A) 330%
B) 225%
C) 200%
D) 300%
Correct Answer: D
Solution :
Kinetic energy is related to momentum through the relation \[K=\frac{{{p}^{2}}}{2m}\] or \[\frac{{{K}_{2}}}{{{K}_{1}}}=\frac{p_{2}^{2}}{p_{1}^{2}}\] ?(i) Given, \[{{p}_{1}}=p,{{p}_{2}}=p+\frac{p\times 100}{100}=p+p=2p,\] \[{{K}_{1}}=K\] \[\therefore \] \[\frac{{{K}_{2}}}{K}=\frac{{{(2p)}^{2}}}{{{p}^{2}}}=\frac{4{{p}^{2}}}{{{p}^{2}}}\] Increase in kinetic energy \[\frac{{{K}_{2}}-K}{K}=\frac{4{{p}^{2}}-{{p}^{2}}}{{{p}^{2}}}=3\] Hence, percentage increase in kinetic energy \[\left( \frac{{{K}_{2}}-K}{K} \right)\times 100=(3\,\times \,100)%\] \[=300%\]
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