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=2 R from the surface of earth will be, :

Solution

Physics sir · 4 steps

At height 2R you are 3R from the centre, so g drops by 9 and the seconds pendulum shrinks by 9.

1. Gravity at height. Gravity follows the inverse square of the distance from the earth's centre. At h = 2R that distance is R + 2R = 3R, three times the surface value.

   g' = g(R/(R+h))² = g/9 = π²/9 m s⁻²

2. Seconds pendulum. A seconds pendulum has time period T = 2 s, one second for each swing. From T = 2π√(L/g') we get L = (g' T²)/4π².
3. Put in numbers. L = ((π²/9) × 4)/4π² = 1/9 m. As a check, on the surface the same formula gives L = 1 m, and the height divides it by 9.
4. So. The length of the seconds pendulum at h = 2R is 1/9 m, option (a).

The idea to keep. For a fixed period, pendulum length is proportional to g. So find g at the new place first, measuring distance from the earth's centre, and scale the length by the same factor.

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