The rotational motion of a body is caused by:

$A$ force $\vec{F}=(2 \hat{i}+3 \hat{j}-5 \hat{k}) \, N$ acts at a point $\vec{r}_1=(2 \hat{i}+4 \hat{j}+7 \hat{k}) \, m$. The torque of the force about the point $\vec{r}_2=(\hat{i}+2 \hat{j}+3 \hat{k}) \, m$ is ............. $N m$.

If $\vec{F} = (4\hat{i} - 10\hat{j})$ and $\vec{r} = (5\hat{i} - 3\hat{j})$,then calculate the torque $\vec{\tau} = \vec{r} \times \vec{F}$. (in $\hat{k}$)

$A$ 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 resultant of the system in the figure is a force of $8 \ N$ parallel to the given forces,acting through point $R$. The value of $PR$ is equal to . . . . . .

$A$ wheel of radius $20\, cm$ has forces applied to it as shown in the figure. The net torque produced by the forces $4\, N$ at $A$,$8\, N$ at $B$,$6\, N$ at $C$,and $9\, N$ at $D$ at the angles indicated is: