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JEE Main 2019 (11 Jan, Shift 2) · Torque, Couple & Equilibrium of Rigid Bodies

Torque, Couple & Equilibrium of Rigid Bodies question from JEE Main 2019 (11 Jan, Shift 2)

Q31JEE Main 2019 · 11 JAN · SHIFT 2Easy

Torque, Couple & Equilibrium of Rigid Bodies

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 5m, the angle between the force and the position vector is (in radians):

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Answer: (b)

  1. τ=rFsin⁡θ\tau = rF\sin\theta
  2. 2.5=5×1×sin⁡θ⇒sin⁡θ=122.5 = 5\times 1\times\sin\theta \Rightarrow \sin\theta = \frac{1}{2}
  3. So the answer is (b) θ=π6\theta = \frac{\pi}{6}.
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