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Question

A particle, which is constrained to move along the x axis, is subjected to a force from the origin as F(x)=kx+ax3. Here k and a are positive constants. For x=0, the functional form of the potential energy U(x) of particle is

A
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B
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C
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D
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Solution

The correct option is D
Slope of potential energy vs x graph is force.
Given,
F(x)=kx+ax3;F(x)=0;
for x=0 x=ka;
So, slope is zero at x=0 x=ka
U=Fdx
U=(kxax3)dx
U=(2kx2ax4)4+C
As x increases beyond ka potential energy contiuously reduces
only option possible is d

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