An object suspended by a wire stretches it by 10 mm. When object is immersed in a liquid the elongation in wire reduces by 5/3mm. The ratio of relative density of the object and liquid is

Given elongation $e_1=10\text{mm}$

$e_2=\left(10-\frac{5}{3}\right)\text{mm}$

Elongation $e_2\text{=}\frac{25}{3}\text{mm}$

We know Young's modulus $Y=\frac{Fl}{Ae}\mathrm{...}\left(1\right)$

$\Rightarrow e_1\propto \text{weight}\mathrm{...}\text{(2)}$

$\Rightarrow e_2\propto \text{Apparent}\text{weight}\mathrm{...}\text{(3)}$

equations$\frac{\left(2\right)}{\left(3\right)}\Rightarrow \frac{10}{10-\frac{5}{3}}=\frac{Mg}{Mg-\frac{Mg\rho }{\sigma }}$

$\Rightarrow \frac{5}{5}\left(\frac{2}{2-\frac{1}{3}}\right)=\frac{1}{1-\frac{\rho }{6}}$

$\Rightarrow \frac{6}{5}=\frac{1}{1-\frac{\rho }{6}}$

$\Rightarrow 1-\frac{\rho }{6}=\frac{5}{6}$

$\Rightarrow \frac{\rho }{\sigma }=1-\frac{5}{6}=\frac{1}{6}$