Rotational Motion

JEE Main level, single correct, medium

124 of 280

Question 124 of 280

The magnitude of torque on a particle of mass 1 kg1 \mathrm{~kg} is 2.5 Nm about the origin. If the force acting on it is 1 N,1 \mathrm{~N}, and the distance of the particle from the origin is 5 m,5 \mathrm{~m}, the angle between the force and the position vector is (in radians):

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Answer: (A) π6\frac{\pi}{6}

Torque about the origin =τ⃗=r→×F→=\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}

=rFsin⁡θ⇒2.5=1×5sin⁡θ=r F \sin \theta \Rightarrow 2.5=1 \times 5 \sin \theta

sin⁡θ=0.5=12\sin \theta=0.5=\frac{1}{2}

⇒θ=π6\Rightarrow \theta=\frac{\pi}{6}

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