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Oscillations question

2023 · Q174
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Oscillations question

2023 · Q174

NEETPhysicsOscillationsMCQ+4 / −1

A simple pendulum oscillating in air has a period of 3 s\sqrt{3} \mathrm{~s}3​ s. If it is completely immersed in non-viscous liquid, having density (14)th \left(\frac{1}{4}\right)^{\text {th }}(41​)th  of the material of the bob, the new period will be :-

  1. A
    232 \sqrt{3}23​ s
  2. B
    23 s\frac{2}{\sqrt{3}} \mathrm{~s}3​2​ s
  3. C
    2 s2 \mathrm{~s}2 s
  4. D
    32 s\frac{\sqrt{3}}{2} \mathrm{~s}23​​ s
View written solutionFree

Correct answer: C

$$ \begin{aligned} & \mathrm{T}_{\text {air }}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}=\sqrt{3} ~\mathrm{sec} \\ & \ln \text { Liquid } \rightarrow g_{\text {net }}=g\left(1-\frac{\rho}{\sigma}\right) \\ & \rho=\text { density of liquid } \\ & \sigma=\text { density of material of bob } \\ & \text { so } \mathrm{T}_{\text {Liq }}=2 \pi \sqrt{\left(\frac{\ell}{g_{\text {net }}}\right)}=2 \pi \sqrt{\frac{\ell}{g\left(1-\frac{\rho}{\sigma}\right)}} \\ & \mathrm{T}_{\text {Liq }}=\frac{\mathrm{T}_{\text {air }}}{\sqrt{1-\frac{\rho}{\sigma}}}=\frac{\sqrt{3}}{\sqrt{1-\frac{1}{4}}}=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}=2 ~\mathrm{sec} \end{aligned} $$

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