Question

A piece of metal weight 46 gm in air, when it is immersed in the liquid of specific gravity 1.24 at 27ºC it weighs 30 gm. When the temperature of liquid is raised to 42ºC the metal piece weight 30.5 gm, specific gravity of the liquid at 42ºC 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 ${27}^{o}C$ is
=46-30=16=$V_1\times 1.24{\rho }_1\times g$ ....(i)
Loss of weight at ${42}^{o}C$ is
=46-30.5=15.5=$V_2\times 1.2{\rho }_1\times g$ ....(ii)
Now dividing (i) by (ii), we get $\frac{16}{15.5}=\frac{V_1}{V_2}\times \frac{1.24}{1.2}$
But $\frac{V_2}{V_1}=1+3\alpha (t_2-t_1)=\frac{15.5\times 1.24}{16\times 1.2}=1.001042$
$\Rightarrow 3\alpha ({42}^o-{27}^o)=0.001042\Rightarrow \alpha =2.316\times {10}^{-5}/{}^{o}C$.