Find out the increase in moment of inertia $I$ of a uniform rod (coefficient of linear expansion $\alpha$ ) about its perpendicular bisector when its temperature is slightly increased by $\Delta T$.
Let the mass and length of a uniform rod be $M$ and $l$ respectively.
Moment of inertia of the rod about its perpendicular bisector. $(I)=\frac{M l^2}{12}$
$$\begin{aligned} &\text { Increase in length of the rod when temperature is increased by } \Delta T \text {, is given by }\\ &\begin{aligned} \Delta l & =l \cdot \alpha \Delta T \quad \text{... (i)}\\ \therefore \text { New moment of inertia of the } \operatorname{rod}(I) & =\frac{M}{12}(l+\Delta l)^2=\frac{M}{12}\left(l^2+\Delta l^2+2 I \Delta l\right) \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { As change in length } \Delta l \text { is very small, therefore, neglecting }(\Delta l)^2 \text {, we get }\\ &\begin{aligned} I^{\prime} & =\frac{M}{12}\left(l^2+2 l \Delta l\right) \\ & =\frac{M l^2}{12}+\frac{M I \Delta l}{6}=l+\frac{M I \Delta l}{6} \end{aligned} \end{aligned}$$
$\therefore$ Increase in moment of inertia $$ \begin{aligned} & \Delta I=l-I=\frac{M l \Delta l}{6}=2 \times\left(\frac{M l^2}{12}\right) \frac{\Delta l}{l} \\ & \Delta I=2 \cdot I \alpha \Delta T \quad \text{[Using Eq. (i)]} \end{aligned}$$
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 snow balls and throw around. Explain the formation of ball out of crushed ice or snow in the light of $p-T$ diagram of water.
Refer to the $\mathrm{p}-T$ diagram of water and double headed arrow. Increasing pressure at $0^{\circ} \mathrm{C}$ and 1 atm takes ice into liquid state and decreasing pressure in liquid state at $0^{\circ} \mathrm{C}$ 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.
100 g of water is supercooled to $-10^{\circ} \mathrm{C}$. 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? $\left[S_w=1 \mathrm{cal} / \mathrm{g} /{ }^{\circ} \mathrm{C}\right.$ and $\left.L_{\text {Fusion }}^w=80 \mathrm{cal} / \mathrm{g}\right]$
Given, mass of water $(m)=100$
Change in temperature $\Delta T=0-(-10)=10^{\circ} \mathrm{C}$
Specific heat of water $\left(S_w\right)=1 \mathrm{cal} / \mathrm{g} /{ }^{\circ} \mathrm{C}$
Latent heat of fusion of water $L_{\text {fusion }}^w=80 \mathrm{cal} / \mathrm{g}$
Heat required to bring water in super cooling from $-10^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$,
$$\begin{aligned} Q & =m s_w \Delta T \\ & =100 \times 1 \times 10=1000 \mathrm{cal} \end{aligned}$$
$$\begin{aligned} &\text { Let } m \text { gram of ice be melted. }\\ &\begin{array}{ll} \therefore & Q=m L \\ \text { or } & m=\frac{Q}{L}=\frac{1000}{80}=12.5 \mathrm{~g} \end{array} \end{aligned}$$
As small mass of ice is melted, therefore the temperature of the mixture will remain 0$$^\circ$$C.
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 (at room temperature) to bring mixture to a comfortable temperature. Suddenly Ramesh had to attend to something which would take some times, say 5-10 min before he could take bath. Now, he had two options (i) fill the remaining bucket completely by cold water and then attend to the work, (ii) 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
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.
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 (both components) would change in such a way that difference between their length $B$ remain constant. If $\alpha_{\text {iron }}=1.2 \times 10^{-5} / \mathrm{K}$ and $\alpha_{\text {brass }}=1.8 \times 10^{-5} / \mathrm{K}$, what should we take as length of each strip?
According to question $l_{\text {iron }}-l_{\text {brass }}=10 \mathrm{~cm}=$ constant at all temperatures
Let $l_0$ be length at temperature $0^{\circ} \mathrm{C}$ and $l$ be the length after change in temperature of $\Delta t$.
Now, we can write $\quad l_{\text {iron }}-l_{\text {brass }}=10 \mathrm{~cm}$ at all temperatures
$$\begin{aligned} l_{\text {iron }}\left(1+\alpha_{\text {iron }} \Delta t\right)-l_{\text {brass }}\left(1+\alpha_{\text {brass }} \Delta t\right) & =10 \mathrm{~cm} \\ l_{\text {iron }} \alpha_{\text {iron }} & =l_{\text {brass }} \alpha_{\text {brass }} \\ \therefore\quad \frac{l_{\text {iron }}}{l_{\text {brass }}} & =\frac{1.8}{1.2}=\frac{3}{2} \\ \therefore \quad \frac{1}{2} l_{\text {brass }} & =10 \mathrm{~cm} \\ \Rightarrow \quad l_{\text {brass }} & =20 \mathrm{~cm} \text { and } l_{\text {iron }} 30 \mathrm{~cm} \end{aligned}$$