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JEE Main 2022 (29 Jul, Shift 2) · Torque, Couple & Equilibrium of Rigid Bodies

Torque, Couple & Equilibrium of Rigid Bodies question from JEE Main 2022 (29 Jul, Shift 2)

Q18JEE Main 2022 · 29 JUL · SHIFT 2Easy

Torque, Couple & Equilibrium of Rigid Bodies

The torque of a force 5i^+3j^−7k^5\hat{i}+3\hat{j}-7\hat{k} about the origin is τ\tau. If the force acts on a particle whose position vector is 2i^+2j^+k^2\hat{i}+2\hat{j}+\hat{k}, then the value of τ\tau will be

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Answer: (c)

  1. τ⃗=r⃗×F⃗\vec{\tau} = \vec{r}\times\vec{F} with r⃗=(2,2,1)\vec{r} = (2,2,1), F⃗=(5,3,−7)\vec{F} = (5,3,-7).
  2. τx=2(−7)−1(3)=−17\tau_x = 2(-7)-1(3) = -17, τy=−[2(−7)−1(5)]=19\tau_y = -[2(-7)-1(5)] = 19, τz=2(3)−2(5)=−4\tau_z = 2(3)-2(5) = -4
  3. So the answer is (c) −17i^+19j^−4k^-17\hat{i}+19\hat{j}-4\hat{k}.
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