Potential energy of a particle moving along x -axis is given by U=\frac{x^{3}}{3}-\frac{9 x^{2}}{2}+20 x. Find out position of stable equilibrium state.

JEE Advanced 2020 Full Test 3 Paper 1

Question : 17 of 54

Marks: +1, -0

Potential energy of a particle moving along x -axis is given by U=‌

−‌

9x2

+20x. Find out position of stable equilibrium state.

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