**Problem (IIT JEE 2008):**
An ideal gas is expanding such that $PT^2=\text{constant}$. The coefficient of volume expansion of the gas is,

- $1/T$
- $2/T$
- $3/T$
- $4/T$

**Solution:**
The coefficient of volume expansion is defined as,
\begin{align}
\label{bga:eqn:1}
& \gamma=\frac{1}{V}\frac{\mathrm{d}V}{\mathrm{d}T}.
\end{align}
The ideal gas equation, $PV=nRT$, gives $P=nRT/V$. Substitute $P$ in $PT^2=\text{constant}$ to get,
\begin{align}
\label{bga:eqn:2}
& V=kT^3,
\end{align}
for some constant $k$. Differentiate second equation with respect to $T$ and substitute in first equation to get $\gamma=3/T$.

**Problem (IIT JEE 2004):**
A cube of coefficient of linear expansion $\alpha_s$ is floating in a bath containing a liquid of coefficient of volume expansion $\gamma_l$. When the temperature is raised by $\Delta T$, the depth up to which the cube is submerged in the liquid remains the same. Find the relation between $\alpha_s$ and $\gamma_l$ showing all the steps.

**Problem (IIT JEE 2003):**
Two rods, one of aluminium and the other made of steel, having initial lengths $l_1$ and $l_2$ are connected together to form a single rod of length $l_1+l_2$. The coefficients of linear expansion for aluminium and steel are $\alpha_a$ and $\alpha_s$, respectively. If the length of each rod increases by the same amount when their temperature are raised by $t\;\mathrm{{}^oC}$, then find the ratio $\frac{l_1}{l_1+l_2}$,

- $\frac{\alpha_s}{\alpha_a}$
- $\frac{\alpha_a}{\alpha_s}$
- $\frac{\alpha_s}{\alpha_a+\alpha_s}$
- $\frac{\alpha_a}{\alpha_a+\alpha_s}$

**Solution:**
After thermal expansion, length of the aluminium and the steel rod become,
\begin{align}
& l_1^\prime=l_1(1+\alpha_at),\\
& l_2^\prime=l_2(1+\alpha_st).
\end{align}
Since length of each rod is increased by the same amount, we get,
\begin{align}
&l_1^\prime-l_1=l_2^\prime-l_2.
\end{align}
Substitute $l_1^\prime$ and $l_2^\prime$ from first and second equation into third equation to get, $l_1\alpha_a=l_2\alpha_s$. Hence,
\begin{align}
\frac{l_1}{l_1+l_2}=\frac{l_1}{l_1+(l_1\alpha_a/\alpha_s)}=\frac{\alpha_s}{\alpha_a+\alpha_s}. \nonumber
\end{align}

**Problem (IIT JEE 1999): **
A bimetallic strip is formed out of two identical strips - one of copper and the other of brass. The coefficients of linear expansion of the two metals are $\alpha_C$ and $\alpha_B$. On heating, the temperature of the strip goes up by $\Delta T$ and the strip bends to form an arc of radius of curvature $R$. Then, $R$ is,

- proportional to $\Delta T$.
- inversely proportional to $\Delta T$.
- proportional to $|\alpha_B -\alpha_C|$.
- inversely proportional to $|\alpha_B -\alpha_C|$.

**Solution:**
Let $d$ and $l_0$ be width and initial length of the two strips. The length of the strips after heating becomes,
\begin{align}
l_C&=l_0(1+\alpha_C\Delta T), \nonumber\\
l_B&=l_0(1+\alpha_B\Delta T).\nonumber
\end{align}
The strips bend as shown in the figure. From figure, the length of the two strips are
\begin{align}
l_C&=R\theta, \nonumber\\
l_B&=(R+d)\theta. \nonumber
\end{align}
Divide to get,
\begin{align}
\label{jqa:eqn:1}
\frac{R+d}{R}&=\frac{l_B}{l_C}=\frac{1+\alpha_B\Delta T}{1+\alpha_C\Delta T}=(1+\alpha_B\Delta T)(1+\alpha_C\Delta T)^{-1}\nonumber\\
&\approx (1+\alpha_B\Delta T)(1-\alpha_C\Delta T)\approx 1+(\alpha_B-\alpha_C)\Delta T,
\end{align}
where we neglected higher order terms. Simplify above equation to get,
\begin{align}
\label{jga:eqn:2}
R=\frac{d}{(\alpha_B-\alpha_C)\Delta T}.
\end{align}
The bending shown in figure assume $\alpha_B > \alpha_C$ (which is generally the case for brass and copper). If $\alpha_B < \alpha_C$, bending of strip will be in opposite direction. To take care of both cases, above equation is written as,
\begin{align}
R=\frac{d}{|\alpha_B-\alpha_C|\,\Delta T}.\nonumber
\end{align}