We would like to make a vessel whose volume does not change with temperature (take a hint from the problem above). We can use brass and iron $\left(\beta_{\text {vbrass }}=6 \times 10^{-5} / K\right.$ and $\left.\beta_{\text {viron }}=3.55 \times 10^{-5} / \mathrm{K}\right)$ to create a volume of 100 cc . How do you think you can achieve this?
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 $0^{\circ} \mathrm{C}$
$V_{i,}, V_b$ be the volume of iron and brass vessel at $\Delta \theta^{\circ} \mathrm{C}$,
$\gamma_i, \gamma_b$ be the coefficient, of volume expansion of iron and brass.
$$\text { As per question, } \quad V_{i o}-V_{b o}=100 \mathrm{cc}=V_i-V_b\quad \text{... (i)}$$
$$\text { Now, } \quad V_i=V_{i 0}\left(1+\gamma_i \Delta \theta\right)$$
$$\begin{aligned} V_b & =V_{b o}\left(1+\gamma_b \Delta \theta\right) \\ V_i-V_b & =\left(V_{i o}-V_{b o}\right)+\Delta \theta\left(V_{i o} \gamma_i-V_{b o} \gamma_b\right) \end{aligned}$$
Since, $V_i-V_b=$ constant.
So, $$V_{i o} \gamma_i=V_{b o} \gamma_b$$
$$\begin{aligned} & \Rightarrow \quad \frac{V_{i o}}{V_{b o}}=\frac{\gamma_b}{\gamma_i}=\frac{\frac{3}{2} \beta_b}{\frac{3}{2} \beta_i}=\frac{\beta_b}{\beta_i}=\frac{6 \times 10^{-5}}{3.55 \times 10^{-5}}=\frac{6}{3.55} \\ & \frac{V_{i o}}{V_{b o}}=\frac{6}{3.55}\quad \text{.... (ii)} \end{aligned}$$
$$\begin{aligned} &\text { Solving Eqs. (i) and (ii), we get }\\ &\begin{aligned} V_{i o} & =244.9 \mathrm{cc} \\ V_{b o} & =144.9 \mathrm{cc} \end{aligned} \end{aligned}$$
Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of $57^{\circ} \mathrm{C}$ is drunk. You can take body (tooth) temperature to be $37^{\circ} \mathrm{C}$ and $\alpha=1.7 \times 10^{-5} /{ }^{\circ} \mathrm{C}$ bulk modulus for copper $=140 \times 10^9 \mathrm{~N} / \mathrm{m}^2$
Given, decrease in temperature $(\Delta t)=57-37=20^{\circ} \mathrm{C}$
Coefficient of linear expansion $(\alpha)=1.7 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
Bulk modulus for copper $(B)=140 \times 10^9 \mathrm{~N} / \mathrm{m}^2$
Coefficient of cubical expansion $(\gamma)=3 \alpha=5.1 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
Let initial volume of the cavity be $V$ and its volume increases by $\Delta V$ due to increase in temperature.
$$\begin{array}{ll} \therefore & \Delta V=\gamma V \Delta t \\ \Rightarrow & \frac{\Delta V}{V}=\gamma \Delta t\quad \text{... (i)} \end{array}$$
Thermal stress produced $=B \times$ Volumetric strain
$$\begin{aligned} & =B \times \frac{\Delta V}{V}=B \times \gamma \Delta t \\ & =140 \times 10^9 \times\left(5.1 \times 10^{-5} \times 20\right) \\ & =1,428 \times 10^8 \mathrm{~N} / \mathrm{m}^2 \end{aligned}$$
This is about $10^3$ times of atmospheric pressure.
A rail track made of steel having length 10 m is clamped on a railway line at its two ends (figure). On a summer day due to rise in temperature by $20^{\circ} \mathrm{C}$. It is deformed as shown in figure. Find $x$ (displacement of the centre) if $\alpha_{\text {steel }}=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}$.
Consider the diagram,
$$\begin{aligned} &\text { Applying Pythagorus theorem in right angled triangle in figure. }\\ &\begin{aligned} \left(\frac{L+\Delta L}{2}\right)^2 & =\left(\frac{L}{2}\right)^2+x^2 \\ \Rightarrow \quad x & =\sqrt{\left(\frac{L+\Delta L}{2}\right)^2-\left(\frac{L}{2}\right)^2} \\ & =\frac{1}{2} \sqrt{(L+\Delta L)^2-L^2} \\ & =\frac{1}{2} \sqrt{\left(L^2+\Delta L^2+2 L \Delta L\right)-L^2} \\ & =\frac{1}{2} \sqrt{\left(\Delta L^2+2 L \Delta L\right)} \end{aligned} \end{aligned}$$
As increase in length $\Delta L$ is very small, therefore, neglecting $(\Delta L)^2$, we get
$$\begin{aligned} x & =\frac{1}{2} \times \sqrt{2 L \Delta L} \quad \text{... (i)}\\ \text{But}\quad \Delta L & =L \alpha \Delta t \quad \text{... (ii)} \end{aligned}$$
$$\begin{aligned} &\text { Substituting value of } \Delta L \text { in Eq. (i) from Eq. (ii) }\\ &\begin{aligned} x & =\frac{1}{2} \sqrt{2 L \times L \alpha \Delta t}=\frac{1}{2} L \sqrt{2 \alpha \Delta t} \\ & =\frac{10}{2} \times \sqrt{2 \times 1.2 \times 10^{-5} \times 20} \\ & =5 \times \sqrt{4 \times 1.2 \times 10^{-4}} \\ & =5 \times 2 \times 1.1 \times 10^{-2}=0.11 \mathrm{~m}=11 \mathrm{~cm} \end{aligned} \end{aligned}$$
A thin rod having length $L_0$ at $0^{\circ} \mathrm{C}$ and coefficient of linear expansion $\alpha$ has its two ends maintained at temperatures $\theta_1$ and $\theta_2$, respectively. Find its new length.
Consider the diagram
$$\theta=\frac{\theta_1+\theta_2}{2}$$
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,
$$\left(\frac{d Q}{d t}\right)=\frac{K A\left(\theta_1-\theta\right)}{\left(L_0 / 2\right)}=\frac{K A\left(\theta-\theta_2\right)}{\left(L_0 / 2\right)}$$
where, $K$ is coefficient of thermal conductivity of the rod.
or $\Rightarrow$ $\theta_1-\theta=\theta-\theta_2$
or $\Rightarrow$ $\theta=\frac{\theta_1+\theta_2}{2}$
Using relation, $L=L_0(1+\alpha \theta)$
or $L=L_0\left[1+\alpha\left(\frac{\theta_1+\theta_2}{2}\right)\right]$
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 $\sigma=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \mathrm{~K}^4$ is known as Stefan's constant. A nuclear weapon may be thought of as a ball of radius 0.5 m . When detonated, it reaches temperature of $10^6 \mathrm{~K}$ and can be treated as a black body.
(a) Estimate the power it radiates.
(b) If surrounding has water at $30^{\circ}$, how much water can $10 \%$ of the energy produced evaporate in 1 s ?
$$\left[S_w=4186.0 \mathrm{~J} / \mathrm{kg} \mathrm{K} \text { and } L_v=22.6 \times 10^5 \mathrm{~J} / \mathrm{kg}\right]$$
(c) If all this energy $U$ is in the form of radiation, corresponding momentum is $p=U / c$. How much momentum per unit time does it impart on unit area at a distance of 1 km ?
$$\begin{aligned} & \text { Given, } \sigma=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \mathrm{~kg} \\ & \text { Radius, }=R=0.5 \mathrm{~m}, T=10^6 \mathrm{~K} \end{aligned}$$
$$\begin{aligned} &\text { (a) Power radiated by Stefan's law }\\ &\begin{aligned} P & =\sigma A T^4=\left(4 \pi R^2\right) T^4 \\ & =\left(5.67 \times 10^{-4} \times 4 \times(3.14) \times(0.5)^2 \times\left(10^6\right) 4\right. \\ & =1.78 \times 10^{17} \mathrm{~J} / \mathrm{s}=1.8 \times 10^{17} \mathrm{~J} / \mathrm{s} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { (b) Energy available per second, } U=1.8 \times 10^{17} \mathrm{~J} / \mathrm{s}=18 \times 10^{16} \mathrm{~J} / \mathrm{s}\\ &\begin{aligned} \text { Actual energy required to evaporate water } & =10 \% \text { of } 1.8 \times 10^{17} \mathrm{~J} / \mathrm{s} \\ & =1.8 \times 10^{16} \mathrm{~J} / \mathrm{s} \end{aligned} \end{aligned}$$
Energy used per second to raise the temperature of $m \mathrm{~kg}$ of water from $30^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$ and then into vapour at $100^{\circ} \mathrm{C}$
$$\begin{aligned} & =m s_w \Delta \theta+m L_v=m \times 4186 \times(100-30)+m \times 22.6 \times 10^5 \\ & =2.93 \times 10^5 m+22.6 \times 10^5 \mathrm{~m}=25.53 \times 10^5 \mathrm{~m} \mathrm{~J} / \mathrm{s} \end{aligned}$$
As per question, $25.53 \times 10^5 \mathrm{~m}=1.8 \times 10^{16}$
or $$m=\frac{1.8 \times 10^{16}}{25.33 \times 10^5}=7.0 \times 10^9 \mathrm{~kg}$$
$$\begin{aligned} &\text { (c) Momentum per unit time, }\\ &p=\frac{U}{C}=\frac{U}{C}=\frac{1.8 \times 10^{17}}{3 \times 10^8}=6 \times 10^8 \mathrm{~kg}-\mathrm{m} / \mathrm{s}^2 \quad\left[\begin{array}{l} P=\text { momentum } \\ V=\text { energy } \\ C=\text { velocity of Light } \end{array}\right] \end{aligned}$$
$$\begin{aligned} &\text { Momentum per unit time per unit }\\ &\text { area } p=\frac{p}{4 \pi R^2}=\frac{6 \times 10^8}{4 \times 3.14 \times\left(10^3\right)^2}\\ &\Rightarrow \quad d=47.7 \mathrm{~N} / \mathrm{m}^2 \quad\left[4 \pi R^2=\text { Surface area }\right] \end{aligned}$$