Question

A simple pendulum has a time period $T$ in vaccum. Its time period when it is completely immersed in a liquid of density one-eight of the density of material of the bob is

• A. $\sqrt{\frac{7}{9}}T$
• 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 $\sqrt{\frac{8}{7}}T$

In vacuum, $T=2\pi \sqrt{\frac{l}{g}}$
Let $V$ be the volume of bob and $\rho$ be the density of the bob.
Density of liquid,$\sigma =\frac{1}{8}\rho$
Upthrust on body is, $F_B=V\sigma g$, since body is completely inside hence volume of fluid displaced$=V$

$\Rightarrow$Effective weight of the bob inside liquid is,
$=\text{weight}-\text{upthrust}$
$W^{\prime }=W-F_B$
$\Rightarrow m{g}^{\prime }=mg-V\left(\frac{\rho }{8}\right)g$
here $g^{\prime }$ is $g_{\text{eff}}$
Mass of the bob, $m=V\times \rho g$

$\Rightarrow V\rho g^{\prime }=V\rho g-V\left(\frac{\rho }{8}\right)g$
$\Rightarrow g^{\prime }=\frac{7g}{8}$
So, time period of oscillation the bob inside the liquid:
$\Rightarrow T^{\prime }=2\pi \sqrt{\frac{l}{\frac{7g}{8}}}$
$\Rightarrow T^{\prime }=2\pi \sqrt{\frac{l}{g}}\times \sqrt{\frac{8}{7}}$
$\therefore T^{\prime }=\sqrt{\frac{8}{7}}T$