A force `vec(F)=4hati-10 hatj` acts on a body at a point having position vector `-5hati-3hatj` relative to origin of co-ordinates on the axis of rotation. The torque acting on the body is

Options

A. `38 hatk`
B. `-25 hatk`
C. `62 hatk`
D. none of these

Explanation

To find the torque acting on the body, we will use the formula for torque, which is given by:

\[ \vec{\tau} = \vec{r} \times \vec{F} \]

where:
- \(\vec{\tau}\) is the torque,
- \(\vec{r}\) is the position vector,
- \(\vec{F}\) is the force vector.

### Step-by-step Solution:

1. **Identify the vectors**:
- The force vector is given as:
\[ \vec{F} = 4 \hat{i} - 10 \hat{j} \]
- The position vector is given as:
\[ \vec{r} = -5 \hat{i} - 3 \hat{j} \]

2. **Set up the cross product**:
- We need to calculate the cross product \(\vec{r} \times \vec{F}\):
\[ \vec{\tau} = (-5 \hat{i} - 3 \hat{j}) \times (4 \hat{i} - 10 \hat{j}) \]

3. **Use the determinant method to calculate the cross product**:
- We can express the cross product using a determinant:
\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & -3 & 0 \\ 4 & -10 & 0 \end{vmatrix} \]

4. **Calculate the determinant**:
- Expanding the determinant, we have:
\[ \vec{\tau} = \hat{i} \begin{vmatrix} -3 & 0 \\ -10 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} -5 & 0 \\ 4 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} -5 & -3 \\ 4 & -10 \end{vmatrix} \]

5. **Calculate each of the 2x2 determinants**:
- The first determinant:
\[ \begin{vmatrix} -3 & 0 \\ -10 & 0 \end{vmatrix} = (-3)(0) - (0)(-10) = 0 \]
- The second determinant:
\[ \begin{vmatrix} -5 & 0 \\ 4 & 0 \end{vmatrix} = (-5)(0) - (0)(4) = 0 \]
- The third determinant:
\[ \begin{vmatrix} -5 & -3 \\ 4 & -10 \end{vmatrix} = (-5)(-10) - (-3)(4) = 50 + 12 = 62 \]

6. **Combine the results**:
- Thus, we have:
\[ \vec{\tau} = 0 \hat{i} - 0 \hat{j} + 62 \hat{k} = 62 \hat{k} \]

7. **Final result**:
- The torque acting on the body is:
\[ \vec{\tau} = 62 \hat{k} \]