$A$ force $\vec{F} = 4\hat{i} - 3\hat{j} + 4\hat{k} \, N$ acts on a particle at a position $\vec{r} = 3\hat{i} + 2\hat{j} + 3\hat{k}$ from the origin. The torque acting on the particle is:

The figure shows three forces $\overrightarrow{F}_{1}, \overrightarrow{F}_{2}$ and $\overrightarrow{F}_{3}$ acting along the sides of an equilateral triangle. If the total torque acting at point $O$ (the center of the triangle) is zero,then the magnitude of $\overrightarrow{F}_{3}$ is:

$A$ tap can be easily opened or closed with the help of two fingers because:

$A$ couple produces:

Find the torque of a force $\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}$ acting at the point $\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}$.

Find the torque about the origin when a force of $3 \hat{j} \text{ N}$ acts on a particle whose position vector is $2 \hat{k} \text{ m}$.