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 find the coefficient of kinetic friction (μ) between the object and the rough inclined plane, we can follow these steps:

### Step 1: Understand the Problem
We have two inclined planes: one is smooth and the other is rough. The time taken to slide down the rough inclined plane (T1) is twice that of the smooth inclined plane (T2). We need to find the coefficient of kinetic friction (μ) for the rough inclined plane.

### Step 2: Set Up the Equations of Motion
For both inclined planes, the distance traveled (S) is the same. The time taken to slide down an inclined plane can be expressed in terms of acceleration (A) as:
\[ T = \sqrt{\frac{2S}{A}} \]

### Step 3: Determine the Accelerations
1. **For the smooth inclined plane (A2)**:
- The only force acting down the incline is the component of gravitational force:
\[ A_2 = g \sin \theta \]

2. **For the rough inclined plane (A1)**:
- The net force acting down the incline is the gravitational component minus the frictional force:
\[ A_1 = g \sin \theta - \mu g \cos \theta \]

### Step 4: Relate the Times
Given that \( T_1 = 2T_2 \), we can write:
\[ \frac{T_1}{T_2} = 2 \]
Using the relationship for time:
\[ \frac{\sqrt{\frac{2S}{A_1}}}{\sqrt{\frac{2S}{A_2}}} = 2 \]
This simplifies to:
\[ \frac{A_2}{A_1} = \frac{1}{4} \]

### Step 5: Substitute the Accelerations
Substituting the expressions for \( A_1 \) and \( A_2 \):
\[ \frac{g \sin \theta}{g \sin \theta - \mu g \cos \theta} = \frac{1}{4} \]

### Step 6: Simplify the Equation
Canceling \( g \) from both sides:
\[ \frac{\sin \theta}{\sin \theta - \mu \cos \theta} = \frac{1}{4} \]

### Step 7: Substitute the Angle
Since \( \theta = 45^\circ \):
\[ \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \]
Substituting these values:
\[ \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}}} = \frac{1}{4} \]

### Step 8: Solve for μ
Cross-multiplying gives:
\[ 4 \left(\frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} \]
Distributing:
\[ 4 \cdot \frac{1}{\sqrt{2}} - 4\mu \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]
Rearranging:
\[ 4 \cdot \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 4\mu \cdot \frac{1}{\sqrt{2}} \]
This simplifies to:
\[ \frac{3}{\sqrt{2}} = 4\mu \cdot \frac{1}{\sqrt{2}} \]
Dividing both sides by \( \frac{1}{\sqrt{2}} \):
\[ 3 = 4\mu \]
Finally, solving for μ:
\[ \mu = \frac{3}{4} \]

### Final Answer
The coefficient of kinetic friction (μ) between the object and the rough plane is:
\[ \mu = \frac{3}{4} \]