Question

A simple pendulum oscillating in air has period $T$. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is $\frac{1}{16}$th of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is :

• A. $4T \sqrt{\frac{1}{15}}$
• B. $2T \sqrt{\frac{1}{10}}$
• C. $4T \sqrt{\frac{1}{14}}$
• D. $2T \sqrt{\frac{1}{14}}$

Answer 1

Answer: (A)

For a simple pendulum T = $2\pi \sqrt{\frac{L}{g_{eff}}}$
situation 1 : when pendulum is in air $\rightarrow \, g_{ef}$f = g
situation 2 : when pendulum is in liquid
$\rightarrow \, \, g_{eff} \, = \, g \, \bigg(1 - \frac{\rho_{liquid}}{\rho_{body}}\bigg) \, = \, g\bigg( 1 - \frac{1}{16}\bigg) \, = \, \frac{15 g}{16}$
$So, \frac{T'}{T} = \frac{2\pi \sqrt{\frac{L}{15g / 16}}}{2\pi \sqrt{\frac{L} {g}}} \Rightarrow T' = \frac{4T}{\sqrt{15}}$
Option (1)