A simple pendulum is making oscillations with its bob immersed in a liquid of density n times less than the density of the bob. What is its period

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$2 π \sqrt{\frac{l}{n g}}$

$2 π \sqrt{\frac{l}{(1 - \frac{1}{n}) g}}$

$2 π \sqrt{\frac{l n}{g}}$

$2 π \sqrt{\frac{l}{(n - 1) g}}$

answer is B.

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Detailed Solution

Time period of pendulum when it is immersed in a liquid density $' ρ'$

$T = 2 π \sqrt{\frac{l}{g (1 - \frac{ρ}{σ})}}$

$ρ$ = density of liquid, $σ$ = density of the bob

Given $ρ = \frac{σ}{n}; T = 2 π \sqrt{\frac{l}{g (1 - \frac{1}{n})}}$

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