A box of mass `8 kg ` placed on a rough inclined plane of inclened `theta` its downward motion can be prevented by applying an upward pull `F` and it can be made to slide upward appliying a force `2F` .The coefficient of friction between the box and the inclined plane is

Options

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

Explanation

To solve the problem, we will analyze the forces acting on the box placed on the inclined plane and use the information given about the forces required to prevent downward motion and to slide the box upward.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Box:**
- The weight of the box (W) is given by \( W = mg = 8g \) N, where \( g \) is the acceleration due to gravity.
- The component of the weight acting down the incline is \( W_{\text{down}} = mg \sin \theta \).
- The component of the weight acting perpendicular to the incline is \( W_{\text{normal}} = mg \cos \theta \).
- The normal force \( N \) acting on the box is equal to \( mg \cos \theta \).

2. **Preventing Downward Motion:**
- When an upward force \( F \) is applied, the box does not slide down. The frictional force \( f \) acts upward and can be expressed as \( f = \mu N = \mu mg \cos \theta \).
- The equation of motion when the box is at rest (not sliding down) can be written as:
\[ F + f = mg \sin \theta \]
- Substituting for \( f \):
\[ F + \mu mg \cos \theta = mg \sin \theta \quad \text{(1)} \]

3. **Sliding Upward:**
- When a force \( 2F \) is applied to slide the box upward, the frictional force now acts downward.
- The equation of motion in this case is:
\[ 2F - f = mg \sin \theta \]
- Again substituting for \( f \):
\[ 2F - \mu mg \cos \theta = mg \sin \theta \quad \text{(2)} \]

4. **Solving the Equations:**
- From equation (1), we can express \( F \):
\[ F = mg \sin \theta - \mu mg \cos \theta \]
- Substitute \( F \) into equation (2):
\[ 2(mg \sin \theta - \mu mg \cos \theta) - \mu mg \cos \theta = mg \sin \theta \]
- Simplifying this gives:
\[ 2mg \sin \theta - 2\mu mg \cos \theta - \mu mg \cos \theta = mg \sin \theta \]
\[ 2mg \sin \theta - mg \sin \theta = 3\mu mg \cos \theta \]
\[ mg \sin \theta = 3\mu mg \cos \theta \]
- Dividing both sides by \( mg \) (assuming \( mg \neq 0 \)):
\[ \sin \theta = 3\mu \cos \theta \]

5. **Finding the Coefficient of Friction:**
- Rearranging gives:
\[ \mu = \frac{\sin \theta}{3 \cos \theta} = \frac{1}{3} \tan \theta \]

### Final Answer:
The coefficient of friction \( \mu \) between the box and the inclined plane is:
\[ \mu = \frac{1}{3} \tan \theta \]