Class JEE Physics ALL Q #1194
COMPETENCY BASED
APPLY
4 Marks 2026 JEE Main 2026 (Online) 21st January Morning Shift MCQ SINGLE
A potential energy curve U vs x is shown. If $F_{AB}, F_{BC}, F_{CD}, F_{DE}$ are forces in respective regions, arrange magnitudes in decreasing order.
(A) $F_{BC}>F_{AB}>F_{CD}>F_{DE}$
(B) $F_{BC}>F_{AB}>F_{DE}>F_{CD}$
(C) $F_{AB}>F_{BC}>F_{DE}>F_{CD}$
(D) $F_{BC}>F_{DE}>F_{AB}>F_{CD}$
Prev Next

AI Tutor Explanation

Powered by Gemini

Step-by-Step Solution

  1. The force is given by the negative gradient of the potential energy: $F = -\frac{dU}{dx}$.
  2. In region AB, the slope is positive and relatively large. Therefore, $F_{AB}$ is negative and has a magnitude proportional to the slope.
  3. In region BC, the potential energy is constant, so the slope is zero. Therefore, $F_{BC} = 0$.
  4. In region CD, the slope is negative and small. Therefore, $F_{CD}$ is positive and has a small magnitude.
  5. In region DE, the slope is positive and has a moderate magnitude. Therefore, $F_{DE}$ is negative and has a moderate magnitude.
  6. Comparing the magnitudes, we have: $|F_{BC}| = 0$, $|F_{AB}|$ is large, $|F_{CD}|$ is small, and $|F_{DE}|$ is moderate.
  7. Therefore, the correct order is $F_{AB} > F_{DE} > F_{CD}$ and $F_{BC}=0$.
  8. Combining these, we get $F_{BC} > F_{AB} > F_{DE} > F_{CD}$ is incorrect. $F_{BC} > F_{AB} > F_{DE} > F_{CD}$ is incorrect. $F_{AB}>F_{BC}>F_{DE}>F_{CD}$ is incorrect. $F_{BC}>F_{DE}>F_{AB}>F_{CD}$ is incorrect.
  9. Since $F_{BC} = 0$, we need to compare the magnitudes of $F_{AB}, F_{DE}, F_{CD}$. From the graph, the magnitude of the slope in region AB is the largest, followed by DE, and then CD. Therefore, $|F_{AB}| > |F_{DE}| > |F_{CD}|$. Since $F_{BC} = 0$, the correct order is $|F_{AB}| > |F_{DE}| > |F_{CD}| > |F_{BC}|$. However, the question asks for the decreasing order of magnitudes, and the options do not include $F_{BC}=0$. The closest option is (C) if we consider the absolute values of the forces. However, since $F_{BC} = 0$, it should be the smallest.
  10. Re-evaluating the slopes: AB has a steep positive slope, BC has zero slope, CD has a shallow negative slope, and DE has a moderate positive slope. Thus, $|F_{AB}|$ is large, $F_{BC} = 0$, $|F_{CD}|$ is small, and $|F_{DE}|$ is moderate. Therefore, the correct order of magnitudes is $|F_{AB}| > |F_{DE}| > |F_{CD}| > |F_{BC}| = 0$. None of the options match this exactly. However, if we consider the closest option, it would be (C), assuming that the question meant to compare the absolute values of the forces.
  11. The correct order should be $F_{AB} > F_{DE} > F_{CD} > F_{BC}$. However, none of the options match this. The closest option is (C), but it has $F_{BC}$ in the middle.
  12. The correct order of magnitudes is $|F_{AB}| > |F_{DE}| > |F_{CD}| > |F_{BC}| = 0$. None of the options are correct.

Correct Answer: None of the options are correct. The correct order should be |F_AB| > |F_DE| > |F_CD| > |F_BC| = 0<\/strong>

|KEY:NONE|||

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the concept of potential energy and its relation to force (F = -dU/dx) to interpret the given potential energy curve and determine the relative magnitudes of the forces in different regions.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding the relationship between potential energy and force, which is a conceptual understanding rather than recall of facts or execution of a procedure.
Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. The question assesses the student's ability to apply the concepts of potential energy and force to a given scenario, rather than simply recalling definitions or formulas.

More from this Chapter

MCQ_SINGLE
An air-filled capacitor of capacitance $C$ is filled with a dielectric $(k=3)$ of width $\frac{d}{3}$, where $d$ is the separation between the plates. The new capacitance is:
MCQ_SINGLE
Find dimensions of $\frac{A}{B}$ if $(P+\frac{A^{2}}{B})+\frac{1}{2}\rho V^{2}=$ constant, where $P\rightarrow$ pressure, $\rho\rightarrow$ density, $V$ speed.
MCQ_SINGLE
An ideal solenoid is kept with its axis vertical. A current $I_{0}$ is flowing in the solenoid. A charge q is thrown vertically downward inside the solenoid. The acceleration of the charged particle is:
NUMERICAL
Given $E=10~V$, $R=10$, $L=10~mH$, $\frac{N}{e}=10^{4}$. Find energy density when current is $-\frac{1}{e}$ of max. If value is $a \frac{\pi}{e^{2}}$. Find $a$ .
MCQ_SINGLE
Find the volume flow rate in the Venturi meter shown below in which water is flowing. Given $\frac{A}{a}=2$, $A=\sqrt{3}m^{2}$, difference in levels is 5 cm and $\rho=1000~kg~m^{-3}$.
View All Questions