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A simple pendulum oscillating in air has a period of $\sqrt3$ s. If it is completely immersed in non-viscous liquid, having density $\left(\dfrac{1}{4}\right)^{th}$ of the material of the bob, the new period will be:-
A
$2\sqrt3$ s
B
$\dfrac{2}{\sqrt3}$ s
C
2s
D
$\dfrac{\sqrt3}{2}$ s
Detailed Solution
$T_{air}=2\pi\sqrt{\dfrac{\ell}{g}}=\sqrt3$ sec. In the liquid, $g_{net}=g\left(1-\dfrac{\rho}{\sigma}\right)$ where $\rho$ is the density of the liquid and $\sigma$ is the density of the bob material. $T_{Liq}=2\pi\sqrt{\dfrac{\ell}{g_{net}}}=\dfrac{T_{air}}{\sqrt{1-\rho/\sigma}}=\dfrac{\sqrt3}{\sqrt{1-1/4}}=\dfrac{\sqrt3}{\sqrt3/2}=2$ sec.