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-eighth of the density of the material of the bob is

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

Answer 1

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

FBD of bob in vacuum and in liquid


Upthrust = force of buoyancy
$={\rho }_l{V}_{l}g=\frac{\rho }{8}Vg$
(density of liquid ${\rho }_l=\frac{\rho }{8}$
volume displaced $V_l$ = volume of bob $V$)

Effective weight of bob in liquid
$=\rho Vg-\frac{\rho }{8}Vg=\frac{7}{8}\rho Vg$
$=\frac{7}{8}mg$

$g_{eff}=\frac{7g}{8}$

Therefore time period in liquid $T^{\prime }=2\pi \sqrt{\frac{l}{7g/8}}=\sqrt{\frac{8}{7}}T$
where $T$ is time period in vacuum.