Question

A simple pendulum has a time period 'T' in vacuum. Its time period when it is completely immersed in a liquid of density one-eight of density of material of bob is :

• A. $\sqrt{\frac{7}{8}}T$
• B. $\frac{5}{8}T$
• C. $\frac{3}{8}T$
• D. $\sqrt{\frac{8}{7}}T$

Answer 1

The correct option is D $\sqrt{\frac{8}{7}}T$

Given,

Density of liquid / density of bob =$\frac{d_L}{d_B}=\frac{1}{8}$

Time period of simple pendulum, $T\propto \frac{1}{\sqrt{g}}$

(Where, $g=$ acceleration of gravity)

Weight in liquid = weight in vaccume – buoyancy force.

$\left(d_{B}V\right)a=\left(d_{B}V\right)g-\left(d_{L}V\right)g$ (Where, $V=$ volume of bob)

Due to buoyancy, downward acceleration become, $a=g[1-\frac{d_L}{d_B}]$

Time period in vacuume / time period in liquid =$\frac{T}{T_2}$

$\frac{T}{T_2}=\sqrt{\frac{g[1-\frac{d_L}{d_B}]}{g}}$

$\frac{T}{T_2}=\sqrt{1-\frac{d_B}{8.d_B}}=\sqrt{\frac{7}{8}}$

$T_2=\sqrt{\frac{8}{7}}T$

Hence, time period in liquid is $T_2=\sqrt{\frac{8}{7}}T$