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Questions  

A particle has a position vector given by r=2i^+j^-2k  m and velocity vector v=i^-j^+k  m/s.  The angular velocity (rad/s) of the particle with respect to the origin at this instant is 

a
i^+4j^+3k9
b
-i^+4j^+3k9
c
-i^-4j^+3k9
d
Zero

detailed solution

Correct option is B

ω→=r→×v→r2

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