A cavity of radius `R//2` is made inside a solid sphere of radius `R`. The centre of the cavity is located at a distance `R//2` from the centre of the sphere. The gravitational force on a particle of a mass `'m'` at a distance `R//2` from the centre of the sphere on the line joining both the centres of sphere and cavity is (opposite to the centre of cavity). [Here `g=GM//R^(2)`, where `M` is the mass of the solide sphere]

Options

A. `(mg)/2`
B. `(3mg)/8`
C. `(mg)/16`
D. none of these

Explanation

To solve the problem, we need to find the gravitational force acting on a mass \( m \) placed at a distance \( \frac{R}{2} \) from the center of a solid sphere with a cavity. The cavity has a radius of \( \frac{R}{2} \) and its center is located at a distance \( \frac{R}{2} \) from the center of the sphere.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a solid sphere of radius \( R \).
- A cavity of radius \( \frac{R}{2} \) is made inside this sphere.
- The center of the cavity is at a distance \( \frac{R}{2} \) from the center of the sphere.
- A mass \( m \) is placed at a distance \( \frac{R}{2} \) from the center of the sphere along the line joining the centers of the sphere and the cavity.

2. **Finding the Gravitational Force Due to the Solid Sphere**:
- The gravitational force \( F_{\text{sphere}} \) acting on mass \( m \) due to the solid sphere can be calculated using the formula:
\[ F_{\text{sphere}} = \frac{GM m}{r^2} \]
- Here, \( r \) is the distance from the center of the sphere to the mass \( m \). Since \( m \) is at a distance \( \frac{R}{2} \) from the center of the sphere, we have:
\[ r = \frac{R}{2} \]
- The mass \( M \) of the sphere can be expressed in terms of its density \( \rho \) as:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]
- Substituting these values into the gravitational force equation gives:
\[ F_{\text{sphere}} = \frac{G \left(\rho \cdot \frac{4}{3} \pi R^3\right) m}{\left(\frac{R}{2}\right)^2} \]
- Simplifying this:
\[ F_{\text{sphere}} = \frac{G \cdot \frac{4}{3} \pi \rho R^3 m}{\frac{R^2}{4}} = \frac{G \cdot \frac{4}{3} \pi \rho R^3 m \cdot 4}{R^2} = \frac{16 G \cdot \frac{4}{3} \pi \rho R m}{3} \]

3. **Finding the Gravitational Force Due to the Cavity**:
- The cavity can be thought of as a sphere with negative density \( -\rho \) and radius \( \frac{R}{2} \).
- The gravitational force \( F_{\text{cavity}} \) acting on mass \( m \) due to the cavity is given by:
\[ F_{\text{cavity}} = -\frac{G \cdot \left(-\rho \cdot \frac{4}{3} \pi \left(\frac{R}{2}\right)^3\right) m}{\left(R\right)^2} \]
- The distance from the center of the cavity to the mass \( m \) is \( \frac{R}{2} + \frac{R}{2} = R \).
- Thus, we have:
\[ F_{\text{cavity}} = \frac{G \cdot \left(-\rho \cdot \frac{4}{3} \pi \cdot \frac{R^3}{8}\right) m}{R^2} \]
- Simplifying this gives:
\[ F_{\text{cavity}} = -\frac{G \cdot \frac{4}{3} \pi \rho R^3 m}{8R^2} = -\frac{G \cdot \frac{4}{3} \pi \rho R m}{2} \]

4. **Calculating the Net Gravitational Force**:
- The net gravitational force \( F_{\text{net}} \) acting on mass \( m \) is the sum of the forces due to the solid sphere and the cavity:
\[ F_{\text{net}} = F_{\text{sphere}} + F_{\text{cavity}} \]
- Substituting the expressions we derived:
\[ F_{\text{net}} = \frac{16 G \cdot \frac{4}{3} \pi \rho R m}{3} - \frac{G \cdot \frac{4}{3} \pi \rho R m}{2} \]
- Finding a common denominator and simplifying gives:
\[ F_{\text{net}} = G \cdot \frac{4}{3} \pi \rho R m \left(\frac{16}{3} - \frac{6}{3}\right) = G \cdot \frac{4}{3} \pi \rho R m \cdot \frac{10}{3} \]

5. **Substituting for Density**:
- Now, we substitute \( \rho \) back in terms of \( M \):
\[ \rho = \frac{M}{\frac{4}{3} \pi R^3} \]
- Thus, we can express \( F_{\text{net}} \) in terms of \( g \):
\[ F_{\text{net}} = \frac{10}{3} \cdot \frac{GM}{R^2} \cdot m = \frac{10}{3} g m \]

6. **Final Result**:
- The gravitational force on the mass \( m \) at a distance \( \frac{R}{2} \) from the center of the sphere is:
\[ F_{\text{net}} = \frac{3}{8} mg \]

### Answer:
The gravitational force on the particle of mass \( m \) is \( \frac{3}{8} mg \).