Q 11-07-088JEE MainJEE Main 2023 (25 Jan, Shift 1)Medium
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 $100\ \text{g}$. The time period of the motion of the particle will be (approximately) (take $g=10\ \text{m s}^{-2}$, radius of earth $=6400\ \text{km}$)
Answer: (B) 1 hour 24 minutes
Inside the earth $g'=g\dfrac rR$, so the motion is SHM with $\omega^2=\dfrac gR$:
$$T=2\pi\sqrt{\frac Rg}=2\pi\sqrt{6.4\times10^5}=2\pi\times800\approx5024\ \text{s}\approx84\ \text{min}$$
i.e. about 1 hour 24 minutes (independent of the mass).
Solution by Sreeraj P, M.Sc Physics