A body of mass `5xx10^(-3)` kg is launched upon a rough inclined plane making an angle of `30^(@)` with the horizontal. Obtain the coefficient of friction between the body and the plane if the time of ascent is half of the time of descent.

Options

A. 0.346
B. 0.921
C. 1.926
D. 2.912

Answer

To solve the problem, we need to analyze the motion of a body of mass \( m = 5 \times 10^{-3} \) kg on a rough inclined plane at an angle \( \theta = 30^\circ \). The time of ascent \( t_1 \) is half of the time of descent \( t_2 \), i.e., \( t_2 = 2t_1 \). We will derive the coefficient of friction \( \mu \) between the body and the plane.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Body:**
- When the body is moving up the incline, the forces acting on it are:
- Weight \( mg \) acting downwards.
- Normal force \( N \) acting perpendicular to the incline.
- Frictional force \( F_f \) acting down the incline (opposite to the direction of motion).

- The components of the weight are:
- \( mg \cos \theta \) perpendicular to the incline.
- \( mg \sin \theta \) parallel to the incline.

2. **Free Body Diagram for Ascent:**
- The net force when moving up is given by:
\[ F_{\text{net}} = -F_f - mg \sin \theta = ma_1 \]
- The frictional force \( F_f = \mu N = \mu mg \cos \theta \).
- Substituting \( N = mg \cos \theta \):
\[ -\mu mg \cos \theta - mg \sin \theta = ma_1 \]
- Dividing through by \( m \):
\[ -\mu g \cos \theta - g \sin \theta = a_1 \]
- Rearranging gives:
\[ a_1 = -\mu g \cos \theta - g \sin \theta \]

3. **Free Body Diagram for Descent:**
- The net force when moving down is given by:
\[ F_{\text{net}} = mg \sin \theta - F_f = ma_2 \]
- Again substituting for \( F_f \):
\[ mg \sin \theta - \mu mg \cos \theta = ma_2 \]
- Dividing through by \( m \):
\[ g \sin \theta - \mu g \cos \theta = a_2 \]

4. **Using the Relationship Between Times:**
- We know \( t_2 = 2t_1 \). Using the equations of motion:
- For ascent:
\[ s = u t_1 - \frac{1}{2} a_1 t_1^2 \]
- For descent:
\[ s = u t_2 - \frac{1}{2} a_2 t_2^2 \]
- Since the distance \( s \) is the same for both ascent and descent, we can equate the two equations.

5. **Equating Distances:**
- Substitute \( t_2 = 2t_1 \) into the descent equation:
\[ s = u(2t_1) - \frac{1}{2} a_2 (2t_1)^2 \]
- This simplifies to:
\[ s = 2ut_1 - 2a_2 t_1^2 \]
- Setting the two equations for \( s \) equal gives:
\[ u t_1 - \frac{1}{2} a_1 t_1^2 = 2ut_1 - 2a_2 t_1^2 \]
- Rearranging leads to:
\[ \frac{1}{2} a_1 t_1^2 = ut_1 - 2a_2 t_1^2 \]

6. **Finding the Ratio of Accelerations:**
- From the previous equations, we can derive:
\[ \frac{a_1}{a_2} = \frac{2t_1^2}{t_1^2} = 4 \]
- Substituting \( a_1 \) and \( a_2 \):
\[ \frac{-\mu g \cos \theta - g \sin \theta}{g \sin \theta - \mu g \cos \theta} = 4 \]

7. **Solving for Coefficient of Friction \( \mu \):**
- Cross-multiplying gives:
\[ -\mu g \cos \theta - g \sin \theta = 4(g \sin \theta - \mu g \cos \theta) \]
- Rearranging leads to:
\[ 5\mu g \cos \theta = 3g \sin \theta \]
- Dividing by \( g \) (assuming \( g \neq 0 \)):
\[ 5\mu \cos \theta = 3\sin \theta \]
- Thus:
\[ \mu = \frac{3 \sin \theta}{5 \cos \theta} = \frac{3}{5} \tan \theta \]
- For \( \theta = 30^\circ \):
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
- Therefore:
\[ \mu = \frac{3}{5} \cdot \frac{1}{\sqrt{3}} = \frac{3}{5\sqrt{3}} \approx 0.346 \]

### Final Answer:
The coefficient of friction \( \mu \) between the body and the inclined plane is approximately \( 0.346 \).