A smooth block is released at rest on a `45^(@)` incline and then slides a distance `d`. The time taken to slide is n times as much to slide on rough incline than on a smooth incline. The coefficient of friction is

Options

A. `mu_(k) = 1- ( 1)/( n^(2))`
B. `mu_(k) = sqrt( 1 - ( 1)/ ( n^(2)))`
C. `mu_(s)=1-(1)/( n^(2))`
D. `mu_(s) = sqrt( 1 - ( 1)/ ( n^(2)))`

Answer

To solve the problem, we need to analyze the motion of a block sliding down both a smooth incline and a rough incline. We will derive the expressions for the time taken to slide down each incline and then relate them to find the coefficient of friction.

### Step-by-step Solution:

1. **Understanding the Forces on the Incline:**
- For a smooth incline (no friction), the only force acting on the block along the incline is the component of gravitational force:
\[ F_{\text{smooth}} = mg \sin \theta \]
- For a rough incline, the frictional force also acts against the motion. The net force acting on the block is:
\[ F_{\text{rough}} = mg \sin \theta - \mu mg \cos \theta \]

2. **Setting Up the Equations of Motion:**
- For the smooth incline, using \(d = \frac{1}{2} a t^2\):
\[ d = \frac{1}{2} g \sin \theta \cdot t_1^2 \quad \Rightarrow \quad t_1^2 = \frac{2d}{g \sin \theta} \]
- For the rough incline:
\[ d = \frac{1}{2} (g \sin \theta - \mu g \cos \theta) t_2^2 \quad \Rightarrow \quad t_2^2 = \frac{2d}{g \sin \theta - \mu g \cos \theta} \]

3. **Relating the Times:**
- According to the problem, the time taken on the rough incline is \(n\) times that on the smooth incline:
\[ t_2 = n t_1 \]
- Squaring both sides gives:
\[ t_2^2 = n^2 t_1^2 \]

4. **Substituting the Expressions for \(t_1^2\) and \(t_2^2\):**
- Substitute the expressions we derived:
\[ \frac{2d}{g \sin \theta - \mu g \cos \theta} = n^2 \cdot \frac{2d}{g \sin \theta} \]
- Cancel \(2d\) from both sides (assuming \(d \neq 0\)):
\[ \frac{1}{g \sin \theta - \mu g \cos \theta} = \frac{n^2}{g \sin \theta} \]

5. **Cross-Multiplying:**
- Cross-multiplying gives:
\[ g \sin \theta = n^2 (g \sin \theta - \mu g \cos \theta) \]
- Rearranging leads to:
\[ g \sin \theta = n^2 g \sin \theta - n^2 \mu g \cos \theta \]
- Simplifying:
\[ n^2 \mu g \cos \theta = (n^2 - 1) g \sin \theta \]

6. **Solving for the Coefficient of Friction \(\mu\):**
- Dividing both sides by \(g\) (assuming \(g \neq 0\)):
\[ n^2 \mu \cos \theta = (n^2 - 1) \sin \theta \]
- Thus, we can express \(\mu\) as:
\[ \mu = \frac{(n^2 - 1) \sin \theta}{n^2 \cos \theta} \]

7. **Substituting \(\theta = 45^\circ\):**
- Since \(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\):
\[ \mu = \frac{(n^2 - 1) \cdot \frac{1}{\sqrt{2}}}{n^2 \cdot \frac{1}{\sqrt{2}}} = \frac{n^2 - 1}{n^2} \]

### Final Answer:
\[ \mu = \frac{n^2 - 1}{n^2} \]