If R is the radius of the earth and the acceleration due to gravity on the surface of earth is $g=\pi^{2} \mathrm{~m} / \mathrm{s}^{2}$, then the length of the second's pendulum at a height $\mathrm{h}=2 \mathrm{R}$ from the surface of earth will be,:

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Elasticity & SHM

If R is the radius of the earth and the acceleration due to gravity on the surface of earth is

g = π^{2} m / s^{2}

, then the length of the second's pendulum at a height

h = 2 R

from the surface of earth will be,:

Answer: (B)

At height

h = 2 R

from the surface, the distance from the Earth's centre is

R + 2 R = 3 R

. Since

g \propto 1/ r^{2}

g^{'} = \frac{g}{( 3 R / R ) ^{2}} = \frac{g}{9} = \frac{π ^{2}}{9} m/s^{2}

For a second's pendulum, the time period is

T = 2

s. Using

T = 2 π \frac{ℓ}{g ^{'}}

2 = 2 π \frac{9 ℓ}{π ^{2}} ⟹ 1 = 3 ℓ ⟹ ℓ = \frac{1}{9} m

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option