Question

A piece of metal weighs $45g$ in air and $25g$ 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 $27g$. 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
$V_{30}=\frac{\text{loss of weight}}{\text{specific gravity}\times g}$
$=\frac{45-25g}{1.5\times g}=13.33c{m}^3$
Similarly, volume of metal at $40^{\circ }C$ is
$V_{40}=\frac{40-27g}{1.25\times g}$
$=14.40c{m}^3$
Now, $V_{40}=V_{30}1+\gamma t_2-t_1$
Or $\gamma =\frac{V_{40}-V_{30}}{V_{30}{t}_2-t_1}$
$=\frac{14.40-13.33}{13.3340-30}$
$=8.03\times 10^{-3}{/}^{\circ }C$
$\therefore$ Coefficient of linear expansion of the metal is
$a=\frac{\gamma }{3}=\frac{8.03\times 10^{-3}}{3}$
$\approx 2.6\times 10^{-3}{/}^{\circ }C$