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JIPMER 2012
Exam
JIPMER
Year
2012
Subject
Physics
Chapter
Mechanical Properties of Fluids

JIPMER 2012 · Physics · Mechanical Properties of Fluids

A soap bubble of radius r is blown up to form a bubble of radius 2...

A soap bubble of radius rr is blown up to form a bubble of radius 2r2 r under isothermal conditions. If TT is the surface tension of soap solution, the energy spent in the blowing

Options

  1. A3πr23 \pi r^2
  2. B6πTr26 \pi T r^2
  3. C12πTr212 \pi T r^2
  4. D24πTr224 \pi T r^2

Correct answer

D. 24πTr224 \pi T r^2

Step-by-step solution

Initially area of soap bubble A1=4πr2A_1=4 \pi r^2 Under isothermal condition radius becomes 2r2 r, Then, area A2=4π(2r)2A_2=4 \pi(2 r)^2 =4π⋅4r2=4 \pi \cdot 4 r^2 =16πr2=16 \pi r^2 Increase in surface area ΔA=2(A2−A1)\Delta A=2\left(A_2-A_1\right)=2\left(16 \pi r^2-4 \pi r^2\right)=24πr2=24 \pi r^2 Energy spent W=T×ΔAW=T \times \Delta A =T⋅24πr2=T \cdot 24 \pi r^2 W=24πTr2 JW=24 \pi T r^2 \mathrm{~J}
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