at speeds 19 ms

$^{-1}$ scaring away predators of all kinds. Assume specific heat of two chemicals and the spray to be same as that of water

$[4.19 \times 10^3 J (kgK)^{-1}]$ and initial temperature of chemicals to be

$20^o$ C. How many times does the pressure increase?

0%
1.5
0%
2.3
0%
1.9
0%
1.28

Explanation

$\displaystyle \frac{P_1}{T_1} = \frac{P_2}{T_2}$
or

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{373}{293} = 1.28$

The correct graph between PV and P of one mol of gas at constant temperature will be

Explanation

For an ideal gas

$PV=nRT$

Thus for a constant temperature,

$PV=constant$

Thus

$PV$ does not change even when P changes.

A gas is filled in a container at any temperature and at pressure 76 cm of Hg. If at the same temperature the mass of gas is increased by 50% then the resultant pressure will be

0%
38 cm of Hg
0%
76 cm of Hg
0%
114 cm of Hg
0%
152 cm of Hg

Explanation

$PV=nRT$

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

$P=\dfrac{m}{M}\dfrac{RT}{V}$ also given that P=76cm of Hg

Given that mass of the gas is increased by 50%

Now

$P_{1}=\left(\dfrac{m+\dfrac{m}{2}}{M}\right)\dfrac{RT}{V}$

$=\dfrac{3}{2}\dfrac{m}{M}\dfrac{RT}{V}$

$=\dfrac{3}{2}P$

$\dfrac{3}{2}\times76$

$114cm\ of\ Hg$

In the gas equation

$PV= RT$ , V is the volume of

0%
1 mol of gas
0%
1 g of gas
0%
gas
0%
1 litre of gas

Explanation

For an Ideal Gas,

$PV=nRT$

Here

$V$ is the volume of n moles of gas.

Thus for

$PV=(1)RT$ ,

$V$ is the volume of 1 mol of gas.

An enclosure of volume 3 litre contains 16g of oxygen, 7g of nitrogen and 11g of carbondioxide at 27

$^o$ C. The pressure exerted by the mixture is approximately

0%
9 atmosphere
0%
8.3 atmosphere
0%
3 atmosphere
0%
1 atmosphere

Explanation

Total number of moles = number of moles of oxygen + number of moles of nitrogen + number of moles of carbon-di-oxide

$\dfrac { 16 }{ 32 } +\dfrac { 7 }{ 14 } +\dfrac { 11 }{ 44 } =1mole$

Molecular weight of Carbon-di-oxide = 44g

Molecular weight of Oxygen =32g

Molecular weight of nitrogen=14g

Using ideal gas equation,

$P=\dfrac { nRT }{ V } =\dfrac { 1 \times8.3 \times300 }{ (3\times{ 10 }^{ -3 }) } =8.3atm(nearly)$

Hence, Option B is correct.

The pressure of a gas in a container is 10

$^{-11}$ pascal at 27

$^o$ C. The number of molecules per unit volume of vessel will be

0%
$6 \times 10^{23} cm^{-3}$
0%
$2.68 \times 10^{19} cm^{-3}$
0%
$2.5 \times 10^{6} cm^{-3}$
0%
$2400 cm^{-3}$

Explanation

$10^5 Pa \equiv 6.023 \times 10^{23}$ (equivalence equation according to information given in question)

$10^{-11} Pa \equiv 6.023 \times 10^7$

$22400 cc \Rightarrow 6.023 \times 10^7$

$1 cc \Rightarrow \displaystyle \frac{6.023 \times 10^7}{2400} =2400 cm^{-3}$

Equal masses of N

$_2$ and O

$_2$ gases are filled in vessels 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

Moles of

$N_2=n_A\dfrac{m}{28}$

Moles of

$O_2=n_B\dfrac{m}{32}$

Given that

$V_B=2V_A$

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

$\implies \dfrac{P_A}{P_B}=\dfrac{n_A}{n_B}\dfrac{V_B}{V_A}=\dfrac{16}{7}$

The dimension of universal gas constant R are

0%
$M^2 L^2 T^{-2}$
0%
$ML^2 T^{-2} \theta^{-1}$
0%
$M^2 L^2 T^{-2} \theta^{-2}$
0%
$MLT^{-2} \theta^{-2}$

Explanation

using ideal gas equation

$R=\dfrac { PV }{ nT }$

dimensions of pressure

$M{ L }^{ -1 }{ T }^{ -2 }$

dimensions of Volume

${L}^{3}$

number of moles is dimensionless

dimensions of Temperature

$\theta$

so, dimensions of R

$\dfrac { (M{ L }^{ -1 }{ T }^{ -2 })({ L }^{ 3 }) }{ \theta } =M{ L }^{ 2 }{ T }^{ -2 }{ \theta }^{ -1 }$

Hence,Option B is correct

In outer space there are 10 molecules per cm

$^3$ on an average and the temperature there is 3K. The average pressure of this light gas is

0%
$10^5 Nm^{-2}$
0%
$5 \times 10^{-14} Nm^{-2}$
0%
$0.4 \times 10^{-16} Nm^{-2}$
0%
$4.14 \times 10^{-16} Nm^{-2}$

Explanation

$P= \displaystyle \frac{nKT}{V} = \frac{10 \times 1.38 \times 10^{-23} \times 3}{10^{-6}}$

$= 4.14 \times 10^{-16} Nm^{-2}$

A cylinder contains 2kg of air at a pressure of 10

$^5$ Pa. If 2 kg more air is pumped into it, keeping the temperature constant, the pressure will be

0%
$10^{10} Pa$
0%
$2 \times 10^5 Pa$
0%
$10^5 Pa$
0%
$0.5 \times 10^5 Pa$

Explanation

Since both volume and Temperature is constant,

$\dfrac { P }{ n } =constant=\dfrac { P }{ m } \\ n=no.of\quad moles=\dfrac { mass\quad of\quad air(m) }{ Molecular\quad weight\quad of\quad water(M) }$

$\dfrac { { 10 }^{ 5 } }{ 2 } =\dfrac { P }{ (2+2) } \\ \Rightarrow P=2\times{ 10 }^{ 5 }Pa$

Hence, Option B is correct

One mole of a gas at a pressure 2 Pa and temperature 27

$^o$ C is heated till both pressure and volume are doubled. What is the temperature of the gas?

0%
1200 K
0%
900 K
0%
600 K
0%
300 K

Explanation

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

$\dfrac{PV}{nR}$

Given, that pressure and volume are doubled i.e,

$P_1=2P$ ,

$V_1=2V$ .

$T_1=\dfrac{P_1V_1}{nR}$

$=\dfrac{\left( 2P \times 2V\right)}{nR}$

$=4T$

Given that T=27

$^o$ C i.e, 300K

$T_1=4\times 300$

$=1200K$

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

0%
Kinetic energy
0%
Momentum
0%
Density
0%
Speed

The temperature of water at the surface of a deep lake is

$2^{\circ}C$ . The temperature expected at the bottom is

0%
$0^{\circ}$
0%
$2^{\circ}$
0%
$4^{\circ}$
0%
$-6^{\circ}$

Explanation

Water exhibits an anomalous behaviour; it contracts on heating between

$0 ^{ o }{ C }$ and

$4 ^{ o }{ C }$ . When the surface of the lake cools down above

$4 ^{ o }{ C }$ the cool water at the surface flows to the bottom because of its greater density. But when the surface temperature reaches to

$2 ^{ o }{ C }$ , the water near the surface is less dense than the warmer water below. Hence the downword flow ceases, and the water near the surface remains colder than that at the bottom.
So the temperature expected at the bottom is

$4 ^{ o }{ C }$
option (C) is the correct answer.

The pressure of a gas filled in a closed vessel increases by 0.4%. When temperature is increases by 1

$^o$ C the initial temperature of the gas is

0%
250 $^o$ C
0%
250 K
0%
250 $^o$ F
0%
2500 $^o$ C

Explanation

Since volume is constant i.e number of moles is constant.

$P_1$ and

$T_1$ are the initial pressure and temperature.

According to the question,

${ P }_{ 2 }=1.004{ P }_{ 1 }\\ { T }_{ 2 }={ T }_{ 1 }+1$

$\therefore \dfrac { { P }_{ 1 } }{ { T }_{ 1 } } =\dfrac{P_2}{T_2}=\dfrac { (1+0.004){ P }_{ 1 } }{ { T }_{ 2 } } \\ \Rightarrow { T }_{ 2 }=1.004{ T }_{ 1 }=({ T }_{ 1 }+1)$

$0.004T_1 = 1$

$\\ \Rightarrow { T }_{ 1 }=250K$

Hence, Option B is correct

CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 6 - 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 6 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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