Verified Solution • 2025–26 Session moderate

QJEEM22Phfce7: JEE Main 2022 Physics - Class 12 JEEM22Phfce7 Solution: Sixty four conducting drops each of radius 0.02 m and each…

Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5 μC are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be :

Multiple Choice Options

  • A 1:04
  • B 4:01
  • C 1:08
  • D 8:01

Step-by-Step Verified Solution

Surface charge density of a spherical conductor is given by,

σ=q4πr2

When all smaller drops combine, the radius of the bigger drop is given by

43πR3=64(43πr3)\frac{4}{3} \pi R^3=64\left(\frac{4}{3} \pi r^3\right)

R3=64r3R=4rσ′=64q4π(4r)2=4σ

 Hence, σ′σ=4:1

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring