Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is 100 g. The time period of the motion of the particle will be (approximately)
(Take g = 10 m s⁻² , radius of earth = 6400 km)

Solution

Physics sir · 4 steps

Inside the earth, gravity grows with distance from the centre, and that linear restoring force is exactly SHM.

1. Gravity inside earth. At a distance r from the centre, only the mass inside radius r pulls on the particle. For uniform density this gives g' = gr/R, always pointing towards the centre.
2. Restoring force. So the force is F = -mg/R r, which is proportional to displacement. That is SHM with ω² = g/R. The mass m cancels, so the 100 g plays no role.

   T = 2π√(R/g)

3. Put in numbers. R/g = (6.4 × 10⁶)/10 = 6.4 × 10⁵ s², so √(R/g) = 800 s. Then T = 2π × 800 ≈ 5027 s ≈ 83.8 min.
4. So. 83.8 minutes is about 1 hour 24 minutes, which is option (b).

The idea to keep. For a tunnel through a uniform earth, the period is 2π√(R/g) ≈ 84 minutes, whatever the mass. It is the same as the period of a satellite skimming the surface.

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