27 similar drops of mercury are maintained at 10V 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.

4 Views|Posted 4 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Alok Kumar Singh | Contributor-Level 10

4 months ago

With the help of conservation of volume, we can write

$27 * \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 3 r . . . . . . . (1)$

With the help of conservation of charge, we can write

Q = 27 q.......(2)

Potential energy of single drop = U₁ = $\frac{q^{2}}{8 π ε_{0} r}$

Potential energy of bigger drop = $U_{2} = \frac{Q^{2}}{8 π ε_{0} R} = \frac{27 * 27 * q^{2}}{8 π ε_{0} (3 r)} = 243 (\frac{q^{2}}{8 π ε_{0} r}) = 243 U_{1}$

$\Rightarrow \frac{U_{2}}{U_{1}} = 243$

Upvote

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K

Colleges

1.2K

Exams

6.9L

Reviews

1.8M

Answers

Learn more about...

Physics Electrostatic Potential and Capacitance 2025

View Exam Details

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

See what others like you are asking & answering