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JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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The torque of a force F=2i^−3j^+5k^\mathbf{F}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}} acting at a point whose position vector r=3i^−3j^+5k^\mathbf{r}=3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}} about the origin is
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Correct option: C
−5j^−3k^-5 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}
The correct answer is C (−5j^−3k^-5 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) because the torque is calculated by taking the cross product of the position vector and the force vector.

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