The period of a simple pendulum is given by $T=2\pi \sqrt{\frac{l}{g}}$. In water, the effective gravity on the bob is reduced due to buoyancy. The net downward acceleration is $g_{\text{eff}}=g-\frac{F_{\text{b}}}{m}=g-\frac{{\rho }_{\text{fluid}}Vg}{{\rho }_{\text{bob}}V}=g(1-\frac{{\rho }_{\text{fluid}}}{{\rho }_{\text{bob}}})$. Given ${\rho }_{\text{bob}}=\frac{4}{3}\times 1000\mathrm{kg}/m^3$ and ${\rho }_{\text{fluid}}=1000\mathrm{kg}/m^3$, the ratio is $\frac{{\rho }_{\text{fluid}}}{{\rho }_{\text{bob}}}=\frac{1000}{4000/3}=\frac{3}{4}$. Therefore, $g_{\text{eff}}=g(1-\frac{3}{4})=\frac{g}{4}$. The period in water becomes $t=2\pi \sqrt{\frac{l}{g/4}}=2\pi \sqrt{\frac{4l}{g}}=2\times 2\pi \sqrt{\frac{l}{g}}=2t_0$.

Final Answer: t = 2t₀