A simple pendulum with an iron bob has a time | vedclass

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(C) 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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$T \sqrt{\frac{12}{11}}$

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(C) 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).The effective acceleration due to gravity $g_{	ext{eff}}$ is given by $g_{	ext{eff}} = g(1 - \frac{ho_L}{ho_S})$,where $ho_L$ is the density of the liquid and $ho_S$ is the density of the solid (iron).Given $ho_L = \frac{1}{12} ho_S$,we have $\frac{ho_L}{ho_S} = \frac{1}{12}$.Thus,$g_{	ext{eff}} = g(1 - \frac{1}{12}) = g(\frac{11}{12})$.The new time period $T'$ is $T' = 2\pi \sqrt{\frac{l}{g_{	ext{eff}}}} = 2\pi \sqrt{\frac{l}{g(\frac{11}{12})}}$.$T' = T \sqrt{\frac{12}{11}}$.

$A$ simple pendulum with an iron 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 $\frac{1}{12}$th that of iron,then the new time period will be:

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