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

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the situation of a particle released from one end of a frictionless tunnel dug along a chord of the Earth at a perpendicular distance of \( R/2 \) from the center of the Earth. We will derive the expressions for the pressing force by the particle on the wall of the tunnel and the acceleration of the particle as it moves through the tunnel.

### Step 1: Understanding the Geometry
The tunnel is dug at a distance of \( R/2 \) from the center of the Earth. The distance of the particle from the center of the Earth is denoted as \( x \). The gravitational force acting on the particle varies with depth, and we need to find expressions for the forces acting on the particle.

### Step 2: Gravitational Force Inside the Earth
Inside the Earth, the gravitational force \( g' \) at a distance \( x \) from the center can be given by:
\[ g' = g \left(1 - \frac{d}{R}\right) \]
where \( d \) is the depth below the surface. Since the depth \( d \) can be expressed as \( R - x \), we have:
\[ g' = g \left(1 - \frac{R - x}{R}\right) = g \left(\frac{x}{R}\right) \]

### Step 3: Pressing Force on the Wall
The pressing force \( N \) exerted by the particle on the wall of the tunnel can be derived from the component of the gravitational force acting perpendicular to the wall. The angle \( \theta \) can be determined from the geometry of the situation, where:
\[ \sin \theta = \frac{R/2}{x} \]
Thus, the normal force can be expressed as:
\[ N = mg' \sin \theta \]
Substituting \( g' \) and \( \sin \theta \):
\[ N = mg \left(\frac{x}{R}\right) \left(\frac{R/2}{x}\right) = \frac{mg}{2} \]
This shows that the pressing force \( N \) is constant and does not depend on \( x \).

### Step 4: Acceleration of the Particle
The net force acting on the particle in the direction of the tunnel is given by:
\[ F = mg' \cos \theta \]
Using \( g' = g \left(\frac{x}{R}\right) \) and \( \cos \theta = \frac{\sqrt{x^2 - (R/2)^2}}{x} \):
\[ F = mg \left(\frac{x}{R}\right) \left(\frac{\sqrt{x^2 - (R/2)^2}}{x}\right) = \frac{mg \sqrt{x^2 - (R/2)^2}}{R} \]
The acceleration \( a \) of the particle can be expressed as:
\[ a = \frac{F}{m} = \frac{g \sqrt{x^2 - (R/2)^2}}{R} \]

### Step 5: Analyzing the Acceleration Expression
The expression for acceleration shows that it varies with \( x \) and is zero when \( x = R/2 \). The acceleration is a parabolic function of \( x \).

### Conclusion
The pressing force by the particle on the wall of the tunnel is constant, while the acceleration of the particle varies parabolically with \( x \). Therefore, the correct options for the graphs representing these relationships are:
- Pressing Force: Constant (Option B)
- Acceleration: Parabolic (Option C)