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 velocit attained by the particle in the subsequent motion is `sqrt((nGM)/(R))` (assuming tunnel to be frictionless). Find n

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy of the particle remains constant since the tunnel is frictionless.

### Step-by-step Solution:

1. **Identify Initial and Final States:**
- The particle is projected with an initial velocity \( v_0 = \sqrt{\frac{2GM}{R}} \) from one end of the tunnel.
- The maximum velocity will occur at the center of the Earth, where the gravitational potential energy is at its minimum.

2. **Calculate Initial Energy:**
- The initial kinetic energy \( KE_i \) of the particle is given by:
\[ KE_i = \frac{1}{2} m v_0^2 = \frac{1}{2} m \left(\frac{2GM}{R}\right) = \frac{mGM}{R} \]
- The gravitational potential energy \( PE_i \) at the surface of the Earth is:
\[ PE_i = -\frac{GMm}{R} \]
- Therefore, the total initial energy \( E_i \) is:
\[ E_i = KE_i + PE_i = \frac{mGM}{R} - \frac{GMm}{R} = 0 \]

3. **Calculate Final Energy at the Center:**
- At the center of the Earth, the potential energy \( PE_f \) is:
\[ PE_f = -\frac{3GMm}{2R} \]
- Let \( v_f \) be the maximum velocity at the center. The kinetic energy \( KE_f \) at this point is:
\[ KE_f = \frac{1}{2} m v_f^2 \]
- The total final energy \( E_f \) is:
\[ E_f = KE_f + PE_f = \frac{1}{2} m v_f^2 - \frac{3GMm}{2R} \]

4. **Apply Conservation of Energy:**
- Since total energy is conserved:
\[ E_i = E_f \]
- Substituting the values:
\[ 0 = \frac{1}{2} m v_f^2 - \frac{3GMm}{2R} \]
- Rearranging gives:
\[ \frac{1}{2} m v_f^2 = \frac{3GMm}{2R} \]

5. **Solve for Maximum Velocity \( v_f \):**
- Dividing both sides by \( m \) and multiplying by 2:
\[ v_f^2 = \frac{3GM}{R} \]
- Taking the square root:
\[ v_f = \sqrt{\frac{3GM}{R}} \]

6. **Express Maximum Velocity in the Given Form:**
- We are given that the maximum velocity can also be expressed as \( v_f = \sqrt{\frac{nGM}{R}} \).
- By comparing the two expressions:
\[ \sqrt{\frac{3GM}{R}} = \sqrt{\frac{nGM}{R}} \]
- Thus, we can equate the coefficients:
\[ n = 3 \]

### Final Answer:
The value of \( n \) is \( 3 \).