The torque of the force $\vec{F} = (2\hat{i} - 3\hat{j} + 4\hat{k}) \text{ N}$ acting at the point $\vec{r} = (3\hat{i} + 2\hat{j} + 3\hat{k}) \text{ m}$ about the origin is:

$A$ uniform cube of side $a$ and mass $m$ rests on a rough horizontal table. $A$ horizontal force $F$ is applied normal to one of the faces at a point that is directly above the centre of the face,at a height $\frac{3a}{4}$ above the base. The minimum value of $F$ for which the cube begins to tilt about the edge is (assume that the cube does not slide):

$A$ triangular plate is shown in the figure. $A$ force $\overrightarrow{F} = 4 \hat{i} - 3 \hat{j}$ is applied at point $P$. The torques at point $P$ with respect to point $O$ and point $Q$ are:

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

Four equal and parallel forces are acting on a rod (as shown in the figure) at distances of $20 \, cm, 40 \, cm, 60 \, cm$,and $80 \, cm$ respectively from one end of the rod. Under the influence of these forces,the rod -

$A$ 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: