A tunnel is dug inside the earth across one of its diameters. Radius of earth is `R` and its mass is `M`. A particle is projected inside the tunnel with velocity `sqrt((2GM)/(R))` from one of its ends then maximum velocity attained by the particle in the subsequent motion is (assuming tunnel to be frictionless)

Options

A. `A. sqrt((GM)/(R))`
B. `B. sqrt((3GM)/(R))`
C. `C. sqrt((2GM)/(R))`
D. `D. sqrt((5GM)/(R))`

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The particle is projected into a frictionless tunnel inside the Earth, and we need to find the maximum velocity it attains during its motion.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The particle is projected with an initial velocity \( v_0 = \sqrt{\frac{2GM}{R}} \) from the surface of the Earth.

2. **Potential Energy at the Surface**:
- The gravitational potential energy \( U \) at the surface of the Earth is given by:
\[ U_{\text{surface}} = -\frac{GMm}{R} \]
- Here, \( M \) is the mass of the Earth, \( m \) is the mass of the particle, and \( R \) is the radius of the Earth.

3. **Kinetic Energy at the Surface**:
- The kinetic energy \( K \) of the particle when projected is:
\[ K_{\text{surface}} = \frac{1}{2} mv_0^2 = \frac{1}{2} m \left(\sqrt{\frac{2GM}{R}}\right)^2 = \frac{1}{2} m \cdot \frac{2GM}{R} = \frac{GMm}{R} \]

4. **Total Mechanical Energy at the Surface**:
- The total mechanical energy \( E_{\text{surface}} \) at the surface is the sum of kinetic and potential energy:
\[ E_{\text{surface}} = K_{\text{surface}} + U_{\text{surface}} = \frac{GMm}{R} - \frac{GMm}{R} = 0 \]

5. **Potential Energy at the Center**:
- At the center of the Earth, the potential energy \( U_{\text{center}} \) is given by:
\[ U_{\text{center}} = -\frac{3}{2} \frac{GMm}{R} \]

6. **Kinetic Energy at the Center**:
- Let the maximum velocity at the center be \( v_{\text{max}} \). The kinetic energy at the center is:
\[ K_{\text{center}} = \frac{1}{2} mv_{\text{max}}^2 \]

7. **Total Mechanical Energy at the Center**:
- The total mechanical energy \( E_{\text{center}} \) at the center is:
\[ E_{\text{center}} = K_{\text{center}} + U_{\text{center}} = \frac{1}{2} mv_{\text{max}}^2 - \frac{3}{2} \frac{GMm}{R} \]

8. **Setting Total Energies Equal**:
- Since mechanical energy is conserved, we set \( E_{\text{surface}} = E_{\text{center}} \):
\[ 0 = \frac{1}{2} mv_{\text{max}}^2 - \frac{3}{2} \frac{GMm}{R} \]

9. **Solving for Maximum Velocity**:
- Rearranging gives:
\[ \frac{1}{2} mv_{\text{max}}^2 = \frac{3}{2} \frac{GMm}{R} \]
- Dividing both sides by \( m \) and multiplying by 2:
\[ v_{\text{max}}^2 = \frac{3GM}{R} \]
- Taking the square root:
\[ v_{\text{max}} = \sqrt{\frac{3GM}{R}} \]

### Final Answer:
The maximum velocity attained by the particle in the subsequent motion is:
\[ v_{\text{max}} = \sqrt{\frac{3GM}{R}} \]