(d) When the bob is immersed in water its effective weight
$= \left( mg - \frac{m}{\rho}g \right) = mg\left( \frac{\rho - 1}{\rho} \right)$
$\therefore g_{eff} = g\left( \frac{\rho - 1}{\rho} \right)$
$\frac{T^{'}}{T} = \sqrt{\frac{g}{g_{eff}}} \Rightarrow T^{'} = T\sqrt{\frac{\rho}{(\rho - 1)}}$