A bob of mass 'm' suspended by a thread of length l undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density 1 \over 4 times that of the bob and the length of the thread is increased by 1/3rd of the original length, then the time period of the simple harmonic oscillations will be :-

Simple Harmonic Motion PYQ — JEE Main 2021 · 31 Aug · Shift 2 · Q51 with Solution

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Correct answer: D

For a simple pendulum,

T = 2\pi \sqrt{\frac{l}{g}}

where $l$ is the length and $g$ is the acceleration due to gravity.

When the bob is immersed in a liquid, buoyancy acts upward.

If the density of the bob is $\rho$ and the density of the liquid is $\rho_l = \frac{\rho}{4}$ , then

Buoyant force is

F_b = \rho_l V g = \frac{\rho}{4} \cdot \frac{m}{\rho} \cdot g = \frac{mg}{4}

So the effective weight becomes

mg_{\text{eff}} = mg - \frac{mg}{4} = \frac{3mg}{4}

Hence,

g_{\text{eff}} = \frac{3g}{4}

The thread length is increased by $\frac{1}{3}$ of the original length:

l' = l + \frac{l}{3} = \frac{4l}{3}

Now the pendulum oscillates with effective gravity $g_{\text{eff}} = \frac{3g}{4}$ and new length $l' = \frac{4l}{3}$ .

Thus,

T' = 2\pi \sqrt{\frac{l'}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{\frac{4l}{3}}{\frac{3g}{4}}}

Simplify:

T' = 2\pi \sqrt{\frac{4l}{3} \cdot \frac{4}{3g}} = 2\pi \sqrt{\frac{16l}{9g}} = \frac{4}{3} \cdot 2\pi \sqrt{\frac{l}{g}}

Therefore,

T' = \frac{4}{3}T

So the correct option is D.