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(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.When the bob is immersed in a liquid,the effective acceleration du

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$\frac{4}{3} T$

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(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.When the bob is immersed in a liquid,the effective acceleration due to gravity $g_{	ext{eff}}$ changes due to the buoyant force.$m g_{	ext{eff}} = m g - F_B$,where $F_B$ is the buoyant force.$F_B = V \sigma g$,where $V$ is the volume of the bob and $\sigma$ is the density of the liquid.Given $\sigma = \frac{ho}{4}$,where $ho$ is the density of the bob,we have $F_B = V \frac{ho}{4} g = \frac{m g}{4}$.Thus,$m g_{	ext{eff}} = m g - \frac{m g}{4} = \frac{3 m g}{4}$,which implies $g_{	ext{eff}} = \frac{3 g}{4}$.The new length of the thread is $l_1 = l + \frac{l}{3} = \frac{4l}{3}$.The new time period $T_1$ is given by $T_1 = 2 \pi \sqrt{\frac{l_1}{g_{	ext{eff}}}} = 2 \pi \sqrt{\frac{4l/3}{3g/4}} = 2 \pi \sqrt{\frac{16l}{9g}} = \frac{4}{3} \left( 2 \pi \sqrt{\frac{l}{g}} ight)$.Therefore,$T_1 = \frac{4}{3} T$.

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