Answer:
So amount of work produced will be [tex]10{-4}J[/tex]
Explanation:
We have given diameter of ammonia bubble is changes from 1 cm to 3 cm
So radius changes from 0.5 cm to 1.5 cm
Surface area of bubble[tex]=4\pi r^2[/tex]
So change in area of bubble [tex]=4\pi (0.015^2-0.005^2)=8\times 3.14\times (0.015^2-0.005^2)=0.00251m^2[/tex]
Surface tension of ammonia = 0.04 N/m
So work done will be [tex]Work\ done=surface\ tension\times change\ in\ area=0.04\times 0.00251= 10^{-4}J[/tex]
