$A$ particle of mass $m$ is moving along a trajectory given by $x = x_0 + a \cos \omega_1 t$ and $y = y_0 + b \sin \omega_2 t$. The torque acting on the particle about the origin at $t = 0$ is:

$A$ force $\overrightarrow{F} = (\hat{i} + 2\hat{j} + 3\hat{k}) \text{ N}$ acts at a point $\vec{r}_1 = (4\hat{i} + 3\hat{j} - \hat{k}) \text{ m}$. The magnitude of torque about the point $\vec{r}_2 = (\hat{i} + 2\hat{j} + \hat{k}) \text{ m}$ is $\sqrt{x} \text{ N-m}$. The value of $x$ is $........$

The torque of a force $\overrightarrow{F} = -3\hat{i} + \hat{j} + 5\hat{k}$ acting at the point $\overrightarrow{r} = 7\hat{i} + 3\hat{j} + \hat{k}$ is:

The rotational effect is produced by

$A$ 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:

How is the direction of torque determined?