Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A soap bubble of radius $r$ is blown up to form a bubble of radius $2 r$ under isothermal conditions. If $T$ is the surface tension of soap solution, the energy spent in the blowing
PhysicsMechanical Properties of FluidsJIPMERJIPMER 2012
Options:
  • A $3 \pi r^2$
  • B $6 \pi T r^2$
  • C $12 \pi T r^2$
  • D $24 \pi T r^2$
Solution:
2331 Upvotes Verified Answer
The correct answer is: $24 \pi T r^2$
Initially area of soap bubble
$A_1=4 \pi r^2$
Under isothermal condition radius becomes $2 r$, Then, area $A_2=4 \pi(2 r)^2$
$=4 \pi \cdot 4 r^2$
$=16 \pi r^2$
Increase in surface area
$\Delta A=2\left(A_2-A_1\right)$
$=2\left(16 \pi r^2-4 \pi r^2\right)$
$=24 \pi r^2$
Energy spent
$W=T \times \Delta A$
$=T \cdot 24 \pi r^2$
$W=24 \pi T r^2 \mathrm{~J}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.