Question

A soap bubble is blown to a diameter of 7 cm. If 36960 erg of work is done in blowing it further, find the new radius. The surface tension of soap solution is 40 dyne/cm.

Answer 1

$$\begin{gathered} Dearstudent, \\ Workdoneinblowingasoapbubble=T\times 8\mathrm{\pi }({{\mathrm{R}}^2}_{\mathrm{final}}-{{\mathrm{R}}^2}_{\mathrm{initial}}) \\ {\mathrm{R}}_{\mathrm{FINAL}}=? \\ {\mathrm{R}}_{\mathrm{initial}}=3.5\mathrm{cm} \\ \mathrm{T}=40\mathrm{dyne}/\mathrm{cm} \\ \mathrm{Work}\mathrm{done}=36960\mathrm{erg} \\ \mathrm{Putting}\mathrm{these}\mathrm{values}\mathrm{in}\mathrm{formula}\mathrm{we}\mathrm{get}, \\ 36960=40\times 8\mathrm{\pi }({{\mathrm{R}}^2}_{\mathrm{final}}-(3.5{)}^2) \\ \Rightarrow {{\mathrm{R}}^2}_{\mathrm{final}}=36.78-12.25 \\ \Rightarrow {\mathrm{R}}_{\mathrm{final}}=4.95\mathrm{cm} \\ \mathrm{Regards} \end{gathered}$$