A given object takes n times as much time to slido down a `45^(@)` rough inclined plane as it takes to slide down a perfectly smooth `45^(@)` inclined plane. The coefficient of kinetic friction between the object and the inclined plane is given by:

Options

A. `(1-1//n^(2))`
B. `1//(1-n^(2))`
C. `sqrt((1-1//n^(2)))`
D. `sqrt({1//(1-n^(2))})`

Answer

To solve the problem, we need to analyze the motion of an object sliding down both a smooth and a rough inclined plane at a 45-degree angle. We will derive the coefficient of kinetic friction based on the time taken to slide down each plane.

### Step-by-Step Solution:

1. **Understanding the Forces on a Smooth Inclined Plane:**
- For a smooth inclined plane, the only force acting parallel to the incline is the component of gravitational force.
- The gravitational force acting down the incline is given by:
\[ F_{\text{smooth}} = mg \sin(45^\circ) = mg \frac{1}{\sqrt{2}} \]
- The acceleration \( a \) of the object down the smooth incline is:
\[ a = \frac{F_{\text{smooth}}}{m} = g \frac{1}{\sqrt{2}} \]

2. **Calculating Time on the Smooth Incline:**
- Using the equation of motion \( S = ut + \frac{1}{2} a t^2 \) where \( S = D \) (the distance down the incline), \( u = 0 \) (initial velocity), and \( a = g/\sqrt{2} \):
\[ D = \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) t^2 \]
- Rearranging gives:
\[ t^2 = \frac{2D \sqrt{2}}{g} \quad \Rightarrow \quad t = \sqrt{\frac{2D \sqrt{2}}{g}} \]

3. **Understanding the Forces on a Rough Inclined Plane:**
- For the rough inclined plane, the forces acting on the object include the gravitational force and the frictional force.
- The frictional force is given by:
\[ F_{\text{friction}} = \mu mg \cos(45^\circ) = \mu mg \frac{1}{\sqrt{2}} \]
- The net force acting down the incline is:
\[ F_{\text{rough}} = mg \sin(45^\circ) - F_{\text{friction}} = mg \frac{1}{\sqrt{2}} - \mu mg \frac{1}{\sqrt{2}} = mg \frac{1 - \mu}{\sqrt{2}} \]
- The acceleration \( a' \) down the rough incline is:
\[ a' = \frac{F_{\text{rough}}}{m} = g \frac{1 - \mu}{\sqrt{2}} \]

4. **Calculating Time on the Rough Incline:**
- Using the same equation of motion:
\[ D = \frac{1}{2} a' t'^2 \]
- Substituting \( a' \):
\[ D = \frac{1}{2} \left(g \frac{1 - \mu}{\sqrt{2}}\right) t'^2 \]
- Rearranging gives:
\[ t'^2 = \frac{2D \sqrt{2}}{g(1 - \mu)} \quad \Rightarrow \quad t' = \sqrt{\frac{2D \sqrt{2}}{g(1 - \mu)}} \]

5. **Relating the Times:**
- We know that \( t' = n t \):
\[ \sqrt{\frac{2D \sqrt{2}}{g(1 - \mu)}} = n \sqrt{\frac{2D \sqrt{2}}{g}} \]
- Squaring both sides:
\[ \frac{2D \sqrt{2}}{g(1 - \mu)} = n^2 \frac{2D \sqrt{2}}{g} \]
- Canceling common terms:
\[ \frac{1}{1 - \mu} = n^2 \quad \Rightarrow \quad 1 - \mu = \frac{1}{n^2} \]

6. **Finding the Coefficient of Kinetic Friction:**
- Rearranging gives:
\[ \mu = 1 - \frac{1}{n^2} \]

### Final Answer:
The coefficient of kinetic friction \( \mu \) between the object and the inclined plane is:
\[ \mu = 1 - \frac{1}{n^2} \]