The excess pressure inside a spherical drop of water is three times that of another drop of water. The ratio of their surface area is

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Aicharya · Accepted Answer

Excess pressure $p=\frac{2 T}{r}$ $\frac{p_{1}}{p_{2}}=\frac{r_{2}}{r_{1}}=3$ $\begin{array}{l} 	herefore r_{2}=3 r_{1} \ \frac{A_{1}}{A_{2}}=\frac{\pi r_{1}^{2}}{\pi r_{2}^{2}}=\left(\frac{r_{1}}{r_{2}}ight)^{2}=\frac{1}{(3)^{2}}=\frac{1}{9} \end{array}$ .

The excess pressure inside a spherical drop of water is three times that of another drop of water. The ratio…

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