A) 44%
B) 66%
C) 80%
D) 30%
Correct Answer: A
Solution :
Energy \[E=\frac{{{p}^{2}}}{2m}\] If m is constant, then \[E\propto {{p}^{2}}\] \[\Rightarrow \] \[\frac{{{E}_{2}}}{{{E}_{1}}}={{\left[ \frac{{{p}_{2}}}{{{p}_{1}}} \right]}^{2}}\] \[={{\left[ \frac{1.2p}{p} \right]}^{2}}\] \[=1.44\] \[\Rightarrow \] \[{{E}_{2}}=1.44{{E}_{1}}\] \[={{E}_{1}}+0.44{{E}_{1}}\]or \[{{E}_{2}}={{E}_{1}}+44%\]of \[{{E}_{1}}\]ie, the kinetic energy will increase by 44%.
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