The change in surface energy when a big spher | vedclass

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(B) The volume of the big drop is equal to the total volume of $n$ small droplets: $\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$. From this, we

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$4 \pi R^2(n^{1/3}-1) T$

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(B) The volume of the big drop is equal to the total volume of $n$ small droplets: $\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$. From this, we get $r^3 = \frac{R^3}{n}$, which implies $r = \frac{R}{n^{1/3}}$. The surface area of the big drop is $A = 4 \pi R^2$. The total surface area of $n$ small droplets is $A' = n 	imes 4 \pi r^2$. Substituting $r = R n^{-1/3}$, we get $A' = n 	imes 4 \pi (R n^{-1/3})^2 = n 	imes 4 \pi R^2 n^{-2/3} = 4 \pi R^2 n^{1/3}$. The change in surface area is $\Delta A = A' - A = 4 \pi R^2 n^{1/3} - 4 \pi R^2 = 4 \pi R^2 (n^{1/3} - 1)$. The change in surface energy is $\Delta U = T 	imes \Delta A = 4 \pi R^2 (n^{1/3} - 1) T$.

The change in surface energy when a big spherical drop of radius $R$ is split into $n$ spherical droplets of radius $r$ is ($T=$ surface tension).

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