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 = π² ms⁻²)

Solution

Physics sir · 3 steps

At height 2R you are 3R from the centre, so g drops to one-ninth and the pendulum shortens by the same factor.

1. Seconds pendulum. A seconds pendulum has a period of T = 2 s, one second for each swing. So its length is L = (g T²)/4π², using whatever g is at that place.
2. Gravity at height. At height h = 2R the distance from the centre is R + 2R = 3R. Since g ∝ 1/r², gravity becomes g' = g(R/3R)² = g/9 = π²/9 m s⁻².

   L = (g' T²)/4π² = (π²/9 × 2²)/4π² = 1/9 m

3. So. On the surface this pendulum would be 1 m long. At height 2R it must be 1/9 m long, which is option (d).

The idea to keep. For a fixed period, pendulum length is proportional to g. Above the surface, use g' = g(R/(R+h))² and measure from the centre of the earth, not from the surface.

Answer from the NTA final answer key. Solution by PrepWiser.