MCQ Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 12

One likes to sit under sunshine in the winter season, because

0%
The air surrounding the body is hot by which body gets heat
0%
We get energy by the sun
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We get heat by conduction by sun
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None of the above

Two rods of the same length and cross-section are joined along the length. Thermal conductivities of the first and second rod are

$K_1$ and

$K_2$ . The temperature of the free ends of the first and second rods are maintained at

$\theta_1$ and

$\theta_2$ respectively. The temperature of the common junction is :

0%
$\dfrac{\theta_1+\theta_2}{2}$
0%
$\dfrac{K_1K_2}{K_1+K_2}(\theta_1+\theta_2)$
0%
$\dfrac{K_1\theta_1+K_2\theta_2}{K_1+K_2}$
0%
$\dfrac{K_2\theta_1+K_1\theta_2}{K_1+K_2}$

Explanation

At steady-state, the rate of heat flow for both blocks will be the same :
i.e.

$\dfrac{K_1A(\theta_1-\theta)}{l_1}= \dfrac{K_2A(\theta-\theta_2)}{l_2}\,(given\,l_1=l_2)$

$\Rightarrow K_1A(\theta_1-\theta)= K_2A(\theta-\theta_2)\,\Rightarrow \theta=\dfrac{K_1\theta_1+K_2\theta_2}{K_1+K_2}$
1719864_1813808_ans_b62110dcab364d23b040a917da4bddc1.jpg

1719864_1813808_ans_b62110dcab364d23b040a917da4bddc1.jpg

Two cylinders P and Q have the same length and diameter and are made of different materials having thermal conductivities in the ratio

$2: 3$ . These two cylinders are combined to make a cylinder. One end of P is kept at

$100 ^\circ C$ and another end of Q at

$0 ^\circ C$ . The temperature at the interface of P and Q is

0%
$30 ^\circ C$
0%
$40 ^\circ C$
0%
$50 ^\circ C$
0%
$60 ^\circ C$

Explanation

Temperature of interface

$\theta=\dfrac{K_1\theta_1+K_1\theta_1}{K_1+K_2}$ ,
where

$K_1$ =

$2K$ and

$K_2$ =

$3K$

$\Bigg(\therefore \dfrac{K_1}{K_2}=\dfrac{2}{3}\Bigg)$

$\Rightarrow \theta=\dfrac{2K \times 100+3K \times 0}{2K+3K}=\dfrac{200 \ K}{5 \ K}=40 ^\circ C$

A thermos flask is polished well

0%
To make attractive
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For shining
0%
To absorb all radiations from outside
0%
To reflect all radiations from outside

Three rods of the same dimension have thermal conductivities

3K,2K3K,2K. The temperature of their junction is

0%
$60 ^\circ C$
0%
$70 ^\circ C$
0%
$50 ^\circ C$
0%
$35 ^\circ C$

Explanation

Let the temperature of junction be

θθ then according to the following figure.

$H=H_1+H_2$

$\dfrac{3K \times A(100-\theta)}{l}=\dfrac{2KA(\theta-50)}{l}+\dfrac{2KA(\theta-20)}{l}$ \

$300-3 \theta=3 \theta -120$

$\theta=70$

1731945_1815566_ans_60b541f907f540d98edb1f2c48a99cc6.jpg

The layers of the atmosphere are heated through

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Convection
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Conduction
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Radiation
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(b) and (c) both

A body of length

$1m$ having a cross-sectional area

$0.75m^2$ has heat flow through it at the rate of

$6000 \ Joule/sec$ . Then find the temperature difference if

$K=200Jm^{-1}K^{-1}$

0%
$20 ^\circ C$
0%
$40 ^\circ C$
0%
$80 ^\circ C$
0%
$100 ^\circ C$

Explanation

$\dfrac{Q}{t}=\dfrac{KA \Delta \theta}{l}\,\Rightarrow 6000=\dfrac{200 \times 0.75 \times \Delta \theta}{1}$ .

$\therefore \Delta \theta=\dfrac{6000 \times 1}{2000 \times 0.75}=40 ^\circ C$

The temperature below which a gas should be cooled, before it can be liquified by pressure only is termed as

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The dew point
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The freezing point
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The saturation point
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The critical point

At constant volume the specific heat of a gas is

$\dfrac{3R}{2}$ , then the value of

$\gamma$ will be

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$\dfrac{3}{2}$
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$\dfrac{5}{2}$
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$\dfrac{5}{3}$
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None of these

At a given volume and temperature , the pressure of a gas

0%
Varies inversely as its mass
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Varies inversely as the square of its mass
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Varies linearly as its mass
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Is independent of its mass

At NTP the mass of one litre of air is

$1.293\,gm$ . The value of specific gas constant will be

0%
$0.29\,J/K-gm$
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$4.2\,J/K-gm$
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$8.3\,J/K-gm$
0%
$16.5\,J-gm$

For matter to exist simultaneously in gas and liquid phases

0%
The temperature must be $0\,K$
0%
The temperature must be less than $0^{\circ}\,C$
0%
The temperature must be less than the critical temperature
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The temperature must be less than the reduced temperature

The following sets of values for

$C_V$ and

$C_P$ of a gas has been reported by different students. The units are cal/gm-mole-K. Which of these sets is most reliable

0%
$C_V = 3 , C_P = 5$
0%
$C_V = 4 , C_P = 6$
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$C_V = 3 , C_P = 2$
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$C_V = 3 , C_P = 4.2$

Beryllium has roughly one-half the specific heat of water

$(H_2O)$ . Rank the quantities of energy input required to produce the following changes from the largest to the smallest. In your ranking, note any cases of equality.

0%
raising the temperature of 1 kg of $H_2O$ from $20^0C$ to $26^0C$
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raising the temperature of 2 kg of $H_2O$ from $20^0C$ to $23^0C$
0%
raising the temperature of 2 kg of $H_2O$ from $1^0C$ to $4^0C$
0%
raising the temperature of 2 kg of $H_2O$ from $1^0C$ to $4^0C$
0%
raising the temperature of 2 kg of $H_2O$ from $21^0C$ to $2^0C$

Explanation

Rankings (e) >(a) =(b) =(c) > (d). We think of the product

$mc\Delta T$ in each case, with c =1 for water and about 0.5 for beryllium: (a) 1. 1. 6 =6, (b) 2. 1. 3 =6, (c) 2. 1. 3 =6, (d) 2(0.5)3 =3, (e) >6 because a large quantity of energy input is required to melt the ice.

Ethyl alcohol has about one-half the specific heat of water. Assume equal amounts of energy are transferred by heat into equal-mass liquid samples of alcohol and water in separate insulated containers. The water rises in temperature by

$25^0C$ . How much will the alcohol rise in temperature?

0%
It will rise by $12^0C$ .
0%
It will rise by $25^0C$ .
0%
It will rise by $50^0C$ .
0%
It depends on the rate of energy transfer.
0%
It will not
rise in temperature.

Choose the correct alternative:
Water expands on reducing its temperature below _______

$^{\circ} C$ .

0%
$0$
0%
$4$
0%
$8$
0%
$12$

A 90 cm long barometer tube contains some air above the mercury. The reading is 74.5 cm when the true pressure is 76 cm at the temperature

$15^{0} C$ . If the reading is observed to be 75.8 cm on a day when the temperature is

$5^{0}C$ , then the true pressure is:

0%
77.38 cm of Hg
0%
75.8 cm of Hg
0%
74 cm of Hg
0%
80 cm of Hg

Explanation

Pressure = Actual pressure - Observed pressure

Volume = A(Barometer length - observed length)

$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \\ \Rightarrow \dfrac { (76-74.5)(90-74.5) }{ 288 } =\dfrac { (P-75.8)(90-75.8) }{ 278 } \\ \Rightarrow P=77.38 \ \ cm \ \ of \ Hg$

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity

$K$ and

$2K$ and thickness

$x$ and

$4x$ respectively are

$T_{2}$ and

$T_{1}$ (

$T_{2} > T_{1}$ ). The rate of heat transfer through the slab in a steady state is

$\displaystyle \left [\frac{A(T_{2}-T_{1})K}{x} \right ]f$ , where

$f$ equals to :

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

Explanation

Let T be the temperature of the interface.

$\dfrac{dQ}{dt}=KA\dfrac{dT}{dx}$
Since the rate of flow of heat across the interface is same,

$KA\dfrac {(T_2-T)}{x}=2KA \dfrac{T-T_1}{4x}$

$\Rightarrow T=\dfrac{2T_2+T_1}{3}$
Substituting the value of T in

$\dfrac{dQ}{dt}$ for slab of thickness

$4x=2K\times A(\dfrac{T-T_1}{dx})$

$=2KA\times \dfrac{( \dfrac{2T_2+T_1}{3}-T_1)}{4x}$

$=\dfrac{KA}{3x} \times (T_2-T_1)$
Comparing this with the given expression

$\dfrac{KA}{x} \times (T_2-T_1)f$
We get

$f=\dfrac{1}{3}$

Two plates

$A$ and

$B$ of equal surface area are placed one on top of the other to form a composite plate of the same surface area. The thickness of

$A$ and

$B$ are

$4.0$ cm and

$6.0$ cm respectively. The temperature of the exposed surface of plate

$A$ is

$-10$

$^{o}$ C and that of the exposed surface of plate B is

$10$

$^{0}$ C. Neglect heat loss from the edges of the composite plate, the temperature of the contact surface is

$T_{1}$ if the plates A and B are made of the same material and

$T_{2}$ if their thermal conductivities are in the ratio

$2:3$ then

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$T_{1}= -4^{0}C$
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$T_{1}= -2^{0}C$
0%
$T_{2}= -3^{0}C$
0%
$T_{2}= 0^{0}C$

Explanation

We use the resistance approach for this problem.

The heat resistance is given by,

$R = \dfrac{L}{KA}$

If the two plates are made of the same plate,

$\dfrac{{K}_{a} A ({T}_{1} + 10)}{4} = - \dfrac{{K}_{b} A ({T}_{1} - 10)}{6}$

Solving this for

${T}_{1}$ gives, it equal to -2

Similarly, if the two plates are made of materials with thermal conductivities in the ratio, 2:3,

$\dfrac{{K}_{a} A ({T}_{1} + 10)}{4} = - \dfrac{{K}_{b} A ({T}_{1} - 10)}{6}$

With

$\dfrac{{K}_{a}}{{K}_{b}} = \dfrac{2}{3}$

Solving this, gives

${T}_{2} = 0$

A closed container of volume 0.02 m $^3$ contains a mixture of neon and argon gases at a temperature of 27 $^{0}$ C and at a pressure of $1\times 10^{5}N/m^{2}$ . The total mass of the mixture is 28 g. If the gram molecular weights of neon and argon are 20 and 40 respectively, the masses of the individual gases in the container are respectively(assuming them to be ideal) [R = 8.314 J/mol K]

0%
16 gm, 12 gm
0%
4 gm, 24 gm
0%
6 gm, 22 gm
0%
12 gm, 16 gm

Explanation

$n=\frac { PV }{ RT } \\ n=\frac { 1\times { 10 }^{ 5 }\times 0.02 }{ 8.314\times 300 } \\ n=0.802\\ { n }_{ 1 }{ +n }_{ 2 }=0.802\\ { 20n }_{ 1 }{ +20n }_{ 2 }=16\quad \left( Multiplying\quad by\quad 20 \right) \\ { w }_{ 1 }{ +w }_{ 2 }=28\\ { n }_{ 1 }{ M }_{ neon }{ +n }_{ 2 }{ M }_{ argon }=28\\ 20{ n }_{ 1 }{ +40n }_{ 2 }=28\\ Solving\quad to\quad two\quad equations\quad we\quad get,\\ 20{ n }_{ 2 }=12\\ { n }_{ 2 }=0.6\\ \therefore { w }_{ 2 }=0.6\times 40=24gms\\ { w }_{ 1 }=28-24=4gms$

One end of a thermally insulated rod is kept at a temperature

$T_{1}$ and the other at

$T_{2}$ . The rod is composed of two sections of lengths

$l_{1}$ and

$l_{2}$ and thermal conductivities

$K_{1}$ and

$K_{2}$ respectively. The temperature at the interface of two sections is :

0%
$\displaystyle \frac{K_{2}l_{1}T_{1}+K_{1}l_{2}T_{2}}{K_{2}l_{1}+K_{1}l_{2}}$
0%
$\displaystyle \frac{K_{1}l_{2}T_{1}+K_{2}l_{1}T_{2}}{K_{1}l_{2}+K_{2}l_{1}}$
0%
$\displaystyle \frac{K_{1}l_{1}T_{1}+K_{2}l_{2}T_{2}}{K_{1}l_{1}+K_{2}l_{2}}$
0%
$\displaystyle \frac{K_{2}l_{2}T_{1}+K_{1}l_{1}T_{2}}{K_{1}l_{1}+K_{2}l_{2}}$

Explanation

Let

$T$ be the temp at the interface.

The rate of flow of heat remains same across the interface.

The equation for heat flow in conduction is:

$\dfrac {dQ}{dT} = -KA\dfrac {dT}{dx}$ where, A is the cross section area across which heat flows.

$\Rightarrow K_1A \dfrac {(T_1-T)}{l_1} =K_2A \dfrac {(T-T_2)}{l_2}$

$\Rightarrow \dfrac {(T_1-T)}{\dfrac {l_1}{K_1}} = \dfrac {(T-T_2)}{\dfrac {l_2}{K_2}}$

$\Rightarrow \dfrac {T_1-T}{T-T_2}=\dfrac {l_1K_2}{l_2K_1}$

$\Rightarrow T(K_1l_2+K_2l_1)= K_1l_2T_1+K_2l_1T_2$

$\Rightarrow T= \dfrac {K_1l_2T_1+K_2l_1T_2}{K_1l_2+K_2l_1}$

A smooth vertical tube with two different cross-sections is open at both ends. They are fitted with pistons of different areas of cross- section and each piston moves within a particular section. One mole of a gas enclosed between the pistons which are tied with non-stretchable threads. The difference in cross-sectional area of pistons

$10\mathrm{c}\mathrm{m}^{2}$ The mass of gas confined between pistons is 5kg. The outside pressure is 1 atmosphere

$=10^{5}N/m^{2}$ . By how many degrees must the gas between pistons be heated to shift the piston by 5 cm?

Given

$R=8.3$ .

0%
$0.9^{\mathrm{o}}\mathrm{C}$
0%
$1.2^{\mathrm{o}}\mathrm{C}$
0%
$1.5^{\mathrm{o}}\mathrm{C}$
0%
$0.5^{\mathrm{o}}\mathrm{C}$

Explanation

Let us consider

$s_1$ and

$s_2$ as the cross-section area of the lower and upper piston respectively.

Let

$P_o$ be the atmospheric pressure and P be the gas pressure.At equilibrium, the downward force and upward forces balances,

i.e.,

$P_o + Ps_1 + mg = Ps_2 + P_{o}s_{1}$ (

$Force = Pressure \times Area$ }

$P = P_{o} + \dfrac{mg}{s_1 - s_2}$ -- (1)

Let

$l_1$ and

$l_2$ be the length of string in lower and upper section.

Initial volume of the gas =

$l_1 s_1 + l_2 s_2$

When the pistons shifts by a distance

$x$ ,

Final volume =

$(l_1 - x)s_1 + (l_2 + x)s_2$

We have

$PV = RT$ . Applying this before heating and after heating the gas,

$P(l_1 s_1 + l_2 s_2) = RT$ -- (2)

$P[(l_1 -x)s_1 + (l_2 + x)s_2] = R (T + \Delta T)$

$P[(l_1 s_1 + l_2 s_2)+ (s_2 - s_1)x]= R (T + \Delta T)$

$[P(l_1 s_1 + l_2 s_2)+ P(s_2 - s_1)x]= R (T + \Delta T)$

$P(l_1 s_1 + l_2 s_2)+ P \Delta S x= R (T + \Delta T)$

Substituting for

$P(l_1 + l_2)$ from equation 2,

$RT+ P \Delta S x= R (T + \Delta T)$

Substituting for

$P$ from equation 1,

$RT+ P_{o} + \dfrac{mg}{s_1 - s_2} \Delta S x= R (T + \Delta T)$

$\implies \Delta T = \dfrac{1}{R} [P_o + mgx)$

$= \dfrac {1}{8.3} [10^5 \times 5 \times 10^{-2} \times 10 \times 0^{-4} + 5 \times 9.81 \times 5 \times 10^{-2} ]$

$= 0.9$

A horizontal uniform glass tube of 100cm length is sealed at both ends contains 10 cm mercury column in the middle the temperature and pressure of air on either side of mercury column are respectively 31 $^{0}$ C and 76cm of mercury if the air column at one end is kept at 0 $^{0}$ C and the other end at 273 $^{0}$ C the pressure of air which is at 0 $^{0}$ C is (in cm of Hg )

0%
$76$
0%
$88.2$
0%
$102.4$
0%
$122$

Explanation

Let

$x$ be the displacement of the mercury column on the side whose temperature is changed to

$0^{\circ}C$ . New pressure on both sides must be same so that the mercury column remains in equilibrium.

$\dfrac{PV}{T}$ is constant for air column.

$\implies \dfrac{P_{0}(45)}{273+31}=\dfrac{P_1 (45-x)}{273+0}$

and

$\dfrac{P_0 (45)}{273+31}=\dfrac{P_1(45+x)}{273+273}$

Therefore,

$x=15\ cm$ and thus

$P=102.4\ cm\ of\ Hg$

A closed hollow insulated cylinder is filled with gas at 0 $^{0}$ C and also contains an insulated piston of negligible weight and negligible thickness at the the middle point. The gas at one side of the piston is heated to 100 $^{0}$ C . If the piston moves 5cm, the length of the hollow cylinder is

0%
13.65 cm
0%
27.3 cm
0%
64.6 cm
0%
54.6 cm

Explanation

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

For section with constant temperature,

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

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

Similarly for section whose temperature increased,

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

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

$\implies L=64.6cm$

Two identical containers each of volume V $_{0}$ are joined by a small pipe. The containers contain identical gases at temperature T $_{0}$ and pressure P $_{0}$ . One container is heated to temperature 2T $_{0}$ while maintaining the other at the same temperature. The common pressure of the gas is P and n is the number of moles of gas in container at temperature 2T $_{0}$ .

0%
$P=2P_{0}$
0%
$P=\dfrac{4}{3}P_{0}$
0%
$n=\dfrac{2P_{0}V_{0}}{3RT_{0}}$
0%
$n=\dfrac{3P_{0}V_{0}}{2RT_{0}}$

Explanation

Since the total number of moles in the system remains the same before and after heating,

$\dfrac{P_0V_0}{T_0}+\dfrac{P_0V_0}{T_0}=\dfrac{PV_0}{T_0}+\dfrac{PV_0}{2T_0}$

$\implies P=\dfrac{4}{3}P_0$

Heat energy always flows from a body at low temperature to a body at high temperrature

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

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

0%
rms speed of a molecule
0%
density
0%
temperature
0%
average of magnitude of momentum

Explanation

Ideal gas equation

$\rightarrow PV=nRT$

Given that

$n_1V_1P$ are same.

So,

$T$ i.e, temperature is same for all ideal gas.

Hence, the answer is Temperature.

A metal ball immersed in water weighs

$w_1$ at

$5^{\circ}C\;and\;w_2$ at

$50^{\circ}C$ . The coefficient of cubical expansion of metal is less than that of water. Then

0%
$w_1\,>\,w_2$
0%
$w_1\,<\,w_2$
0%
$w_1=w_2$
0%
Insufficient information

Explanation

Weight of immersed ball = mg -buoyancy force

Buoyancy force is equal to weight of liquid displaced.

now when we increase the temperature both will expand but expansion in water will be more,hence the mass of volume displaced will decrease.

Hence,

$w_1<w_2$ .

A black body is at

$(727)^{0}C$ . It emits energy at a rate which is proportional to:

0%
$(727)^{4}$
0%
$(727)^{2}$
0%
$(1000)^{4}$
0%
$(1000)^{2}$

Explanation

$E= \sigma T^{4}\therefore E \infty (727+273)^{4}\Rightarrow E \infty \left ( 1000 \right )^{4}$

Three rods

$A,B$ and

$C$ have the same dimensions. Their thermal conductivities

$k_A, \space k_B$ and

$k_C$ respectively.

$A$ and

$B$ are placed end to end, with the free ends kept at a certain temperature difference.

$C$ is placed separately with its ends kept at same temperature difference. The two arrangements conduct heat at the same rate

$k_C$ equal to

0%
$k_A+k_B$
0%
$\displaystyle \frac{k_Ak_B}{k_A+k_B}$
0%
$\displaystyle \frac{1}{2}(k_A+k_B)$
0%
$\displaystyle 2\frac{k_Ak_B}{k_A+k_B}$

Explanation

Given that three rods A, B and C have same dimensions. Their thermal conductivities

$k_{A}$ ,

$k_{B}$ and

$k_{C}$ respectively. A and B are placed end to end,
with the free ends kept at a certain temperature difference. C is placed separately with its ends kept at same temperature difference.
Heat flow through A and B rods at thermal equilibrium is equal

${ H }_{ A }={ H }_{ B }=\dfrac { { k }_{ A }a(T-{ T }_{ 1 }) }{ L } =\dfrac { { k }_{ B }a({ T }_{ 2 }-T) }{ L } \\ \quad \quad \quad \quad \quad ({ k }_{ A }+{ k }_{ B })T=\quad { k }_{ A }{ T }_{ 1 }+{ k }_{ B }{ T }_{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad T\quad =\dfrac { { k }_{ A }{ T }_{ 1 }+{ k }_{ B }{ T }_{ 2 } }{ { k }_{ A }+{ k }_{ B } }$
where T is the temperature of the common end Of A and B,

$T_{1}$ and

$T_{2}$ are temperatures at the other ends of A and B,

$a$ is area and

$L$ is the length of the each rod
Heat flow through C

${ H }_{ C }=\dfrac { { k }_{ C }a({ T }_{ 2 }-{ T }_{ 1 }) }{ L }$
If the rate of heat flow for the above two arrangements are equal

${ H }_{ A }={ H }_{ B }={ H }_{ C }$

$\dfrac { { k }_{ A }a(\dfrac { { k }_{ A }{ T }_{ 1 }+{ k }_{ B }{ T }_{ 2 } }{ { k }_{ A }+{ k }_{ B } } -{ T }_{ 1 }) }{ L } =\dfrac { { k }_{ C }a({ T }_{ 2 }-{ T } _{ 1 }) }{ L } \\ \dfrac { { k }_{ A }({ k }_{ A }{ T }_{ 1 }+{ k }_{ B }{ T }_{ 2 }-{ k }_{ A }{ T }_{ 1 }-{ k }_{ B }{ T }_{ 1 }) }{ { k }_{ A }+{ k }_{ B } } ={ k } _{ C }({ T }_{ 2 }-{ T }_{ 1 })\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \dfrac { { k }_{ A }{ k }_{ B } }{ { k }_{ A }+{ k }_{ B } } ={ k }_{ C }$
option (B) is the correct answer.

During an experiment, an ideal gas is found to obey a condition

$\dfrac{P^{2}}{\rho }$

$=$ constant [

$\rho =$ density of the gas]. The gas is initially at temperature T, pressure P and density

$\rho$ . The gas expands such that density changes to

$\dfrac{\rho}{2}$

0%
The pressure of the gas changes to $\sqrt{2}P$
0%
The temperature of the gas changes to $\sqrt{2}T$
0%
The graph of the above process on the P-T diagram is parabola.
0%
The graph of the above process on the P-T diagram is hyperbola.

Explanation

Equation of process

$\Rightarrow \dfrac{P^{2}}{\rho }=$ constant

$= C$ .... (1)
Equation of state

$\Rightarrow \dfrac{P}{\rho }=\dfrac{R}{M}T$ .... (2)
From 1 and 2,

$PT=$ constant

$\Rightarrow$ C is false, D is true.
Now,

$\rho$ -changes to

$\dfrac{\rho }{2}\Rightarrow$ P changes to

$\dfrac{P}{\sqrt{2}}$ from equation (1)

$\Rightarrow$ A is false.
Hence, T changes to

$\sqrt{2}T \Rightarrow$ B is true

When a liquid is heated in copper vessel its coefficient of apparent expansion is

$6\times 10^{-6}/^oC$ . When the same liquid is heated is heated in a steel vessel its coefficient of apparent expansion is

$24\times 10^{-6}/^oC$ . If the coefficient of linear expansion for copper is

$18\times 10^{-6}/^oC$ , the coefficient of linear expansion for steel is:

0%
$20\times 10^{-6}/^oC$
0%
$24\times 10^{-6}/^oC$
0%
$34\times 10^{-6}/^oC$
0%
$12\times 10^{-6}/^oC$

Explanation

Here,

$\gamma_{real}=\gamma_{app}+\gamma_{container}$

So,

$\gamma_{app(Cu)}+\gamma_{Cu}=\gamma_{app(S)}+\gamma_{S}$

$6\times 10^{-6}+3(18\times 10^{-6})=24\times 10^{-6}+3\alpha_S$ where

$(\gamma=3\alpha)$

$\alpha_S=12\times 10^{-6} /^oC$

The value of

$C_p-C_v$ is 1.09 R for a gas sample in state A and is 1.00 R in state R. Let

$T_A, T_B$ denote the temperature and

$p_A$ and

$p_B$ denote the pressure of the states A and B respectively. Then

0%
$p_A < p_B$ and $T_A > T_B$
0%
$p_A > p_B$ and $T_A > T_B$
0%
$p_A = p_B$ and $T_A < T_B$
0%
$p_A > p_B$ and $T_A < T_B$

Two slabs

$A$ and

$B$ of equal surface area are placed one over the other such that their surfaces are completely in contact. The thickness of slab

$A$ is twice that of

$B$ . The coefficient of thermal conductivity or slab

$A$ is twice that of

$B$ . The first surface of slab

$A$ is maintained at

$100$ , while the second surface of slab

$B$ is maintained at

$25$ . The temperature at the contact of their surfaces is

0%
$62.5$
0%
$45$
0%
$55$
0%
$85$

Explanation

The temperature at the contact of the surface

$=\dfrac { { K }_{ 1 }{ d }_{ 2 }{ \theta }_{ 1 }+{ K }_{ 2 }{ d }_{ 1 }{ \theta }_{ 2 } }{ { K }_{ 1 }{ d }_{ 2 }+{ K }_{ 2 }{ d }_{ 1 } }$

$=\dfrac { 2{ K }_{ 2 }{ d }_{ 2 }\times 100+2{ d }_{ 2 }\times { K }_{ 2 }\times 25 }{ 2{ K }_{ 2 }{ d }_{ 2 }+{ K }_{ 2 }2{ d }_{ 2 } }$

$=\dfrac { 200+50 }{ 4 } =62.5$

A thermocol box has a total wall area (including the lid) of

$1.0{ m }^{ 2 }$ and wall thickness of

$3cm$ . It is filled with ice at

${ 0 }^{ o }C$ . If the average temperature outside the box is

${ 30 }^{ o }C$ throughout the day, the amount of ice that melts in one day is
(Use

${ K }_{ thermocol }=0.03W/mK,{ L }_{ fusion(ice) }=3.00\times { 10 }^{ 5 }J/kg\quad$

0%
$1kg$
0%
$2.88kg$
0%
$25.92kg$
0%
$8.64kg$

Explanation

Given :

$A=I{ m }^{ 2 }$

$L=3\times { 10 }^{ 5 } \ J/kg$

$l=3cm=3\times { 10 }^{ -2 }m$

$t=24\times 60\times 60 \ s$

$\Delta \theta =30-0={ 30 }^{ o }C$
Using

$\cfrac { Q }{ t } =\cfrac { KA }{ l } \Delta \theta \quad$
or

$\cfrac { mL }{ t } =\cfrac { KA }{ l } \Delta \theta \quad$
Putting the values,

$\cfrac { m\times 3\times { 10 }^{ 5 } }{ 24\times 60\times 60 } =\cfrac { 0.03\times 1 }{ 3\times { 10 }^{ -2 } } \times 30$

$\implies \ m=8.641kg$

One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is

$P_0$ . Choose the correct option(s) from the following.

0%
Internal energies at A and B are the same
0%
Work done by the gas in process AB is $P_oV_o$ ln $4$
0%
Pressure at C is $\dfrac{P_o}{4}$
0%
Temperature at C is $\dfrac{T_o}{4}$

Explanation

From the graph given in the question, we can see that temperature at states Aand C are same,

i.e,

$T_A=T_B$

Internal energy

$=\dfrac{t}{2}nRT=\dfrac{t}{2}RT$ ( Here

$n=1$ )

$\Rightarrow$ Here,

$f=$ degree of freedom

$n=$ moles

$R=$ universal gas constant.

As,

$f$ and

$R$ are constant and

$T_A=T_B,$

So, internal energy of the gas at A and B are same .

Hence, option A is correct.

For the process AB, temperature is constant throughout i.e, the process is isothermal.

Work done by a gas in an isothermal process

$=nRT\iota \eta \left(\dfrac{v_2}{v_1}\right)$

$=\left(1\right) \left(R\right) \left(T_0\right) \iota \eta \left(\dfrac{v_B}{v_A}\right)$ [ Hence

$n=1$ ]

$=RT_0\iota \eta \left(\dfrac{4v_0}{v_0}\right)$ [ On applying,

$Pv=nRT$ at

$'A',$ we will get

$RT_0=p_0v_0$ ]

$=P_0v_0 \iota \eta 4.$

So, option B is correct.

Hence, both option A and B are correct.

An insulated cylindrical tube of an air-conditioner's condenser contains a hot fluid. Temperature of fluid is

${500}^{o}C$ and outside temperature is

${40}^{o}C$ . Hot fluid tube is very thin and is covered with

$3$ layers of different insulating materials. Cross-section of the tube is as shown in figure. Given

${ r }_{ 1 }=1cm;{ r }_{ 2 }=2cm;{ r }_{ 3 }=8cm;{ r }_{ 4 }=64cm$

${ K }_{ A }=1W/{ m }^{ -1 }\: { _{ }^{ o }{ C } }^{ -1 };{ K }_{ B }=2W/{ m }^{ -1 }\: { _{ }^{ o }{ C } }^{ -1 };{ K }_{ C }=3W/{ m }^{ -1 }\: { _{ }^{ o }{ C } }^{ -1 }$
Heat loss per unit length (in watts) of tube will be:

0%
$\cfrac { 460 }{ \ln { (2) } }$
0%
$\cfrac { 460 }{ \pi \ln { (2) } }$
0%
$\cfrac { \pi \times 460 }{ \ln { (2) } }$
0%
$\cfrac {2 \pi \times 460 }{ 3\ln { (2) } }$

Explanation

To solve this question let's have some general discussion.

Top view of cylinder only one layer is A (fig i)

Thermal resistance

$\dfrac{l}{KA}$

Thermal resistance of element with thickness dr of radius r

$\int { dR=\int _{ { r }_{ 1 } }^{ { r }_{ 2 } }{ \dfrac { dr }{ K2\pi rL } } }$

$R=\dfrac { 1 }{ 2\pi KL } ln\left( \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \right)$ (1)

Now in fig(ii) there are three layers and these layers are in series.

Resistance of layer A using formula(1)

$=\dfrac { 1 }{ 2\pi { K }_{ A }L } ln\left( \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \right)$ (2)

similarly resistance of layer B

$=\dfrac { 1 }{ 2\pi { K }_{ B }L } ln\left( \dfrac { { r }_{ 3 } }{ { r }_{ 2 } } \right)$ (3)

Similarly resistance of layer C

$=\dfrac { 1 }{ 2\pi { K }_{ C }L } ln\left( \dfrac { { r }_{ 4 } }{ { r }_{ 3 } } \right)$

Requivalent

$=R_A+R_B+R_C$

$=\dfrac { 1 }{ 2\pi { K }_{ A }L } ln\left( \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \right) +\dfrac { 1 }{ 2\pi { K }_{ B }L } ln\left( \dfrac { { r }_{ 3 } }{ { r }_{ 2 } } \right) +\dfrac { 1 }{ 2\pi { K }_{ C }L } ln\left( \dfrac { { r }_{ 4 } }{ { r }_{ 3 } } \right)$

$H=\dfrac { \left( { T }_{ 1 }-{ T }_{ 4 } \right) }{ \dfrac { ln\left( \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \right) }{ 2\pi { K }_{ A }L } +\dfrac { ln\left( \dfrac { { r }_{ 3 } }{ { r }_{ 2 } } \right) }{ 2\pi { K }_{ B }L } +\dfrac { ln\left( \dfrac { { r }_{ 4 } }{ { r }_{ 3 } } \right) }{ 2\pi { K }_{ C }L } }$

$H=\dfrac { 2\pi L\times \left( { T }_{ 1 }-{ T }_{ 4 } \right) }{ \dfrac { ln\left( \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \right) }{ { K }_{ A } } +\dfrac { ln\left( \dfrac { { r }_{ 3 } }{ { r }_{ 2 } } \right) }{ { K }_{ B } } +\dfrac { ln\left( \dfrac { { r }_{ 4 } }{ { r }_{ 3 } } \right) }{ { K }_{ C } } }$

$=\dfrac { 2\pi L\times 460 }{ \dfrac { { ln }2 }{ 1 } +\dfrac { ln4 }{ 2 } +\dfrac { ln8 }{ 3 } }$

$H=\dfrac { 2\pi L\times 460 }{ ln2+ln2+ln2 }$

$H=\left( \dfrac { 2 }{ 3 } \right) \dfrac{\pi L\times 460}{ln2}$ W

$\dfrac{H}{L}=\dfrac{2}{3}\dfrac{\pi \times460}{ln2}$

A shining metallic ball with a small black spot on its surface is heated to a very high temperature and then quickly taken to a dark room. Then:

0%
Both appear equally bright
0%
The spot appears brighter than the ball
0%
The spot appears darker than the ball
0%
Both are invisible in the dark room

Two bodies

$A$ and

$B$ have thermal emissivities of

$0.01$ and

$0.81$ , respectively. The outer surface areas of the two bodies are the same. The two bodies emit total radiant power of the same rate. Wavelength

$\lambda_{B}$ corresponding to maximum spectral radiancy in the radiation from

$B$ shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from

$A$ , by

$1.00\mu m$ . If the temperature of

$B$ is

$5802\ K$ .

0%
The temperature of $B$ is $1934 K$
0%
$\lambda_{B} = 1.5\mu m$
0%
The temperature of $B$ is $11604 K$
0%
The temperature of $B$ is $2901 K$

Explanation

According to question -

$P_A=P_B$

$\Rightarrow \sigma { e }_{ A }{ A }_{ A }{ T }_{ A }^{ 4 }=\sigma { e }_{ B }{ A }_{ B }{ T }_{ B }^{ 4 }$ [outer areas of both bodies are same]

$\Rightarrow { e }_{ A }{ T }_{ A }^{ 4 }={ e }_{ B }{ T }_{ B }^{ 4 }$

$\Rightarrow \left(\dfrac{T_A}{T_B}\right)^4=\dfrac{e_B}{e_A}=81$

$\Rightarrow \dfrac{T_A}{T_B}=3$

$T_A=3T_B$

$\Rightarrow T_B=\dfrac{T_A}{3}=\dfrac{5802}{3}=1934K$ (A)

Now, by wien's law -

$\Rightarrow \lambda_B=\dfrac{b}{T_B}=\dfrac{\left(0.3cm\;kdvin\right)}{1934k}=0.000155cm$

$=1.5\mu m.$

Hence, the answer is

$1.5\mu m$ and temperature of B is

$1934K.$

An ideal gas is taken from the state A(pressure

$P_0$ , volume

$V_0$ ) to the state B (pressure

$P_0/2$ , volume

$2V_0$ ) along a straight line path in the P-V diagram. Select the correct statement(s) from the following.

0%
The work done by the gas in the process A to B exceeds the work that would be done by it if the system were taken from A to B along the isotherm
0%
In the T-V diagram, the path AB becomes a part of the parabola
0%
In the P-T diagram, the path AB becomes a part of hyperbola
0%
In going from A to B, the temperature T of the gas first increases to maximum value and then decreases

Explanation

Work done by the gas

$=$ area under the graph on volume axis.

Clearly, from the graph, we can see that area under the isotherm is less than the area under straight line AB process.

So, option (A) is correct.

Equation of AB :

$P=\left(\dfrac{-P_0}{2V_0}\right)V+\dfrac{3P_0}{2}$

$\Rightarrow \dfrac{nRT}{V}-\left(\dfrac{-P_0}{2V_0}\right)V+\dfrac{3P_0}{2}$ [ Applying

$PV=nRT$ ]

$\Rightarrow T=\left(\dfrac{-P_0}{2nRV_0}\right)V^2+\dfrac{3P_0}{2}V \equiv y=-Ax^2+Bx$

On drawing various isotherms across AB, we can see that temperature first increases then decreases.

Hence, this question has multiple answers.

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

0%
$PV=\Big(\dfrac{5}{32}\Big)RT$
0%
$PV = 5 RT$
0%
$PV = \Big(\dfrac{5}{2}\Big)RT$
0%
$PV = \Big(\dfrac{5}{16}\Big)RT$

Explanation

Given,

$m=5g$

$M=32g$ (molecular mass of the oxygen molecule)

Number of moles,

$n=\dfrac{m}{M}$

$n=\dfrac{5}{32}$

From the equation of gas,

$PV=nRT$

$PV=(\dfrac{5}{32})RT$

The movement of water being heated in a pot on a stove is an example of

0%
conduction
0%
convection
0%
radiation
0%
condensation

In the given figure conainer

$A$ holds an ideal gas at a pressure of

$5.0 \times 10^{5}\ Pa$ and a temperature of

$300\ K.$ It is connected by the tin tube (and a closed valve) to container

$B$ , with four times the volume of

$A$ . Container

$B$ hold same ideal gas at a pressure of

$1.0\ \times 10^{5}\ Pa$ and a temperature of

$400\ K$ . The valve is opened to allow the pressure to equalize, but the temperature of each container kept constant at its initial value. The final pressure in the two containers will be close to :

0%
$2.0 \times 10^{5}\ Pa$
0%
$2.5 \times 10^{5}\ Pa$
0%
$3.0 \times 10^{5}\ Pa$
0%
$3.5 \times 10^{5}\ Pa$

Temperature of a body increases from

$-73^{\circ}C$ to

$127^{\circ}C$ . The radiant energy per second will become.

0%
$2\ times$
0%
$256\ times$
0%
$16\ times$
0%
$32\ times$

$10g$ of ice at

$-{20}^{o}C$ is added to

$10g$ of water at

${50}^{o}C$ . Specific heat of water

$=1cal/g ^{o}C$ , specific heat of ice

$=0.5cal/g^{o}C$ . Latent heat of ice

$=80cal/g$ . Then resulting temperature is -

0%
$-{20}^{o}C$
0%
${15}^{o}C$
0%
${0}^{o}C$
0%
${50}^{o}C$

A rod of length

$l$ with thermally insulated lateral surface consists of material whose heat conductivity coefficient varies with temperature as

$k = \alpha/ T$ , where

$\alpha$ is a constant. The ends of the rod are kept at temperature

$T_{1}$ and

$T_{2}$ . Find the function

$T(x)$ , where

$x$ is the distance from the end whose temperature is

$T_{1}$ , and the heat flow density.

0%
$T_{(x)} = 2T_{1}\left (\dfrac {T_{2}}{T_{1}}\right )^{xf}; H = \dfrac {\alpha}{l}ln \left (\dfrac {T_{2}}{T_{1}}\right )$ .
0%
$T_{(x)} = T_{1}\left (\dfrac {T_{2}}{T_{1}}\right )^{xf}; H = \dfrac {\alpha}{l}ln \left (\dfrac {T_{2}}{T_{1}}\right )$ .
0%
$T_{(x)} = 3T_{1}\left (\dfrac {T_{2}}{T_{1}}\right )^{xf}; H = \dfrac {\alpha}{l}ln \left (\dfrac {T_{2}}{T_{1}}\right )$ .
0%
$T_{(x)} = 4T_{1}\left (\dfrac {T_{2}}{T_{1}}\right )^{xf}; H = \dfrac {\alpha}{l}ln \left (\dfrac {T_{2}}{T_{1}}\right )$ .

A rod of length

$l$ and cross section area A has a variable thermal conductivity given by

$k=\alpha T$ where

$\alpha$ is a positive constant and

$T$ is temperature in Kelvin. Two ends of the rod are maintained at temperature

$T_1$ and

$T_2$

$(T_1>T_2)$ . Heat current flowing through the rod will be

0%
$\dfrac{A \alpha (T_1^2-T_2^2)}{l}$
0%
$\dfrac{A \alpha (T_1^2+T_2^2)}{l}$
0%
$\dfrac{A \alpha (T_1^2+T_2^2)}{3l}$
0%
$\dfrac{A \alpha (T_1^2-T_2^2)}{2l}$

A pond has an ice layer of thickness 3 cm. If K of ice is 0.005 CGS units, surface temperature of surroundings is

$-20^\circ C$ , density of ice is

$0.9 gm/cc$ , the time taken for the thickness to increase by 1 cm is

0%
30 min
0%
35 min
0%
42 min
0%
60 min

The solar constant for the earth is

$S$ . The surface temperature of the sun is

$T\ K$ . The sun subtends an angle

$\theta$ at the earth.

0%
$S\propto T^{4}$
0%
$S\propto T^{2}$
0%
$S\propto \theta^{2}$
0%
$S\propto \theta$

Explanation

Let

$R =$ radius of the sun,

$d =$ distance of the earth from the sun.
Power radiated by the sun

$= (4\pi R^{2})\sigma T^{4} = P$ .
Energy received per unit area per second normally on the earth

$= S = \dfrac {P}{4\pi d^{2}} = \dfrac {4\pi R^{2}\sigma T^{4}}{4\pi d^{2}} = (\sigma T^{4})\left (\dfrac {R}{d}\right )^{2} = \dfrac {1}{4}\sigma T^{4}\left (\dfrac {2R}{d}\right )^{2}$

Angle subtended by the sun at the earth

$= \theta = \dfrac {2R}{d}$ .
or,

$S = \dfrac {\sigma}{4}T^{4}\theta^{2}$

Hence (A) and (C) are correct.

Three rods of a same material same area of cross-section but different lengths 10 cm, 20 cm, and 30 cm are connected at a point as shown. What is a temperature of junction O?

0%
19.2
0%
16.4
0%
11.5
0%
22

Explanation

Let the temperature at point 0 be

$T_0$

so,

$\dfrac{T_A -T_O}{l_1} = \dfrac{T_O - T_B}{l_2} + \dfrac{T_O - T_C}{l_3}$

$\dfrac{30-T_O}{30} = \dfrac{T_O -10}{10} + \dfrac{T_O - 20}{20}$

Solving this

$T_O = 19.2$

960238_1039086_ans_866bf3aaa2c34bdaaf5a86a132e413d5.png

CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 12 - 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 12 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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