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Question

The given plot shows the variation of the potential energy (U) of interaction between two particles with respect to the distance (r) separating them. Then,


A
B and D are equilibrium points
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B
C is a point of unstable equilibrium
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C
The force of interaction between the two particles is attractive between points C and D and repulsive between D and E
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D
The force of interaction between the particles is repulsive between points E and F.
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Solution

The correct option is D The force of interaction between the particles is repulsive between points E and F.
At B and D, dUdr0, so they are not equilibrium points.
At C, dUdr=0 and potential energy is minimum. So, it is a stable equilibrium position.
Now, F=dUdr= − (Slope of Ur curve)
In the region CD, the slope of the graph is positive, therefore by the relation, F is negative i.e, force is attractive in nature.
In the region DE, the slope of the graph is again positive, therefore by the relation, F is negative i.e, force is attractive in nature in the region DE too.
But in the region EF, the slope of the graph is negative which means F is positive i.e, force is repulsive in nature.

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