A bob of mass 'm' suspended by a thread of length l undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density 1/4 times that of the bob and the length of the thread is increased by 1/3ʳᵈ of the original length, then the time period of the simple harmonic oscillations will be :-

Solution

Physics sir · 3 steps

In a liquid, buoyancy weakens gravity, so use an effective g and a new length in the same formula.

1. Effective gravity. The bob displaces liquid of density 1/4ρ, so the upthrust is 1/4mg. The net downward force is mg - 1/4mg = 3/4mg, so the bob behaves as if g_(eff) = 3/4g.
2. New length. Increasing l by one third gives l' = l + l/3 = 4/3l.

   T' = 2π√(l'/g_(eff)) = 2π√((4/3l)/(3/4g)) = 2π√(16/9·l/g) = 4/3× 2π√(l/g)

3. So. Since T = 2π√(l/g), the new period is T' = 4/3T, option (d).

The idea to keep. For a pendulum bob in a liquid, replace g by g(1 - ρ_(liquid)/ρ_(bob)). Then change l and g together inside T = 2π√(l/g).

Answer from the NTA final answer key. Solution by PrepWiser.