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A piece of metal weighs $45 \mathrm{~g}$ in air and $25 \mathrm{~g}$ in a liquid of density $1.5 \times 10^{3} \mathrm{~kg}-\mathrm{m}^{-3}$ kept at $30^{\circ} \mathrm{C}$. When the temperature of the liquid is raised to $40^{\circ} \mathrm{C}$, the metal piece weights $27 \mathrm{~g}$. the density of liquid of $40^{\circ} \mathrm{C}$ is $1.25 \times 10^{3} \mathrm{~kg}-\mathrm{m}^{-3}$. the coefficient of linear expansion of metal is
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Verified Answer
The correct answer is:
$2.6 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
Volume of the metal at $30^{\circ} \mathrm{C}$,
$$
\begin{aligned}
V_{30} &=\frac{\text { loss of weight }}{\text { specific gravity } \times g} \\
&=\frac{(45-25) g}{1.5 \times g}=13.33 \mathrm{~cm}^{3}
\end{aligned}
$$
Similarly, volume of metal at $40^{\circ} \mathrm{C}$
$$
V_{40}=\frac{(45-27) g}{1.25 \times g}=14.40 \mathrm{~cm}^{3}
$$
Now, $V_{40}=V_{30}\left[1+\gamma\left(t_{2}-t_{1}\right)\right]$
$$
\text { or } \begin{aligned}
& \gamma=\frac{V_{40}-V_{30}}{V_{30}\left(t_{2}-t_{1}\right)}=\frac{14.40-13.33}{13.33(40-30)} \\
&=8.03 \times 10^{-3} /{ }^{\circ} \mathrm{C}
\end{aligned}
$$
$\therefore \quad$ Coefficient of linear expansion of the metal
$$
\alpha=\frac{\gamma}{3}=\frac{8.03 \times 10^{-3}}{3} \approx 2.6 \times 10^{-3} /{ }^{\circ} \mathrm{C}
$$
$$
\begin{aligned}
V_{30} &=\frac{\text { loss of weight }}{\text { specific gravity } \times g} \\
&=\frac{(45-25) g}{1.5 \times g}=13.33 \mathrm{~cm}^{3}
\end{aligned}
$$
Similarly, volume of metal at $40^{\circ} \mathrm{C}$
$$
V_{40}=\frac{(45-27) g}{1.25 \times g}=14.40 \mathrm{~cm}^{3}
$$
Now, $V_{40}=V_{30}\left[1+\gamma\left(t_{2}-t_{1}\right)\right]$
$$
\text { or } \begin{aligned}
& \gamma=\frac{V_{40}-V_{30}}{V_{30}\left(t_{2}-t_{1}\right)}=\frac{14.40-13.33}{13.33(40-30)} \\
&=8.03 \times 10^{-3} /{ }^{\circ} \mathrm{C}
\end{aligned}
$$
$\therefore \quad$ Coefficient of linear expansion of the metal
$$
\alpha=\frac{\gamma}{3}=\frac{8.03 \times 10^{-3}}{3} \approx 2.6 \times 10^{-3} /{ }^{\circ} \mathrm{C}
$$
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