A) \[25%\]
B) \[50%\]
C) \[100%\]
D) \[125%\]
Correct Answer: D
Solution :
The relation between momentum \[(p)\] and kinetic energy \[(K)\] is \[p=\sqrt{2m\,\,K}\] When momentum is increased by \[50%\], then \[150=\sqrt{2mK'}\] ? (i) and \[100p=\sqrt{2mK}\] ? (ii) From Eqs. (i) and (ii), we get \[\frac{150p}{100p}=\frac{\sqrt{K'}}{\sqrt{K}}\] \[\Rightarrow \] \[{{\left( \frac{150}{100} \right)}^{2}}=\frac{K'}{K}\] \[\Rightarrow \] \[\frac{K'}{K}-1=\frac{9}{4}-1\] \[\Rightarrow \] \[\frac{K'-K}{K}\times 100%=\frac{5}{4}\times 100=125%\]
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