A pendulum has time period $T$ in air. When it is made to oscillate in water, it acquired a time period $T' = \sqrt 2 T$. The density of the pendulum bob is equal to (density of water $= 1$)
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(b) The effective acceleration of a bob in water
$ = g' = g\,\left( {1 - \frac{\sigma }{\rho }} \right)$ where $\sigma$ and $\rho$ are the density of water and the bob respectively. Since the period of oscillation of the bob in air and water are given as $T = 2\pi \sqrt {\frac{l}{g}} $ and $T' = 2\pi \sqrt {\frac{l}{{g'}}} $
$\therefore $ $\frac{T}{{T'}} = \sqrt {\frac{{g'}}{g}} = \sqrt {\frac{{g(1 - \sigma /\rho )}}{g}} $$ = \sqrt {1 - \frac{\sigma }{\rho }} = \sqrt {1 - \frac{1}{\rho }} $
Putting $\frac{T}{{T'}} = \frac{1}{{\sqrt 2 }}$. We obtain, $\frac{1}{2} = 1 - \frac{1}{\rho }$
$ \Rightarrow \rho = 2$
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