A simple pendulum with a brass 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 1/8 that of brass, the time period of the same pendulum will be

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

$\frac{8}{7} T$

$\frac{8^{2}}{7^{2}} T$

answer is A.

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Detailed Solution

Due to buoyancy, the effective acceleration due to gravity decreases from g to $g^{'} = g - g / 8 = 7 g / 8$ . Since $T = 2 π \sqrt{\frac{l}{g}}$ . The new time period will be
$T^{'} = 2 π \sqrt{\frac{l}{g^{'}}} = 2 π \sqrt{\frac{l}{7 g / 8}}$
Thus $\frac{T^{'}}{T} = \sqrt{\frac{g}{7 g / 8}} = \sqrt{\frac{8}{7}}$ or $T^{'} = \sqrt{\frac{8}{7}} T$ .

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