A solid sphere is rolling on a horizontal surface such that speed of the centre of sphere is u. Mase of the sphere is m and radius R. Magnitude of angular momentum about the point of contact is found to be `(7)/(n)mvR`/. Find the value of n

Explanation

To solve the problem, we need to find the value of \( n \) given that the angular momentum \( L \) about the point of contact of a rolling solid sphere is expressed as \( \frac{7}{n} mvR \).

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a solid sphere rolling without slipping on a horizontal surface.
- The speed of the center of the sphere is \( u \).
- The mass of the sphere is \( m \) and its radius is \( R \).

2. **Angular Momentum about the Point of Contact**:
- The angular momentum \( L \) about the point of contact can be calculated using the formula:
\[ L = R \times P + I_{cm} \omega \]
- Here, \( R \) is the position vector from the point of contact to the center of mass, \( P = mv \) is the linear momentum of the center of mass, \( I_{cm} \) is the moment of inertia of the sphere about its center, and \( \omega \) is the angular velocity.

3. **Calculating Linear Momentum**:
- The linear momentum \( P \) of the center of mass is:
\[ P = mv \]
- Since the sphere is rolling, the speed \( v \) at the center of mass is \( u \).

4. **Calculating Moment of Inertia**:
- The moment of inertia \( I_{cm} \) for a solid sphere about its center is given by:
\[ I_{cm} = \frac{2}{5} m R^2 \]

5. **Calculating Angular Velocity**:
- The angular velocity \( \omega \) is related to the linear speed \( u \) by:
\[ \omega = \frac{u}{R} \]

6. **Position Vector**:
- The position vector \( R \) from the point of contact to the center of mass is equal to the radius \( R \).

7. **Calculating Angular Momentum**:
- Substituting the values into the angular momentum formula:
\[ L = R \times mv + I_{cm} \omega \]
- The first term becomes:
\[ R \times mv = R \cdot mv \cdot \hat{k} \quad \text{(where \( \hat{k} \) is the unit vector perpendicular to the plane)} \]
- The second term becomes:
\[ I_{cm} \omega = \frac{2}{5} m R^2 \cdot \frac{u}{R} = \frac{2}{5} m R u \]

8. **Combining Terms**:
- The total angular momentum \( L \) about the point of contact is:
\[ L = mvR + \frac{2}{5} m R u = mvR + \frac{2}{5} mvR = \left(1 + \frac{2}{5}\right) mvR = \frac{7}{5} mvR \]

9. **Finding \( n \)**:
- According to the problem statement, the angular momentum is given as \( \frac{7}{n} mvR \).
- Comparing both expressions:
\[ \frac{7}{5} mvR = \frac{7}{n} mvR \]
- Therefore, we can equate:
\[ n = 5 \]

### Final Answer:
The value of \( n \) is \( 5 \).