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JEE Main 2025 (22 Jan, Shift 2) · Torque, Couple & Equilibrium of Rigid Bodies

Torque, Couple & Equilibrium of Rigid Bodies question from JEE Main 2025 (22 Jan, Shift 2)

Q6JEE Main 2025 · 22 JAN · SHIFT 2Easy

Torque, Couple & Equilibrium of Rigid Bodies

The torque due to the force (2i^+j^+2k^)(2\hat{i}+\hat{j}+2\hat{k}) about the origin, acting on a particle whose position vector is (i^+j^+k^)(\hat{i}+\hat{j}+\hat{k}), would be

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Answer: (b)

  1. τ⃗=r⃗×F⃗\vec{\tau} = \vec{r}\times\vec{F} with r⃗=(1,1,1)\vec{r} = (1,1,1), F⃗=(2,1,2)\vec{F} = (2,1,2).
  2. τx=1(2)−1(1)=1\tau_x = 1(2)-1(1) = 1, τy=−[1(2)−1(2)]=0\tau_y = -[1(2)-1(2)] = 0, τz=1(1)−1(2)=−1\tau_z = 1(1)-1(2) = -1
  3. So the answer is (b) i^−k^\hat{i}-\hat{k}.
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