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(C) Let $S$ be the surface tension of water. The energy of the bigger drop is $E = S \cdot 4\pi R^2$. The energy of $n$ small drops is $n \cdot S \cdo

Vedclass · Accepted Answer

$64$

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(C) Let $S$ be the surface tension of water. The energy of the bigger drop is $E = S \cdot 4\pi R^2$. The energy of $n$ small drops is $n \cdot S \cdot 4\pi r^2$. The energy loss is $\Delta U = n(S \cdot 4\pi r^2) - S \cdot 4\pi R^2 = 3E$. Substituting $E = S \cdot 4\pi R^2$,we get $n(S \cdot 4\pi r^2) - S \cdot 4\pi R^2 = 3(S \cdot 4\pi R^2)$. $n(S \cdot 4\pi r^2) = 4(S \cdot 4\pi R^2) \implies n r^2 = 4R^2$. Since the volume is conserved,$\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$,so $R^3 = nr^3$,which means $n = (R/r)^3$. Substituting $n$ in $nr^2 = 4R^2$: $(R/r)^3 \cdot r^2 = 4R^2 \implies R^3/r = 4R^2 \implies R/r = 4$. Thus,$n = (4)^3 = 64$.

When a big drop of water is formed from $n$ small drops of water,the energy loss is $3E$,where $E$ is the energy of the bigger drop. The radius of the bigger drop is $R$ and that of the smaller drop is $r$. Then the value of $n$ is:

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