$A$ couple produces:

Moment of a force of magnitude $20 \, N$ acting along the positive $x$-direction at point $(3 \, m, 0, 0)$ about the point $(0, 2, 0)$ (in $N \cdot m$) is ...........

The moment of the force,$\overrightarrow{F} = 4\hat{i} + 5\hat{j} - 6\hat{k}$ at point $(2, 0, -3)$,about the point $(2, -2, -2)$,is given by

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:

If $\vec{F}$ is the force acting on a particle having position vector $\vec{r}$ and $\vec{\tau}$ is the torque of this force about the origin,then

What does torque measure in terms of producing rotational motion?