The metallic bob of simple pendulum has the relative density 5 ....

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

Then, ${g}^{'} = \dfrac{mg - {F}_{B}}{m}$ $= \dfrac{{ρ}_{B}Vg - {ρ}_{W}Vg}{{ρ}_{B}V}$

=\left(\dfrac{{ρ}_{B} - {ρ}_{w}}{{ρ}_{B}}\right)g = \dfrac{5 - 1}{5} \times g$$= \dfrac{4}{5}g$

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

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

$⇒ {T}^{'} = T\sqrt{\dfrac{5}{4}} = \dfrac{10}{2}\sqrt{5}$

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

Hence, value of $x = 5.