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At a given instant of time the position vector of a particle moving in a circle with a velocity $3 \hat i- 4 \hat j + 5 \hat k$ is $\hat i+9 \hat j -3 \hat k$. Its angular velocity at that time is :
\[\begin {array} {1 1} (a)\;\frac{(13 \hat i+29 \hat j-31 \hat k)}{\sqrt {146}} & \quad (b)\;\frac{(13 \hat i-29 \hat j-31 \hat k)}{146} \\ (c)\;\frac{(13 \hat i+29 \hat j-31 \hat k)}{\sqrt {146}} & \quad (d)\;\frac{(13 \hat i+29 \hat j+31 \hat k)}{146} \end {array}\]
jeemain
eamcet
physics
2005
q2
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Nov 7, 2013
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meena.p
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