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Question

A simple pendulum oscillating in air has a period of $\sqrt{3} \mathrm{~s}$. If it is completely immersed in non-viscous liquid, having density $\left(\frac{1}{4}\right)^{\text {th }}$ of the material of the bob, the new period will be :-

$2 \sqrt{3}$ s
$\frac{2}{\sqrt{3}} \mathrm{~s}$
$2 \mathrm{~s}$
$\frac{\sqrt{3}}{2} \mathrm{~s}$

Solution

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