A) 150%
B) 200%
C) 225%
D) 300%
Correct Answer: D
Solution :
\[E=\frac{{{P}^{2}}}{2m}\] Þ \[{{E}_{2}}={{E}_{1}}{{\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)}^{2}}\]\[={{E}_{1}}{{\left( \frac{2P}{P} \right)}^{2}}\] Þ \[{{E}_{2}}=4E=E+3E\]\[=E+300%\] of E
You need to login to perform this action.
You will be redirected in
3 sec
