Types of Equilibrium — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Types of Equilibrium MCQs with step-by-step solutions (8 questions). Part of Work, Power and Energy. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Types of Equilibrium · easy · theory
In STABLE equilibrium, the potential energy of the body is at a:
A. Minimum ✓ Correct
B. Constant value everywhere
C. Point of inflection
D. Maximum
Solution: Stable: U minimum (displacement raises U, restoring force pulls back). Unstable: U maximum. Neutral: U constant.
Q2 — Types of Equilibrium · medium · theory
The mathematical conditions for STABLE equilibrium at x₀ are:
A. dU/dx = 0 and d²U/dx² < 0
B. dU/dx > 0
C. d²U/dx² = 0 only
D. dU/dx = 0 and d²U/dx² > 0 ✓ Correct
Solution: Equilibrium needs zero force (dU/dx = 0); a positive second derivative makes it a minimum ⇒ stable.
Q3 — Types of Equilibrium · easy · numerical
For U(x) = x² (J), the equilibrium at x = 0 is:
A. Stable ✓ Correct
B. Unstable
C. Not an equilibrium
D. Neutral
Solution: dU/dx = 2x = 0 at x = 0; d²U/dx² = 2 > 0 ⇒ minimum ⇒ stable (this is the spring potential).
Q4 — Types of Equilibrium · medium · numerical
For U(x) = x³ − 3x (J), the STABLE equilibrium position is:
A. x = −1 m
B. x = ±√3
C. x = +1 m ✓ Correct
D. x = 0
Solution: dU/dx = 3x² − 3 = 0 ⇒ x = ±1. d²U/dx² = 6x: positive at x = +1 (stable), negative at x = −1 (unstable).
Q5 — Types of Equilibrium · hard · numerical
For U(x) = x⁴/4 − x²/2 (J), the stable equilibrium positions are:
A. x = ±2
B. x = 0 only
C. x = ±1 (x = 0 is unstable) ✓ Correct
D. None exist
Solution: dU/dx = x³ − x = 0 ⇒ x = 0, ±1. d²U/dx² = 3x² − 1: at 0 it is −1 (max, unstable); at ±1 it is +2 (minima, stable) — a double-well potential.
Q6 — Types of Equilibrium · medium · numerical
For U(x) = 2x³ − 6x (J), the equilibrium at x = +1 m is:
A. Neutral
B. Unstable
C. Not an equilibrium point
D. Stable ✓ Correct
Solution: dU/dx = 6x² − 6 = 0 at x = ±1; d²U/dx² = 12x = +12 > 0 at x = 1 ⇒ U minimum ⇒ stable.
Q7 — Types of Equilibrium · hard · numerical
For U(x) = x² − 4x + 3 (J), the equilibrium position and the force at x = 0 are:
A. x = 1 m; F(0) = zero
B. x = 2 m; F(0) = −4 N
C. x = 2 m; F(0) = +4 N (towards the equilibrium) ✓ Correct
D. x = 4 m; F(0) = +3 N
Solution: F = −dU/dx = −(2x − 4) = 4 − 2x; F = 0 at x = 2 (stable, d²U/dx² = 2 > 0). At x = 0, F = +4 N, pointing towards x = 2.
Q8 — Types of Equilibrium · medium · numerical
A particle experiences a force F(x) = x³ − 4x (N). The equilibrium at x = 0 is:
A. Neutral
B. Not an equilibrium
C. Stable ✓ Correct
D. Unstable
Solution: F(0) = 0 ✓. d²U/dx² = −dF/dx = −(3x² − 4) = +4 > 0 at x = 0 ⇒ stable. (x = ±2 are unstable.) Trap: analyse −dF/dx, not dF/dx.