If R is the radius of the earth and the acceleration due to gravity on the surface of earth is g = π² m/s², then the length of the second's pendulum at a height h = 2R from the surface of earth will be: (a) 1/9 m (b) 8/9 m (c) 2/9 m (d) 4/9 m

Answer: (a). At h = 2R: g' = (g)/((1+2)²) = (π²)/(9) m/s². Seconds pendulum: T = 2 s = 2π√(l/g') ⇒ l = (g')/(π²) l = 1/9 m So the answer is (a) 1/9 m.

If R is the radius of the earth and the acceleration due to gravity… | JEE Main 2024 (1 Feb, Shift 1)

Q38JEE Main 2024 · 1 FEB · SHIFT 1Medium

Acceleration Due to Gravity

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

Show answer and solution

Answer: (a)

At $h = 2R$ : $g' = \frac{g}{(1+2)^2} = \frac{\pi^2}{9}\text{ m/s}^2$ .
Seconds pendulum: $T = 2\text{ s} = 2\pi\sqrt{\frac{l}{g'}} \Rightarrow l = \frac{g'}{\pi^2}$
$l = \frac{1}{9}\text{ m}$
So the answer is (a) $\frac{1}{9}\text{ m}$ .

Report a mistake

Acceleration Due to Gravity question from JEE Main 2024 (1 Feb, Shift 1)

More questions on Acceleration Due to Gravity