$A$ couple produces:

$A$ 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 $........$

Give the formula for the moment of a couple.

$A$ rod of length $l$ is acted upon by a couple as shown in the figure. The moment of the couple is $\tau \text{ Nm}$. If the force at each end of the rod is $F$,then the magnitude of each force is (given $\sin 30^{\circ} = \cos 60^{\circ} = 0.5$):

The magnitude of torque on a particle of mass $1\,kg$ is $2.5\,Nm$ about the origin. If the force acting on it is $1\,N$,and the distance of the particle from the origin is $5\,m$,the angle between the force and the position vector is (in radians)

$A$ 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: