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{40 - 27 \,g}{1.25 \times g}$
$ = 14.40 \, cm^{3}$
Now, $ V_{40 } = V_{30} 1 + \gamma t_{2} - t_{1}$
Or $ \gamma = \frac{V_{40} - V_{30}}{V_{30} t_{2} - t_{1}} $
$ = \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
$ a = \frac{\gamma}{3} = \frac{8.03 \times 10^{-3}}{3}$
$ \approx 2.6 \times 10^{-3} /^{\circ}C$