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JEE Main 201816 AprilOnline
Exam
JEE Main
Year
2018
Date
16 April
Mode
Online
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2018 · 16 April · Online · Physics · Mechanical Properties of Fluids

A small soap bubble of radius 4cm is trapped inside another bubble...

A small soap bubble of radius 4cm is trapped inside another bubble of radius 6cm without any contact. Let P2{P}_{2} be the pressure inside the inner bubble and P0{P}_{0}, the pressure outside the outer bubble. Radius of another bubble with pressure difference P2−P0{P}_{2} - {P}_{0} between its inside and outside would be:A small soap bubble of radius 4cm is trapped inside another bubble of radius 6cm without any contact. Let P _ 2 be the

Options

  1. A

    2.4cm2 .4 cm

  2. B12cm12cm
  3. C4.8cm4 .8 cm
  4. D6cm6cm

Correct answer

A.

2.4cm2 .4 cm

Step-by-step solution

A small soap bubble of radius 4cm is trapped inside another bubble of radius 6cm without any contact. Let P _ 2 be the

P2−P1=4Tr1....(1)P1−P0=4Tr2.....(2)adding(1) and (2)P2−P0=4Tr1+4Tr2P2=P0+4Tr1+4Tr2P2=P0+4T6+4T4\begin{array}{ll}{P}_{2} - {P}_{1} = \dfrac{4T}{{r}_{1}}....( 1 ) \\ {P}_{1} - {P}_{0} = \dfrac{4T}{{r}_{2}} ..... ( 2 ) \\ adding ( 1 ) \text{ and } ( 2 ) \\ {P}_{2} - {P}_{0} = \dfrac{4T}{{r}_{1}} + \dfrac{4T}{{r}_{2}} \\ {P}_{2} = {P}_{0} + \dfrac{4T}{{r}_{1}} + \dfrac{4T}{{r}_{2}} \\ {P}_{2} = {P}_{0} + \dfrac{4T}{6} + \dfrac{4T}{4}\end{array}

for another bubble of radius r

P2=P0+4Tr{P}_{2} = {P}_{0} + \dfrac{4T}{r}

⇒1r=16+14⇒ \dfrac{1}{r} = \dfrac{1}{6} + \dfrac{1}{4}

⇒r=2.4cm⇒ r = 2.4cm
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