$A$ force of $-F \hat{k}$ acts on $O$,the origin of the coordinate system. The torque about the point $(1, -1)$ is

$A$ force $F = 2.0\,N$ acts on a particle $P$ in the $xz-$ plane. The force $F$ is parallel to the $x-$ axis. The particle $P$ (as shown in the figure) is at a distance $3\,m$ from the origin,and the line joining $P$ with the origin makes an angle of $30^\circ$ with the $x-$ axis. The magnitude of the torque on $P$ with respect to the origin $O$ (in $N-m$) is:

The torque of the force $\vec{F} = (2\hat{i} - 3\hat{j} + 4\hat{k}) \text{ N}$ acting at the point $\vec{r} = (3\hat{i} + 2\hat{j} + 3\hat{k}) \text{ m}$ about the origin is:

$A$ door is hinged at one end and is free to rotate about a vertical axis (figure). Does its weight cause any torque about this axis? Give a reason for your answer.

Why is the handle of a door or window placed at the end opposite to the hinged side?

The torque due to the force $\vec{F} = (2 \hat{i} + \hat{j} + 2 \hat{k})$ about the origin,acting on a particle whose position vector is $\vec{r} = (\hat{i} + \hat{j} + \hat{k})$,is: