Rotation Ques 8
8 The magnitude of torque on a particle of mass $1 \mathrm{~kg}$ is 2.5 $\mathrm{N}-\mathrm{m}$ about the origin. If the force acting on it is $1 \mathrm{~N}$ and the distance of the particle from the origin is $5 \mathrm{~m}$, then the angle between the force and the position vector is (in radian)
(Main 2019, 11 Jan II)
(a) $\frac{\pi}{8}$
(b) $\frac{\pi}{4}$
(c) $\frac{\pi}{3}$
(d) $\frac{\pi}{6}$
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Answer:
Correct Answer: 8.( d )
Solution:
- Given,
$ \begin{aligned} & m=1 \mathrm{~kg} \\ & |\tau|=2.5 \mathrm{~N}-\mathrm{m}, F=1 \mathrm{~N} \text { and } r=5 \mathrm{~m} \end{aligned} $
We know that, torque $\quad|\tau|=r F \sin \theta$
$ \begin{array}{ll} \Rightarrow & 2.5=5 \times 1 \times \sin \theta \\ \Rightarrow & \sin \theta=\frac{1}{2} \\ \text { or } & \theta=\frac{\pi}{6} \mathrm{rad} \end{array} $
BETA
