JEE Mains · Physics · STD 11 - 13. oscillations
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 :-
- A \({T}\)
- B \(\frac{3}{2} {T}\)
- C \(\frac{3}{4} {T}\)
- D \(\frac{4}{3} {T}\)
Answer & Solution
Correct Answer
(D) \(\frac{4}{3} {T}\)
Step-by-step Solution
Detailed explanation
\({T}=2 \pi \sqrt{\ell / {g}}\) When bob is immersed in liquid \({mg}_{{eff}}={mg}-\) Buoyant force \({mg}_{{eff}} ={mg}-{v} \sigma {g} \quad(\sigma=\text { density of liquid })\) \(={mg}-{v} \frac{\rho}{4} {g}\) \(={mg}-\frac{{mg}}{4}=\frac{3 {mg}}{4}\)…
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