A body is pushed up on a rough inclined plane making an angle `30^@` to the horizontal. If its time of ascent on the plane is half the time of its descent, find coefficient of friction between the body and the incined plane.

Options

A. `(sqrt3)/(5)`
B. `sqrt(3//5)`
C. `3/(sqrt(5))`
D. `3//5`

Answer

To solve the problem, we will analyze the motion of a body on a rough inclined plane at an angle of \(30^\circ\) to the horizontal. We need to find the coefficient of friction (\(\mu\)) given that the time of ascent is half the time of descent.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Body:**
- The weight of the body is \(mg\) acting vertically downward.
- The normal force \(N\) acts perpendicular to the inclined plane.
- The frictional force \(f_k\) acts opposite to the direction of motion.

2. **Resolve the Weight into Components:**
- The component of weight parallel to the incline:
\[ F_{\text{parallel}} = mg \sin(30^\circ) = \frac{mg}{2} \]
- The component of weight perpendicular to the incline:
\[ F_{\text{perpendicular}} = mg \cos(30^\circ) = mg \cdot \frac{\sqrt{3}}{2} \]

3. **Normal Force Calculation:**
- The normal force \(N\) is equal to the perpendicular component of the weight:
\[ N = mg \cos(30^\circ) = mg \cdot \frac{\sqrt{3}}{2} \]

4. **Frictional Force:**
- The kinetic frictional force is given by:
\[ f_k = \mu N = \mu \left(mg \cdot \frac{\sqrt{3}}{2}\right) \]

5. **Equation of Motion for Ascent:**
- When the body is moving upward, the net force equation is:
\[ mA = f_k + mg \sin(30^\circ) \]
- Substituting the values:
\[ mA = \mu \left(mg \cdot \frac{\sqrt{3}}{2}\right) + \frac{mg}{2} \]
- Dividing by \(m\):
\[ A = \mu \left(g \cdot \frac{\sqrt{3}}{2}\right) + \frac{g}{2} \]

6. **Equation of Motion for Descent:**
- When the body is moving downward, the net force equation is:
\[ mA' = mg \sin(30^\circ) - f_k \]
- Substituting the values:
\[ mA' = \frac{mg}{2} - \mu \left(mg \cdot \frac{\sqrt{3}}{2}\right) \]
- Dividing by \(m\):
\[ A' = \frac{g}{2} - \mu \left(g \cdot \frac{\sqrt{3}}{2}\right) \]

7. **Relating Times of Ascent and Descent:**
- Let \(t_1\) be the time of ascent and \(t_2\) be the time of descent. Given that \(t_1 = \frac{1}{2} t_2\), we can use the equations of motion:
\[ s = \frac{1}{2} A t_1^2 \quad \text{(for ascent)} \]
\[ s = \frac{1}{2} A' t_2^2 \quad \text{(for descent)} \]

8. **Setting Up the Equation:**
- From the above equations, we can express \(s\) in terms of \(A\) and \(A'\):
\[ \frac{1}{2} A \left(\frac{1}{2} t_2\right)^2 = \frac{1}{2} A' t_2^2 \]
- Simplifying gives:
\[ A \cdot \frac{t_2^2}{4} = A' t_2^2 \]
- Dividing both sides by \(t_2^2\):
\[ \frac{A}{4} = A' \]

9. **Substituting for \(A\) and \(A'\):**
- From the equations for \(A\) and \(A'\):
\[ \frac{\mu g \sqrt{3}}{2} + \frac{g}{2} = 4 \left(\frac{g}{2} - \frac{\mu g \sqrt{3}}{2}\right) \]
- Simplifying gives:
\[ \mu g \sqrt{3} + g = 8 - 4 \mu g \sqrt{3} \]
- Rearranging terms:
\[ 5 \mu g \sqrt{3} = 7 \]
- Solving for \(\mu\):
\[ \mu = \frac{7}{5g \sqrt{3}} \]

10. **Final Calculation:**
- Since \(g\) cancels out, we get:
\[ \mu = \frac{7}{5\sqrt{3}} \]

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