Question

A simple pendulum oscillating in air has a period of $1$ minute. The bob of the pendulum is completely immersed in a non-viscous liquid. If the density of the liquid is ${\left(\frac{3}{4}\right)}^{th}$ of the density of the material of the body and the bob is inside the liquid at all time, its period of oscillation in the liquid will be

• A. $60\text{s}$
• B. $100\text{s}$
• C. $120\text{s}$
• D. $140\text{s}$

Answer 1

The correct option is C $120\text{s}$

Given that,
Time period of the pendulum in air, $T=1\text{min}=60\text{s}$
Let the volume of the bob is $V$, length of string is $L$, density is $\rho$, mass is $m$ and density of the liquid is $\sigma$.

The time period of simple pendulum is given by $T=2\pi \sqrt{\frac{L}{g_{eff}}}$
When the pendulum is in air, $g_{eff}=g$
$\Rightarrow T=2\pi \sqrt{\frac{L}{g}}=60\text{s}.......\left(1\right)$

When the pendulum is immersed in a liquid a thrust will oppose the weight


$\Rightarrow m{g}_{eff}=mg-\text{Thrust}$
$g_{eff}=g-\frac{V\sigma g}{V\rho }$

$\Rightarrow g_{eff}=g-\frac{\sigma }{\rho }g$
$\Rightarrow g_{eff}=g\{1-\frac{\sigma }{\rho }\}$
It is given that $\sigma =\frac{3}{4}\rho \Rightarrow \frac{\sigma }{\rho }=\frac{3}{4}$
$g_{eff}=g\{1-\frac{3}{4}\}\Rightarrow g_{eff}=\frac{g}{4}$

$\Rightarrow$ Time period $T^{\prime }$ of the pendulum in the liquid is.
$T^{\prime }=2\pi \sqrt{\frac{L}{g_{eff}}}$
$\Rightarrow T^{\prime }=2\pi \sqrt{\frac{4L}{g}}$
$\Rightarrow T^{\prime }=2\times 2\pi \sqrt{\frac{L}{g}}$
Using (1) we get
$T^{\prime }=2\times T$
$\Rightarrow T^{\prime }=2\times T$
$\Rightarrow T^{\prime }=120\text{s}$

Hence, option (c) is correct.