Explanation

(b) The effective acceleration of a bob in water
$= g^{'} = g\left( 1 - \frac{\sigma}{\rho} \right)$ where $\sigma\ $and$\ \rho$ are the density of water and the bob respectively. Since the period of oscillation of the bob in air and water are given as
$T = 2\pi\sqrt{\frac{l}{g}}\ and\ T^{'} = 2\pi\sqrt{\frac{l}{g^{'}}}$
$\therefore\frac{T}{T^{'}} = \sqrt{\frac{g^{'}}{g}} = \sqrt{\frac{g(1 - \sigma/\rho)}{g}} = \sqrt{1 - \frac{\sigma}{\rho}} = \sqrt{1 - \frac{1}{\rho}}$
Putting $\frac{T}{T^{'}} = \frac{1}{\sqrt{2}}.$ We obtain, $\frac{1}{2} = 1 - \frac{1}{\rho} \Rightarrow \rho = 2$