Suppose a smooth tunnel is dug along a straight line joining two points on the surface of the earth and a particle is dropped from rest at its one end. Assume that mass of earth is uniformly distributed over its volume. Then

Options

A. The particle will emerge from the other end with velocity `(GM_(e))/(2R_(e))`, where `M_(e)` and `R_(e)` are earth's mass and radius, respectively.
B. The particle will come to rest at the centre of the tunnel because at this position, the particle is closest to the earth's centre.
C. Potential energy of the particle will be equal to zero at centre of the tunnel if it is along a diameter.
D. Acceleration of the particle will be proportional to its distance from the mid-point of the tunnel.

Answer

A, B, C

Explanation

When the particle is dropped in the tunnel at its one end, it starts to execute simple harmonic motion with mean position as the centre of the tunnel. Hence, acceleration of the particle is directly proportional to its distance from he centre of the tunnel. It means option d is correct.
The particle has zero velocity at one end because it is dropped from rest at that end, therefore ends of the tunnel are its extreme positions. Hence at the other end, the velocity will extreme positions. Hence at the other end, the velocity will become equal to zero. It means optioin a is incorrect.
Velocity of a particle executing S.H.M. is maximum at its mean position. Hence at the centre of the tunnel, its velocity will be maximum possible. Hence, option b is incorrect.
When the particle moves from extreme position to centre of the tunel, its velocity increases. It means `KE` increases or `PE` decreases. But initial potential energy becomes more negative and it is least at the mid-point of the tunnel because `KE` is maximum there. It means the gravitational PE can never be equal to zero. Hence, option c is also incorrect.