A tunnel ois dug along a chord of the earth a perpendicular distance `R/2` from the earth's centre. The wall of the tunnel may be assumed to be frictionless. Find the force exerted by the eall on a particle of mass m when it is at a distance x from the centre of the tunnel.

Explanation

To find the force exerted by the wall of the tunnel on a particle of mass \( m \) when it is at a distance \( x \) from the center of the tunnel, we can follow these steps:

### Step 1: Understand the Setup
We have a tunnel dug along a chord of the Earth, at a perpendicular distance of \( R/2 \) from the center of the Earth. We need to find the gravitational force acting on a mass \( m \) located at a distance \( x \) from the center of the tunnel.

### Step 2: Define the Geometry
Let \( R \) be the radius of the Earth. The distance from the center of the Earth to the tunnel is \( R/2 \). The distance from the center of the tunnel to the mass \( m \) is \( d \), which can be expressed in terms of \( x \) and \( R/2 \) using the Pythagorean theorem:
\[ d = \sqrt{(R/2)^2 + x^2} \]

### Step 3: Calculate the Mass of the Sphere
The mass \( M' \) of the sphere of radius \( d \) (the sphere that would enclose the mass \( m \)) is given by:
\[ M' = \frac{4}{3} \pi d^3 \rho \]
where \( \rho \) is the density of the Earth. The total mass \( M \) of the Earth is:
\[ M = \frac{4}{3} \pi R^3 \rho \]

### Step 4: Relate the Masses
Using the ratio of the masses, we can express \( M' \) in terms of \( M \):
\[ \frac{M'}{M} = \frac{d^3}{R^3} \]
Thus,
\[ M' = M \cdot \frac{d^3}{R^3} \]

### Step 5: Calculate the Gravitational Force
The gravitational force \( F \) acting on the mass \( m \) due to the mass \( M' \) is given by Newton's law of gravitation:
\[ F = \frac{G m M'}{d^2} \]
Substituting \( M' \):
\[ F = \frac{G m \left(M \cdot \frac{d^3}{R^3}\right)}{d^2} = \frac{G m M d}{R^3} \]

### Step 6: Determine the Direction of the Force
The force \( F \) acts towards the center of the Earth. Since we are interested in the component of force acting along the direction of the tunnel, we need to find the angle \( \theta \) formed with the vertical. Using trigonometry:
\[ \cos \theta = \frac{R/2}{d} \]

### Step 7: Substitute \( \cos \theta \) into the Force Equation
The effective force acting along the tunnel is:
\[ F_{\text{tunnel}} = F \cos \theta = \frac{G m M d}{R^3} \cdot \frac{R/2}{d} = \frac{G m M R}{2 R^3} = \frac{G m M}{2 R^2} \]

### Final Result
Thus, the force exerted by the wall of the tunnel on the particle of mass \( m \) when it is at a distance \( x \) from the center of the tunnel is:
\[ F = \frac{G m M}{2 R^2} \]