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
Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' coalesce to form a big drop. The radius of the big drop is
PhysicsMechanical Properties of FluidsMHT CETMHT CET 2020 (12 Oct Shift 2)
Options:
  • A $\left(r_{1}-r_{2}\right)^{\frac{1}{2}}$
  • B $\left(r_{1}+r_{2}\right)^{\frac{1}{2}}$
  • C $\left(r_{1}^{2}+r_{2}^{2}\right)^{\frac{1}{2}}$
  • D $\left(r_{1}^{2}-r_{2}^{2}\right)^{\frac{1}{2}}$
Solution:
1983 Upvotes Verified Answer
The correct answer is: $\left(r_{1}^{2}+r_{2}^{2}\right)^{\frac{1}{2}}$
Under isothermal conditions the surface tension remains constant. If $r$ is the radius of the digger drop, then Final surface energy $=$ Initial surface energy $\therefore 8 \pi r^{2} T=8 \pi r_{1}^{2} T+8 \pi r_{2}^{2} T$
$\therefore r^{2}=r_{1}^{2}+r_{2}^{2}$
$\therefore r=\left(r_{1}^{2}+r_{2}^{2}\right)^{\frac{1}{2}}$

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.