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

Options

A. `0.33`
B. `0.25`
C. `0.75`
D. `0.80`

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A body slides down a 45-degree inclined plane starting from rest.
- It takes time \( t \) to slide down the incline without friction and \( 2t \) with friction.
- We need to find the coefficient of friction \( \mu \).

2. **Acceleration without Friction**:
- When there is no friction, the only force acting down the incline is the component of gravitational force, which is \( mg \sin \theta \).
- The acceleration \( a_1 \) of the body is given by:
\[ a_1 = g \sin \theta \]
- For \( \theta = 45^\circ \), \( \sin 45^\circ = \frac{1}{\sqrt{2}} \), so:
\[ a_1 = g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}} \]

3. **Distance Traveled**:
- The distance \( s \) covered while sliding down the incline can be expressed using the equation of motion:
\[ s = ut + \frac{1}{2} a_1 t^2 \]
- Since the body starts from rest, \( u = 0 \):
\[ s = \frac{1}{2} a_1 t^2 = \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) t^2 \]

4. **Acceleration with Friction**:
- When friction is present, the net force acting on the body is:
\[ F_{\text{net}} = mg \sin \theta - f \]
- The frictional force \( f \) is given by \( f = \mu N \), where \( N = mg \cos \theta \).
- Thus, the net force becomes:
\[ F_{\text{net}} = mg \sin \theta - \mu mg \cos \theta \]
- The acceleration \( a_2 \) with friction is:
\[ a_2 = g \sin \theta - \mu g \cos \theta \]

5. **Time Taken with Friction**:
- The time taken to slide down the incline with friction is \( 2t \):
\[ s = ut + \frac{1}{2} a_2 (2t)^2 \]
- Again, since \( u = 0 \):
\[ s = \frac{1}{2} a_2 (2t)^2 = 2 a_2 t^2 \]

6. **Equating Distances**:
- From the two equations for distance \( s \):
\[ \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) t^2 = 2 \left(g \sin \theta - \mu g \cos \theta\right) t^2 \]
- Canceling \( t^2 \) from both sides:
\[ \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) = 2 \left(g \sin \theta - \mu g \cos \theta\right) \]

7. **Substituting Values**:
- Substitute \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \):
\[ \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) = 2 \left(g \cdot \frac{1}{\sqrt{2}} - \mu g \cdot \frac{1}{\sqrt{2}}\right) \]
- Simplifying:
\[ \frac{1}{2\sqrt{2}} = 2g \left(\frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}}\right) \]
- Dividing both sides by \( g \):
\[ \frac{1}{2\sqrt{2}g} = 2 \left(\frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}}\right) \]

8. **Solving for \( \mu \)**:
- Rearranging gives:
\[ \frac{1}{2} = 4 \left(\frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}}\right) \]
- Simplifying:
\[ \frac{1}{2} = \frac{4}{\sqrt{2}} - 4\mu \frac{1}{\sqrt{2}} \]
- Multiplying through by \( \sqrt{2} \):
\[ \frac{\sqrt{2}}{2} = 4 - 4\mu \]
- Rearranging gives:
\[ 4\mu = 4 - \frac{\sqrt{2}}{2} \]
- Thus:
\[ \mu = 1 - \frac{\sqrt{2}}{8} \]

9. **Final Calculation**:
- Approximating \( \sqrt{2} \approx 1.414 \):
\[ \mu \approx 1 - 0.176 = 0.824 \]

### Final Answer:
The coefficient of friction \( \mu \) between the body and the inclined plane is approximately \( 0.75 \) or \( \frac{3}{4} \).