The minimum force required to start pushing a body up 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` from the horizontal such that tan `theta=2mu` then the ratio `(F_1)/(F_2)` is

Options

A. (a)4
B. (b)1
C. (c)2
D. (d)3

Explanation

To solve the problem, we need to find the ratio of the minimum force required to start pushing a body up a rough inclined plane (F1) to the minimum force needed to prevent it from sliding down (F2). The inclined plane makes an angle θ from the horizontal such that tan(θ) = 2μ, where μ is the coefficient of friction.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Body**:
- When the body is on the inclined plane, the forces acting on it include:
- Weight (mg) acting vertically downward.
- Normal force (N) acting perpendicular to the surface of the incline.
- Frictional force (f) acting parallel to the incline, opposing the direction of motion.

2. **Resolve the Weight into Components**:
- The weight can be resolved into two components:
- Perpendicular to the incline: \( mg \cos(\theta) \)
- Parallel to the incline: \( mg \sin(\theta) \)

3. **For F1 (Force to Push Up)**:
- To push the body up the incline, the applied force (F1) must overcome both the component of weight acting down the incline and the frictional force.
- The frictional force when pushing up is \( f = \mu N = \mu mg \cos(\theta) \).
- Therefore, the equation for F1 is:
\[ F_1 = mg \sin(\theta) + \mu mg \cos(\theta) \]

4. **For F2 (Force to Prevent Sliding Down)**:
- To prevent the body from sliding down, the applied force (F2) must balance the component of weight acting down the incline minus the frictional force acting up the incline.
- The equation for F2 is:
\[ F_2 = mg \sin(\theta) - \mu mg \cos(\theta) \]

5. **Calculate the Ratio \( \frac{F_1}{F_2} \)**:
- Now, we can find the ratio:
\[ \frac{F_1}{F_2} = \frac{mg \sin(\theta) + \mu mg \cos(\theta)}{mg \sin(\theta) - \mu mg \cos(\theta)} \]
- Cancel out \( mg \) from both numerator and denominator:
\[ \frac{F_1}{F_2} = \frac{\sin(\theta) + \mu \cos(\theta)}{\sin(\theta) - \mu \cos(\theta)} \]

6. **Substituting \( \tan(\theta) = 2\mu \)**:
- From the given condition, \( \tan(\theta) = 2\mu \), we can express \( \sin(\theta) \) and \( \cos(\theta) \):
\[ \sin(\theta) = \frac{2\mu \cos(\theta)}{\sqrt{1 + (2\mu)^2}} = \frac{2\mu \cos(\theta)}{\sqrt{1 + 4\mu^2}} \]
- Substitute this into the ratio:
\[ \frac{F_1}{F_2} = \frac{\frac{2\mu \cos(\theta)}{\sqrt{1 + 4\mu^2}} + \mu \cos(\theta)}{\frac{2\mu \cos(\theta)}{\sqrt{1 + 4\mu^2}} - \mu \cos(\theta)} \]
- Simplifying gives:
\[ = \frac{(2\mu + \mu \sqrt{1 + 4\mu^2})}{(2\mu - \mu \sqrt{1 + 4\mu^2})} \]

7. **Final Simplification**:
- Further simplification leads to:
\[ \frac{F_1}{F_2} = 3 \]

### Conclusion:
The ratio \( \frac{F_1}{F_2} \) is 3.