A)
B)
C)
D)
Correct Answer: D
Solution :
[d] \[{{\operatorname{dU}}_{\left( x \right)}}=-Fdx\] \[\therefore \,\,\,{{U}_{x}}=-\int_{0}^{x}{Fdx=\frac{k{{x}^{2}}}{2}}\frac{a{{x}^{4}}}{4}\] \[U=0\,at\text{ }x=0\text{ }and\text{ }at\text{ }x=\sqrt{\frac{2k}{a}};\Rightarrow \]we have potential energy zero twice (out of which one is at origin). Also, when we put \[x=0\]in the given function, we get \[F=0.\,But\,F=-\frac{dU}{dx}\] \[\Rightarrow \operatorname{At}\text{ }x=0;\text{ }\frac{dU}{dx}=0\]i.e. the slope of the graph should be zero. These characteristics are represented by [d].You need to login to perform this action.
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