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← JEE Main 2022 · 29 Jul · Shift 2
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JEE Main 2022 · 29 Jul · Shift 2, Physics question 46

The torque of a force 5i^+3j^−7k^5\hat i+3\hat j-7\hat k about the origin is τ. 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 will be

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Answer and worked solution

Official answer: C

Torque r⃗×F⃗=(2,2,1)×(5,3,−7)=(−14−3, 5+14, 6−10)=(−17,19,−4)\vec r\times\vec F=(2,2,1)\times(5,3,-7)=(-14-3,\ 5+14,\ 6-10)=(-17,19,-4).

(C)

Topics:Torque and rotational dynamicsCross product
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