JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Elasticity & SHM

A bob of mass '

m

' suspended by a thread of length I undergoes simple harmonic oscillations with time period

T

. If the bob is immersed in a liquid that has density

\frac{1}{4}

times that of the bob and the length of the thread is increased by

1/ 3^{rd}

of the original length, then the time period of the simple harmonic oscillations will be :-

Answer: (D)

Solution

Initial time period:

T = 2 π \frac{l}{g}

When immersed in liquid, effective gravity:

g_{e f f} = g (1 - \frac{ρ _{l i q u i d}}{ρ _{b o b}}) = g (1 - \frac{1}{4}) = \frac{3 g}{4}

New length:

l^{'} = l + \frac{l}{3} = \frac{4 l}{3}

New time period:

T^{'} = 2 π \frac{l ^{'}}{g _{e f f}} = 2 π \frac{4 l /3}{3 g /4} = 2 π \frac{16 l}{9 g} = \frac{4}{3} \times 2 π \frac{l}{g} = \frac{4}{3} T

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option