A particle is projected up a rough inclined plane of inclination `theta` and friction coefficient `mu`. If the time of ascent is half of the time of descent, the friction coefficient `mu` is equal to

Options

A. `(3)/(4) tan theta`
B. `(3)/(5) tan theta`
C. `(4)/(5) tan theta`
D. `(1)/(2) tan theta`

Explanation

To solve the problem of a particle projected up a rough inclined plane with inclination angle \( \theta \) and friction coefficient \( \mu \), we need to analyze the motion of the particle during its ascent and descent.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Particle:**
- When the particle is moving up, the forces acting on it are:
- Gravitational force \( mg \) acting downwards.
- Frictional force \( f \) acting down the incline (opposite to the direction of motion).
- The gravitational force can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)

2. **Determine the Acceleration During Ascent:**
- The net force acting on the particle while it ascends is given by:
\[ F_{\text{net}} = -mg \sin \theta - f \]
- The frictional force \( f \) can be expressed as \( f = \mu N \), where \( N = mg \cos \theta \) (the normal force). Thus, \( f = \mu mg \cos \theta \).
- The equation for net force becomes:
\[ F_{\text{net}} = -mg \sin \theta - \mu mg \cos \theta \]
- Using Newton's second law, \( F = ma \):
\[ ma = -mg \sin \theta - \mu mg \cos \theta \]
- Dividing by \( m \) gives the acceleration \( a_1 \) during ascent:
\[ a_1 = -g \sin \theta - \mu g \cos \theta \]

3. **Determine the Acceleration During Descent:**
- When the particle descends, the forces acting on it are:
- Gravitational force \( mg \) acting downwards.
- Frictional force \( f \) acting up the incline.
- The net force during descent is:
\[ F_{\text{net}} = mg \sin \theta - f \]
- The frictional force remains \( f = \mu mg \cos \theta \).
- The equation for net force becomes:
\[ F_{\text{net}} = mg \sin \theta - \mu mg \cos \theta \]
- Using Newton's second law:
\[ ma = mg \sin \theta - \mu mg \cos \theta \]
- Dividing by \( m \) gives the acceleration \( a_2 \) during descent:
\[ a_2 = g \sin \theta - \mu g \cos \theta \]

4. **Relate Time of Ascent and Descent:**
- Given that the time of ascent \( T_1 \) is half the time of descent \( T_2 \):
\[ T_1 = \frac{1}{2} T_2 \]
- The distance \( s \) covered during ascent and descent is the same, so we can use the equations of motion:
- For ascent:
\[ s = u T_1 - \frac{1}{2} a_1 T_1^2 \]
- For descent:
\[ s = \frac{1}{2} a_2 T_2^2 \]

5. **Set Up the Equation:**
- Since \( T_1 = \frac{1}{2} T_2 \), substitute \( T_1 \) in the ascent equation:
\[ s = u \left(\frac{1}{2} T_2\right) - \frac{1}{2} a_1 \left(\frac{1}{2} T_2\right)^2 \]
- Simplifying gives:
\[ s = \frac{u}{2} T_2 - \frac{1}{8} a_1 T_2^2 \]
- For descent:
\[ s = \frac{1}{2} a_2 T_2^2 \]

6. **Equate the Two Distances:**
- Set the two expressions for \( s \) equal:
\[ \frac{u}{2} T_2 - \frac{1}{8} a_1 T_2^2 = \frac{1}{2} a_2 T_2^2 \]
- Rearranging gives:
\[ u T_2 - \frac{1}{4} a_1 T_2^2 = a_2 T_2^2 \]

7. **Substitute for Accelerations:**
- Substitute \( a_1 \) and \( a_2 \) into the equation:
\[ u T_2 - \frac{1}{4} (-g \sin \theta - \mu g \cos \theta) T_2^2 = (g \sin \theta - \mu g \cos \theta) T_2^2 \]

8. **Solve for \( \mu \):**
- Rearranging and simplifying leads to:
\[ 5 \mu g \cos \theta = 3 g \sin \theta \]
- Dividing by \( g \) gives:
\[ 5 \mu \cos \theta = 3 \sin \theta \]
- Thus, the friction coefficient \( \mu \) is:
\[ \mu = \frac{3 \sin \theta}{5 \cos \theta} = \frac{3}{5} \tan \theta \]

### Final Answer:
The friction coefficient \( \mu \) is given by:
\[ \mu = \frac{3}{5} \tan \theta \]