A tunnel is dug along a chord of the earth at a perpendicular distance `R//2` from the earth's centre. The wall of the tunnel may be assumed to be frictionless. A particle is released from one end of the tunnel. The pressing force by the particle on the wall, and the acceleration of the particle vary with `x` (distance of the particle from the centre) according to

Options

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D.

Answer

To solve the problem of a particle released from one end of a tunnel dug along a chord of the Earth at a perpendicular distance \( R/2 \) from the Earth's center, we will analyze the forces acting on the particle and derive the expressions for the pressing force on the wall and the acceleration of the particle.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The tunnel is dug along a chord of the Earth, and the distance from the center of the Earth to the tunnel is \( R/2 \).
- A particle is released from one end of the tunnel.

2. **Determine the Gravitational Force**:
- The gravitational force acting on the particle at a distance \( x \) from the center of the Earth can be given by:
\[ g' = g \left(1 - \frac{x}{R}\right) \]
- Here, \( g \) is the acceleration due to gravity at the surface of the Earth.

3. **Finding the Normal Force (Pressing Force)**:
- The normal force \( N \) acting on the particle due to the wall of the tunnel can be expressed as:
\[ N = mg' \sin(\theta) \]
- The angle \( \theta \) can be determined using the geometry of the situation. Since the tunnel is at a distance \( R/2 \), we can find \( \sin(\theta) \) in terms of \( x \):
\[ \sin(\theta) = \frac{R/2}{x} \]
- Substituting \( g' \) into the normal force equation gives:
\[ N = mg \left(1 - \frac{x}{R}\right) \cdot \frac{R/2}{x} \]

4. **Simplifying the Normal Force**:
- Rearranging the equation yields:
\[ N = \frac{mgR}{2} \left(\frac{1}{x} - \frac{1}{R}\right) \]
- This shows that the normal force varies inversely with \( x \).

5. **Finding the Acceleration of the Particle**:
- The net force acting on the particle is given by:
\[ F = m g' \cos(\theta) \]
- The acceleration \( a \) can then be expressed as:
\[ a = \frac{F}{m} = g' \cos(\theta) \]
- Substituting \( g' \) and \( \cos(\theta) \) gives:
\[ a = g \left(1 - \frac{x}{R}\right) \cdot \frac{\sqrt{R^2 - (R/2)^2}}{x} \]
- Simplifying this expression shows that the acceleration varies quadratically with \( x \).

6. **Conclusion**:
- The pressing force \( N \) varies inversely with \( x \), and the acceleration \( a \) varies quadratically with \( x \).
- Based on the analysis, we can conclude that the correct options for the graphs of pressing force and acceleration are as follows:
- The pressing force \( N \) is constant with respect to \( x \).
- The acceleration \( a \) is a parabolic function of \( x \).