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(D) For a conservative force,$F = -dU/dx$. Equilibrium occurs where $F = 0$. In the given $F-x$ graph,the force is zero at points $A$ and $C$.An equil

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$C$

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(D) For a conservative force,$F = -dU/dx$. Equilibrium occurs where $F = 0$. In the given $F-x$ graph,the force is zero at points $A$ and $C$.An equilibrium point is unstable if a small displacement from the equilibrium position results in a force that pushes the body further away from the equilibrium point.For unstable equilibrium,the potential energy $U$ must be a maximum,which implies $d^2U/dx^2  0$.At point $A$,the slope of the $F-x$ graph $(dF/dx)$ is negative.At point $C$,the slope of the $F-x$ graph $(dF/dx)$ is positive.Therefore,at point $C$,the body is in unstable equilibrium.

The force $F$ acting on a body varies with its position $x$ as shown in the figure. At which point is the body in unstable equilibrium?

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