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$\frac{\pi}{6}$
- B$\frac{\pi}{3}$
- C$\frac{\pi}{8}$
- D$\frac{\pi}{4}$
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$A$ rod of length $l$ is acted upon by a couple as shown in the figure. The moment of the couple is $\tau \text{ Nm}$. If the force at each end of the rod is $F$,then the magnitude of each force is (given $\sin 30^{\circ} = \cos 60^{\circ} = 0.5$):
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View SolutionIf $\vec{F}$ 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
MediumAIPMT 2009
View SolutionLet $\vec{F}$ 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:
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