A) \[2v\]
B) \[>2v\]
C) \[<2v\]
D) between \[2v\] and \[4v\]
Correct Answer: B
Solution :
\[\because \] \[E={{W}_{0}}+\frac{1}{2}mv_{\max }^{2}\] \[\Rightarrow \] \[{{v}_{\max }}=\sqrt{\frac{2(hf-{{W}_{0}})}{m}}\] If frequency becomes 4f then \[v=\sqrt{\frac{2(h\times 4f-{{W}_{0}})}{m}}=2\sqrt{\frac{2\left( hf-\frac{{{W}_{0}}}{4} \right)}{m}}\] \[\Rightarrow \] \[v>2v\]You need to login to perform this action.
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