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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 :-
${T}$
$\frac{3}{2} {T}$
$\frac{3}{4} {T}$
$\frac{4}{3} {T}$
Solution
${T}=2 \pi \sqrt{\ell / {g}}$
When bob is immersed in liquid
${mg}_{{eff}}={mg}-$ Buoyant force
${mg}_{{eff}} ={mg}-{v} \sigma {g} \quad(\sigma=\text { density of liquid })$
$={mg}-{v} \frac{\rho}{4} {g}$
$={mg}-\frac{{mg}}{4}=\frac{3 {mg}}{4}$
$\therefore {g}_{{eff}} =\frac{3 {g}}{4}$
${T}_{1}= 2 \pi \sqrt{\frac{\ell_{1}}{{g}_{{eff}}}} \quad \ell_{1}=\ell+\frac{\ell}{3}=\frac{4 \ell}{3}, \quad \ell_{{eff}}=\frac{3 {g}}{4}$
By solving
${T}_{1}=\frac{4}{3} 2 \pi \sqrt{\ell / {g}}$
${T}_{1}=\frac{4 {T}}{3}$
