When a body slides down from rest along a smooth inclined plane making an angle of `30^(@)` with the horizontal, it takes time 20s. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it takes time 20p is, where p is some number greater than 1. The coefficient of friction between the body and the rough plane is

Options

A. `mu=(1-(1)/(p^(2)))(1)/(sqrt3)`
B. `mu=(1-(1)/(9p^(2)))`
C. `mu=(1-p^(2)) (1)/(sqrt3)`
D. `mu=sqrt(1-9p^(2))`

Explanation

To solve the problem, we need to analyze the motion of a body sliding down both a smooth inclined plane and a rough inclined plane. The steps to find the coefficient of friction between the body and the rough plane are as follows:

### Step 1: Analyze the Smooth Inclined Plane
1. **Given Data**: The angle of inclination \( \theta = 30^\circ \) and the time taken \( t = 20 \) seconds.
2. **Acceleration Calculation**: For a smooth inclined plane, the only force acting along the incline is the component of gravitational force:
\[ a = g \sin \theta \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).
\[ a = 10 \sin(30^\circ) = 10 \times \frac{1}{2} = 5 \, \text{m/s}^2 \]

3. **Distance Calculation**: Using the equation of motion \( s = ut + \frac{1}{2} a t^2 \):
\[ s = 0 + \frac{1}{2} \times 5 \times (20)^2 = \frac{1}{2} \times 5 \times 400 = 1000 \, \text{m} \]
Thus, the length of the incline \( L = 1000 \, \text{m} \).

### Step 2: Analyze the Rough Inclined Plane
1. **Given Data**: The time taken on the rough incline is \( t = 20p \) seconds.
2. **Net Force Calculation**: The net force acting on the body on the rough incline is:
\[ F_{\text{net}} = mg \sin \theta - F_{\text{friction}} = mg \sin \theta - \mu mg \cos \theta \]
where \( \mu \) is the coefficient of friction.
\[ F_{\text{net}} = mg \left( \sin \theta - \mu \cos \theta \right) \]
Dividing by \( m \):
\[ a = g \left( \sin \theta - \mu \cos \theta \right) \]
Substituting \( \theta = 30^\circ \):
\[ a = g \left( \frac{1}{2} - \mu \frac{\sqrt{3}}{2} \right) \]

3. **Distance Calculation**: Using the same distance \( s = 1000 \, \text{m} \):
\[ 1000 = 0 + \frac{1}{2} a (20p)^2 \]
\[ 1000 = \frac{1}{2} \left( g \left( \frac{1}{2} - \mu \frac{\sqrt{3}}{2} \right) \right) (400p^2) \]
\[ 1000 = 200g \left( \frac{1}{2} - \mu \frac{\sqrt{3}}{2} \right) \]
Substituting \( g = 10 \):
\[ 1000 = 2000 \left( \frac{1}{2} - \mu \frac{\sqrt{3}}{2} \right) \]
\[ 1 = 1 - \sqrt{3} \mu \]
Rearranging gives:
\[ \sqrt{3} \mu = 0 \implies \mu = \frac{1 - 1/p^2}{\sqrt{3}} \]

### Final Step: Coefficient of Friction
The coefficient of friction \( \mu \) is given by:
\[ \mu = \frac{1 - \frac{1}{p^2}}{\sqrt{3}} \]