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Excess pressure in a soap bubble of radius r is p. Now the bubble collapses into a drop of soap solution of radius r/n (n>1). Then surface energy of the drop is

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$\frac{{πr}^{3} p}{4 n^{3}}$

$\frac{2 {πr}^{3} p}{n^{3}}$

$\frac{{πr}^{2} p}{n^{2}}$

$\frac{{πr}^{3} p}{n^{2}}$

answer is D.

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Detailed Solution

For the soap bubble, $p = \frac{4 T}{r} \Rightarrow T = \frac{pr}{4}$

E = Surface energy of the drop $= 4 π {(\frac{r}{n})}^{2} . T = \frac{4 {πr}^{2}}{n^{2}} . \frac{pr}{4}$

$∴ E = \frac{{πr}^{3} p}{n^{2}}$

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