A)
B)
C)
D)
Correct Answer: D
Solution :
\[F=\frac{-dU}{dx}\Rightarrow dU=-F\,\,dx\] \[\Rightarrow U=-\int_{0}^{x}{(-Kx\,+\,a{{x}^{3}})dx}\]\[=\frac{k{{x}^{2}}}{2}-\frac{a{{x}^{4}}}{4}\] \[\therefore \] We get U = 0 at x = 0 and x = \[\sqrt{2k/a}\] and also U = negative for \[x>\sqrt{2k/a}\]. So F = 0 at x = 0 i.e. slope of U ? x graph is zero at x = 0.You need to login to perform this action.
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