The minimum force required to start pushing a body up a rough (frictional coefficient `mu`) inclined plane is `F_(1)` while the minimum force needed to prevent it from sliding down is `F_(2)`. If the inclined plane makes an angle `theta` with the horizontal such that `tan theta =2mu` then the ratio `(F_(1))/(F_(2))` is .

Options

A. 4
B. 1
C. 2
D. 3

Explanation

To solve the problem, we need to determine the ratio of the forces \( F_1 \) and \( F_2 \) required to push a body up and prevent it from sliding down a rough inclined plane. Given that \( \tan \theta = 2\mu \), we can derive the required expressions step by step.

### Step-by-Step Solution

1. **Identify Forces Acting on the Body**:
- When the body is pushed up the incline, the forces acting on it are:
- The applied force \( F_1 \) (up the incline)
- The gravitational force component \( mg \sin \theta \) (down the incline)
- The frictional force \( f \) (acting down the incline, opposing the motion)
- The normal force \( N \) (perpendicular to the incline)

2. **Free Body Diagram for Pushing Up**:
- For the case when the body is pushed up:
\[ F_1 = f + mg \sin \theta \]
- The frictional force \( f \) can be expressed as:
\[ f = \mu N \]
- The normal force \( N \) is given by:
\[ N = mg \cos \theta \]
- Therefore, the frictional force becomes:
\[ f = \mu mg \cos \theta \]
- Substituting this into the equation for \( F_1 \):
\[ F_1 = \mu mg \cos \theta + mg \sin \theta \]

3. **Free Body Diagram for Preventing Sliding Down**:
- For the case when the body is prevented from sliding down:
\[ F_2 = mg \sin \theta - f \]
- Again substituting for \( f \):
\[ F_2 = mg \sin \theta - \mu mg \cos \theta \]

4. **Expressing the Ratio \( \frac{F_1}{F_2} \)**:
- Now we can find the ratio:
\[ \frac{F_1}{F_2} = \frac{\mu mg \cos \theta + mg \sin \theta}{mg \sin \theta - \mu mg \cos \theta} \]
- Canceling \( mg \) from the numerator and denominator gives:
\[ \frac{F_1}{F_2} = \frac{\mu \cos \theta + \sin \theta}{\sin \theta - \mu \cos \theta} \]

5. **Substituting \( \tan \theta = 2\mu \)**:
- From \( \tan \theta = 2\mu \), we have:
\[ \sin \theta = 2\mu \cos \theta \]
- Substituting this into the ratio:
\[ \frac{F_1}{F_2} = \frac{\mu \cos \theta + 2\mu \cos \theta}{2\mu \cos \theta - \mu \cos \theta} \]
- Simplifying the expression:
\[ \frac{F_1}{F_2} = \frac{3\mu \cos \theta}{\mu \cos \theta} = 3 \]

### Final Result
Thus, the ratio \( \frac{F_1}{F_2} \) is \( 3 \).