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NEET PHYSICSEasy

What would be the torque about the origin when a force 3j^ N3\hat{j} \text{ N} acts on a particle whose position vector is 2k^ m2\hat{k} \text{ m}?

A

6j^ N-m6\hat{j} \text{ N-m}

B

−6i^ N-m-6\hat{i} \text{ N-m}

C

6k^ N-m6\hat{k} \text{ N-m}

D

6i^ N-m6\hat{i} \text{ N-m}

Step-by-Step Solution

Torque is defined as the cross product of the position vector and the force vector: τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}. Given: Position vector, r⃗=2k^ m\vec{r} = 2\hat{k} \text{ m} Force vector, F⃗=3j^ N\vec{F} = 3\hat{j} \text{ N} τ⃗=(2k^)×(3j^)\vec{\tau} = (2\hat{k}) \times (3\hat{j}) τ⃗=6(k^×j^)\vec{\tau} = 6(\hat{k} \times \hat{j}) Since k^×j^=−i^\hat{k} \times \hat{j} = -\hat{i}, we get: τ⃗=−6i^ N-m\vec{\tau} = -6\hat{i} \text{ N-m}

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