Question

A piece of metal weighs $45 g$ in air and $25 g$ in a liquid of density $1.5 \times 10^{3} kg - m ^{-3}$ kept at $30^{\circ} C$. When the temperature of the liquid is raised to $40^{\circ} C$, the metal piece weighs $27 g$. The density of liquid at $40^{\circ} C$ is $1.25 \times 10^{3} kg - m ^{-3}$. The coefficient of linear expansion of metal is

• A. $1.3 \times 10^{-3} /{ }^{\circ} C$
• B. $5.2 \times 10^{-3} /{ }^{\circ} C$
• C. $2.6 \times 10^{-3} /{ }^{\circ} C$
• D. $0.26 \times 10^{-3} /{ }^{\circ} C$

Answer 1

Answer: (C)

The volume of the metal at $30^{\circ} C$ is
$V_{30} =\frac{\text { loss of weight }}{\text { specific gravity } \times g} $
$=\frac{(45-25) g}{1.5 \times g}=13.33 \,cm ^{3}$
Similarly, volume of metal at $40^{\circ} C$ is
$V_{40}=\frac{(45-27) g}{1.25 \times g}=14.40 \,cm ^{3} $
Now, $V_{40} =V_{30}\left[1+\gamma\left(t_{2}-t_{1}\right)\right]$
or $\gamma =\frac{V_{40}-V_{30}}{V_{30}\left(t_{2}-t_{1}\right)} $
$=\frac{14.40-13.33}{13.33(40-30)} $
$=8.03 \times 10^{-3} /{ }^{\circ} C$
$\therefore $ Coefficient of linear expansion of the metal is
$\alpha=\frac{\gamma}{3} =\frac{8.03 \times 10^{-3}}{3}$
$ \approx 2.6 \times 10^{-3} /{ }^{\circ} C$