A) 50%
B) 100%
C) 125%
D) 150%
Correct Answer: C
Solution :
[c] \[{{P}_{2}}=\frac{3}{2}{{p}_{1}}\] \[\therefore \] \[{{v}_{2}}=\frac{3}{2}\,\,{{v}_{1}}\] We have, \[E\,\,\infty \,{{v}_{2}}\] \[\therefore \] \[{{E}_{2}}=\frac{9}{4}\,{{v}^{2}}_{1}\] Increase in K.E. \[=\frac{({{E}_{2}}-{{E}_{1}})}{{{E}_{1}}}\times 100\] =125%You need to login to perform this action.
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