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 $$m=\frac{1.8 \times 10^{16}}{25.33 \times 10^5}=7.0 \times 10^9 \mathrm{~kg}$$
$$\begin{aligned} &\text { (c) Momentum per unit time, }\\ &p=\frac{U}{C}=\frac{U}{C}=\frac{1.8 \times 10^{17}}{3 \times 10^8}=6 \times 10^8 \mathrm{~kg}-\mathrm{m} / \mathrm{s}^2 \quad\left[\begin{array}{l} P=\text { momentum } \\ V=\text { energy } \\ C=\text { velocity of Light } \end{array}\right] \end{aligned}$$
$$\begin{aligned} &\text { Momentum per unit time per unit }\\ &\text { area } p=\frac{p}{4 \pi R^2}=\frac{6 \times 10^8}{4 \times 3.14 \times\left(10^3\right)^2}\\ &\Rightarrow \quad d=47.7 \mathrm{~N} / \mathrm{m}^2 \quad\left[4 \pi R^2=\text { Surface area }\right] \end{aligned}$$