A) \[\sqrt{\frac{hv}{(2m)}}\]
B) \[\sqrt{\frac{hv}{m}}\]
C) \[\sqrt{\frac{2hv}{m}}\]
D) \[2\sqrt{\frac{hv}{m}}\]
Correct Answer: C
Solution :
As \[\frac{1}{2}mv_{\max }^{2}=hv\] \[\Rightarrow \] \[v_{\max }^{2}=\frac{2hv}{m}\] \[\therefore \] \[{{v}_{\max }}=\sqrt{\frac{2hv}{m}}\]
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