JEE Main 2018PhysicsMechanical Properties of FluidsHardMCQ

JEE Main 2018 — Mechanical Properties of Fluids Question with Solution

JEE Main 2018 (16 Apr Online)

Question

A small soap bubble of radius 4cm is trapped inside another bubble of radius 6cm without any contact. Let P2 be the pressure inside the inner bubble and P0, the pressure outside the outer bubble. Radius of another bubble with pressure difference P2-P0 between its inside and outside would be:

Choose an option

Show full solutionCorrect option: A
Correct answer
A

2.4 cm

Step-by-step explanation



P2−P1=4Tr1      ....(1)P1−P0=4Tr2  .....(2)adding (1) & (2)P2−P0=4Tr1+4Tr2P2=P0+4Tr1+4Tr2P2=P0+4T6+4T4

for another bubble of radius r

P2= P0+4Tr

⇒   1r=16+14

⇒   r=2.4 cm

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About this question

This is a previous-year question from JEE Main 2018, 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.