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

Options

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

Explanation

To solve the problem step by step, we will analyze the motion of the body on the inclined plane both with and without friction.

### Step 1: Understanding the Problem
The body starts from rest and slides down a \(45^\circ\) inclined plane. It takes time \(t\) to slide down the incline with friction and \(t/2\) to slide down the same distance without friction.

### Step 2: Setting Up the Equations
1. **Without Friction:**
- The acceleration \(a\) of the body down the incline without friction is given by:
\[ a = g \sin \theta = g \sin 45^\circ = \frac{g}{\sqrt{2}} \]
- Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
- Since the body starts from rest (\(u = 0\)):
\[ s = \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) \left(\frac{t}{2}\right)^2 = \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) \left(\frac{t^2}{4}\right) = \frac{gt^2}{8\sqrt{2}} \]

2. **With Friction:**
- The acceleration \(a'\) of the body down the incline with friction is:
\[ a' = g \sin \theta - f = g \sin 45^\circ - \mu g \cos 45^\circ = \frac{g}{\sqrt{2}} - \mu \frac{g}{\sqrt{2}} = \frac{g(1 - \mu)}{\sqrt{2}} \]
- Using the same equation of motion:
\[ s = \frac{1}{2} a' t^2 = \frac{1}{2} \left(\frac{g(1 - \mu)}{\sqrt{2}}\right) t^2 \]

### Step 3: Equating the Distances
Since the distance \(s\) is the same in both cases, we can set the two equations equal to each other:
\[ \frac{gt^2}{8\sqrt{2}} = \frac{1}{2} \left(\frac{g(1 - \mu)}{\sqrt{2}}\right) t^2 \]

### Step 4: Simplifying the Equation
1. Cancel \(g\) and \(t^2\) from both sides (assuming \(g\) and \(t\) are not zero):
\[ \frac{1}{8\sqrt{2}} = \frac{1 - \mu}{2\sqrt{2}} \]
2. Multiply both sides by \(2\sqrt{2}\):
\[ \frac{2}{8} = 1 - \mu \]
\[ \frac{1}{4} = 1 - \mu \]
3. Rearranging gives:
\[ \mu = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 5: Conclusion
The coefficient of friction \(\mu\) between the body and the inclined plane is:
\[ \mu = \frac{3}{4} \]

### Final Answer
The correct option is \( \frac{3}{4} \).

---