Practice portal › Oscillations › The Simple Pendulum

A bob of mass m on a thread of length l oscillates with time period T. The bob is then immersed in a liquid whose density is 1/4 of the density of the bob, and the thread is lengthened by 1/3 of its original length. The new time period of the small oscillations is

Asked in JEE Main 31st Aug 2nd Shift 2021 · Pendulum in a liquid

Answer: (2) 4/3 T

Step-by-step solution

Buoyancy reduces the effective gravity to g'=g(1-σ/ρ)=g(1-1/4)=(3g)/4.

The new length is l'=l+l/3=(4l)/3.

T'=2π√(4l/3)/(3g/4)=2π√(16l)/(9g)=4/3 T.

Why the other options are wrong

More The Simple Pendulum questionsAll The Simple Pendulum questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer