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(C) The time period of a simple pendulum in air is given by $T = 2\pi \sqrt{\frac{l}{g}}$.When the bob is immersed in a liquid,it experiences an upwar

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$\frac{T}{\sqrt{1 - \rho}}$

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(C) The time period of a simple pendulum in air is given by $T = 2\pi \sqrt{\frac{l}{g}}$.When the bob is immersed in a liquid,it experiences an upward buoyant force $F_{up} = V \sigma g$,where $V$ is the volume of the bob and $\sigma$ is the density of the liquid.The effective weight of the bob becomes $mg' = mg - F_{up} = V \rho_0 g - V \sigma g$,where $\rho_0$ is the density of the bob material.Thus,the effective acceleration due to gravity $g'$ is $g' = g \left(1 - \frac{\sigma}{\rho_0}\right)$.Given the ratio $\rho = \frac{\sigma}{\rho_0}$,we have $g' = g(1 - \rho)$.The new time period $T'$ is $T' = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{g(1 - \rho)}}$.Comparing this with the original time period $T$,we get $T' = \frac{T}{\sqrt{1 - \rho}}$.

The period of small oscillation of a simple pendulum is $T$. The ratio of the density of the liquid to the density of the material of the bob is $\rho$ (where $\rho < 1$). When the bob is immersed in the liquid,the new time period of small oscillation will be:

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