A solid sphere is released from the state of rest on an inclined plane. Mass of the sphere is m and its radius is r. Inclined plane makes an angle with the horizontal. If minimum coefficient of friction required for pure rolling is `(n)/(7)tan theta` then calculate value of n.

Explanation

To solve the problem, we need to analyze the motion of a solid sphere rolling down an inclined plane. Here are the step-by-step calculations:

### Step 1: Analyze Forces Acting on the Sphere
When the sphere is released on the inclined plane, the forces acting on it are:
- The gravitational force acting downwards: \( mg \)
- The component of gravitational force acting along the incline: \( mg \sin \theta \)
- The normal force acting perpendicular to the incline: \( N = mg \cos \theta \)
- The frictional force \( F \) acting up the incline.

### Step 2: Apply Newton's Second Law Along the Incline
Using Newton's second law along the incline, we can write:
\[ mg \sin \theta - F = ma \]
where \( a \) is the linear acceleration of the sphere.

### Step 3: Relate Linear and Angular Acceleration
For a solid sphere rolling without slipping, the relationship between linear acceleration \( a \) and angular acceleration \( \alpha \) is given by:
\[ a = \alpha r \]
where \( r \) is the radius of the sphere.

### Step 4: Apply Torque Equation
The torque \( \tau \) due to the frictional force \( F \) about the center of mass is:
\[ \tau = F \cdot r \]
According to the rotational dynamics, this torque is also equal to the moment of inertia \( I \) times the angular acceleration \( \alpha \):
\[ F \cdot r = I \alpha \]
For a solid sphere, the moment of inertia about its center of mass is:
\[ I = \frac{2}{5} m r^2 \]
Substituting \( \alpha = \frac{a}{r} \) into the torque equation gives:
\[ F \cdot r = \frac{2}{5} m r^2 \cdot \frac{a}{r} \]
This simplifies to:
\[ F = \frac{2}{5} m a \]

### Step 5: Substitute Friction Force into the Equation of Motion
Now, substituting \( F = \frac{2}{5} m a \) into the equation from Step 2:
\[ mg \sin \theta - \frac{2}{5} m a = ma \]
Rearranging gives:
\[ mg \sin \theta = ma + \frac{2}{5} m a \]
\[ mg \sin \theta = \frac{7}{5} m a \]
From this, we can solve for \( a \):
\[ a = \frac{5}{7} g \sin \theta \]

### Step 6: Calculate the Friction Force
Now substituting \( a \) back into the expression for friction:
\[ F = \frac{2}{5} m a = \frac{2}{5} m \left( \frac{5}{7} g \sin \theta \right) = \frac{2}{7} mg \sin \theta \]

### Step 7: Relate Friction to the Coefficient of Friction
The maximum static friction force can be expressed as:
\[ F \leq \mu N \]
where \( N = mg \cos \theta \). Therefore:
\[ \frac{2}{7} mg \sin \theta \leq \mu mg \cos \theta \]
Cancelling \( mg \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{2}{7} \sin \theta \leq \mu \cos \theta \]
Rearranging gives:
\[ \mu \geq \frac{2}{7} \tan \theta \]

### Step 8: Identify the Minimum Coefficient of Friction
According to the problem, the minimum coefficient of friction required for pure rolling is given as:
\[ \mu = \frac{n}{7} \tan \theta \]
Setting the two expressions for \( \mu \) equal gives:
\[ \frac{n}{7} \tan \theta = \frac{2}{7} \tan \theta \]
This implies:
\[ n = 2 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{2} \]