A) \[4\times {{10}^{-3}}\]
B) \[1.6\times {{10}^{-5}}\]
C) \[4\times {{10}^{-6}}\]
D) 4
Correct Answer: A
Solution :
: K.E of rotation\[=10\text{ }J\]or \[\frac{1}{2}I{{\omega }^{2}}=10\] Or \[{{\omega }^{2}}=\frac{2\times 10}{I}\] Or \[{{\omega }^{2}}=\frac{2\times 10}{8\times {{10}^{-7}}}=\frac{1}{4\times {{10}^{-8}}}\]or \[\omega =\frac{1}{2\times {{10}^{-4}}}\] Angular momentum \[(L)=I\omega \] \[=(8\times {{10}^{-7}})\times \frac{1}{(2\times {{10}^{-4}})}\] \[=4\times {{10}^{-3}}\]S.I. units
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