JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Elasticity & SHM

If the time period of a two meter long simple pendulum is

2 s

, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :

Answer: (C)

Solution

The time period of a simple pendulum is

T = 2 π \frac{ℓ}{g}

.

Given

T = 2

s and

ℓ = 2

2 = 2 π \frac{2}{g}

1 = π \frac{2}{g}

\frac{1}{π ^{2}} = \frac{2}{g}

g = 2 π^{2} m/s^{2}

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option