A)
B)
C)
D)
Correct Answer: C
Solution :
Energy density (E) \[=\frac{I}{v}=2{{\pi }^{2}}\rho {{n}^{2}}{{A}^{2}}\] \[{{v}_{\max }}=\omega A=2\pi nA\]Þ \[E\propto {{({{v}_{\max }})}^{2}}\] i.e., graph between E and \[{{v}_{\max }}\] will be a parabola symmetrical about E axis.You need to login to perform this action.
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