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: Component formula
$\tau_x = yF_z - zF_y$, $\tau_y = zF_x - xF_z$, $\tau_z = xF_y - yF_x$.

Step 2: Substitute
$\tau_x = (1)(-1) - (-1)(2) = 1$. $\tau_y = (-1)(3) - (1)(-1) = -2$. $\tau_z = (1)(2) - (1)(3) = -1$.

Step 3: Magnitude
$\sqrt{1^2 + (-2)^2 + (-1)^2} = \sqrt{6}$ N m.

Step 4: Check
As a test, the dot product $\vec{\tau}\cdot\vec{r} = 1 - 2 + 1 = 0$, as it must be, since torque is perpendicular to $\vec{r}$.

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