d
(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}}}}} $
$\frac{{T'}}{T} = \sqrt {\frac{\rho}{{\rho -1}}} $