A simple pendulum has a time period ‘T’ in vacuum. Its time period when it is completely immersed in a liquid of density one-eighth of the density of the material of the bob is

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

$\sqrt{\frac{5}{8}} T$

$\sqrt{\frac{3}{8}} T$

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

answer is D.

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Detailed Solution

$\begin{array}{l} T = 2 π \sqrt{\frac{ℓ}{g}} \\ T' = 2 π \sqrt{\frac{ℓ}{g (1 - \frac{d_{ℓ}}{d_{s}})}} = 2 π \sqrt{\frac{ℓ}{g (1 - \frac{1}{8})}} \\ T^{1} = \sqrt{\frac{8}{7}} T \end{array}$

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