A) 40%
B) 44%
C) 66%
D) 50%
Correct Answer: B
Solution :
The kinetic energy \[E=\frac{1}{2}m{{v}^{2}}=\frac{1}{2m}{{(mv)}^{2}}\] \[=\frac{{{p}^{2}}}{2m}\] ( \[\because \] \[p=mv\]) \[\therefore \] \[E\propto {{p}^{2}}\] \[\Rightarrow \] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{p_{1}^{2}}{p_{2}^{2}}\] ??(i) Given, \[{{E}_{1}}=E,\] \[{{p}_{1}}=p,\] \[{{p}_{2}}=p+\frac{20p}{100}=\frac{6p}{5}\] So, from Eq. (i) \[\frac{E}{{{E}_{2}}}=\frac{{{p}^{2}}}{{{(6p/5)}^{2}}}\] \[\Rightarrow \] \[\frac{E}{{{E}_{2}}}=\frac{25}{36}\] \[\Rightarrow \] \[\frac{{{E}_{2}}}{E}=\frac{36}{25}\] \[\therefore \] \[\frac{{{E}_{2}}-E}{E}=\frac{36-25}{25}=\frac{11}{25}\] So, percentage increase \[=\left( \frac{{{E}_{2}}-E}{E}\times 100 \right)\] \[=\frac{11}{25}\times 100=44%\]
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