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JEE Main 2022 (25 Jun, Shift 1) · Torque, Couple & Equilibrium of Rigid Bodies

Torque, Couple & Equilibrium of Rigid Bodies question from JEE Main 2022 (25 Jun, Shift 1)

Q19JEE Main 2022 · 25 JUN · SHIFT 1Easy

Torque, Couple & Equilibrium of Rigid Bodies

If force F⃗=3i^+4j^−2k^\vec{F} = 3\hat{i}+4\hat{j}-2\hat{k} acts on a particle position vector 2i^+j^+2k^2\hat{i}+\hat{j}+2\hat{k} then, the torque about the origin will be:

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Answer: (b)

  1. τ⃗=r⃗×F⃗\vec{\tau} = \vec{r}\times\vec{F} with r⃗=(2,1,2)\vec{r} = (2,1,2), F⃗=(3,4,−2)\vec{F} = (3,4,-2).
  2. τx=1(−2)−2(4)=−10\tau_x = 1(-2)-2(4) = -10, τy=−[2(−2)−2(3)]=10\tau_y = -[2(-2)-2(3)] = 10, τz=2(4)−1(3)=5\tau_z = 2(4)-1(3) = 5
  3. So the answer is (b) −10i^+10j^+5k^-10\hat{i}+10\hat{j}+5\hat{k}.
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