Assume that a frictionless tunnel is made in the earth along its diameter, a particle in projected in the tunnel from the surface of the earth with an initial speed `v_(0)=sqrt(gR)`, where g is the acceleration due to gravity on the earth's surface & R is the radius of the earth, if time taken by the particle to reach the centre of the earth is `(pi)/(n)sqrt((4R)/(g))` the value of `n` is?

Explanation

To solve the problem, we need to analyze the motion of a particle projected into a frictionless tunnel through the Earth. The particle is projected with an initial speed \( v_0 = \sqrt{gR} \), where \( g \) is the acceleration due to gravity and \( R \) is the radius of the Earth. We are tasked with finding the value of \( n \) in the expression for the time taken by the particle to reach the center of the Earth, given as \( \frac{\pi}{n} \sqrt{\frac{4R}{g}} \).

### Step-by-Step Solution:

1. **Understand the Energy Conservation**:
The total mechanical energy (kinetic + potential) of the particle remains constant. At the surface, the total energy is:
\[ E_{\text{initial}} = \frac{1}{2} m v_0^2 - \frac{GMm}{R} \]
where \( v_0 = \sqrt{gR} \) and \( G \) is the gravitational constant.

2. **Substituting Values**:
We know that \( g = \frac{GM}{R^2} \). Therefore, we can write:
\[ E_{\text{initial}} = \frac{1}{2} m (gR) - \frac{GMm}{R} \]
Simplifying this gives:
\[ E_{\text{initial}} = \frac{1}{2} mgR - mg = \frac{1}{2} mgR - mg = \frac{1}{2} mgR - \frac{2}{2} mg = \frac{1}{2} mgR - \frac{2mg}{2} \]
\[ E_{\text{initial}} = \frac{1}{2} mgR - mg = \frac{1}{2} mgR - \frac{2mgR}{2R} = \frac{(1 - 2)mgR}{2} \]

3. **At the Center of the Earth**:
At the center, the potential energy is:
\[ U_{\text{center}} = -\frac{3}{2} \frac{GMm}{R} \]
The kinetic energy at the center is:
\[ E_{\text{final}} = \frac{1}{2} mv^2 - \frac{3}{2} \frac{GMm}{R} \]

4. **Equating Energies**:
Setting \( E_{\text{initial}} = E_{\text{final}} \):
\[ \frac{1}{2} mgR - mg = \frac{1}{2} mv^2 - \frac{3}{2} \frac{GMm}{R} \]

5. **Finding the Velocity at the Center**:
After simplifying, we can find the velocity \( v \) at the center:
\[ mgR - 2mg = mv^2 - 3 \frac{GMm}{R} \]
After some algebra, we can derive that:
\[ v = \sqrt{2gR} \]

6. **Using Simple Harmonic Motion**:
The motion of the particle in the tunnel can be modeled as simple harmonic motion. The angular frequency \( \omega \) is given by:
\[ \omega = \sqrt{\frac{g}{R}} \]

7. **Finding the Time to Reach the Center**:
The time \( t \) to reach the center from the surface is given by:
\[ t = \frac{\pi}{2\omega} = \frac{\pi}{2} \sqrt{\frac{R}{g}} \]

8. **Relating to Given Expression**:
We need to relate this to the given expression:
\[ t = \frac{\pi}{n} \sqrt{\frac{4R}{g}} \]
Setting the two expressions for time equal gives:
\[ \frac{\pi}{2} \sqrt{\frac{R}{g}} = \frac{\pi}{n} \sqrt{\frac{4R}{g}} \]

9. **Solving for \( n \)**:
Canceling \( \pi \) and simplifying:
\[ \frac{1}{2} = \frac{2}{n} \]
Thus, \( n = 4 \).

### Final Answer:
The value of \( n \) is \( 4 \).