Q: 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 material of the bob is

(a) $ \displaystyle \sqrt{\frac{7}{8}} T $

(b) $ \displaystyle \sqrt{\frac{5}{8}} T $

(c) $ \displaystyle \sqrt{\frac{3}{8}} T $

(d) $ \displaystyle \sqrt{\frac{8}{7}} T $

Ans: (d)

$ \displaystyle T = 2\pi \sqrt{\frac{l}{g}} $

In a liquid ;

$ \displaystyle T’ = 2\pi \sqrt{\frac{l}{g(1-\frac{\rho}{\sigma})}} $

Where ρ = density of liquid

and σ = density of solid

$ \displaystyle T’ = 2\pi \sqrt{\frac{l}{g(1-\frac{\sigma /8}{\sigma})}} $

$ \displaystyle T’ = 2\pi \sqrt{\frac{l}{g(1- \frac{1}{8})}} $

$ \displaystyle T’ = 2\pi \sqrt{\frac{l}{g}.\frac{8}{7}} $

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