JEE Main 2026 · 21 January · morning · Single correct · +4 / −1
Potential energy ( V ) versus distance ( x ) is given by the graph. Rank various regions as per the magnitudes of the force ( F ) acting on a particle from high to low.
- (a) F_(BC)>F_(AB)>F_(DE)>F_(CD)
- (b) F_(BC)>F_(CD)>F_(DE)>F_(AB)
- (c) F_(CD)>F_(AB)>F_(BC)>F_(DE)
- (d) F_(CD)>F_(DE)>F_(AB)>F_(BC)
Answer and solution
Answer (a) F_(BC)>F_(AB)>F_(DE)>F_(CD)
Solution
Physics sir · 3 steps
Force is minus the slope of the V–x graph, so just compare how steep each straight piece is.
- Force from slope. For a conservative force, F = -dV/dx. So the size of the force in any region equals the magnitude of the slope of the V–x graph there, not the height of V.
- Read the points. From the grid: A (0, 0), B (1, 1), C (2, 4), D (4, 4), E (6, 3). Each region is a straight line, so its slope is constant.
Region Δ V / Δ x |F| AB (1 - 0)/(1 - 0) = 1 1 BC (4 - 1)/(2 - 1) = 3 3, largest CD (4 - 4)/(4 - 2) = 0 0, smallest DE (3 - 4)/(6 - 4) = -1/2 0.5 - So. From high to low: F_(BC) > F_(AB) > F_(DE) > F_(CD), which is option (a).
The idea to keep. On a potential energy graph, steepness gives force and the sign of the slope gives its direction. A flat part means zero force, however high the potential is.
Answer from the NTA final answer key. Solution by PrepWiser.