Q. A simple pendulum has a time period $T$ in vacuum. Its time period when it is completely immersed in a liquid of density one-eight of the density of material of the bob is
A) $\sqrt{\frac{7}{8}}T$
B) $\sqrt{\frac{5}{8}}T$
C) $\sqrt{\frac{3}{8}}T$
D) $\sqrt{\frac{8}{7}}T$
Explanation
(d) In vacuum$,\ T = 2\pi\sqrt{\frac{l}{g}}$
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Let $V$ be the volume and $T$ be the density of the mass of the bob.
Net downward force acting on the bob in side the liquid = Weight $-$ upthrust
$= V\rho g - V.\frac{T}{8}g = \frac{7}{8}V\rho g$
$i.e.,$ effective value of $g$ is $\frac{7}{8}g$
So, time period of the bob inside the liquid
$\therefore T_{l} = 2\pi\sqrt{\frac{l}{(7/8)g}} = 2\pi\sqrt{\frac{l}{g}} \times \sqrt{\frac{8}{7}} = \sqrt{\frac{8}{7}}T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.