JEE Physics 2026 Question with Solution

Given: A potential energy versus position graph with regions AB, BC, CD, DE.

Find: The ranking of the magnitudes of force in these regions.

The force is related to potential energy by

$$F = -\frac{dV}{dx}$$

So, the magnitude of force is equal to the magnitude of the slope of the $V$-$x$ graph:

$$|F| = \left|\frac{dV}{dx}\right|$$

From the graph, the approximate coordinates are:

$$A=(0,0),\; B=(1,1),\; C=(2,4),\; D=(4,4),\; E=(6,3)$$

Now calculate slopes in each region:

For AB,

$$m_{AB}=\frac{1-0}{1-0}=1$$

Hence,

$$|F_{AB}|=1$$

For BC,

$$m_{BC}=\frac{4-1}{2-1}=3$$

Hence,

$$|F_{BC}|=3$$

For CD,

$$m_{CD}=\frac{4-4}{4-2}=0$$

Hence,

$$|F_{CD}|=0$$

For DE,

$$m_{DE}=\frac{3-4}{6-4}=-\frac{1}{2}$$

Hence,

$$|F_{DE}|=\frac{1}{2}$$

Therefore, the order from high to low is

$$F_{BC} > F_{AB} > F_{DE} > F_{CD}$$

So, the correct option is C.

Given: The force magnitude must be obtained from the slope of the potential energy graph, not from the height of the graph.

Find: Which region has the largest and smallest $|F|$.

A steeper line means a larger magnitude of slope, and hence a larger magnitude of force. A horizontal segment means zero slope, so zero force magnitude.

• BC is the steepest segment, so it has the largest force magnitude.
• AB is less steep than BC, so its force magnitude is smaller than BC.
• DE slopes downward gently, so its slope magnitude is smaller than AB.
• CD is horizontal, so its force magnitude is zero.

Thus,

$$F_{BC} > F_{AB} > F_{DE} > F_{CD}$$

Therefore, the correct option is C.

The solution states point values that make AB and DE equal, but the graph and the listed options support the stricter ordering above. Hence the defensible answer is C.