Question

The torque due to the force \( \left( 2\hat{i} + \hat{j} + 2\hat{k} \right) \) about the origin, acting on a particle whose position vector is \( \hat{i} + \hat{j} + \hat{k} \), would be:

• A. \( \hat{i} + \hat{k} \)
• B. \( \hat{i} - \hat{k} \)
• C. \( \hat{i} + \hat{j} + \hat{k} \)
• D. \( \hat{j} + \hat{k} \)

Hint:

To compute torque using the cross product: - Use the formula \( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} \). - Remember that \( \hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0 \).

Answer 1

The Correct Option is A

The torque \( \mathbf{\tau} \), resulting from a force \( \mathbf{F} \) applied to a particle at position \( \mathbf{r} \), is defined by the cross product: \[ \mathbf{\tau} = \mathbf{r} \times \mathbf{F} \] Given \( \mathbf{r} = \hat{i} + \hat{j} + \hat{k} \) and \( \mathbf{F} = 2\hat{i} + \hat{j} + 2\hat{k} \), the cross product is computed as: \[ \mathbf{\tau} = (\hat{i} + \hat{j} + \hat{k}) \times (2\hat{i} + \hat{j} + 2\hat{k}) \] Expanding the cross product: \[ \mathbf{\tau} = \hat{i} \times 2\hat{i} + \hat{i} \times \hat{j} + \hat{i} \times 2\hat{k} + \hat{j} \times 2\hat{i} + \hat{j} \times \hat{j} + \hat{j} \times 2\hat{k} + \hat{k} \times 2\hat{i} + \hat{k} \times \hat{j} + \hat{k} \times 2\hat{k} \] Applying cross product properties yields: \[ \mathbf{\tau} = \hat{i} + \hat{k} \] The resulting torque is \( \boxed{\hat{i} + \hat{k}} \).