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A simple pendulum with a brass bob has a period T. The bob is now immersed in a nonviscous liquid and oscillated. If the density of the liquid is 1/8th of brass, the time period of the same pendulum will be

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$\sqrt{\frac{8}{7}} T$

$\frac{8}{7} T$

$\frac{64}{49} T$

answer is A.

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Detailed Solution

$T^{1} = \frac{T}{\sqrt{1 - \frac{d_{I}}{d_{s}}}}$

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