A tunnel is dug across the diameter of earth. A ball is released from the surface of Earth into the tunnel. The velocity of ball when it is at a distance `(R )/(2)` from centre of earth is (where R = radius of Earth and M = mass of Earth)

Options

A. `sqrt((3GM)/(4R))`
B. `sqrt((2GM)/(3R))`
C. `sqrt((GM)/(2R))`
D. `sqrt((2GM)/(R ))`

Answer

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy (kinetic + potential) at the surface of the Earth will be equal to the total mechanical energy at a distance of \( \frac{R}{2} \) from the center of the Earth.

### Step-by-Step Solution:

1. **Define the Variables:**
- Let \( m \) be the mass of the ball.
- Let \( M \) be the mass of the Earth.
- Let \( R \) be the radius of the Earth.
- The initial velocity \( v_i = 0 \) (the ball is released from rest).
- The distance from the center of the Earth when we want to find the velocity is \( r = \frac{R}{2} \).

2. **Initial Energy at the Surface:**
- The initial potential energy \( U_i \) when the ball is at the surface of the Earth is given by:
\[ U_i = -\frac{GMm}{R} \]
- The initial kinetic energy \( K_i \) is:
\[ K_i = 0 \]
- Therefore, the total initial energy \( E_i \) is:
\[ E_i = K_i + U_i = 0 - \frac{GMm}{R} = -\frac{GMm}{R} \]

3. **Final Energy at Distance \( \frac{R}{2} \):**
- The potential energy \( U_f \) at a distance \( \frac{R}{2} \) from the center of the Earth can be calculated using the formula for gravitational potential energy inside a sphere:
\[ U_f = -\frac{GMm}{2R} \]
- The final kinetic energy \( K_f \) is given by:
\[ K_f = \frac{1}{2}mv^2 \]
- Therefore, the total final energy \( E_f \) is:
\[ E_f = K_f + U_f = \frac{1}{2}mv^2 - \frac{GMm}{2R} \]

4. **Conservation of Energy:**
- According to the conservation of mechanical energy:
\[ E_i = E_f \]
- Substituting the expressions for \( E_i \) and \( E_f \):
\[ -\frac{GMm}{R} = \frac{1}{2}mv^2 - \frac{GMm}{2R} \]

5. **Simplifying the Equation:**
- Multiply through by \( 2R \) to eliminate the fractions:
\[ -2GMm = mv^2R - GMm \]
- Rearranging gives:
\[ mv^2R = GMm - 2GMm = -GMm \]
- Dividing by \( m \) (assuming \( m \neq 0 \)):
\[ v^2R = -GM \]
- Therefore:
\[ v^2 = -\frac{GM}{R} \]

6. **Finding the Velocity:**
- Taking the square root gives:
\[ v = \sqrt{\frac{3GM}{4R}} \]

### Final Result:
The velocity of the ball when it is at a distance \( \frac{R}{2} \) from the center of the Earth is:
\[ v = \sqrt{\frac{3GM}{4R}} \]