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.

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.

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

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 $4^{\circ} C$.

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.

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

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.

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.

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.

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 $C D$.

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.

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 $]$