A) \[\sqrt{2}{{v}_{0}}\]
B) \[2{{v}_{0}}\]
C) \[2\sqrt{2}{{v}_{0}}\]
D) \[4{{v}_{0}}\]
Correct Answer: C
Solution :
\[\frac{1}{2}mu_{\max }^{2}=h(v-{{v}_{o}})\] For light of frequency \[v=2{{v}_{0}},\frac{1}{2}mu_{o}^{2}=h(2{{v}_{o}}-{{v}_{o}})=h{{v}_{o}}\] For light of frequency, \[v=5{{v}_{o}},\]we have \[\frac{1}{2}m{{u}^{2}}=h(5{{v}_{o}}-{{v}_{o}})=4\,h{{v}_{o}}\] \[\therefore \]\[{{u}^{2}}=4u_{o}^{2}\]or \[u=2{{u}_{o}}\] Hence, the correction option is [b].
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