Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance `(R//2)` from the earth's centre, where 'R' is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period :

Options

A. `(2piR)/g`
B. `g/(2pir)`
C. `1/(2pi)sqrt(g/r)`
D. `2pisqrt(R/g)`

Answer

To solve the problem, we need to determine the time period of a particle executing simple harmonic motion (SHM) when released in a tunnel dug along a chord of the Earth at a perpendicular distance of \( \frac{R}{2} \) from the Earth's center, where \( R \) is the radius of the Earth.

### Step-by-step Solution:

1. **Understanding the Setup**:
- The tunnel is dug at a distance of \( \frac{R}{2} \) from the center of the Earth.
- When a particle is released in this tunnel, it will experience gravitational force directed towards the center of the Earth.

2. **Gravitational Force Inside the Earth**:
- The gravitational force acting on the particle at a distance \( r \) from the center of the Earth is given by:
\[ F_g = -\frac{G M_e m}{r^2} \]
- However, since the particle is at a distance of \( \frac{R}{2} \), we need to find the gravitational field at that point.

3. **Gravitational Field Inside the Earth**:
- The gravitational field \( E \) at a distance \( r \) from the center of the Earth is given by:
\[ E = \frac{G M_e}{r^2} \]
- For \( r = \frac{R}{2} \):
\[ E = \frac{G M_e}{\left(\frac{R}{2}\right)^2} = \frac{4 G M_e}{R^2} \]

4. **Restoring Force**:
- The restoring force \( F \) acting on the particle can be expressed as:
\[ F = m E = m \cdot \frac{4 G M_e}{R^2} \]
- This force acts towards the center of the Earth, which is characteristic of SHM.

5. **Relating to SHM**:
- In SHM, the restoring force is proportional to the displacement \( x \) from the equilibrium position:
\[ F = -k x \]
- Here, we can identify \( k \) as:
\[ k = \frac{4 G M_e}{R^2} \]

6. **Finding the Time Period**:
- The time period \( T \) of SHM is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
- Substituting for \( k \):
\[ T = 2\pi \sqrt{\frac{m}{\frac{4 G M_e}{R^2}}} = 2\pi \sqrt{\frac{m R^2}{4 G M_e}} \]
- Since \( g = \frac{G M_e}{R^2} \), we can express \( M_e \) in terms of \( g \):
\[ T = 2\pi \sqrt{\frac{R}{g}} \]

7. **Final Result**:
- The time period of the particle executing SHM in the tunnel is:
\[ T = 2\pi \sqrt{\frac{R}{g}} \]

### Conclusion:
Thus, the time period of the particle executing simple harmonic motion in the tunnel is \( T = 2\pi \sqrt{\frac{R}{g}} \).