A block moves down a smooth inclined plane of inclination `theta`. Its velocity on reaching the bottom is v. If it slides down a rough inclined plane of some inclination, its velocity on reaching the bottom is v/n, where n is a number greater than 0. The coefficient of friction is given by -

Options

A. `mu = tan theta(1-1/(n^2))`
B. `mu = cot theta(1-1/(n^2))`
C. `mu = tan theta(1-1/(n^2))^(1//2)`
D. `mu = cot theta(1-1/(n^2))^(1//2)`

Answer

To solve the problem, we need to analyze the motion of a block sliding down two different inclined planes: one smooth and one rough. The goal is to find the coefficient of friction (μ) in terms of the angle of inclination (θ) and the number (n) given in the problem.

### Step-by-Step Solution:

1. **Identify Forces on the Rough Inclined Plane:**
- When the block is sliding down the rough inclined plane, the forces acting on it are:
- The gravitational force component parallel to the incline: \( F_{\text{parallel}} = mg \sin \theta \)
- The gravitational force component perpendicular to the incline: \( F_{\text{perpendicular}} = mg \cos \theta \)
- The frictional force acting opposite to the direction of motion: \( F_{\text{friction}} = \mu N = \mu mg \cos \theta \)

2. **Apply Newton's Second Law:**
- The net force acting along the incline can be expressed as:
\[ F_{\text{net}} = mg \sin \theta - \mu mg \cos \theta \]
- According to Newton's second law, this net force is also equal to the mass times acceleration (a):
\[ ma = mg \sin \theta - \mu mg \cos \theta \]
- Dividing through by \( m \):
\[ a = g \sin \theta - \mu g \cos \theta \]

3. **Use Kinematic Equation for Velocity:**
- The block starts from rest and slides down a distance \( L \) along the incline. Using the kinematic equation:
\[ v^2 = u^2 + 2aL \]
- Since the initial velocity \( u = 0 \):
\[ v^2 = 2aL \]
- Substituting for \( a \):
\[ v^2 = 2(g \sin \theta - \mu g \cos \theta)L \]
- This simplifies to:
\[ v^2 = 2gL(\sin \theta - \mu \cos \theta) \]

4. **Velocity on the Smooth Incline:**
- For the smooth incline, the block reaches the bottom with velocity \( v \):
\[ v^2 = 2gL \sin \theta \]

5. **Relate the Two Velocities:**
- When sliding down the rough incline, the velocity is given as \( \frac{v}{n} \):
\[ \left(\frac{v}{n}\right)^2 = 2gL(\sin \theta - \mu \cos \theta) \]
- Substituting \( v^2 = 2gL \sin \theta \):
\[ \frac{(2gL \sin \theta)}{n^2} = 2gL(\sin \theta - \mu \cos \theta) \]

6. **Cancel Common Terms:**
- Dividing both sides by \( 2gL \):
\[ \frac{\sin \theta}{n^2} = \sin \theta - \mu \cos \theta \]

7. **Rearranging to Solve for μ:**
- Rearranging gives:
\[ \mu \cos \theta = \sin \theta - \frac{\sin \theta}{n^2} \]
- Factor out \( \sin \theta \):
\[ \mu \cos \theta = \sin \theta \left(1 - \frac{1}{n^2}\right) \]
- Finally, solving for \( \mu \):
\[ \mu = \frac{\sin \theta \left(1 - \frac{1}{n^2}\right)}{\cos \theta} \]
- This simplifies to:
\[ \mu = \tan \theta \left(1 - \frac{1}{n^2}\right) \]

### Final Answer:
The coefficient of friction \( \mu \) is given by:
\[ \mu = \tan \theta \left(1 - \frac{1}{n^2}\right) \]