Five particles each of mass 'm' are kept at five verties of a regular pentagon . A sixth particles of mass 'M' is kept at center of the pentagon 'O'.Distance between'M' and is 'm' is 'a'. Find
(a) net force on 'M'
(b) magnitude of net force on 'M' if any one particle is removed from one of the verties.

Explanation

To solve the problem, we need to analyze the forces acting on the mass \( M \) located at the center of the pentagon due to the five particles of mass \( m \) located at the vertices.

### Step-by-Step Solution:

**(a) Finding the net force on \( M \):**

1. **Identify the Setup:**
- We have five particles, each of mass \( m \), located at the vertices of a regular pentagon.
- A sixth particle of mass \( M \) is located at the center \( O \) of the pentagon.
- The distance from \( O \) to any vertex (where mass \( m \) is located) is given as \( a \).

2. **Calculate the Gravitational Force:**
- The gravitational force between the mass \( M \) at the center and any mass \( m \) at the vertex can be calculated using Newton's law of gravitation:
\[ F = \frac{G \cdot M \cdot m}{a^2} \]
- Here, \( G \) is the gravitational constant.

3. **Direction of Forces:**
- Each mass \( m \) exerts a gravitational force on \( M \) directed towards itself. Since the pentagon is symmetrical, the forces exerted by the particles will have equal magnitudes but different directions.

4. **Net Force Calculation:**
- The angle between the lines connecting the center \( O \) to any two adjacent vertices of the pentagon is \( 72^\circ \) (since \( 360^\circ / 5 = 72^\circ \)).
- Due to the symmetry of the pentagon, the horizontal components of the forces from the five masses will cancel out, and the vertical components will also cancel out.
- Therefore, the net force on \( M \) due to the five masses is:
\[ F_{\text{net}} = 0 \]

**(b) Finding the magnitude of net force on \( M \) if one particle is removed:**

1. **Removing One Particle:**
- Let's assume we remove one of the particles, say particle \( B \). Now, we have four particles left at the vertices.

2. **Calculate the Remaining Forces:**
- The remaining particles \( A, C, D, \) and \( E \) will still exert gravitational forces on \( M \).
- The forces from particles \( A, C, D, \) and \( E \) will not cancel out completely because of the asymmetry introduced by removing one particle.

3. **Resultant Force Calculation:**
- The net force will now be the vector sum of the forces due to the four remaining particles.
- The force due to each remaining particle is still given by:
\[ F = \frac{G \cdot M \cdot m}{a^2} \]
- The resultant force will be directed towards the center of mass of the remaining particles, which will be in the direction of the vertex opposite to the removed particle.

4. **Magnitude of Net Force:**
- Since we have four forces acting at angles, we can find the resultant force using vector addition. However, for simplicity, we can state that the net force \( F_{\text{net}} \) on \( M \) when one particle is removed will be:
\[ F_{\text{net}} = \frac{G \cdot M \cdot m}{a^2} \]
- This force will be directed towards the remaining particles.

### Final Answers:
- (a) The net force on \( M \) is \( 0 \).
- (b) The magnitude of the net force on \( M \) if one particle is removed is \( \frac{G \cdot M \cdot m}{a^2} \).