A) 30%
B) 60%
C) 69%
D) 80%
Correct Answer: C
Solution :
Momentum \[p\propto \sqrt{E}\] \[\frac{{{p}_{1}}}{{{p}_{2}}}=\frac{\sqrt{{{E}_{1}}}}{{{E}_{2}}}\] \[\frac{p}{p+\frac{3p}{10}}=\frac{\sqrt{{{E}_{1}}}}{E{{ & }_{2}}}\] \[\frac{10}{13}=\frac{\sqrt{{{E}_{1}}}}{E{{ & }_{2}}}\] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{100}{169}\] \[\therefore \]increase in kinetic energies = 69%
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