The time taken by a body to slide down a rough `45^(@)` inclined plane is twice that required to slide down a smooth `45^(@)` inclined plane. The coefficient of kinetic friction between the object and rough plane is given by

Options

A. `(1)/(3)`
B. `(3)/(4)`
C. `sqrt((3)/(4))`
D. `sqrt((2)/(3))`

Explanation

To solve the problem, we need to analyze the motion of a body sliding down both a smooth and a rough inclined plane at an angle of \(45^\circ\). We will derive the coefficient of kinetic friction between the object and the rough plane based on the given information.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have two inclined planes: one smooth and one rough, both inclined at an angle of \(45^\circ\).
- The time taken to slide down the rough plane (\(t_1\)) is twice that of the smooth plane (\(t_2\)):
\[ t_1 = 2t_2 \]

2. **Setting Up the Forces:**
- For the smooth inclined plane, the only force causing acceleration is the component of gravity along the incline:
\[ F_{\text{smooth}} = mg \sin(45^\circ) = mg \cdot \frac{1}{\sqrt{2}} \]
- The acceleration (\(a_2\)) on the smooth plane is:
\[ a_2 = g \sin(45^\circ) = g \cdot \frac{1}{\sqrt{2}} \]

3. **For the Rough Inclined Plane:**
- On the rough inclined plane, the forces acting on the body include the gravitational component down the incline and the frictional force opposing the motion:
\[ F_{\text{rough}} = mg \sin(45^\circ) - f_k \]
- The frictional force \(f_k\) is given by:
\[ f_k = \mu mg \cos(45^\circ) = \mu mg \cdot \frac{1}{\sqrt{2}} \]
- Therefore, the net force on the rough plane is:
\[ F_{\text{net}} = mg \sin(45^\circ) - \mu mg \cos(45^\circ) = mg \cdot \frac{1}{\sqrt{2}} - \mu mg \cdot \frac{1}{\sqrt{2}} \]
- The acceleration (\(a_1\)) on the rough plane is:
\[ a_1 = \frac{F_{\text{net}}}{m} = g \cdot \frac{1}{\sqrt{2}} (1 - \mu) \]

4. **Relating Time and Acceleration:**
- The time taken to slide down an incline is inversely related to the square root of the acceleration:
\[ t \propto \frac{1}{\sqrt{a}} \]
- Therefore, we can write:
\[ \frac{t_1}{t_2} = \sqrt{\frac{a_2}{a_1}} \]
- Substituting the expressions for \(a_1\) and \(a_2\):
\[ \frac{t_1}{t_2} = \sqrt{\frac{g \cdot \frac{1}{\sqrt{2}}}{g \cdot \frac{1}{\sqrt{2}} (1 - \mu)}} = \sqrt{\frac{1}{1 - \mu}} \]

5. **Substituting the Time Relation:**
- Since \(t_1 = 2t_2\), we substitute this into our equation:
\[ 2 = \sqrt{\frac{1}{1 - \mu}} \]
- Squaring both sides gives:
\[ 4 = \frac{1}{1 - \mu} \]
- Rearranging leads to:
\[ 1 - \mu = \frac{1}{4} \]
- Thus:
\[ \mu = 1 - \frac{1}{4} = \frac{3}{4} \]

6. **Final Answer:**
- The coefficient of kinetic friction between the object and the rough plane is:
\[ \mu = \frac{3}{4} \]