Question

A piece of metal weight 46 gm in air, when it is immersed in the liquid of specific gravity 1.24 at ${27}^{\circ }C$ it weighs 30 gm. When the temperature of liquid is raised to ${42}^{\circ }C$ the metal piece weight 30.5 gm, specific gravity of the liquid at ${42}^{\circ }C$ is 1.20, then the linear expansion of the metal will be

• A. 3.316 × 10^-5/ºC
• B. 2.316 × 10^-5/ºC
• C. 4.316 × 10^-5/ºC
• D. None of these

Answer 1

The correct option is B

2.316 × 10^-5/ºC

Loss of weight at ${27}^{\circ }C$ is

= 46 - 30 = 16 = V₁ × 1.24 r_l × g ...(i)

Loss of weight at ${42}^{\circ }C$ is

= 46 - 30.5 = 15.5 = $V_2$ × 1.2 $r_1$ × g ...(ii)

Now dividing (i) by (ii) , we get $\frac{16}{15.5}=\frac{V_1}{V_2}\times \frac{1.24}{1.2}$

But $\frac{V_2}{V_1}=1+3\alpha (t_2-t_1)=\frac{15.5\times 1.24}{16\times 1.2}=1.001042$

$\Rightarrow 3\alpha ({42}^{\circ }-{27}^{\circ })=0.001042\Rightarrow \alpha =2.316\times {10}^{-}5{/}^{\circ }C$