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
$V_{30}=\frac{\text{ loss of weight }}{\text{ specific gravity }\times g}$
$=\frac{(45-25)g}{1.5\times g}=13.33c{m}^3$
Similarly, volume of metal at $40^{\circ }C$ is
$V_{40}=\frac{(45-27)g}{1.25\times g}=14.40c{m}^3$
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
$\alpha =\frac{\gamma }{3}=\frac{8.03\times 10^{-3}}{3}$
$\approx 2.6\times 10^{-3}/{}^{\circ }C$