The metallic bob of simple pendulum has the relative density \(5\). The time period of this pendulum is \(10\,s\). If the metallic bob is immersed in water, then the new time period becomes \(5 \sqrt{ x } s\). The value of \(x\) will be.

\(=\frac{\rho_{B} V g-\rho_{\pi} V g}{\rho_{B} V}\)

\(=\left(\frac{\rho_{B}-\rho_{\pi}}{\rho_{B}}\right) g \quad T=2 \pi \sqrt{\frac{\ell}{g}}\)

\(=\frac{5-1}{5} \times g\)

\(=\frac{4}{5} g\)

\(\frac{T^{\prime}}{T^{\prime}}=\sqrt{\frac{g}{g^{\prime}}}=\sqrt{\frac{g}{5} g}=\sqrt{\frac{5}{4}}\)

\(T^{\prime}=T \sqrt{\frac{5}{4}}=\frac{10}{2} \sqrt{5}\)

\(T^{\prime}=5 \sqrt{5}\)