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:

$A$ 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:

Let $\overrightarrow{F}$ be the force acting on a particle having position vector $\overrightarrow{r}$,and let $\overrightarrow{\tau}$ be the torque of this force about the origin. Then:

$A$ wheel of radius $R$ with an axle of radius $R/2$ is shown in the figure and is free to rotate about a frictionless axis through its centre and perpendicular to the page. Three forces are exerted as shown in the figure. The magnitude of the net torque acting on the system is nearly (in $FR$)

What is the torque of a force $\vec{F} = 3\hat{i} + 7\hat{j} + 4\hat{k}$ about the origin,if the force acts on a particle whose position vector is $\vec{r} = 2\hat{i} + 2\hat{j} + 1\hat{k}$?

$A$ force $\vec{F} = 4\hat{i} + 3\hat{j} + 4\hat{k}$ is applied at the intersection point of the plane $x = 2$ and the $x$-axis. The magnitude of the torque of this force about the point $(2, 3, 4)$ is .......... . (Round off to the nearest integer)