MCQ
A simple pendulum oscillating in air has period $T.$ The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is $\frac {1}{16}$ of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is
  • A
    $2T\sqrt {\frac {1}{10}}$
  • B
    $2T\sqrt {\frac {1}{14}}$
  • $4T\sqrt {\frac {1}{15}}$
  • D
    $4T\sqrt {\frac {1}{14}}$

Answer

Correct option: C.
$4T\sqrt {\frac {1}{15}}$
c
For a simple pendulum $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}_{\mathrm{err}}}}$

Situation $1$: when pendulum is in air $\rightarrow g_{\text {eff }}=g$

Situation $2$ :when pendulum is in liquid

$\rightarrow g_{e f}=g\left(1-\frac{\rho_{\text {liquid }}}{\rho_{\text {body }}}\right)=g\left(1-\frac{1}{16}\right)=\frac{15 g}{16}$

So, $\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\frac{2 \pi \sqrt{\frac{\mathrm{L}}{15 \mathrm{g} / 16}}}{2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}}$

$\Rightarrow \mathrm{T}^{\prime}=\frac{4 \mathrm{T}}{\sqrt{15}}$

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