d
\({T}=2 \pi \sqrt{\ell / {g}}\)

When bob is immersed in liquid

\({mg}_{{eff}}={mg}-\) Buoyant force

\({mg}_{{eff}} ={mg}-{v} \sigma {g} \quad(\sigma=\text { density of liquid })\)

\(={mg}-{v} \frac{\rho}{4} {g}\)

\(={mg}-\frac{{mg}}{4}=\frac{3 {mg}}{4}\)

\(\therefore {g}_{{eff}} =\frac{3 {g}}{4}\)

\({T}_{1}= 2 \pi \sqrt{\frac{\ell_{1}}{{g}_{{eff}}}} \quad \ell_{1}=\ell+\frac{\ell}{3}=\frac{4 \ell}{3}, \quad \ell_{{eff}}=\frac{3 {g}}{4}\)

By solving

\({T}_{1}=\frac{4}{3} 2 \pi \sqrt{\ell / {g}}\)

\({T}_{1}=\frac{4 {T}}{3}\)