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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.When the bob is immersed in a liquid,it experiences an upward buoya

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$\frac{2T}{\sqrt{3}}$

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.When the bob is immersed in a liquid,it experiences an upward buoyant force (upthrust).Let $ho$ be the density of the metal and $\sigma$ be the density of the liquid. Given $\sigma = \frac{ho}{4}$.The effective acceleration due to gravity $g_{	ext{eff}}$ is given by $g_{	ext{eff}} = g(1 - \frac{\sigma}{ho})$.Substituting $\sigma = \frac{ho}{4}$,we get $g_{	ext{eff}} = g(1 - \frac{1}{4}) = \frac{3}{4}g$.The new time period $T'$ is $T' = 2\pi \sqrt{\frac{l}{g_{	ext{eff}}}} = 2\pi \sqrt{\frac{l}{\frac{3}{4}g}} = 2\pi \sqrt{\frac{4l}{3g}}$.This simplifies to $T' = \sqrt{\frac{4}{3}} 	imes (2\pi \sqrt{\frac{l}{g}}) = \frac{2}{\sqrt{3}}T$.

$A$ simple pendulum with a metallic bob has a time period $T$. The bob is now immersed in a non-viscous liquid and oscillated. If the density of the liquid is $1/4$ that of the metal,the time period of the same pendulum will be ...........

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