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)

$A$ couple produces:

$A$ force $\vec{F}$ acts on a particle with position vector $\vec{r}$. The torque $\vec{\tau}$ about the origin due to this force is given by $\vec{\tau} = \vec{r} \times \vec{F}$. Which of the following is true?

$A$ 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 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 ....... $Nm$.

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?