A) 150%
B) 200%
C) 225%
D) 300%
Correct Answer: D
Solution :
\[KE=\frac{{{p}^{2}}}{2m}\Rightarrow K{{E}_{2}}=K{{E}_{1}}{{\left( \frac{{{p}_{2}}}{{{p}_{1}}} \right)}^{2}}=K{{E}_{1}}{{\left( \frac{2p}{p} \right)}^{2}}\] \[\Rightarrow \] \[K{{E}_{2}}=4K{{E}_{1}}\] \[=K{{E}_{1}}+3K{{E}_{1}}=K{{E}_{1}}+300%\]of\[K{{E}_{1}}\]
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