Force acting on a particale is `(2hat(i)+3hat(j))N`. Work done by this force is zero, when the particle is moved on the line `3y+kx=5`. Here value of k is `(`Work done `W=vec(F).vec(d))`

Options

A. 2
B. 4
C. 6
D. 8

Explanation

To solve the problem, we need to find the value of \( k \) such that the work done by the force \( \vec{F} = 2\hat{i} + 3\hat{j} \) is zero when the particle is moved along the line defined by the equation \( 3y + kx = 5 \).

### Step-by-Step Solution:

1. **Understand the Work Done Formula**:
The work done \( W \) by a force \( \vec{F} \) when moving a particle along a path \( \vec{d} \) is given by the dot product:
\[ W = \vec{F} \cdot \vec{d} \]
In differential form, this can be expressed as:
\[ dW = \vec{F} \cdot d\vec{s} \]
where \( d\vec{s} = dx \hat{i} + dy \hat{j} \).

2. **Substitute the Force Vector**:
Given \( \vec{F} = 2\hat{i} + 3\hat{j} \), we can write:
\[ dW = (2\hat{i} + 3\hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = 2dx + 3dy \]
Since the work done is zero, we have:
\[ 2dx + 3dy = 0 \]

3. **Differentiate the Line Equation**:
The line equation is given as \( 3y + kx = 5 \). To find the relationship between \( dx \) and \( dy \), we differentiate this equation:
\[ 3dy + kdx = 0 \]
Rearranging gives:
\[ dy = -\frac{k}{3}dx \]

4. **Substitute \( dy \) in the Work Done Equation**:
Now, substitute \( dy \) into the work done equation:
\[ 2dx + 3\left(-\frac{k}{3}dx\right) = 0 \]
This simplifies to:
\[ 2dx - kdx = 0 \]
Factoring out \( dx \):
\[ (2 - k)dx = 0 \]

5. **Solve for \( k \)**:
For the equation \( (2 - k)dx = 0 \) to hold true for any \( dx \) (i.e., \( dx \neq 0 \)), we must have:
\[ 2 - k = 0 \implies k = 2 \]

### Final Answer:
Thus, the value of \( k \) is \( 2 \).