Verified Solution • 2025–26 Session moderate

QJEEM18Ph35b9: JEE Main 2018 Physics - Class 11 JEEM18Ph35b9 Solution: A small soap bubble of radius 4 cm is trapped inside another…

A small soap bubble of radius 4 cm is trapped inside another bubble of radius 6 cm 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 :

Multiple Choice Options

  • A 12 cm
  • B 2.4 cm
  • C 6 cm
  • D 4.8 cm

Step-by-Step Verified Solution

Pressure difference inside the inner bubble,

p2 −- p1 = 4Tr2{{4T} \over {{r_2}}} b . . . . . (1)

JEE Main 2018 (Online) 16th April Morning Slot Physics - Properties of Matter Question 232 English Explanation

And for outer bubble

p1 −- p0 = 4Tr1{{4T} \over {{r_1}}} . . . . . . . (2)

∴   \therefore\,\,\, p2 −- p0 = 4T (1r2+1r1)\left( {{1 \over {{r_2}}} + {1 \over {{r_1}}}} \right)

⇒\Rightarrow    \,\,\, p2 −- p0 = 4Tr{{4T} \over r}

Here r is the radius of the bubble.

∴   \therefore\,\,\, 1r=1r2+1r1{1 \over r} = {1 \over {{r_2}}} + {1 \over {{r_1}}}

⇒\Rightarrow    \,\,\, r = r1r2r1+r2{{{r_1}{r_2}} \over {{r_1} + {r_2}}}

= 4×64+60{{4 \times 6} \over {4 + 60}}

= 2.4 cm

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring