Question

A simple pendulum of length `l` is making oscillations with its bob immersed in a liquid of density $\eta$ times less than the density of the bob. What is it's time period of oscillation?

• A. $2\pi \sqrt{\frac{l}{\eta g}}$
• B. $2\pi \sqrt{\frac{l}{(1-\frac{1}{\eta })g}}$
• C. $2\pi \sqrt{\frac{l\eta }{g}}$
• D. $2\pi \sqrt{\frac{l}{(\eta -1)g}}$

Answer 1

The correct option is B $2\pi \sqrt{\frac{l}{(1-\frac{1}{\eta })g}}$

As discussed in theory part,

$2\pi \sqrt{\frac{l}{g(1-\frac{\rho }{\sigma })}}$, where

$\rho =$ density of the liquid
$\sigma =$ density of the bob

Its given, $\rho =\frac{\sigma }{\eta }$

Using the above equations, we can find out that
$T=2\pi \sqrt{\frac{l}{(1-\frac{1}{\eta })g}}$