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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 $1 / 3^{\text {rd }}$ of the original length, then the time period of the simple harmonic oscillations will be :-
Correct Option: , 4
$\mathrm{T}=2 \pi \sqrt{\ell / \mathrm{g}}$
When bob is immersed in liquid $\mathrm{mg}_{\text {eff }}=\mathrm{mg}-$ Buoyant force
$\mathrm{mg}_{\text {eff }}=\mathrm{mg}-\mathrm{v} \sigma \mathrm{g}$ $(\sigma=$ density of liquid $)$
$=m g-V \frac{\rho}{4} g$
$=\mathrm{mg}-\frac{\mathrm{mg}}{4}=\frac{3 \mathrm{mg}}{4}$
$\therefore \mathrm{g}_{\text {eff }}=\frac{3 \mathrm{~g}}{4}$
$\mathrm{T}_{1}=2 \pi \sqrt{\frac{\ell_{1}}{\mathrm{~g}_{\mathrm{eff}}}} \quad \ell_{1}=\ell+\frac{\ell}{3}=\frac{4 \ell}{3}, \quad \ell_{\mathrm{eff}}=\frac{3 \mathrm{~g}}{4}$
By solving
$\mathrm{T}_{1}=\frac{4}{3} 2 \pi \sqrt{\ell / \mathrm{g}}$
$\mathrm{T}_{1}=\frac{4 \mathrm{~T}}{3}$
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