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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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A particle moves in a potential energy region given as U=x2−14x+12U=x^2-14 x+12. Its state of equilibrium will be-
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Correct option: A
x=7x=7
Option A (x=7x=7) is correct because at equilibrium, the force (derivative of potential energy) is zero.

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