Let $\vec{F}$ be the force acting on a particle having position vector $\vec{r}$ and $\vec{T}$ be the torque of this force about the origin. Then:

$A$ particle of mass $m$ is moving along a trajectory given by $x = x_0 + a \cos \omega_1 t$ and $y = y_0 + b \sin \omega_2 t$. The torque acting on the particle about the origin at $t = 0$ 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 ....... $Nm$.

$A$ wheel of radius $R$ with an axle of radius $R/2$ is as shown in the figure. It 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: a force $F$ at an angle of $45^{\circ}$ to the tangent of the outer rim,a force $F$ tangential to the inner axle,and a force $2F$ tangential to the outer rim. The magnitude of the net torque acting on the system is nearly ............. $FR$.

$A$ 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?

Show that the moment of a couple does not depend on the point about which you take the moments.