The time period of a simple pendulum is $T$. The pendulum is oscillated with its bob immersed in a liquid of density $\sigma$. If the density of the bob is $\rho$ and viscous effect is neglected, the time period of the pendulum will be

Let $m$ be the mass of the bob and $V$ its volume. Its mass $m=\rho V$. The weight of the ball is

$W=mg=\rho Vg$

The volume of the liquid displaced $=V$. If $\sigma$ is the density of the liquid, the weight of the liquid displaced is the upthrust $U$ it experiences.

$U=V\sigma g$

The net downward force acting on the bob is
$$\begin{gathered} F=W-U \\ F=(\rho -\sigma )Vg=\left(\frac{\rho -\sigma }{\rho }\right)mg (\because m=\rho V) \\ \text{ or } m{g}^{\text{'}}=\left(\frac{\rho -\sigma }{\rho }\right)mg \\ \text{ or } g^{\text{'}}=\left(\frac{\rho -\sigma }{\rho }\right)g \end{gathered}$$
This is the effective acceleration due to gravity.
$$\begin{gathered} \therefore T^{\text{'}}=2\pi \sqrt{\frac{l}{g^{\text{'}}}}=2\pi \sqrt{\frac{\rho l}{(\rho -\sigma )g}} \\ ={\left(\frac{\rho }{\rho -\sigma }\right)}^{1/2}T \left(\because T=2\pi \sqrt{\frac{l}{g}}\right) \end{gathered}$$
Hence the correct choice is (1).