A) \[{{x}^{2}}\]
B) \[{{x}^{3}}\]
C) \[\ln \,x\]
D) \[{{e}^{x}}\]
Correct Answer: B
Solution :
\[\frac{dv}{dt}=v\frac{dv}{dx}=-k{{x}^{2}}\] \[\Rightarrow \] \[vdv=-k{{x}^{2}}dx\] Let \[{{v}_{1}}\]and \[{{v}_{2}}\]be the velocities of the particle at locations \[{{x}_{1}}\]and \[{{x}_{2}}k,\] respectively. Then \[\frac{v_{2}^{2}-v_{1}^{2}}{2}=-k\left[ \frac{x_{2}^{3}-x_{1}^{3}}{3} \right]\] \[\therefore \] Loss in \[KE\propto {{x}^{3}}\]
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