Potential energy of a particle of mass 1 kg moving along x-axis is given by $U = (\frac{x^{3}}{3} - 4 x + 12)$ joule where x is in metres. For this force field :
(i) x =Zm and x= - 2mare positions of equilibrium
(ii) x = 2 m is the position of stable equilibrium
(iii) x = - 2 m is the position of unstable equilibrium
(iv) potential energy is minimum at x = 2m

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(i, ii, iii, iv)

(ii, iv)

(i, ii)

(i, ii, iii)

answer is D.

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Detailed Solution

At positions of equilibrium F $= 0, i.e., \frac{dU}{dX} = 0$
So, $\frac{dU}{dX} = \frac{3 x^{2}}{3} - 4 (1) = 0$
X = ±2m
At position of stable equilibrium $\frac{d^{2} U}{{dX}^{2}} = + ve$
As $\frac{d^{2} U}{{dX}^{2}} = 2 x$ which is positive at x =- 2 m
At position of unstable equilibrium $\frac{d^{2} U}{{dX}^{2}} = - ve$
For minimum potential energy $\frac{dU}{dX} = 0 and \frac{d^{2} U}{{dX}^{2}} = + ve$