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. ![JEE Main 2026 Physics, Work Power Energy — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Work_Power_Energy/6983c17c8bfdbb243563955c/question_1__q_6983c17c8bfdbb243563955c__cdn-question-pool.getmarks.app__21S1_q_43_1__ad633b37fd_final_ppt_sync.png)

Potential energy (V) versus… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Mechanics

2026mcqmedium

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.

JEE Main 2026 Physics, Work Power Energy — question figure

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Options

A
$F_{C D}>F_{A B}>F_{B C}>F_{D E}$
B
$F_{C D}>F_{D E}>F_{A B}>F_{B C}$
C
$F_{B C}>F_{C D}>F_{D E}>F_{A B}$
D
$F_{B C}>F_{A B}>F_{D E}>F_{C D}$

Did you get this right?

Solution

Force is related to potential energy by $F = -\frac{dV}{dx}$ .

Magnitude of force $|F| = \left|\frac{dV}{dx}\right|$ = |slope of V-x graph|

From graph coordinates: A(1, 0), B(1.5, 1), C(2, 4), D(4, 4), E(5, 3)

$|F_{AB}| = \frac{1-0}{1.5-1} = 2$

$|F_{BC}| = \frac{4-1}{2-1.5} = 6$

$|F_{CD}| = \frac{4-4}{4-2} = 0$

$|F_{DE}| = \frac{|3-4|}{5-4} = 1$

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

Your note

JEE Main Mechanics in other years

Mechanics 2025 Mechanics 2024 Mechanics 2023 Mechanics 2022 Mechanics 2021 Mechanics 2020 Mechanics 2019 Mechanics 2018 Mechanics 2017 Mechanics 2016 Mechanics 2015 Mechanics 2014

Keep practicing Mechanics

Work through every JEE Main Mechanics PYQ, year by year.

Browse all Physics topics

Options

$F_{C D}>F_{A B}>F_{B C}>F_{D E}$

$F_{C D}>F_{D E}>F_{A B}>F_{B C}$

$F_{B C}>F_{C D}>F_{D E}>F_{A B}$

$F_{B C}>F_{A B}>F_{D E}>F_{C D}$

Solution

Force is related to potential energy by $F = -\frac{dV}{dx}$ .

Magnitude of force $|F| = \left|\frac{dV}{dx}\right|$ = |slope of V-x graph|

From graph coordinates: A(1, 0), B(1.5, 1), C(2, 4), D(4, 4), E(5, 3)

$|F_{AB}| = \frac{1-0}{1.5-1} = 2$

$|F_{BC}| = \frac{4-1}{2-1.5} = 6$

$|F_{CD}| = \frac{4-4}{4-2} = 0$

$|F_{DE}| = \frac{|3-4|}{5-4} = 1$

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

Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics

Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics