A simple pendulum doing small oscillations at a place R height above earth surface has time period of T₁=4 s. T₂ would be it's time period if it is brought to a point which is at a height 2 R from earth surface. Choose the correct relation [R= radius of earth] :

Solution

Physics sir · 3 steps

Measure distance from the centre of the earth, not from the surface, and T grows in step with it.

1. g above the surface. At a distance r from the centre of the earth, g = GM/r². A height h above the surface means r = R + h.
2. Period depends on r. The pendulum period is T = 2π√(L/g), and g ∝ 1/r². So T ∝ √r² = r: the period grows in direct proportion to the distance from the centre.

   T₂/T₁ = r₂/r₁ = (R + 2R)/(R + R) = 3R/2R = 3/2

3. So. 2T₂ = 3T₁, which gives T₂ = 6 s. This is the relation 3T₁ = 2T₂, option (a).

The idea to keep. Above the earth, g ∝ 1/r², so a pendulum's period T ∝ r. Always add R to the height before you take any ratio.

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