The position vectors of two particles of a rigid body are $(3, 0, 0) \ m$ and $(0, 3, 0) \ m$. Forces of $(0, 1, 0) \ N$ and $(0, -1, 0) \ N$ act on these particles respectively. The torque of the couple is ....... $Nm$.

Four forces are applied to a wheel of radius $10\,cm$ as shown in the figure. The net torque produced by all these forces about $C$ is

The vector sum of a system of non-collinear forces acting on a rigid body is given to be nonzero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero,does this mean that it is necessarily zero about any arbitrary point?

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

$A$ 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:

$A$ wheel of radius $R$ with an axle of radius $R/2$ is shown in the figure and is free to rotate about a frictionless axis through its centre and perpendicular to the page. Three forces are exerted as shown in the figure. The magnitude of the net torque acting on the system is nearly (in $FR$)