JEE Main 2023PhysicsMechanical Properties of FluidsMediumMCQ

JEE Main 2023 — Mechanical Properties of Fluids Question with Solution

JEE Main 2023 (29 Jan Shift 1)

Question

Surface tension of a soap bubble is 2.0×10-2 N m-1 . Work done to increase the radius of soap bubble from 3.5 cm to 7 cm will be : [Take π=227 ]

Choose an option

Show full solutionCorrect option: C
Correct answer
C18.48×10-4 J

Step-by-step explanation

As we know,

Work done = change in surface energy of soap bubble

Therefore,

Work done =2T4πr22-4πr12

=8πTr22-r12

=8×227×0.0272-3.52×10-4 J

=18.48×10-4J

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About this question

This is a previous-year question from JEE Main 2023, covering the Mechanical Properties of Fluids chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.