A) 25%
B) 50%
C) 100%
D) 125%
Correct Answer: D
Solution :
The relation between momentum (p) and kinetic energy (K) is \[p=\sqrt{2mK}\] When momentum is increased by 50%, then \[150\,p=\sqrt{2mK}\] ?(i) and \[100\,p\sqrt{2mK}\] ?(ii) From Eqs. (i) and (ii), we get \[\frac{150p}{100p}=\frac{\sqrt{K}}{\sqrt{K}}\] \[\Rightarrow \] \[{{\left( \frac{150}{100} \right)}^{2}}=\frac{K}{K}\] \[\Rightarrow \] \[\frac{K}{K}-1=\frac{9}{4}-1\] \[\Rightarrow \] \[\frac{K-K}{K}\times 100%=\frac{5}{4}\times 100=125%\]
You need to login to perform this action.
You will be redirected in
3 sec
