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Question

A force F=2^i+3^j^k acts at a point (0,0,2)then the magnitude of torque about origin is:

A
6
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B
7.21
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C
52
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D
None of these
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Solution

The correct options are
B 7.21
C 52
Position vector r=2^k
Force is given as F=2^i+3^j^k
Torque about origin τ=r×F
τ=2^k×(2^i+3^j^k)=6^i+4^j
Magnitude of torque |τ|=62+42=52 Nm=7.21 Nm

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