Question

The excess pressure inside a spherical drop of water is four times that of another drop

Answer 1

Answer:

The excess pressure inside a spherical drop of water equals:

$p = \frac{2 \gamma }{r}$

$where \: \gamma - surface \: tension \: of \: water \\ r - radius \: of \: the \: drop \\ so \: if \: we \: have \: 2 \: drop \: with \: excess \: pressures \: 4p1 = p2 \\ 4 \frac{2 \gamma }{r1} = \frac{2 \gamma }{r2} \\ mass \: of \: sperical \: drop = m = \frac{4}{3}\pi {r}^{2}p \\ p - density \: of \: water \\ r - radius \: of \: the \: drop \\ \frac{m1}{m2} = \frac{r13}{r23} {( \frac{r1}{r2} })^{3} = {4}^{2} = 64$