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Question:

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 1/16th of the material of the bob. If the bob is inside the liquid all the time, its period of oscillation in this liquid is:

4T√(1/15)

T√(1/14)

2T√(1/14)

2T√(1/10)

Solution:

Correct option is A. 4T√(1/15)
For simple pendulum
T=2π√(L/geff)
Situation 1: when pendulum is in air → geff = g
Situation 2: when pendulum is in liquid
geff = g(1 - ρliquid/ρbody) = g(1 - 1/16) = (15g)/16
So, T'/T = 2π√(L/(15g/16)) / 2π√(L/g) → T' = 4T√(1/15)