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

Explanation

To solve the problem, we will analyze the motion of a block sliding down both a smooth incline and a rough incline, and we will derive the coefficient of friction based on the time taken to slide down each incline.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Angle of incline, \( \theta = 45^\circ \)
- Distance slid down the incline, \( d \)
- Time taken on the smooth incline, \( t \)
- Time taken on the rough incline, \( T = n \cdot t \) (where \( n \) is a given factor)

2. **Calculate Acceleration on the Smooth Incline:**
- For a smooth incline, the acceleration \( a_1 \) is given by:
\[ a_1 = g \sin \theta = g \sin 45^\circ = \frac{g}{\sqrt{2}} \]

3. **Use the Equation of Motion for the Smooth Incline:**
- Using the second equation of motion \( s = ut + \frac{1}{2} a t^2 \) (where initial velocity \( u = 0 \)):
\[ d = 0 \cdot t + \frac{1}{2} a_1 t^2 \Rightarrow d = \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) t^2 \]
- Rearranging gives:
\[ d = \frac{g}{2\sqrt{2}} t^2 \]

4. **Calculate Acceleration on the Rough Incline:**
- For the rough incline, the acceleration \( a_2 \) is given by:
\[ a_2 = g \sin 45^\circ - \mu g \cos 45^\circ = \frac{g}{\sqrt{2}} - \mu \frac{g}{\sqrt{2}} = \frac{g(1 - \mu)}{\sqrt{2}} \]

5. **Use the Equation of Motion for the Rough Incline:**
- The time taken on the rough incline is \( T = n \cdot t \):
\[ d = \frac{1}{2} a_2 T^2 = \frac{1}{2} \left(\frac{g(1 - \mu)}{\sqrt{2}}\right) (n t)^2 \]
- Rearranging gives:
\[ d = \frac{g(1 - \mu)}{2\sqrt{2}} n^2 t^2 \]

6. **Equate the Two Expressions for Distance \( d \):**
- From the smooth incline:
\[ d = \frac{g}{2\sqrt{2}} t^2 \]
- From the rough incline:
\[ d = \frac{g(1 - \mu)}{2\sqrt{2}} n^2 t^2 \]
- Setting these equal gives:
\[ \frac{g}{2\sqrt{2}} t^2 = \frac{g(1 - \mu)}{2\sqrt{2}} n^2 t^2 \]

7. **Cancel Common Terms:**
- Cancel \( \frac{g}{2\sqrt{2}} t^2 \) from both sides (assuming \( t \neq 0 \)):
\[ 1 = (1 - \mu) n^2 \]

8. **Solve for the Coefficient of Friction \( \mu \):**
- Rearranging gives:
\[ 1 - \mu = \frac{1}{n^2} \]
- Thus, we find:
\[ \mu = 1 - \frac{1}{n^2} \]

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