Electrostatics

JEE Main level, numerical answer, medium

191 of 311

Question 191 of 311

2727 similar drops of mercury are maintained at 10V10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ___________ times that of a smaller drop.

Show answer and solution

Answer: 243

(27)(43πr3)=43πR3\left(27\right)\left(\dfrac{4}{3}\pi {r}^{3}\right)=\dfrac{4}{3}\pi {R}^{3}
R=3rR=3r
Potential energy of smaller drop U1=35kq2r{U}_{1}=\dfrac{3}{5}\dfrac{k{q}^{2}}{r}
Potential energy of bigger drop : U=35kQ2RU=\dfrac{3}{5}\dfrac{k{Q}^{2}}{R}
U=35k(27q)2RU=\dfrac{3}{5}\dfrac{k{\left(27q\right)}^{2}}{R}
U=35k(27)(27)q23rU=\dfrac{3}{5}k\dfrac{\left(27\right)\left(27\right){q}^{2}}{3r}
U=(27)(27)3(35kq2r)U=\dfrac{\left(27\right)\left(27\right)}{3}\left(\dfrac{3}{5}\dfrac{k{q}^{2}}{r}\right)
U=243U1U=243{U}_{1}

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: Electric field lines simulationDrag the sliders and watch the numbers in this question change.Revise Electrostatics firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

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