A particle moves under the action of a force `F=20hati + 15hatj` along a straight line `3y + alphax =5`, where, `alpha` is a constant. If the work done by the force F is zero the value of `alpha` is .

Options

A. `4/9`
B. `9/4`
C. `3`
D. `4`

Explanation

To solve the problem, we need to find the value of the constant \( \alpha \) such that the work done by the force \( \mathbf{F} = 20 \hat{i} + 15 \hat{j} \) is zero while the particle moves along the line defined by \( 3y + \alpha x = 5 \).

### Step-by-Step Solution:

1. **Understanding the Condition for Zero Work Done:**
The work done by a force is given by the dot product of the force vector and the displacement vector. For the work done to be zero, the force must be perpendicular to the direction of motion. This means that the angle \( \theta \) between the force vector \( \mathbf{F} \) and the direction of motion must be \( 90^\circ \).

2. **Finding the Direction of the Force:**
The force vector is given by:
\[ \mathbf{F} = 20 \hat{i} + 15 \hat{j} \]
The slope of the force vector can be calculated as:
\[ \tan \theta_1 = \frac{F_y}{F_x} = \frac{15}{20} = \frac{3}{4} \]

3. **Finding the Slope of the Line:**
The equation of the line is given by:
\[ 3y + \alpha x = 5 \]
Rearranging this into the slope-intercept form \( y = mx + b \):
\[ y = -\frac{\alpha}{3}x + \frac{5}{3} \]
The slope \( m \) of the line is:
\[ \tan \theta_2 = -\frac{\alpha}{3} \]

4. **Setting Up the Perpendicular Condition:**
For the two vectors (the force and the direction of motion) to be perpendicular, the product of their slopes must equal \(-1\):
\[ \tan \theta_1 \cdot \tan \theta_2 = -1 \]
Substituting the values we found:
\[ \left(\frac{3}{4}\right) \left(-\frac{\alpha}{3}\right) = -1 \]

5. **Solving for \( \alpha \):**
Simplifying the equation:
\[ -\frac{3\alpha}{12} = -1 \]
This simplifies to:
\[ \frac{\alpha}{4} = 1 \]
Therefore:
\[ \alpha = 4 \]

### Conclusion:
The value of \( \alpha \) is \( 4 \).