The time period of a simple pendulum in air i | vedclass

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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 is immersed in a liquid of density $\rho$

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$\sqrt{\frac{2}{5}}$

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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 is immersed in a liquid of density $\rho$ and the bob has a density $\sigma$,the effective acceleration due to gravity $g'$ is given by $g' = g(1 - \frac{\rho}{\sigma})$.Here,$\rho = 1000 \ kg \ m^{-3}$ and $\sigma = \frac{5000}{3} \ kg \ m^{-3}$.Thus,$g' = g(1 - \frac{1000}{5000/3}) = g(1 - \frac{3000}{5000}) = g(1 - \frac{3}{5}) = g(\frac{2}{5})$.The time period in water is $t = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{g(2/5)}} = \sqrt{\frac{5}{2}} \cdot 2\pi \sqrt{\frac{l}{g}} = \sqrt{\frac{5}{2}} T$.Therefore,$\frac{T}{t} = \sqrt{\frac{2}{5}}$.

The time period of a simple pendulum in air is $T$. If the pendulum is immersed in water and executes $SHM$,its time period is $t$. The value of $\frac{T}{t}$ is [density of the bob is $\frac{5000}{3} \ kg \ m^{-3}$ and density of water is $1000 \ kg \ m^{-3}$].

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