Thinking Process

The metallic strip with higher coefficient of linear expansion $(\alpha_{\text {Al }})$ will expand more.

Answer (d) As $\alpha_{\text {Al }}>\alpha$ steal, aluminium will expand more. So, it should have larger radius of curvature. Hence, aluminium will be on convex side.

• (a) remain straight: This option is incorrect because the two metals have different coefficients of thermal expansion. When heated, they will expand by different amounts, causing the strip to bend rather than remain straight.

• (b) get twisted: This option is incorrect because twisting would imply a non-uniform expansion or some form of torsional stress, which is not the case here. The strip will bend uniformly due to the difference in thermal expansion coefficients.

• (c) will bend with aluminium on concave side: This option is incorrect because aluminium has a higher coefficient of thermal expansion than steel. Therefore, aluminium will expand more and will be on the convex side, not the concave side.

Answer (b) As the rod is heated, it expands. No external torque is acting on the system so angular momentum should be conserved.

$$ \begin{aligned} & \quad L=\text { Angular momentum }=I \omega=\text { constant } \\ & \Rightarrow \quad I_1 \omega_1=I_2 \omega_2 \end{aligned} $$

Due to expansion of the rod $I_2>I_1$

$$ \begin{matrix} \Rightarrow & \frac{\omega_2}{\omega_1}=\frac{I_1}{I_2}<1 \\ \Rightarrow & \omega_2<\omega_1 \end{matrix} $$

So, angular velocity (speed of rotation) decreases.

• Option (a) its speed of rotation increases: This is incorrect because when the rod is heated and expands, its moment of inertia increases. Since no external torque is acting on the system, angular momentum must be conserved. An increase in the moment of inertia results in a decrease in angular velocity, not an increase.

• Option (c) its speed of rotation remains same: This is incorrect because the expansion of the rod due to heating changes its moment of inertia. Since angular momentum is conserved and the moment of inertia increases, the angular velocity must decrease to maintain the conservation of angular momentum.

• Option (d) its speed increases because its moment of inertia increases: This is incorrect because an increase in the moment of inertia actually leads to a decrease in angular velocity when angular momentum is conserved. The relationship is inversely proportional, not directly proportional.

Answer (b) It is clear from the graph that lowest point for scale $A$ is $30^{\circ}$ and lowest point for scale $B$ is $0^{\circ}$. Highest point for the scale $A$ is $180^{\circ}$ and for scale $B$ is $100^{\circ}$. Hence, correct relation is

$\Rightarrow \frac{t_{A}-30}{180-30}=\frac{t_{B}-0}{100-0}$

$\Rightarrow \quad \frac{t_{A}-30}{150}=\frac{t_{B}}{100}$

where, LFP $\rightarrow$ Lower fixed point

UFP $\rightarrow$ Upper fixed point

• Option (a): The equation $\frac{t_{A}-180}{100}=\frac{t_{B}}{150}$ is incorrect because it incorrectly assumes that the lower fixed point for scale $A$ is $180^{\circ}$ and the lower fixed point for scale $B$ is $0^{\circ}$. The correct lower fixed point for scale $A$ is $30^{\circ}$, not $180^{\circ}$.

• Option (c): The equation $\frac{t_{B}-180}{150}=\frac{t_{A}}{100}$ is incorrect because it incorrectly assumes that the lower fixed point for scale $B$ is $180^{\circ}$. The correct lower fixed point for scale $B$ is $0^{\circ}$, not $180^{\circ}$.

• Option (d): The equation $\frac{t_{B}-40}{100}=\frac{t_{A}}{180}$ is incorrect because it incorrectly assumes that the lower fixed point for scale $B$ is $40^{\circ}$. The correct lower fixed point for scale $B$ is $0^{\circ}$, not $40^{\circ}$. Additionally, it incorrectly assumes that the upper fixed point for scale $A$ is $180^{\circ}$ without considering the correct lower fixed point for scale $A$.

Thinking Process

Density of water is maximum at $4{ }^{\circ} C$, this is because of anomalous expansion of water.

Answer (a) Let volume of the sphere is $V$ and $\rho$ is its density, then we can write buoyant force

$\begin{aligned} & F=V \rho G \quad(g=\text { acceleration due to gravity }) \\ & \Rightarrow \quad F \propto \rho \quad(\because V \text { and } g \text { are almost constant }) \\ & \Rightarrow \quad \frac{F_{4 C}}{F_{0 C}}=\frac{\rho_{4 C}}{\rho_{0 C}}>1 \quad\left(\because \rho_{4 C}>\rho_{0 C}\right) \\ & \Rightarrow \quad F_{4 C}>F_{0 C} \\ \end{aligned}$

Hence, 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 $4^{\circ} C$: This is incorrect because the density of water is higher at $4^{\circ} C$ than at $0^{\circ} C$. Since buoyant force is directly proportional to the density of the fluid, the buoyant force will be greater in water at $4^{\circ} C$.

• (c) Buoyancy in water at $0^{\circ} C$ will be same as that in water at $4^{\circ} C$: This is incorrect because the density of water changes with temperature. The density of water is maximum at $4^{\circ} C$, so the buoyant force will not be the same at $0^{\circ} C$ and $4^{\circ} C$.

• (d) Buoyancy may be more or less in water at $4^{\circ} C$ depending on the radius of the sphere: This is incorrect because the buoyant force depends on the volume of the displaced fluid and the density of the fluid, not on the radius of the sphere. The density of water at $4^{\circ} C$ is higher than at $0^{\circ} C$, so the buoyant force will always be greater at $4^{\circ} C$ regardless of the radius of the sphere.

Answer (a) As the temperature is increased length of the pendulum increases. We know that time period of pendulum $T=2 \pi \sqrt{\frac{L}{g}}$

$$ \Rightarrow \quad T \propto \sqrt{L} \text {, as } L \text {, increases. } $$

So, time period $(T)$ also increases.

• (b) The period of a pendulum does not decrease as its effective length increases. According to the formula ( T = 2\pi \sqrt{\frac{L}{g}} ), the period ( T ) is directly proportional to the square root of the length ( L ). Therefore, if the length increases, the period must also increase, not decrease.

• (c) The period of a pendulum increases due to the increase in the effective length, not because of the shifting of the center of mass below the center of the bob. The center of mass of the bob remains at the center, and the increase in period is solely due to the increase in length.

• (d) The period of a pendulum does not decrease if the effective length remains the same. The formula ( T = 2\pi \sqrt{\frac{L}{g}} ) shows that the period is dependent on the length ( L ). If the length does not change, the period will not decrease. Additionally, the center of mass shifting above the center of the bob is not a factor in the period of a pendulum.

Answer (a) We know that as temperature increases vibration of molecules about their mean position increases hence, kinetic energy associated with random motion of molecules increases.

• (b) Kinetic energy of orderly motion of molecules: Heat is not associated with the orderly motion of molecules. Orderly motion refers to a uniform or structured movement, which is characteristic of macroscopic objects rather than the microscopic random motion of molecules that constitutes heat.

• (c) Total kinetic energy of random and orderly motion of molecules: Heat specifically refers to the kinetic energy due to random motion of molecules. The inclusion of orderly motion in this option makes it incorrect because heat does not account for the kinetic energy from orderly motion.

• (d) Kinetic energy of random motion in some cases and kinetic energy of orderly motion in other: Heat is consistently associated with the random motion of molecules. It does not switch between random and orderly motion depending on the case. This option is incorrect because it suggests a variability that does not exist in the definition of heat.

Answer (d) Let the radius of the sphere is $R$. As the temperature increases radius of the sphere increases as shown.

$ \text { Original volume } V_0=\frac{4}{3} \pi R^{3} $

Coefficient of linear expansion $=\alpha$

$\therefore \quad$ Coefficient of volume expansion $=3 \alpha$

$\therefore \quad \frac{1}{V} \frac{d V}{d T}=3 \alpha \Rightarrow d V=3 V \alpha d t=4 \pi R^{3} \alpha \Delta T$

$=$ Increase in the volume

• Option (a): $2 \pi R \alpha \Delta T$

  • This option is incorrect because it suggests a linear relationship between the increase in volume and the temperature change. However, the volume expansion of a sphere is a cubic relationship, not linear. The correct factor should involve $R^3$ rather than $R$.

• Option (b): $\pi R^{2} \alpha \Delta T$

  • This option is incorrect because it suggests a quadratic relationship between the increase in volume and the temperature change. The volume expansion of a sphere is a cubic relationship, not quadratic. The correct factor should involve $R^3$ rather than $R^2$.

• Option (c): $4 \pi R^{3} \alpha \Delta T / 3$

  • This option is incorrect because it underestimates the increase in volume by a factor of 3. The coefficient of volume expansion is $3\alpha$, and when applied to the original volume $\frac{4}{3} \pi R^3$, the correct increase in volume should be $4 \pi R^3 \alpha \Delta T$, not $\frac{4 \pi R^3 \alpha \Delta T}{3}$.

Thinking Process

In this problem the cooling will be in the form of radiations that is according to Stefan’s law. Since, emissive power directly proportional to surface. Here, for given volume, sphere have least surface area and circular plate of greatest surface area.

Answer (c) Consider the diagram where all the three objects are heated to same temperature $T$. We know that density, $\rho=\frac{\text { mass }}{\text { volume }}$ as $\rho$ is same for all the three objects hence, volume will also be same.

Sphere

Cube

Plate

As thickness of the plate is least hence, surface area of the plate is maximum.

We know that, according to Stefan’s law of heat loss $H \alpha A T^{4}$ where, $A$ is surface area of for object and $T$ is temperature.

Hence,

$$ \begin{aligned} & H_{\text {sphere }}: H_{\text {cube }}: H_{\text {plate }} \\ = & A_{\text {sphere }}: A_{\text {cube }}: A_{\text {plate }} \end{aligned} $$

As $A_{\text {plate }}$ is maximum.

Hence, the plate will cool fastest.

As, the sphere is having minimum surface area hence, the sphere cools slowest.

• Option (a): Plate will cool fastest and cube the slowest

  • Incorrect because the sphere has a smaller surface area compared to the cube. Since the rate of cooling is proportional to the surface area, the cube will not cool the slowest; the sphere will.

• Option (b): Sphere will cool fastest and cube the slowest

  • Incorrect because the sphere has the smallest surface area among the three objects. According to Stefan’s law, the object with the smallest surface area will cool the slowest, not the fastest.

• Option (d): Cube will cool fastest and plate the slowest

  • Incorrect because the plate has the largest surface area, which means it will lose heat the fastest according to Stefan’s law. The cube, having a larger surface area than the sphere but smaller than the plate, will not cool the fastest.

Answer $(b, d)$

According to question

$$T_x=T _y \quad(\because \text{x and y are in thermal equilibrium })$$ $$T _x \neq T_z \quad(\because \text{ x is not in thermal equilibrium with z})$$ $$ \text{Clearly}, T _y \neq T _z$$

Hence, $\quad y$ and $z$ are not in thermal equilibrium. (d) Given, $T_{x} \neq T_{y}$ and $T_{x} \neq T_{z}$

We cannot say about equilibrium of $Y$ and $Z$, they may or may not be in equilibrium.

• (a) A system (X) is in thermal equilibrium with (Y) but not with (Z). The systems (Y) and (Z) may be in thermal equilibrium with each other.

  • This option is incorrect because if (X) is in thermal equilibrium with (Y) ((T_x = T_y)) and (X) is not in thermal equilibrium with (Z) ((T_x \neq T_z)), then (T_y \neq T_z). Therefore, (Y) and (Z) cannot be 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.

  • This option is incorrect because if (X) is not in thermal equilibrium with (Y) ((T_x \neq T_y)) and (X) is not in thermal equilibrium with (Z) ((T_x \neq T_z)), there is no information to conclude that (Y) and (Z) must be in thermal equilibrium with each other. They may or may not be in thermal equilibrium.

Answer $(b, c)$

Smaller gulab jamuns are having least surface area hence, they will be heated first.

As in case of smaller gulab jamun heat radiates will be less

Similarly, smaller pizzas are heated before bigger ones because they are of small surface areas.

• Option (a): Both size gulab jamuns will get heated in the same time

  • This option is incorrect because the rate of heating depends on the surface area to volume ratio. Smaller gulab jamuns have a higher surface area to volume ratio compared to bigger ones, which means they will heat up faster.

• Option (d): Bigger pizzas are heated before smaller

  • This option is incorrect because, similar to the gulab jamuns, the rate of heating for pizzas also depends on the surface area to volume ratio. Smaller pizzas have a higher surface area to volume ratio compared to bigger ones, which means they will heat up faster.

Thinking Process

During phase change process, temperature of the system remains constant.

Answer $(a, d)$

During the process $A B$ temperature of the system is $0^{\circ} C$ Hence, it represents phase change that is transformation of ice into water while temperature remains $0^{\circ} C$.

$B C$ represents rise in temperature of water from $0^{\circ} C$ to $100^{\circ} C$ (at $C$ ).

Now, water starts converting into steam which is represent by $C D$.

• Option (b): At point B, water does not start boiling. Instead, point B marks the end of the phase change from ice to water, and the temperature of the water begins to rise from 0°C to 100°C.

• Option (c): At point C, not all the water gets converted into steam. Point C represents the temperature at which water starts to boil (100°C), but the phase change from water to steam occurs over the region from C to D.

Answer $(b, c, d)$

When hot milk spread on the table heat is transferred to the surroundings by conduction, convection and radiation.

According to Newton’s law of cooling temperature of the milk falls off exponentially. Heat also will be transferred from surroundings to the milk but will be lesser than that of transferred from milk to the surroundings.

• Option (a) is incorrect because the rate of cooling is not constant. According to Newton’s law of cooling, the rate of cooling depends on the temperature difference between the milk and the surroundings, which decreases over time. Therefore, the rate of cooling decreases as the milk approaches the temperature of the surroundings.

Answer From the 1st observation $\alpha=\frac{\Delta l}{l \Delta T} \Rightarrow \alpha=\frac{4 \times 10^{-4}}{2 \times 10}=2 \times 10^{-5}{ }^{\circ} C^{-1}$

For 2nd observation

$$ \begin{aligned} \Delta l & =\alpha l \Delta T \\ & =2 \times 10^{-5} \times 1 \times 10=2 \times 10^{-4} m \neq 4 \times 10^{-4} m \text { (Wrong) } \end{aligned} $$

For 3rd observation $\Delta l=\alpha l \Delta T$

$$ =2 \times 10^{-5} \times 2 \times 20=8 \times 10^{-4} m \neq 2 \times 10^{-4} m(\text { Wrong }) $$

For 4th observation

$$ \Delta l=\alpha l \Delta T $$

$$ =2 \times 10^{-5} \times 3 \times 10=6 \times 10^{-4} m=6 \times 10^{-4} m $$

[i.e., observed value (Correct)]

Thinking Process

According to Kirchhoff’s law, good radiator are good absorbers.

Answer Due to difference in conductivity, metals having high conductivity compared to wood. On touch with a finger, heat from the surrounding flows faster to the finger from metals and so one feels the heat.

Similarly, when one touches a cold metal the heat from the finger flows away to the surroundings faster.

Answer Let $Q$ be the value of temperature having same value an Celsius and Fahrenheit scale. Now, we can write

$\frac{{ }^{\circ} F-32}{180} =\frac{{ }^{\circ} C}{100} $

$\Rightarrow \text { Let } \frac{F}{} =C=Q $

$\Rightarrow \frac{Q-32}{180} =\frac{Q}{100}=Q=-40^{\circ} C \text { or }-40^{\circ} F$

Answer As copper is a good conductor of heat as compared to steel. The steel utensils with copper bottom absorbs heat more quickly than steel and give it to the food in utensil. As a result, of it, the food in utensil is heated uniformly and quickly.

Thinking Process

As temperature increases length of the rod also increases hence, moment of inertia of the rod also increases.

Answer Let the mass and length of a uniform rod be $M$ and $l$ respectively.

Moment of inertia of the rod about its perpendicular bisector. $(I)=\frac{M l^{2}}{12}$

Increase in length of the rod when temperature is increased by $\Delta T$, is given by

$\begin{aligned} \Delta l & =l \cdot \alpha \Delta T \\ \therefore \text { New moment of inertia of the } rod(I) & =\frac{M}{12}(l+\Delta l)^{2}=\frac{M}{12}(l^{2}+\Delta l^{2}+2 I \Delta l)\end{aligned}$

As change in length $\Delta l$ is very small, therefore, neglecting $(\Delta l)^{2}$, we get

$\therefore$ Increase in moment of inertia

$$ \begin{aligned} I^{\prime} & =\frac{M}{12}(l^{2}+2 l \Delta l) \\ & =\frac{M l^{2}}{12}+\frac{M I \Delta l}{6}=l+\frac{M I \Delta l}{6} \end{aligned} $$

$\Delta I=l-I=\frac{M l \Delta l}{6}=2 \times(\frac{M l^{2}}{12}) \frac{\Delta l}{l}$

$\Delta I=2 \cdot I \alpha \Delta T$

[Using Eq. (i)]

Answer Refer to the $p-T$ diagram of water and double headed arrow. Increasing pressure at $0^{\circ} C$ and $1 atm$ takes ice into liquid state and decreasing pressure in liquid state at $0^{\circ} C$ and $1 atm$ takes water to ice state.

When crushed ice is squeezed, some of it melts, filling up gap between ice flakes upon releasing pressure. This water freezes, binding all ice flakes and making the ball more stable.

Answer Given, mass of water $(m)=100$

Change in temperature $\Delta T=0-(-10)=10^{\circ} C$

Specific heat of water $(S_{w})=1 cal / g /{ }^{\circ} C$

Latent heat of fusion of water $L_{\text {fusion }}^{w}=80 cal / g$

Heat required to bring water in super cooling from $-10^{\circ} C$ to $0^{\circ} C$,

$$ \begin{aligned} Q & =m s_{w} \Delta T \\ & =100 \times 1 \times 10=1000 cal \end{aligned} $$

Let $m$ gram of ice be melted.

$$ \begin{matrix} \therefore & Q=m L \\ \text { or } & m=\frac{Q}{L}=\frac{1000}{80}=12.5 g \end{matrix} $$

As small mass of ice is melted, therefore the temperature of the mixture will remain $0^{\circ} C$.

Note To find the temperature of the mixture we must go through the two steps $(Q=m s$ DT $)$ and $(Q=m L)$, we should not directly apply first one.

Thinking Process

We should apply logic in this problem in the context of Newton’s law of cooling which gives a consequence about rate of fall of temperature of a body with respect to the difference of temperature of body and surroundings.

Answer The first option would have kept water warmer because according to Newton’s law of cooling, the rate of loss of heat is directly proportional to the difference of temperature of the body and the surrounding and in the first case the temperature difference is less, so, rate of loss of heat will be less.

Answer According to question $l_{\text {iron }}-l_{\text {brass }}=10 cm=$ constant at all temperatures

Let $l_0$ be length at temperature $0^{\circ} C$ and $l$ be the length after change in temperature of $\Delta t$.

$$ \begin{aligned} & \text { Now, we can write } \quad l_{\text {iron }}-l_{\text {brass }}=10 cm \text { at all temperatures } \\ & l_{\text {iron }}(1+\alpha_{\text {iron }} \Delta t)-l_{\text {brass }}(1+\alpha_{\text {brass }} \Delta t)=10 cm \\ & l_{\text {ron }} \alpha_{\text {iron }}=l_{\text {brass }} \alpha_{\text {brass }} \\ & \frac{l_{\text {iron }}}{l_{\text {brass }}}=\frac{1.8}{1.2}=\frac{3}{2} \\ & \frac{1}{2} l_{\text {brass }}=10 cm \\ & \Rightarrow \quad l_{\text {brass }}=20 cm \text { and } l_{\text {iron }} 30 cm \end{aligned} $$

Answer In the previous problem the difference in the length was constant. In this problem the difference in volume is constant.

The situation is shown in the diagram.

Let $V_{i o}, V_{b o}$ be the volume of iron and brass vessel at $0^{\circ} C$ $V_{i,}, V_{b}$ be the volume of iron and brass vessel at $\Delta \theta^{\circ} C$,

$\gamma_{i}, \gamma_{b}$ be the coefficient, of volume expansion of iron and brass.

As per question, $\quad V_{i o}-V_{b o}=100 c c=V_{i}-V_{b}$

Now,

Since, $V_{i}-V_{b}=$ constant.

$$ \begin{aligned} V_{i 0}-V_{b o} & =100 c c=V_{i}-V_{b} \\ V_{i} & =V_{i 0}(1+\gamma_{i} \Delta \theta) \\ V_{b} & =V_{b o}(1+\gamma_{b} \Delta \theta) \\ V_{i}-V_{b} & =(V_{i o}-V_{b o})+\Delta \theta(V_{i o} \gamma_{i}-V_{b o} \gamma_{b}) \end{aligned} $$

So,

$$ \begin{aligned} & V_{i o} \gamma_{i}=V_{b o} \gamma_{b} \\ & \frac{V_{i o}}{V_{b o}}=\frac{\gamma_{b}}{\gamma_{i}}=\frac{\frac{3}{2} \beta_{b}}{\frac{3}{2} \beta_{i}}=\frac{\beta_{b}}{\beta_{i}}=\frac{6 \times 10^{-5}}{3.55 \times 10^{-5}}=\frac{6}{3.55} \\ & \frac{V_{i o}}{V_{b o}}=\frac{6}{3.55} \end{aligned} $$

Solving Eqs. (i) and (ii), we get

$$ \begin{aligned} V_{i 0} & =244.9 cc \\ V_{bo} & =144.9 cc \end{aligned} $$

Answer Given, decrease in temperature $(\Delta t)=57-37=20^{\circ} C$

Coefficient of linear expansion $(\alpha)=1.7 \times 10^{-5} /{ }^{\circ} C$

Bulk modulus for copper $(B)=140 \times 10^{9} N / m^{2}$

Coefficient of cubical expansion $(\gamma)=3 \alpha=5.1 \times 10^{-5} /{ }^{\circ} C$

Let initial volume of the cavity be $V$ and its volume increases by $\Delta V$ due to increase in temperature. $\begin{matrix} \therefore & \Delta V=\gamma \vee \Delta t \\ \Rightarrow & \frac{\Delta V}{V}=\gamma \Delta t\end{matrix} $

Thermal stress produced $=B \times$ Volumetric strain

$$ \begin{aligned} & =B \times \frac{\Delta V}{V}=B \times \gamma \Delta t \\ & =140 \times 10^{9} \times(5.1 \times 10^{-5} \times 20) \\ & =1,428 \times 10^{8} N / m^{2} \end{aligned} $$

This is about $10^{3}$ times of atmospheric pressure.

Answer Consider the diagram.

Applying Pythagorus theorem in right angled triangle in figure.

$$ \begin{aligned} (\frac{L+\Delta L}{2})^{2} & =(\frac{L}{2})^{2}+x^{2} \\ x & =\sqrt{(\frac{L+\Delta L}{2})^{2}-(\frac{L}{2})^{2}} \\ & =\frac{1}{2} \sqrt{(L+\Delta L)^{2}-L^{2}} \\ & =\frac{1}{2} \sqrt{(L^{2}+\Delta L^{2}+2 L \Delta L)-L^{2}} \\ & =\frac{1}{2} \sqrt{(\Delta L^{2}+2 L \Delta L)} \end{aligned} $$

As increase in length $\Delta L$ is very small, therefore, neglecting $(\Delta L)^{2}$, we get

But

$$ \begin{aligned} x & =\frac{1}{2} \times \sqrt{2 L \Delta L} \\ \Delta L & =L \alpha \Delta t \end{aligned} $$

Substituting value of $\Delta L$ in Eq. (i) from Eq. (ii)

$$ \begin{aligned} x & =\frac{1}{2} \sqrt{2 L \times L \alpha \Delta t}=\frac{1}{2} L \sqrt{2 \alpha \Delta t} \\ & =\frac{10}{2} \times \sqrt{2 \times 1.2 \times 10^{-5} \times 20} \\ & =5 \times \sqrt{4 \times 1.2 \times 10^{-4}} \\ & =5 \times 2 \times 1.1 \times 10^{-2}=0.11 m=11 cm \end{aligned} $$

Note Here we have assumed $\Delta L$ to be very small so that it can be neglected compared to $L$.

Thinking Process

When temperature of a rod varies linearly, temperature of the middle point of the rod can be taken as mean of temperatures at the two ends.

Answer Consider the diagram

$$ \begin{gathered} \theta_1 \bigcirc \\ \theta=\frac{\theta_1+\theta_2}{2} \end{gathered} $$

Let temperature varies linearly in the rod from its one end to other end. Let $\theta$ be the temperature of the mid-point of the rod. At steady state,

Rate of flow of heat,

$$ (\frac{d Q}{d t})=\frac{K A(\theta_1-\theta)}{(L_0 / 2)}=\frac{K A(\theta-\theta_2)}{(L_0 / 2)} $$

where, $K$ is coefficient of thermal conductivity of the rod.

or $\Rightarrow$

$$ \theta_1-\theta=\theta-\theta_2 $$

or $\Rightarrow$

Using relation,

$$ \theta=\frac{\theta_1+\theta_2}{2} $$

or

$$ \begin{aligned} & L=L_0(1+\alpha \theta) \\ & L=L_0[1+\alpha(\frac{\theta_1+\theta_2}{2})] \end{aligned} $$

Answer Given, $\sigma=5.67 \times 10^{-8} W / m^{2} kg$

Radius, $=R=0.5 m, T=10^{6} K$ (a) Power radiated by Stefan’s law

$$ \begin{aligned} P & =\sigma A T^{4}=(4 \pi R^{2}) T^{4} \\ & =(5.67 \times 10^{-4} \times 4 \times(3.14) \times(0.5)^{2} \times(10^{6}) 4. \\ & =1.78 \times 10^{17} J / s=1.8 \times 10^{17} J / s \end{aligned} $$

(b) Energy available per second, $U=1.8 \times 10^{17} J / s=18 \times 10^{16} J / s$

Actual energy required to evaporate water $=10 %$ of $1.8 \times 10^{17} J / s$

$$ =1.8 \times 10^{16} J / s $$

Energy used per second to raise the temperature of $m kg$ of water from $30^{\circ} C$ to $100^{\circ} C$ and then into vapour at $100^{\circ} C$

$$ \begin{aligned} & =m s_{w} \Delta \theta+m L_{v}=m \times 4186 \times(100-30)+m \times 22.6 \times 10^{5} \\ & =2.93 \times 10^{5} m+22.6 \times 10^{5} m=25.53 \times 10^{5} m J / s \end{aligned} $$

As per question, $25.53 \times 10^{5} m=1.8 \times 10^{16}$

or

$$ m=\frac{1.8 \times 10^{16}}{25.33 \times 10^{5}}=7.0 \times 10^{9} kg $$

(c) Momentum per unit time,

$$ p=\frac{U}{C}=\frac{U}{C}=\frac{1.8 \times 10^{17}}{3 \times 10^{8}}=6 \times 10^{8} kg-m / s^{2} \quad[\begin{matrix} P=\text { momentum } \\ V=\text { energy } \\ C=\text { velocity of Light } \end{matrix} ] $$

Momentum per unit time per unit

$$ \text { area } p=\frac{p}{4 \pi R^{2}}=\frac{6 \times 10^{8}}{4 \times 3.14 \times(10^{3})^{2}} $$

$\Rightarrow \quad d=47.7 N / m^{2} \quad[4 \pi R^{2}=.$ Surface area $]$