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 \( \vec{F} = (2\hat{i} + 3\hat{j}) \, N \) is zero when the particle moves along the line given by the equation \( 3y + kx = 5 \).

### Step-by-Step Solution:

1. **Identify the Force Vector**:
The force acting on the particle is given by:
\[ \vec{F} = 2\hat{i} + 3\hat{j} \]

2. **Equation of the Line**:
The line along which the particle moves is given by:
\[ 3y + kx = 5 \]

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

4. **Displacement Vector**:
The displacement vector \( d\vec{r} \) can be expressed in terms of \( dx \) and \( dy \):
\[ d\vec{r} = dx \hat{i} + dy \hat{j} \]
Substituting \( dy \):
\[ d\vec{r} = dx \hat{i} - \frac{k}{3}dx \hat{j} = dx \left( \hat{i} - \frac{k}{3}\hat{j} \right) \]

5. **Work Done by the Force**:
The work done \( W \) by the force is given by the dot product:
\[ W = \vec{F} \cdot d\vec{r} \]
Substituting the expressions for \( \vec{F} \) and \( d\vec{r} \):
\[ W = (2\hat{i} + 3\hat{j}) \cdot \left( dx \left( \hat{i} - \frac{k}{3}\hat{j} \right) \right) \]
This simplifies to:
\[ W = dx \left( 2 \cdot 1 + 3 \cdot \left(-\frac{k}{3}\right) \right) = dx \left( 2 - k \right) \]

6. **Setting Work Done to Zero**:
For the work done to be zero:
\[ 2 - k = 0 \]
Solving for \( k \):
\[ k = 2 \]

### Final Answer:
The value of \( k \) is \( 2 \).