A) \[\frac{l}{4}\]
B) \[\frac{l}{3}\]
C) \[\frac{l}{2}\]
D) \[I\]
Correct Answer: A
Solution :
\[I\alpha 4{{\alpha }^{2}}\,{{\cos }^{2}}\frac{\phi }{2}\] In the first case, \[\phi =-\frac{2\pi }{3}\] \[\therefore \] \[I\alpha \,4{{a}^{2}}\] In the second case, \[\phi =-\frac{2\pi }{3}\] \[\therefore \] \[I\alpha 4{{a}^{2}}{{\cos }^{2}}\frac{2\pi }{3}\] or \[I\alpha {{a}^{2}}\] \[\frac{I}{I}=\frac{1}{4}\] or \[I=\frac{1}{4}\]
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