Starting from rest a body slides down a `45^(@)` inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of the body and the inclined plane is :

Options

A. 0.2
B. 0.25
C. 0.75
D. 0.5

Explanation

To solve the problem of finding the coefficient of friction between a body and a 45-degree inclined plane, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a body sliding down a 45-degree inclined plane.
- The body starts from rest and takes twice the time to slide down the same distance when friction is present compared to when it is absent.

2. **Case 1: Frictionless Incline**
- When there is no friction, the only force acting on the body is the component of gravitational force along the incline.
- The acceleration \( a_1 \) of the body is given by:
\[ a_1 = g \sin \theta \]
where \( \theta = 45^\circ \). Thus, \( \sin 45^\circ = \frac{1}{\sqrt{2}} \), and we have:
\[ a_1 = g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}} \]

3. **Using the Equation of Motion:**
- The distance \( s \) covered in time \( t \) can be expressed as:
\[ s = ut + \frac{1}{2} a_1 t^2 \]
Since the body starts from rest, \( u = 0 \):
\[ s = \frac{1}{2} a_1 t^2 = \frac{1}{2} \cdot \frac{g}{\sqrt{2}} t^2 \]

4. **Case 2: Incline with Friction**
- When friction is present, the net force acting on the body is:
\[ F = mg \sin \theta - \mu mg \cos \theta \]
- The acceleration \( a_2 \) can be expressed as:
\[ a_2 = g \sin \theta - \mu g \cos \theta \]
Substituting \( \theta = 45^\circ \):
\[ a_2 = g \cdot \frac{1}{\sqrt{2}} - \mu g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}}(1 - \mu) \]

5. **Using the Equation of Motion for Case 2:**
- The time taken to slide down with friction is \( 2t \):
\[ s = u(2t) + \frac{1}{2} a_2 (2t)^2 \]
Again, since \( u = 0 \):
\[ s = \frac{1}{2} a_2 (2t)^2 = 2 a_2 t^2 \]
- Substituting for \( a_2 \):
\[ s = 2 \left(\frac{g}{\sqrt{2}}(1 - \mu)\right) t^2 \]

6. **Equating Distances:**
- Since both cases cover the same distance \( s \):
\[ \frac{1}{2} \cdot \frac{g}{\sqrt{2}} t^2 = 2 \left(\frac{g}{\sqrt{2}}(1 - \mu)\right) t^2 \]
- Simplifying gives:
\[ \frac{1}{2} = 4(1 - \mu) \]
- Rearranging:
\[ 1 - \mu = \frac{1}{8} \]
\[ \mu = 1 - \frac{1}{8} = \frac{7}{8} \]

7. **Final Calculation:**
- The coefficient of friction \( \mu \) is:
\[ \mu = 0.875 \]

### Conclusion:
The coefficient of friction between the body and the inclined plane is \( \mu = 0.875 \).