JEE Main 2023PhysicsMechanical Properties of FluidsEasyMCQ

JEE Main 2023 — Mechanical Properties of Fluids Question with Solution

JEE Main 2023 (01 Feb Shift 1)

Question

A mercury drop of radius 10–3 m is broken into 125 equal size droplets. Surface tension of mercury is 0.45 N m–1 . The gain in surface energy is:

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Show full solutionCorrect option: A
Correct answer
A2.26 × 10–5 J

Step-by-step explanation

As the volume of the mercury will remain same, therefore

43πR3=12543πr3⇒R=5r

Now, increase in surface energy

∆U=125×S×4πr2-S×4πR2

 =0.45×4π×10-3212525-1

=4×0.45×4π×10-6

=2.26×10-5J

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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.