A) 44%
B) 55%
C) 66%
D) 77%
Correct Answer: A
Solution :
\[E=\frac{{{P}^{2}}}{2m}.\] If m is constant then \[E\propto {{P}^{2}}\] Þ\[\frac{{{E}_{2}}}{{{E}_{1}}}={{\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)}^{2}}={{\left( \frac{1.2P}{P} \right)}^{2}}=1.44\] Þ \[{{E}_{2}}=1.44{{E}_{1}}={{E}_{1}}+0.44{{E}_{1}}\] \[{{E}_{2}}={{E}_{1}}+44%\] of \[{{E}_{1}}\] i.e. the kinetic energy will increase by 44%
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