Physics - SHM Question with Solution | TestHub

When the bob is immersed in water its effective weight is $m{g}^{\text{'}}=mg-F_B$, where, $F_B$ is buoyant force.

Then, $g^{\text{'}}=\frac{mg-F_B}{m}$ $=\frac{{\rho }_{\mathrm{B}}\mathrm{Vg}-{\rho }_{\mathrm{W}}\mathrm{Vg}}{{\rho }_{\mathrm{B}}\mathrm{V}}$

$=\left(\frac{{\rho }_B-{\rho }_w}{{\rho }_B}\right)g$ $=\frac{5-1}{5}\times g$$=\frac{4}{5}g$

Now, time period of simple pendulum is given by $T=2\pi \sqrt{\frac{l}{g}}$

Thus, the ratio $\frac{T^{\text{'}}}{T}=\sqrt{\frac{g}{g^{\text{'}}}}=\sqrt{\frac{g}{{\displaystyle \frac{4}{5}}g}}=\sqrt{\frac{5}{4}}$

$\Rightarrow T^{\text{'}}=T\sqrt{\frac{5}{4}}=\frac{10}{2}\sqrt{5}$

Thus, the new time period is $T^{\text{'}}=5\sqrt{5} \mathrm{s}$. 

Hence, value of $x=5$.