A piece of metal weighs $45 \mathrm{~g}$ in air and $25 \mathrm{~g}$ in a liquid of density $1.5 \times 10^3…
A piece of metal weighs $45 \mathrm{~g}$ in air and $25 \mathrm{~g}$ in a liquid of density $1.5 \times 10^3 \mathrm{~kg}^{-3} \mathrm{~m}^{-3}$ kept at $30^{\circ} \mathrm{C}$. When the temperature of the liquid is raised to $40^{\circ} \mathrm{C}$, the metal piece weighs $27 \mathrm{~g}$. The density of liquid at $40^{\circ} \mathrm{C}$ is $1.25 \times 10^3 \mathrm{~kg}-\mathrm{m}^{-3}$. The coefficient of linear expansion of metal is
$1.3 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
$5.2 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
$2.6 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
$0.26 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
Solution
The volume of the metal at $30^{\circ} \mathrm{C}$ is
$\begin{aligned}
V_{30} & =\frac{\text { loss of weight }}{\text { specific gravity } \times g} \\
& =\frac{(45-25) g}{1.5 \times g}=13.33 \mathrm{~cm}^3
\end{aligned}$
Similarly, volume of metal at $40^{\circ} \mathrm{C}$ is
$V_{40}=\frac{(45-27) g}{1.25 \times g}=14.40 \mathrm{~cm}^3$
Now, $\quad V_{40}=V_{30}\left[1+\gamma\left(t_2-t_1\right)\right]$
$\begin{aligned}
\text{or} \quad \gamma & =\frac{V_{40}-V_{30}}{V_{30}\left(t_2-t_1\right)} \\
& =\frac{14.40-13.33}{13.33(40-30)} \\
& =8.03 \times 10^{-3} /{ }^{\circ} \mathrm{C}
\end{aligned}$
$\therefore$ Coefficient of linear expansion of the metal is
$\begin{aligned}
\alpha=\frac{\gamma}{3} & =\frac{8.03 \times 10^{-3}}{3} \\
& \approx 2.6 \times 10^{-3} /{ }^{\circ} \mathrm{C}
\end{aligned}$