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
The surface tension of soap solution is  3.5×10-2 N m-1. The amount of work done required to increase the radius of soap bubble from 10 cm to 20 cm is _____×10-4 J . (take π=227)
PhysicsMechanical Properties of FluidsJEE MainJEE Main 2023 (11 Apr Shift 2)
Solution:
2309 Upvotes Verified Answer
The correct answer is: 264

The amount of work done is equal to the change in the surface energy to increase the radius of the bubble.

The formula to calculate the change in surface energy is given by

 ΔU=2(T×ΔA)   ...1

Substitute the values of the known parameters into equation (1) to calculate the required work done.

U=3.5×10-2×4π(0.2)2-(0.1)2×2 J=3.5×10-2×4×227×0.03×2=2×1.32×10-2=264×10-4 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.