askiitiansexpert soumyajit_iitkanpur
Last Activity: 15 Years ago
Let r be the instantaneous radius of the bubble in the process of enlargement and let the radius increases to r+dr. Then increase in surface area ds = 4*pi*(r+dr)^2 - 4*pi*r^2 = 8*pi*rdr (neglecting 4*pi*(dr)^2 which is small)
Now work done in increase of this surface area dw = T*ds (T surface tension of soap solution)
As tehre are two free surfaces of soap bubble so net work done dw = 16*pi*T*rdr
therefore net work done in increasing diameter D to 2d is W = 16*pi*T ∫D/2 d r dr = 8*pi*T*(d^ 2 - (D^2)/4)