Work, Energy and Power NEET 2012: Stable Equilibrium

What is the correct answer to the Work, Energy and Power NEET 2012 equilibrium question?

Option C is correct for the supplied Work, Energy and Power NEET 2012 Physics question: $r = \frac{2 A}{B}$

The task is to find the distance from the field centre where a particle has stable equilibrium. Its potential energy is:

U (r) = \frac{A}{r^{2}} - \frac{B}{r}, A > 0, B > 0, r > 0

The supplied choices are:

The answer needs two checks: zero force establishes equilibrium; positive potential-energy curvature establishes stability. Finding a distance where the potential energy is zero does not establish either condition.

How do you find the equilibrium distance using the first derivative?

Equilibrium requires zero radial force, so the first derivative of potential energy must vanish. Start with the force relation, not the value of potential energy:

F_{r} = - \frac{d U}{d r}

F_{r} = 0 \Rightarrow \frac{d U}{d r} = 0

Rewrite the fractions as powers before differentiating. This makes the signs easier to track because the second term already carries a minus sign:

U = A r^{- 2} - B r^{- 1}

Apply the power rule separately to each term:

\frac{d}{d r} (A r^{- 2}) = - 2 A r^{- 3}

\frac{d}{d r} (- B r^{- 1}) = + B r^{- 2}

The contribution from the second term is positive because the negative coefficient multiplies the negative exponent. Combining the two contributions gives:

\frac{d U}{d r} = - \frac{2 A}{r^{3}} + \frac{B}{r^{2}}

Set this equal to zero and move the negative term to the other side:

- \frac{2 A}{r^{3}} + \frac{B}{r^{2}} = 0

\frac{B}{r^{2}} = \frac{2 A}{r^{3}}

Multiplying by the cube of the radial distance is valid because the distance is strictly positive: $B r = 2 A$ $r = \frac{2 A}{B}$

This gives option C, but it does not yet prove stability. A stationary potential-energy value is only an equilibrium candidate: the slope test alone does not tell you whether it is a minimum or a maximum.

How does the second derivative prove that option C is stable?

The second derivative is positive at the candidate distance, proving that the potential energy has a local minimum. Follow the official solution by differentiating the first derivative term by term:

\frac{d}{d r} (- 2 A r^{- 3}) = 6 A r^{- 4}

\frac{d}{d r} (B r^{- 2}) = - 2 B r^{- 3}

Combining these terms gives:

\frac{d^{2} U}{d r^{2}} = \frac{6 A}{r^{4}} - \frac{2 B}{r^{3}}

Only at the equilibrium point, the relation already found allows this substitution:

B r = 2 A \Rightarrow B = \frac{2 A}{r}

Use it after differentiating, not before. It is a condition at one position, not an identity valid at every distance.

{\frac{d^{2} U}{d r^{2}} |}_{e q} = \frac{6 A}{r^{4}} - \frac{4 A}{r^{4}} = \frac{2 A}{r^{4}} > 0

Both the constant and the radial distance are positive, so the curvature is positive. A small radial displacement raises the potential energy, and the resulting radial force acts back towards equilibrium.

The official two-stage test therefore confirms stable equilibrium at option C:

r = \frac{2 A}{B}

Why does setting potential energy to zero give option D instead?

Setting potential energy to zero finds a zero-energy position, not an equilibrium position. That wrong physical condition produces option D:

\frac{A}{r^{2}} - \frac{B}{r} = 0

A - B r = 0 \Rightarrow r = \frac{A}{B}

Check the slope at this distance:

{\frac{d U}{d r} |}_{r = A / B} = - \frac{2 A}{(A / B)^{3}} + \frac{B}{(A / B)^{2}} = - \frac{B^{3}}{A^{2}} \neq 0

The force is therefore nonzero. This position cannot be an equilibrium point.

Potential energy has an arbitrary reference zero. Adding a constant changes its zero-energy condition but leaves its derivative, force and equilibrium distance unchanged:

\tilde{U} = U + C, \frac{d \tilde{U}}{d r} = \frac{d U}{d r}

A secondary units check rejects two choices, but it cannot replace the derivative test:

[A] = energy \times {length}^{2}, [B] = energy \times length

[\frac{A}{B}] = length, [\frac{B}{A}] = {length}^{- 1}

Dimensions cannot distinguish C from D. In your error log, label this mistake “wrong physical condition”:

U = 0 instead of \frac{d U}{d r} = 0

Use NEET Score Improvement: Fix Errors Before Your Next Mock to plan a targeted retry. Start that retry by writing the zero-force condition before doing any algebra.

How can you apply the same test to two related questions?

Use the equilibrium position to calculate energy, but use derivatives to judge stability. These are original practice questions, not verified past-year questions. Attempt each before checking its worked answer.

What is the potential energy at stable equilibrium for the same potential?

The equilibrium energy is negative, but the equilibrium remains stable. Original practice question 1: Find the potential energy at stable equilibrium for:

U = \frac{A}{r^{2}} - \frac{B}{r}, A > 0, B > 0, r > 0

Substitute the equilibrium distance already established: $r = \frac{2 A}{B}$

U_{e q} = \frac{A}{4 A^{2} / B^{2}} - \frac{B}{2 A / B} = \frac{B^{2}}{4 A} - \frac{B^{2}}{2 A} = - \frac{B^{2}}{4 A}

Negative potential energy does not mean instability. Its sign depends on the energy reference; positive curvature establishes stability here.

Where is equilibrium for a quadratic potential, and is it stable?

The equilibrium is stable at the constant force divided by the stiffness. Original practice question 2: A particle moves along the x-axis with the following potential; find its equilibrium position and determine its stability:

U (x) = \frac{k x^{2}}{2} - F_{0} x, k > 0, F_{0} > 0

Differentiate, set the slope to zero, then check the curvature:

\frac{d U}{d x} = k x - F_{0}

k x - F_{0} = 0 \Rightarrow x = \frac{F_{0}}{k}

\frac{d^{2} U}{d x^{2}} = k > 0

For a worked numerical check, take these values within this original example:

k = 4 N m^{- 1}, F_{0} = 6 N

x = \frac{6}{4} m = 1.5 m

The curvature is positive:

\frac{d^{2} U}{d x^{2}} = 4 N m^{- 1} > 0

For your next attempt, write this sequence before calculating: differentiate potential energy, set its first derivative to zero, then evaluate its second derivative at the candidate position.

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Frequently asked questions

What is the answer to the Work, Energy and Power NEET 2012 equilibrium question?

Option C, r = 2A/B, is correct for U(r) = A/r² − B/r with A, B and r positive. Setting dU/dr = 0 gives this distance, and the positive second derivative confirms stable radial equilibrium.

How do you check stable equilibrium from potential energy?

First set the first derivative of potential energy to zero, since force is the negative potential-energy gradient. Then evaluate the second derivative at that position; a positive value confirms a local minimum and stable equilibrium. For U(r) = A/r² − B/r, the curvature at r = 2A/B is 2A/r⁴ > 0.

Why is setting U = 0 wrong for finding equilibrium?

Equilibrium requires zero force, so the correct condition is dU/dr = 0, not U = 0. For U(r) = A/r² − B/r, setting U = 0 gives r = A/B, where the force is nonzero. Changing the reference zero of potential energy does not change the force or equilibrium position.

Can potential energy be negative at stable equilibrium?

Yes; the sign of potential energy depends on the chosen reference and does not establish stability. For U(r) = A/r² − B/r with A and B positive, the energy at stable equilibrium is −B²/(4A). Stability follows from the positive curvature at r = 2A/B.