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=tantheta(1-(1)/(n^(2)))`
B. `mu=cottheta(1-(1)/(n^(2)))`
C. `mu=tantheta(1+(1)/(n^(2)))`
D. `mu=cottheta(1+(1)/(n^(2)))`

Answer

To solve the problem, we need to analyze the motion of the block on both the smooth and rough inclined planes, and then derive the expression for the coefficient of friction.

### Step-by-Step Solution:

1. **Understanding the Motion on the Smooth Inclined Plane:**
- For a smooth inclined plane at an angle \( \theta \), the only force causing the block to accelerate down the incline is the component of gravitational force along the incline.
- The gravitational force acting on the block is \( mg \), where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity.
- The component of this force acting down the incline is \( mg \sin \theta \).
- The acceleration \( a \) of the block is given by:
\[ a = g \sin \theta \]

2. **Finding the Velocity at the Bottom of the Smooth Incline:**
- Using the kinematic equation \( v^2 = u^2 + 2as \) (where \( u = 0 \) at the top), we have:
\[ v^2 = 0 + 2(g \sin \theta) S \]
- Thus, the velocity \( v \) at the bottom of the smooth incline is:
\[ v^2 = 2g \sin \theta S \]

3. **Understanding the Motion on the Rough Inclined Plane:**
- For the rough inclined plane, the block experiences friction. The frictional force \( f_k \) acts up the incline and is given by \( f_k = \mu N \), where \( \mu \) is the coefficient of friction and \( N \) is the normal force.
- The normal force \( N \) on the inclined plane is \( mg \cos \theta \).
- The net force acting down the incline is:
\[ F_{\text{net}} = mg \sin \theta - f_k = mg \sin \theta - \mu mg \cos \theta \]
- According to Newton's second law, \( F = ma \):
\[ ma = mg \sin \theta - \mu mg \cos \theta \]
- Dividing through by \( m \):
\[ a = g \sin \theta - \mu g \cos \theta \]

4. **Finding the Velocity at the Bottom of the Rough Incline:**
- The final velocity when the block reaches the bottom of the rough incline is given as \( \frac{v}{n} \).
- Using the same kinematic equation:
\[ \left(\frac{v}{n}\right)^2 = 0 + 2\left(g \sin \theta - \mu g \cos \theta\right) S \]
- Thus, we have:
\[ \frac{v^2}{n^2} = 2\left(g \sin \theta - \mu g \cos \theta\right) S \]

5. **Equating the Two Expressions for \( v^2 \):**
- From the smooth incline, we have:
\[ v^2 = 2g \sin \theta S \]
- Substituting this into the rough incline equation:
\[ \frac{2g \sin \theta S}{n^2} = 2\left(g \sin \theta - \mu g \cos \theta\right) S \]
- Canceling \( 2S \) from both sides (assuming \( S \neq 0 \)):
\[ \frac{g \sin \theta}{n^2} = g \sin \theta - \mu g \cos \theta \]

6. **Solving for the Coefficient of Friction \( \mu \):**
- Rearranging gives:
\[ g \sin \theta - \frac{g \sin \theta}{n^2} = \mu g \cos \theta \]
- Factoring out \( g \sin \theta \):
\[ g \sin \theta \left(1 - \frac{1}{n^2}\right) = \mu g \cos \theta \]
- Dividing both sides by \( g \cos \theta \):
\[ \mu = \frac{\sin \theta \left(1 - \frac{1}{n^2}\right)}{\cos \theta} \]
- This simplifies to:
\[ \mu = \tan \theta \left(1 - \frac{1}{n^2}\right) \]

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