A body takes time `t` to reach the bottom oa a smooth inclined plane of angle `theta` with the horizontal. If the plane is made rought, time taken now is `2t`.The coefficient of friction of the rough surface is

Options

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

Explanation

To solve the problem step by step, we will analyze the motion of the body on both the smooth and rough inclined planes.

### Step-by-Step Solution:

1. **Understanding the Motion on a Smooth Inclined Plane:**
- When the body slides down a smooth inclined plane, the only force acting along the incline is the component of gravitational force.
- The gravitational force acting down the incline is given by:
\[ F_{\text{gravity}} = mg \sin \theta \]
- Since there is no friction, the acceleration \( a_1 \) of the body is:
\[ a_1 = g \sin \theta \]

2. **Using the Equation of Motion:**
- The distance \( s \) traveled down the incline is given by:
\[ s = \frac{1}{2} a_1 t^2 \]
- Substituting \( a_1 \):
\[ s = \frac{1}{2} (g \sin \theta) t^2 \]

3. **Understanding the Motion on a Rough Inclined Plane:**
- When the inclined plane is rough, friction acts against the motion. The frictional force \( f \) is given by:
\[ f = \mu N = \mu mg \cos \theta \]
- The net force acting down the incline is:
\[ F_{\text{net}} = mg \sin \theta - f = mg \sin \theta - \mu mg \cos \theta \]
- The acceleration \( a_2 \) of the body on the rough incline is:
\[ a_2 = g \sin \theta - \mu g \cos \theta \]

4. **Using the Equation of Motion for the Rough Plane:**
- The time taken to reach the bottom of the rough incline is \( 2t \). Therefore, we can write:
\[ s = \frac{1}{2} a_2 (2t)^2 = 2 a_2 t^2 \]
- Substituting \( a_2 \):
\[ s = 2 \left(g \sin \theta - \mu g \cos \theta\right) t^2 \]

5. **Equating the Distances:**
- Since the distance \( s \) is the same for both cases, we can set the two equations for \( s \) equal to each other:
\[ \frac{1}{2} (g \sin \theta) t^2 = 2 \left(g \sin \theta - \mu g \cos \theta\right) t^2 \]

6. **Simplifying the Equation:**
- Dividing both sides by \( t^2 \) (assuming \( t \neq 0 \)):
\[ \frac{1}{2} g \sin \theta = 2 \left(g \sin \theta - \mu g \cos \theta\right) \]
- Rearranging gives:
\[ \frac{1}{2} g \sin \theta = 2g \sin \theta - 2\mu g \cos \theta \]
- Simplifying further:
\[ 2\mu g \cos \theta = 2g \sin \theta - \frac{1}{2} g \sin \theta \]
\[ 2\mu g \cos \theta = \frac{3}{2} g \sin \theta \]

7. **Solving for the Coefficient of Friction \( \mu \):**
- Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ 2\mu \cos \theta = \frac{3}{2} \sin \theta \]
- Therefore:
\[ \mu = \frac{3}{4} \tan \theta \]

### Final Answer:
The coefficient of friction \( \mu \) of the rough surface is:
\[ \mu = \frac{3}{4} \tan \theta \]