One large soap bubble of diameter 'D' breaks | vedclass

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(C) Let the radius of the large bubble be $R = D/2$. The volume of the large bubble is $V = \frac{4}{3}\pi R^3$.Let the radius of each small bubble be

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$6$

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(C) Let the radius of the large bubble be $R = D/2$. The volume of the large bubble is $V = \frac{4}{3}\pi R^3$.Let the radius of each small bubble be $r$. Since the volume remains constant, $V = 64 	imes (\frac{4}{3}\pi r^3)$.Thus, $R^3 = 64r^3$, which implies $R = 4r$ or $r = R/4$.The initial surface area of the large soap bubble (having two surfaces) is $A_i = 2 	imes (4\pi R^2) = 8\pi R^2$.The final surface area of $64$ small bubbles is $A_f = 64 	imes 2 	imes (4\pi r^2) = 512\pi r^2$.Substituting $r = R/4$, we get $A_f = 512\pi (R/4)^2 = 512\pi (R^2/16) = 32\pi R^2$.The change in surface area is $\Delta A = A_f - A_i = 32\pi R^2 - 8\pi R^2 = 24\pi R^2$.The change in surface energy is $\Delta U = T 	imes \Delta A = T 	imes 24\pi R^2$.Substituting $R = D/2$, we get $\Delta U = T 	imes 24\pi (D/2)^2 = T 	imes 24\pi (D^2/4) = 6\pi T D^2$.

One large soap bubble of diameter '$D$' breaks into $64$ smaller bubbles of equal size. If the surface tension of the soap solution is '$T$',what is the change in surface energy (in $\pi T D^2$)?

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