NCERT Exemplar Class 11 Physics Solutions Chapter 11 Thermal Properties Of Matter

NCERT Exemplar Class 11 Physics Solutions Chapter 11: MCQ I

Question:11.1

A bimetallic strip is made of aluminium and steel $({\alpha }_{Al}>{\alpha }_{steel})$. On heating, the strip will
a) remain straight
b) get twisted
c) will bend with aluminium on concave side
d) will bend with steel on concave side

Answer:

The answer is the option (d) Will bend with steel on concave side.
Explanation: Initially, Aluminium and Steel are fixed together in a bimetallic strip and when you heat them, the expansion in steel will be smaller than the expansion in aluminium. Hence, the steel strip will be on the concave side, whereas the aluminium strip will be on the convex side.

Question:11.2

A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly
a) its speed of rotation increases
b) its speed of rotation decreases
c) its speed of rotation remains same
d) its speed increases because its moment of inertia increases

Answer:

When the rod is heated uniformly to raise its temperature slightly, the length of the rod will be increased due to heating. This leads to increase in its moment of inertia (I).
Since angular momentum,L=Iω
To maintain the angular momentum conserved, ω(angular speed ) decreases with an increase in moment of inertia.

Question:11.3

The graph between two temperature scales A and B is shown in the figure. Between upper fixed point and lower fixed point there are 150 equal division on scale A and 100 on scale B. The relationship for conversion between the two scales is given by

$a)\frac{t_A-180}{100}=\frac{t_B}{150}$
$b)\frac{t_A-30}{150}=\frac{t_B}{100}$
$c)\frac{t_B-180}{150}=\frac{t_A}{150}$
$d)\frac{t_B-40}{100}=\frac{t_A}{180}$

Answer:

The answer is the option $b)\frac{t_A-30}{150}=\frac{t_B}{100}$
Explanation: Now, from graph t_A, we know that,
Lower fixed point (LFP) = ${30}^{\circ }$
& upper fixed point (UFP) = ${180}^{\circ }$
Similarly, in case of scale B,
UFP = ${100}^{\circ }$
& LFP = $0^{\circ }$
Thus, the formula,
$\frac{t_A-(LFP)}{(UFP{)}_A-(LFP{)}_A}=\frac{t_B-(LFP{)}_B}{(UFP{)}_B-(LFP{)}_B}$
$\frac{T_A-30}{180-30}=\frac{t_B-0}{100-0}or\frac{t_A-30}{150}=\frac{t_B}{100}$

Question:11.4

An aluminium sphere is dipped into water. Which of the following is true?
a) buoyancy will be less in water at $0^{\circ }C$ than that in water at $4^{\circ }C$
b) buoyancy will be more in water at $0^{\circ }C$ than that in water at $4^{\circ }C$
c) buoyancy in water at $0^{\circ }C$ will be same as that in water at $4^{\circ }C$
d) buoyancy may be more or less in water at $4^{\circ }C$ depending on the radius of the sphere

Answer:

We know that,
Buoyant force (B.F.) =$V^{{}^{\prime }}\rho lg$
Where, V’ = volume of displaced liquid by dipped body
$\rho l$ = density if liquid
Now, at 4⁰C, the density of water is maximum, thus, the density of water and buoyant force will also be maximum at 4⁰C.
$B.F.\propto P_l$
Thus,
$\frac{F_A}{F_0}=\frac{{\rho }_A}{{\rho }_0},where,{\rho }_A>{\rho }_0$
$\frac{F_A}{F_0}>1$or
$F_A>F_0$

Question:11.5

As the temperature is increased, the time period of a pendulum
a) increases as its effective length increases even though its centre of mass still remains at the centre of the bob
b) decreases as its effective length increases even though its centre of mass still remains at the centre of the bob
c) increases as its effective length increases due to shifting of centre of mass below the centre of the bob
d) decreases as its effective length remains same but the centre of mass shifts above the centre of the bob

Answer:

The answer is the option (a) increases as its effective length increases even though its centre of mass still remains at the centre of the bob.
Explanation: Due to linear expansion, the length will increase with temperature,
$T=\sqrt{\frac{L}{g}}orT\propto L$
Hence, the effective length and T increases on increasing temperature.

Question:11.6

Heat is associated with
a) kinetic energy of random motion of molecules
b) kinetic energy of orderly motion of molecules
c) total kinetic energy of random and orderly motion of molecules
d) kinetic energy of random motion in some cases and kinetic energy of orderly motion in other

Answer:

The answer is the option (a) kinetic energy of random motion of molecules.
Explanation: Kinetic energy associated with random motion of molecules increases as vibration of molecules about their mean position increases on increasing temperature.

Question:11.7

The radius of a metal sphere at room temperature T is R, and the coefficient of linear expansion of the metal is $\alpha$. The sphere is heated a little by a temperature $\mathrm{\Delta }T$ so that its new Temperature is $T+\mathrm{\Delta }T$. The increase in the volume of the sphere is approximately
$(a)2\pi R\alpha \mathrm{\Delta }T$
$(b)\pi R^2\alpha \mathrm{\Delta }T$
$(c)\frac{\pi R^3\alpha \mathrm{\Delta }T}{3}$
$(d)4\pi R^3\alpha \mathrm{\Delta }T$

Answer:

The answer is the option
Explanation: Here, we know that,
$\alpha$ = coefficient of linear expansion
$3\alpha$= ϒ = coefficient of cubical expansion
Now, $V_0=\frac{4}{3}\pi R^3$
$\gamma =\frac{\mathrm{\Delta }V}{V\mathrm{\Delta }T}$
Thus, $\mathrm{\Delta }V=\gamma V\mathrm{\Delta }T$
i.e.,$\mathrm{\Delta }V=3\alpha .\frac{4}{3}\pi R^3\mathrm{\Delta }T$
$=4\pi R^3\mathrm{\Delta }T$
Hence, opt (d).

Question:11.8

A sphere, a cube, and a thin circular plate, all of the same material and same mass are initially heated to same high temperature.
a) plate will cool fastest and cube the slowest
b) sphere will cool fastest and cube the slowest
c) plate will cool fastest and sphere the slowest
d) cube will cool fastest and plate the slowest

Answer:

The answer is the option (c) Plate will cool fastest and sphere the slowest.
Explanation: On cooling, loss of heat is directly proportional to surface area exposed to the surrounding, material of object & temperature difference between the surrounding and the body.
Here, opt (c) is verified as the surface area of the sphere is minimum and thus it will cool slowest, while the surface area of circular plate is maximum, and it will cool fastest.

NCERT Exemplar Class 11 Physics Solutions Chapter 11: MCQ II

Question:11.9

Mark the correct options:
a) A system X is in thermal equilibrium with Y but not with Z. System Y and Z may be in thermal equilibrium with each other
b) A system X is in thermal equilibrium with Y but not with Z. Systems Y and Z are not in thermal equilibrium with each other
c) A system X is neither in thermal equilibrium with Y nor with Z. The systems Y and Z must be in thermal equilibrium with each other
d) A system X is neither in thermal equilibrium with Y nor with Z. The system Y and Z may be in thermal equilibrium with each other

Answer:

The answer is the option (b) A system X is in thermal equilibrium with Y but not with Z. System Y and Z are not in thermal equilibrium with each other and option (d) A system X is neither in thermal equilibrium with Y nor with Z. The system Y and Z may be in thermal equilibrium with each other.
Explanation: (a) Here Y and Z are not in thermal equilibrium since,
$T_x=T_y$ & $T_x$ is not equal to $T_z$.
Thus, $T_x=T_y\ne T_z.$hence, it is incorrect.
(b) System Y and Z are not in thermal equilibrium since,
$T_x=T_y$ and $T_x\ne T_z$
Thus, $T_y\ne T_z$. Thus, (b) is correct.
(c) Y and Z are not in thermal equilibrium because,
$T_x\ne T_y$ and $T_x\ne T_z$
Thus, $T_y\ne T_z$. Hence, opt (c) is incorrect.
(d) Here,
$T_x\ne T_y$ & $T_x\ne T_z$.
Here, $T_z$ may be equal to $T_y$. Hence, opt (d) is correct.

Question:11.10

Gulab Jamuns’ (assumed to be spherical) are to be heated in an oven. They are available in two sizes, one twice bigger than the other. Pizzas (assumed to be discs) are also to be heated in oven. They are also in two sizes, one twice big in radius than the other. All four are put together to be heated to oven temperature. Choose the correct option from the following:
a) both size gulab jamuns will get heated in the same time
b) smaller gulab jamuns are heated before bigger ones
c) smaller pizzas are heated before bigger ones
d) bigger pizzas are heated before smaller ones

Answer:

The answer is the option (b) Smaller gulab jamuns are heated before bigger ones and (c) Smaller pizzas are heated before bigger ones.
Explanation: Here, the smaller ones have a small thickness, through which heat to transfer up to depth gets heated before the bigger ones that have larger depth.
Hence, opt (b) & (c).

Question:11.11

Refer to the plot of temperature versus time showing the changes in the state of ice on heating. Which of the following is correct?

a) the region AB represents ice and water in thermal equilibrium
b) at B water starts boiling
c) at C all the water gets converted into steam
d) C to D represents water and steam in equilibrium at boiling point

Answer:

The answer is the option (a) The region AB represents ice and water in thermal equilibrium and (d) C to D all represents water and steam in equilibrium at boiling point.
Explanation: If the temperature of a substance does not change on applying heat continuously, then its state changes.
Here, AB = $0^{\circ }C$ & CD = ${100}^{\circ }C$ , thus,
(i) AB, represents ice and water up to B &
(ii) CD represents water & steam.

Question:11.12

A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct?
a) the rate of cooling is constant till milk attains the temperature of the surrounding
b) the temperature of milk falls off exponentially with time
c) while cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools
d) all three phenomenon, conduction, convection, and radiation are responsible for the loss of heat from milk to the surroundings

Answer:

The answer is the option (b) The temperature of milk falls off exponentially with time, (c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools and (d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk to the surroundings.
Explanation:

1. Loss of heat is proportional to temperature difference with surrounding and body. As milk cools, rate of cooling decreases with time. Thus, opt (a) is incorrect.
2. The heat of milk falls exponentially by Newton’s law of cooling.
3. As compared to heat lost by milk to surrounding while cooling it, a very small amount of heat also flows from surrounding to milk. Hence, it is correct.
4. By conduction, convection, and radiation, when hot milk is spread on a table, it transfers, heat to the surroundings. Hence it is correct.

NCERT Exemplar Class 11 Physics Solutions Chapter 11: Very Short Answer

Question:11.13

Is the bulb of a thermometer made of diathermic or adiabatic wall?
Diathermic walls are used in the bulb of thermometer as they allow to conduct heat through it into mercury of bulb, whereas adiabatic walls does not allow to pass heat through it.
A student records the initial length l, change in temperature $\mathrm{\Delta }T$ and change in length $\mathrm{\Delta }l$ of a rod as follows:

If the first observation is correct, what can you say about observations 2, 3, and 4.

Answer:

Here, it is given that there is a rod and a material that has linear expansion ‘$\alpha$’, remains same.
$\mathrm{\Delta }t$ is same here.
Now, from observation no. 1
$\alpha =\frac{\mathrm{\Delta }L}{L\mathrm{\Delta }t}$
$\alpha =\frac{4\times {10}^{-4}}{2\times 10}=2\times {10}^{-5}/C^0$
Now, for observation no. 2,
$\mathrm{\Delta }L=\alpha L\mathrm{\Delta }t$
$=2\times {10}^{-5}\times 1\times 10$
$=2\times {10}^{-4}m$ viz. not equal to $4\times {10}^{-4}m$
Now, for observation no. 3,
$\mathrm{\Delta }L=\alpha L\mathrm{\Delta }t$
$=2\times {10}^{-5}\times 2\times 20$
$=8\times {10}^{-4}$ viz. not equal to $=2\times {10}^{-4}m$
Now, for the last observation, i.e., no. 4,
$\mathrm{\Delta }L=\alpha L\mathrm{\Delta }t$
$=2\times {10}^{-5}\times 3\times 10$
$=6\times {10}^{-4}=6\times {10}^{-4}m.$

Question:11.14

A student records the initial length l, change in temperature $\mathrm{\Delta }T$ and change in length $\mathrm{\Delta }l$ of a rod as follows:

If the first observation is correct, what can you say about observations 2, 3, and 4.

Answer:

$\mathrm{\Delta }t$ is same here.
Now, from observation no. 1
$\alpha =\frac{\mathrm{\Delta }L}{L\mathrm{\Delta }t}$
$\frac{4\times {10}^{-4}}{2\times 10}=2\times {10}^{-5}{C}^{\circ }-1$
Now, for observation no. 2,
$\mathrm{\Delta }L=\alpha L\mathrm{\Delta }t$
$=2\times {10}^{-5}\times 1\times 10$
$=2\times {10}^{-4}m$ viz. not equal to $4\times {10}^{-4}m$
Now, for observation no. 3,
$\mathrm{\Delta }L=\alpha L\mathrm{\Delta }t$
$=2\times {10}^{-5}\times 2\times 20$
$=8\times {10}^{-4}$ viz. not equal to $=2\times {10}^{-4}m$
Now, for the last observation, i.e., no. 4,
$\mathrm{\Delta }L=\alpha L\mathrm{\Delta }t$
$=2\times {10}^{-5}\times 3\times 10$
$=6\times {10}^{-4}=6\times {10}^{-4}m.$

Question:11.15

Why does a metal bar appear hotter than a wooden bar at the same temperature? Equivalently it also appears cooler than wooden bar if they are both colder than room temperature.

Answer:

The rate of transferring heat in metal is larger than that in wood, as the conductivity of the metal bar is extremely high as compared to the wood.
Also, metal requires exceedingly small quantities of heat than wood to change temperature as the specific heat of metal is exceptionally low as compared to the wood.
Thus, due to larger conductivity and smaller specific heat metal becomes colder and hotter than wood when kept in cold and hot regions, respectively.

Question:11.16

Calculate the temperature which has same numerical value on Celsius and Fahrenheit scale.

Answer:

Let us consider that the required temperature is,
$x^{\circ }C=x^{\circ }F$
Now, we know that,
$\frac{C}{100}=\frac{F-32}{180}$
Thus,
$\frac{x}{5}=\frac{x-32}{9}$
Thus, $5x-160=9x$
-9x +5x = 160
-4x = 160x
Thus,$x=-{40}^{\circ }$
Thus, $-{40}^{\circ }F=-{40}^{\circ }C$

Question:11.17

These days people use steel utensils with copper bottom. This is supposed to be good for uniform heating of food. Explain this effect using the fact that copper is the better conductor.

Answer:

Copper base transfers heat quickly to food from the bottom and also needs a low quantity of heat as compared to steel because, as compared to steel, copper has higher conductivity and low specific heat. Thus, heat is supplied from burner to food in utensil quickly and in a large amount when the base is copper.

NCERT Exemplar Class 11 Physics Solutions Chapter 11: Short Answer

Question:11.18

Find out the increase in moment of inertia I of a uniform rod about its perpendicular bisector when its temperature is slightly increased by $\mathrm{\Delta }T$.

Answer:

Now, we know that moment of inertia (I) of a rod when its axis is along perpendicular bisector it is =
$\frac{1}{12}M{L}^2$
Now,$\mathrm{\Delta }L=\alpha .L\mathrm{\Delta }T$
Thus, $I^{{}^{\prime }}=\frac{1}{12}M(L+\mathrm{\Delta }L{)}^2$
$=\frac{1}{12}M(L^2+\mathrm{\Delta }{L}^2+2L\mathrm{\Delta }L{)}^2$
Since $\mathrm{\Delta }{L}^2$ is a very small term, we will neglect it,
$=\frac{M}{12}(L^2+2L\mathrm{\Delta }L)$
$=\frac{M{L}^2}{12}+\frac{ML\mathrm{\Delta }L}{6}\times \frac{2L}{2L}$
$=\frac{M{L}^2}{12}+\frac{M{L}^2}{12}.\frac{2\mathrm{\Delta }L}{L}$
$I^{{}^{\prime }}=I(1+2\alpha \mathrm{\Delta }T)$
Thus, the new moment of inertia will increase by $2I\alpha \mathrm{\Delta }T$.

Question:11.19

During summers in India, one of the common practice to keep cool is to make ice balls of crushed ice, dip it in flavoured sugar syrup and sip it. For this a stick is inserted into crushed ice and is squeezed in the palm to make it into the ball. Equivalently in winter, in those areas where it snows, people make snowballs and throw around. Explain the formation of ball out of crushed ice or snow in the light P-T diagram of water.

Answer:

From the P-T graph, we know that at decreasing pressure in the liquid state at $0^{\circ }C$ and 1 atm takes water to ice state & increasing pressure from $0^{\circ }C$ to 1 atm takes water to ice state.
If we squeeze crushed ice, some parts of it melt into water at $0^{\circ }C$ and fill the gap between ice flakes. Also, when we squeeze crushed ice, its melting point increases and the water present in the ice flakes freezes binding them together and making the ball more stable.

Question:11.20

100 g of water is supercooled to $-{10}^{\circ }C$. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze?

Answer:

Given: Mass of water = 100g
Ice mixes with water at $-{10}^{\circ }C$
Now, the heat required by -$-{10}^{\circ }C$ ice to $0^{\circ }C$ ice $=ms\mathrm{\Delta }t$
$=100\times 1\times [0-(-10)]$
Thus, Q = 1000cal
Thus,
$m=\frac{Q}{L}$
$=\frac{1000}{80}$
= 12.5 gm
Thus, there is 12.5 gm of water and ice in the mixture and hence its temperature remains $0^{\circ }C$.

Question:11.21

One day in the morning, Ramesh filled up 1/3 bucket of hot water from geyser, to take bath. Remaining 2/3 was to be filled by cold water to bring mixture to a comfortable temperature. Suddenly Ramesh had to attend something which would take sometimes, say 5-10 minutes before he could take bath. Now he has two options:
a) fill the remaining bucket completely by cold water and then attend the work
b) first attend to the work and fill the remaining bucket just before taking bath.
Which option do you think would have kept water warmer? Explain.

Answer:

According to Newton’s law of cooling,
Rate of cooling(loss of heat) is proportional to Temperature difference of body & surrounding.
Thus, for option

1. Since the temperature of the surroundings and water is small, water should be kept warmer, thus, the amount of heat energy lost here is less.
2. Here a large amount of heat energy is lost since the temperature difference between water and surrounding is large.

NCERT Exemplar Class 11 Physics Solutions Chapter 11: Long Answer

Question:11.22

We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 cm. We can use a bimetallic strip made of brass and iron each of different length whose length would change in such a way that difference between their lengths remain constant. If ${\alpha }_{iron}=\frac{1.2\times {10}^{-5}}{K}$ and ${\alpha }_{brass}=\frac{1.8\times {10}^{-5}}{K}$what should we take as the length of each strip?

Answer:

We can use iron and brass to make the required scale. Here, one end will be connected with brass and the other end will be only of iron at a distance of 10 cm at any temperature. Let us consider the initial length of iron and brass to be-
${\alpha }_{iron}=\frac{1.2\times {10}^{-5}}{K}$ &
${\alpha }_{brass}=\frac{1.8\times {10}^{-5}}{K}$
Now, $L_{11}-L_{1B}=10cm$ …… (i)
Thus, $\alpha =\frac{\mathrm{\Delta }L}{L_0\mathrm{\Delta }T}or\frac{L_2-L_1}{L_1\mathrm{\Delta }T}$
Now, $L_2=L_1+L_1\alpha \mathrm{\Delta }T$
$=L_1(1+{\alpha }_B\mathrm{\Delta }T)$
If the rod is heated then, the length will become,
$L_{21}and{L}_{2B}$
Now, $L_{21}-L_{2B}=10cm$
$L_{11}(1+{\alpha }_1\mathrm{\Delta }T)-L_{1B}(1+{\alpha }_B\mathrm{\Delta }T)=10$
$L_{11}+{\alpha }_1{L}_{11}\mathrm{\Delta }T-L_{1B}-L_{1B}{\alpha }_B\mathrm{\Delta }T=10$
Thus, from (i),
$10+({\alpha }_1{L}_{11}-{\alpha }_B{L}_{1B})\mathrm{\Delta }T=10or{\alpha }_1{L}_{11}-{\alpha }_B{L}_{1B}=0$
${\alpha }_1{L}_{11}={\alpha }_B{L}_{1B}$
$\frac{L_{11}}{L_{1B}}=\frac{{\alpha }_B}{{\alpha }_1}$
$=\frac{1.8\times {10}^{-5}}{1.2\times {10}^{-5}}$
$=\frac{3}{2}$
Now, let$L_{11}=3x$ and $L_{1B}=2x$
$L_{11}-L_1=10$
3x – 2x = 10
Thus, x = 10.
Therefore, length of iron rod, (3x) = 30 & length of brass rod, (2x) = 20
Thus, the difference between the second ends will be 10 cm.

Question:11.23

We would like to make a vessel whose volume does not change with temperature. We can use brass and iron $({\beta }_{brass}=\frac{6\times {10}^{-6}}{K}and{\beta }_{iron}=\frac{3.55\times {10}^{-5}}{K})$to create a volume of 100 cc. How do you think you can achieve this?

Answer:

For, the above situation, we need to make a double container whose volume difference will be 100cc.
Let us consider $V_{1i}$& $V_{1b}$ as the initial volumes of the iron and brass container, i.e., $V_{1i}-V_{1b}=100cc$.
After heating by $\mathrm{\Delta }TK$ the difference will be the same, but the new volumes will be V2i &V2b.
Now, $V_{2i}-V_{2b}=100cc$
$\gamma =\frac{\mathrm{\Delta }V}{V\mathrm{\Delta }T}$
Therefore, $V_2-V_1=\gamma V_1\mathrm{\Delta }T$
$V_2=V_1+\gamma V_1\mathrm{\Delta }T$
$=V_1(1+\gamma \mathrm{\Delta }T)$
$V_{2i}=V_{1i}(1+{\gamma }_i\mathrm{\Delta }T)$
$V_{2b}=V_{1b}(1+{\gamma }_b\mathrm{\Delta }T)$
$V_{1i}+V_{1i}{\gamma }_i\mathrm{\Delta }T-(V_{ib}+V_{ib}{\gamma }_b\mathrm{\Delta }T)=100cc.$
$100+(V_{1i}{\gamma }_i-V_{1b}{\gamma }_b)\mathrm{\Delta }T=100$
$V_{1i}{\gamma }_i-V_{1b}{\gamma }_b=0$
$\frac{V_{1i}}{V_{1b}}=\frac{{\gamma }_b}{{\gamma }_i}$
$=\frac{6\times {10}^{-5}}{3.55\times {10}^{-5}}$
$=\frac{6}{3.55}$
$=\frac{120}{71}$
Now, let $V_{1i}=120x,V_{1b}=71x$
& $V_{1i}-V_{1b}=100$
Thus, 120x – 71x = 100
49x = 100
$x=\frac{100}{49}$
= 2.04
Thus, $V_{1i}=120\times 2.04=245cc$
$V_{1b}=71\times 2.04=145cc$

Question:11.24

Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of ${57}^{\circ }C$ is drunk. You can take body temperature to be ${37}^{\circ }C$ and 1.7 x 10^{-5 0}C, bulk modulus for copper =$140\times {10}^{9}N/m^2.$

Answer:

Let $\mathrm{\Delta }T$ be the change in temperature, $\mathrm{\Delta }T=57-37={20}^{\circ }C$
Let $\alpha$ be linear expansion of body, $\alpha =\frac{1.7\times {10}^{-5}}{K}$
& ϒ be the cubical expansion $=3\alpha =3\times 1.7\times {10}^{-5}$
$=\frac{5.1\times {10}^{-5}}{K}$
Let V be the volume of the cavity and $\mathrm{\Delta }V$ be the increase in its volume which is a result of increase in temperature by $\mathrm{\Delta }T$.
$\mathrm{\Delta }V=\gamma V.\mathrm{\Delta }T$
$\frac{\mathrm{\Delta }V}{V}=\gamma \mathrm{\Delta }T$
Now, we know that,
Thermal stress production = B x Volumetric strain
$=B.\frac{\mathrm{\Delta }V}{V}$
$=B.\gamma \mathrm{\Delta }T$
$=140\times {10}^9\times 5.1\times {10}^{-5}\times 20$
$=1.428\times {10}^{8}N{m}^{-2}$
Thus, the stress is $1.01\times {10}^{5}N{m}^{-2}$, viz., ${10}^3$ times of atmospheric pressure.

Question:11.25

A rail track made of steel having length 10 m is clamped on a railway line at its two ends. On a summer day due to rise in temperature by ${20}^{\circ }C$, it is deformed as shown in the figure. Find x if $\alpha steel=\frac{1.2\times {10}^{-5}}{{}^{\circ }C}$.

Answer:

Given data:
$\alpha steel=\frac{1.2\times {10}^{-5}}{{}^{\circ }C}$
L₀ = 10m
& $\mathrm{\Delta }T={20}^{\circ }C$
Now, by using Pythagoras theorem,
$x^2={[(L+\mathrm{\Delta }L)\frac{1}{2}]}^2-{\left(\frac{L}{2}\right)}^2$
$=\frac{1}{4}[L^2+\mathrm{\Delta }{L}^2+2L\mathrm{\Delta }L]-\frac{L^2}{4}$
Thus, $x=\frac{L^2}{4}+\frac{\mathrm{\Delta }{L}^2}{4}+\frac{2L\mathrm{\Delta }L}{4}-\frac{L^2}{4}$ ……… (neglecting $\mathrm{\Delta }{L}^{2}since\mathrm{\Delta }{L}^2<<L$)
$x^2=\frac{2L\mathrm{\Delta }L}{4}$
Thus, $x=\frac{1}{2}\sqrt{2L\mathrm{\Delta }L}$
$\mathrm{\Delta }L=L_c\alpha .\mathrm{\Delta }T$
$=10\times 1.2\times {10}^{-5}\times 20$
$=240\times {10}^{-5}$
Thus, $x=\frac{1}{2}\sqrt{2\times 10\times 240\times {10}^{-5}}$
$=\frac{40}{2}\times \sqrt{0.3}\times {10}^{-2}$
$=10.8\times {10}^{-2}m$
Thus, $x=10.8cm$

Question:11.26

A thin rod having length Lo at $0^{\circ }C$ and coefficient of linear expansion α has its two ends maintained at temperatures ${\theta }_1$and ${\theta }_2$, respectively. Find its new length.

Answer:

From one end to another end the temperature of rod varies from ${\theta }_1$ to ${\theta }_2$
Mean temperature of rod= $\frac{{\theta }_1+{\theta }_2}{2^{\circ }C}$


Here, the rate of flow of heat from A to C to B is equal
Thus, ${\theta }_1>\theta >{\theta }_2$
Thus, $\frac{d\theta }{dt}=\frac{KA({\theta }_1-\theta )}{\frac{L_0}{2}}$
$=\frac{KA(\theta -{\theta }_2)}{\frac{L_0}{2}}$
Here K is the coefficient of thermal conductivity.
Therefore, ${\theta }_1-\theta =\theta -{\theta }_2$
$\theta ={\theta }_1+\frac{{\theta }_2}{2}$
Thus, $L=L_0(1+\alpha \theta )$
$=L_0[1+\alpha \left(\frac{{\theta }_1+{\theta }_2}{2}\right)]$

Question:11.27

According to Stefan’s law of radiation, a black body radiates energy $\sigma T^4$ from its unit surface area every second where T is the surface temperature of the black body and $\sigma =5.67\times {10}^{-8}W/m^2{K}^4$ is known as Stefan’s constant. A nuclear weapon may be thought of as a ball of radius 0.5 m. When denoted, it reaches temperature of 106K and can be treated as a black body.
a) estimate the power it radiates
b) if surrounding has water at ${30}^{\circ }C$, how much water can 10% of the energy produced evaporate in 1 sec?
c) if all this energy U is in the form of radiation, corresponding momentum is $p=\frac{U}{c}$. How much momentum per unit time does it impart on unit area at a distance of 1 km?

Answer:

Given: E = $\sigma T^4$ per sec per sq. m
Thus, total E is equal to radiated from all surface area A per second will be power radiated by nuclear weapon
Thus, P = $\sigma A{T}^4$
Now, $\sigma =5.67\times {10}^{-8}W/m^2{K}^4$ &
$P=5.67\times {10}^{-8}\times (4\times \pi R^2)({10}^6{)}^4$
$=5.67\times 4\times 3.14\times \times 0.5\times 0.5\times {10}^{-8}\times {10}^{24}$ …. (R = 0.5m, T = 106K)
$=5.67\times 3.14\times {10}^{24-8}\times 1$
Thus, $P\cong 18\times {10}^{16}Watt$
$=1.8\times {10}^{17}J/s$ …… (i)
(b) $P=18\times {10}^{16}Watt$ from (i)
Now, for evaporation of water 10% of this power is required.
Thus, $E=\frac{10}{100}\times 18\times {10}^{16}Watt$
$=1.8\times {10}^{17}J/s$
Now, energy required by m kg of water at ${30}^{\circ }C$ to evaporate at ${100}^{\circ }C$ = E required to heat up water from ${30}^{\circ }C+{100}^{\circ }C$+ E required to evaporate water into vapour
$=m{S}_w(T_2-T_1)+mL$
$=m[S_w(T_2-T_1)+L]$
Thus,
$18\times {10}^{16}=m[4180(100-30)+22.6\times {10}^5]$
$m=\frac{18\times {10}^{16}}{25.5\times {10}^5}$
$\cong 7\times {10}^{9}kg$
(c) Now, we know that,
Momentum per unit time $p^{{}^{\prime }}=\frac{U}{c}$
$=\frac{18\times {10}^{16}}{3\times {10}^8}$

Thus,
$p^{{}^{\prime }}=6\times {10}^{8}kgm{s}^{-1}$
Required momentum per unit time
$=\frac{6\times {10}^8}{4\times 3.14\times ({10}^3{)}^2}$
$\frac{6\times {10}^8}{4\times 3.14\times {10}^6}=47.77\frac{kgm{s}^{-1}}{m^2}$