Question

A force of \(3\hat{i}+2\hat{j}-\hat{k}\) N acts on a particle with position vector \(\hat{i}+\hat{j}-\hat{k}\) m. The magnitude of torque of given force is

• A. \(\sqrt{5}\,\text{N}\,m\)
• B. \(\sqrt{8}\,\text{N}\,m\)
• C. \(\sqrt{6}\,\text{N}\,m\)
• D. \(\sqrt{10}\,\text{N}\,m\)

Hint:

Torque is the cross product of position vector and force.

Answer 1

The Correct Option is C

Step 1: Understand the concept
The torque about the origin is \(\vec{\tau} = \vec{r}\times\vec{F}\). Its magnitude is then found from the components.

Step 2: Write the vectors
\(\vec{r} = (1, 1, -1)\) m and \(\vec{F} = (3, 2, -1)\) N.

Step 3: Compute the cross product
\[ \vec{\tau} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ 1 & 1 & -1\\ 3 & 2 & -1\end{vmatrix} = \hat{i}(-1 + 2) - \hat{j}(-1 + 3) + \hat{k}(2 - 3) = \hat{i} - 2\hat{j} - \hat{k} \]

Step 4: Magnitude
\[ |\vec{\tau}| = \sqrt{1 + 4 + 1} = \sqrt{6}\ \text{N m} \]
Option (C).

Final Answer:
The torque magnitude is sqrt 6 N m. This is option (C). \[ \boxed{\text{(C) }\sqrt{6}\ \text{N m}} \]