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