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(B) The time period of a simple pendulum in air is given by $T = 2\pi \sqrt{\frac{l}{g}}$.When the pendulum oscillates in water,it experiences an upwa

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$2$

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(B) The time period of a simple pendulum in air is given by $T = 2\pi \sqrt{\frac{l}{g}}$.When the pendulum oscillates in water,it experiences an upward buoyant force. The effective acceleration due to gravity $g'$ becomes $g' = g \left( 1 - \frac{\sigma}{\rho} \right)$,where $\sigma$ is the density of water and $\rho$ is the density of the bob.The time period in water is $T' = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{g(1 - \sigma/\rho)}}$.Dividing the two equations,we get $\frac{T'}{T} = \sqrt{\frac{g}{g'}} = \frac{1}{\sqrt{1 - \sigma/\rho}}$.Given $T' = \sqrt{2} T$,so $\frac{T'}{T} = \sqrt{2}$.Thus,$\sqrt{2} = \frac{1}{\sqrt{1 - 1/\rho}}$.Squaring both sides,$2 = \frac{1}{1 - 1/\rho}$.$1 - 1/\rho = 1/2$.$1/\rho = 1 - 1/2 = 1/2$.Therefore,$\rho = 2$.

$A$ pendulum has a time period $T$ in air. When it is made to oscillate in water,it acquires a time period $T' = \sqrt{2} T$. The density of the pendulum bob is (density of water $= 1$)

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