Work Energy Power

JEE Main level, single correct, medium

90 of 183

Question 90 of 183

Potential energy as a function of rr is given by U=Ar10−Br5U=\dfrac{A}{{r}^{10}}-\dfrac{B}{{r}^{5}}, where rr is the interatomic distance, AA and BB are positive constants. The equilibrium distance between the two atoms will be :

Show answer and solution

Answer: (C) (2AB)15{\left(\dfrac{2A}{B}\right)}^{\dfrac{1}{5}}

Given U=Ar10−Br5U=\dfrac{A}{{r}^{10}}-\dfrac{B}{{r}^{5}}.
Now, for equilibrium, dUdr=0\dfrac{dU}{dr}=0.
So, differentiating the given equation,
−10Ar11+5Br6=0\dfrac{-10A}{{r}^{11}}+\dfrac{5B}{{r}^{6}}=0
⇒10Ar11=5Br6⇒r5=2AB⇒r=(2AB)15\Rightarrow \dfrac{10A}{{r}^{11}}=\dfrac{5B}{{r}^{6}} \Rightarrow {r}^{5}=\dfrac{2A}{B} \Rightarrow r={\left(\dfrac{2A}{B}\right)}^{\dfrac{1}{5}}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Collisions, centre of mass and rockets simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Work Energy Power (Energy), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.