Question

A piece of metal weighs 46 g in air and 30 g in a liquid of density $1.24\times {10}^{3}kg/m^3$ kept at ${27}^{\circ }C$. When the temperature of the liquid is raised to ${42}^{\circ }C$, the metal piece weighs 30.5 g. The density of the liquid at ${42}^{\circ }C$ is $1.20\times {10}^{3}kg/m^3$. Calculate the coefficient of linear expansion of the metal.

• A. $2.3\times {10}^6{/}^{\circ }C$
• B. $1.2\times {10}^{-5}{/}^{\circ }C$
• C. $1.2\times {10}^5{/}^{\circ }C$
• D. $2.3\times {10}^{-5}{/}^{\circ }C$

Answer 1

The correct option is D $2.3\times {10}^{-5}{/}^{\circ }C$

Let the volume of the metal piece be $V_0$ at ${27}^{\circ }C$ and $V_{\theta }$ at ${42}^{\circ }C$. The density of the liquid at ${27}^{\circ }C$ is ${\rho }_0$ = $1.24\times {10}^{3}kg/m^3$, and the density at ${42}^{\circ }C$ is ${\rho }_{\theta }$ = $1.20\times {10}^{3}kg/m^3$.

The weight of the liquid displaced = apparent loss in the weight of the metal piece when dipped in the liquid. Thus,

$v_0{\rho }_0$ = 46 g - 30 g = 16 g

and $v_{\theta }{\rho }_{\theta }$ = 46 g - 30.5 g = 15.5 g.

Thus,

$\frac{v_{\theta }}{v_0}$ = $\frac{{\rho }_0}{{\rho }_{\theta }}\times \frac{15.5}{16}$

or, $1+3\alpha \Delta \theta$ = $\frac{1.24\times {10}^{3}15.5}{1.20\times {10}^3\times 16}$

or, $1+3\theta ({42}^{\circ }C-{27}^{\circ }C)$ = 1.00104

or, $\alpha$ = $2.3\times {10}^{-5}{/}^{\circ }C$