The time period of a simple pendulum is T . The pendulum is oscillated with its bob immersed in a liquid of density $σ$ . If the density of the bob is p and viscous effect is neglected, the time period of the pendulum will be

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${(\frac{σ}{p - σ})}^{1 / 2} T$

${(\frac{p}{p - σ})}^{1 / 2} T$

${(\frac{σ}{p})}^{1 / 2} T$

${(\frac{p}{σ})}^{1 / 2} T$

answer is A.

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Detailed Solution

This is the effective acceleration due to gravity
$\begin{array}{l} T^{'} = 2 π \sqrt{\frac{1}{g_{eff}}} = 2 π \sqrt{\frac{pl}{(p - σ) g}} = {(\frac{p}{p - σ})}^{1 / 2} T \\ (∵ T = 2 π \sqrt{\frac{l}{g}}) \end{array}$

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