A) 50%
B) 100%
C) 125%
D) 25%
Correct Answer: C
Solution :
Let \[{{P}_{1}}=P\], \[{{P}_{2}}={{P}_{1}}+50%\] of \[{{P}_{1}}\]= \[{{P}_{1}}+\frac{{{P}_{1}}}{2}=\frac{3{{P}_{1}}}{2}\] \[E\propto {{P}^{2}}\]Þ \[\frac{{{E}_{2}}}{{{E}_{1}}}={{\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)}^{2}}={{\left( \frac{3{{P}_{1}}/2}{{{P}_{1}}} \right)}^{2}}=\frac{9}{4}\] Þ\[{{E}_{2}}=2.25E\]\[={{E}_{1}}+1.25{{E}_{1}}\] \\[{{E}_{2}}={{E}_{1}}+125%\ \text{of }{{E}_{1}}\] i.e. kinetic energy will increase by 125%.
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