JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Elasticity & SHM

The length of a seconds pendulum at a height

h = 2 R

from earth surface will be:
(Given:

R =

Radius of earth and acceleration due to gravity at the surface of earth

g = π^{2} m/s^{2}

)

Answer: (D)

Solution

A seconds pendulum has period

T = 2

s. At height

h = 2 R

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

r = R + 2 R = 3 R

. The acceleration due to gravity at this height is

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

.

Using the pendulum formula

T = 2 π \frac{L}{g ^{'}}

2 = 2 π \frac{L}{π ^{2} /9} = 2 π \frac{9 L}{π ^{2}} = 2 π \cdot \frac{3 L}{π} = 6 L

Solving:

L = \frac{1}{3}

, so

L = \frac{1}{9} m

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option