Solve this:

Q18. A simple pendulum has solid bob of relative density 6. It is oscillating inside a non-viscous liquid of relative density 1.2. The time period of small oscillation of this pendulum (assume SHM) is given by

     $$\begin{gathered} \left(1\right) 2\mathrm{\pi }\sqrt{\frac{61}{5g}} \left(2\right) 2\mathrm{\pi }\sqrt{\frac{5l}{4\mathrm{g}}} \\ \left(3\right) 2\mathrm{\pi }\sqrt{\frac{4 l}{5\mathrm{g}}} \left(4\right) 2\mathrm{\pi }\sqrt{\frac{5 l}{6\mathrm{g}}} \end{gathered}$$