The length of a seconds pendulum at a height h = 2R from earth surface will be: (Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = \pi 2 ms - 2)

Gravitation PYQ — JEE Main 2022 · 25 Jul · Shift 2 · Q56 with Solution

View written solutionFree

Correct answer: D

For a simple pendulum,

T = 2\pi \sqrt{\frac{L}{g'}}

where $L$ is the length and $g'$ is the acceleration due to gravity at that location.

A seconds pendulum has time period

T = 2\text{ s}

So,

2 = 2\pi \sqrt{\frac{L}{g'}}

1 = \pi \sqrt{\frac{L}{g'}}

\sqrt{\frac{L}{g'}} = \frac{1}{\pi}

\frac{L}{g'} = \frac{1}{\pi^2}

L = \frac{g'}{\pi^2}

Height above surface is $h = 2R$ . Hence distance from earth's center is

r = R + h = R + 2R = 3R

Using variation of gravity with height,

g' = g\left(\frac{R}{R+h}\right)^2

g' = g\left(\frac{R}{3R}\right)^2 = g\left(\frac{1}{3}\right)^2 = \frac{g}{9}

Given

g = \pi^2\,\text{m s}^{-2}

So,

g' = \frac{\pi^2}{9}

From

L = \frac{g'}{\pi^2}

substitute $g' = \frac{\pi^2}{9}$ :

L = \frac{\pi^2/9}{\pi^2} = \frac{1}{9}\text{ m}

L = \frac{1}{9}\text{ m}

So the correct option is:

D: $\frac{1}{9}$ m