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 , radius of earth = 6400 km)
(Take , radius of earth = 6400 km)
- A.24 hours
- B.1 hour 40 minutes
- C.12 hours
- D.1 hour 24 minutes
Show answer and solution
Correct option: D
Solution
The time period of SHM in the tunnel is T = 2π√(R/g).
T = 2π√(6400×10³/10) = 2π√(640000) = 2π×800 = 5026 seconds.
Converting: 5026/60 ≈ 83.8 minutes ≈ 1 hour 24 minutes.
The restoring force on the particle at depth is due to the mass of earth within radius r, giving F = -mg·x/R, identical to a spring-mass system with ω² = g/R, independent of the particle's mass.
T = 2π√(6400×10³/10) = 2π√(640000) = 2π×800 = 5026 seconds.
Converting: 5026/60 ≈ 83.8 minutes ≈ 1 hour 24 minutes.
The restoring force on the particle at depth is due to the mass of earth within radius r, giving F = -mg·x/R, identical to a spring-mass system with ω² = g/R, independent of the particle's mass.
[JEE Main 2023 · 25 Jan · Shift 1]