A) \[400%\]
B) \[800%\]
C) \[200%\]
D) \[100%\]
Correct Answer: B
Solution :
\[E=\frac{{{L}^{2}}}{2l}\] \[\therefore \] \[E\propto {{L}^{2}}\] \[\frac{{{E}_{2}}}{{{E}_{1}}}={{\left( \frac{{{L}_{2}}}{{{L}_{1}}} \right)}^{2}}\] \[\frac{{{E}_{2}}}{{{E}_{1}}}=\left[ \frac{{{L}_{1}}+200%\,\,of\,\,{{L}_{1}}}{{{L}_{1}}} \right]\] \[={{\left[ \frac{{{L}_{1}}+2{{L}_{1}}}{{{L}_{1}}} \right]}^{2}}={{(3)}^{2}}\] \[{{E}_{2}}=9{{E}_{1}}\] Increment in kinetic energy \[\Delta E={{E}_{2}}-{{E}_{1}}\] \[=9{{E}_{2}}-{{E}_{1}}\] \[\Delta E=8{{E}_{1}}\] \[\frac{\Delta E}{{{E}_{1}}}=8\] or percentage increase\[=800%\]
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