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 1/4 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:- (a) T (b) 3/2 T (c) 3/4 T (d) 4/3 T

Answer: (d). In a liquid, buoyancy reduces the effective gravity: geff = g(1 - (σ)/(ρ)) = g(1-1/4) = 3g/4. New length l' = l + l/3 = 4l/3. T'/T = √l'/lgeff/g = √((4/3)/(3/4)) = 4/3. So the answer is (d) 4/3T.

A bob of mass 'm' suspended by a thread of length l undergoes simple… | JEE Main 2021 (31 Aug, Shift 2)

Q66JEE Main 2021 · 31 AUG · SHIFT 2Medium

Simple Pendulum

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 $\frac{1}{4}$ 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 :-

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Answer: (d)

In a liquid, buoyancy reduces the effective gravity: $g_{eff} = g\left(1 - \frac{\sigma}{\rho}\right) = g\left(1-\frac{1}{4}\right) = \frac{3g}{4}$ .
New length $l' = l + \frac{l}{3} = \frac{4l}{3}$ .
$\frac{T'}{T} = \sqrt{\frac{l'/l}{g_{eff}/g}} = \sqrt{\frac{4/3}{3/4}} = \frac{4}{3}$ .
So the answer is (d) $\frac{4}{3}T$ .

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