Angular velocity of a body moving on a circular track is given by →ω=^i+^j+^k and the radius vector →r=2^i−2^j−2^k at the instant of time ‘t′. Then the magnitude of linear velocity of the body at that instant is.
A
2.01m/s
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B
4.23m/s
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C
6.21m/s
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D
5.65m/s
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Solution
The correct option is D5.65m/s Given →ω=^i+^j+^k and →r=2^i−2^j−2^k ∴→v=→ω×→r =⎡⎢⎣^i^j^k1112−2−2⎤⎥⎦ =4^j−4^k=√42+42=5.65m/s