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
$$\begin{aligned} l_{\text {iron }}\left(1+\alpha_{\text {iron }} \Delta t\right)-l_{\text {brass }}\left(1+\alpha_{\text {brass }} \Delta t\right) & =10 \mathrm{~cm} \\ l_{\text {iron }} \alpha_{\text {iron }} & =l_{\text {brass }} \alpha_{\text {brass }} \\ \therefore\quad \frac{l_{\text {iron }}}{l_{\text {brass }}} & =\frac{1.8}{1.2}=\frac{3}{2} \\ \therefore \quad \frac{1}{2} l_{\text {brass }} & =10 \mathrm{~cm} \\ \Rightarrow \quad l_{\text {brass }} & =20 \mathrm{~cm} \text { and } l_{\text {iron }} 30 \mathrm{~cm} \end{aligned}$$
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 $L=L_0\left[1+\alpha\left(\frac{\theta_1+\theta_2}{2}\right)\right]$