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(D) The excess pressure inside a spherical drop of radius $r$ and surface tension $T$ is given by $P = \frac{2T}{r}$.Let $r_1$ and $r_2$ be the radii

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$1: 27$

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(D) The excess pressure inside a spherical drop of radius $r$ and surface tension $T$ is given by $P = \frac{2T}{r}$.Let $r_1$ and $r_2$ be the radii of the first and second drops,respectively.Given that the excess pressure in the first drop is three times that of the second drop: $P_1 = 3P_2$.Substituting the formula: $\frac{2T}{r_1} = 3 \left( \frac{2T}{r_2} \right)$.This simplifies to $\frac{1}{r_1} = \frac{3}{r_2}$,which means $\frac{r_1}{r_2} = \frac{1}{3}$.The mass of a drop is given by $m = V \rho = \frac{4}{3} \pi r^3 \rho$,where $\rho$ is the density of water.Since both drops are of water,the density $\rho$ is the same for both.Therefore,the ratio of the masses is $\frac{m_1}{m_2} = \frac{\frac{4}{3} \pi r_1^3 \rho}{\frac{4}{3} \pi r_2^3 \rho} = \left( \frac{r_1}{r_2} \right)^3$.Substituting the ratio of radii: $\frac{m_1}{m_2} = \left( \frac{1}{3} \right)^3 = \frac{1}{27}$.

The excess pressure inside a first spherical drop of water is three times that of a second spherical drop of water. Then the ratio of the mass of the first spherical drop to that of the second spherical drop is

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