A diametrical tunnel is dug across the earth. A ball dropped into the tunnel from one side. The velocity of the ball when it reaches the centre of the earth is [Given: gravitational potential at the centre of earth `= -3//2(GM//R)`]

Options

A. `sqrt(R)`
B. `sqrt(gR)`
C. `sqrt(2.5gR)`
D. `sqrt(7.1gR)`

Explanation

To find the velocity of a ball when it reaches the center of the Earth after being dropped into a diametrical tunnel, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Gravitational Potential**:
- The gravitational potential \( V \) at the surface of the Earth is given by:
\[ V = -\frac{GM}{R} \]
- At the center of the Earth, the gravitational potential is given as:
\[ V_{center} = -\frac{3}{2} \frac{GM}{R} \]

2. **Calculate the Change in Gravitational Potential**:
- The change in gravitational potential \( \Delta V \) as the ball moves from the surface to the center is:
\[ \Delta V = V_{initial} - V_{final} = -\frac{GM}{R} - \left(-\frac{3}{2} \frac{GM}{R}\right) \]
- Simplifying this gives:
\[ \Delta V = -\frac{GM}{R} + \frac{3}{2} \frac{GM}{R} = \frac{1}{2} \frac{GM}{R} \]

3. **Relate Change in Potential to Kinetic Energy**:
- The change in gravitational potential energy \( \Delta PE \) is converted into kinetic energy \( KE \) as the ball falls:
\[ \Delta PE = m \Delta V \]
- Thus:
\[ \Delta PE = m \cdot \frac{1}{2} \frac{GM}{R} \]

4. **Express Kinetic Energy**:
- The kinetic energy at the center of the Earth can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
- Setting the change in potential energy equal to the kinetic energy gives:
\[ m \cdot \frac{1}{2} \frac{GM}{R} = \frac{1}{2} mv^2 \]

5. **Cancel Mass and Solve for Velocity**:
- Cancel \( m \) from both sides and multiply by 2:
\[ \frac{GM}{R} = v^2 \]
- Taking the square root gives:
\[ v = \sqrt{\frac{GM}{R}} \]

6. **Relate to Gravitational Acceleration**:
- We know that \( g = \frac{GM}{R} \), so:
\[ v = \sqrt{gR} \]

### Final Answer:
Thus, the velocity of the ball when it reaches the center of the Earth is:
\[ v = \sqrt{gR} \]