$ = 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$