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(D) The excess pressure inside a spherical drop of radius $r$ is given by $P = \frac{2T}{r}$,where $T$ is the surface tension.For drop $A$,$P_A = \fra

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$1: 64$

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(D) The excess pressure inside a spherical drop of radius $r$ is given by $P = \frac{2T}{r}$,where $T$ is the surface tension.For drop $A$,$P_A = \frac{2T}{r_A}$.For drop $B$,$P_B = \frac{2T}{r_B}$.Given that $P_A = 4P_B$,we have $\frac{2T}{r_A} = 4 \left( \frac{2T}{r_B} \right)$.This simplifies to $\frac{1}{r_A} = \frac{4}{r_B}$,or $\frac{r_A}{r_B} = \frac{1}{4}$.The mass of a drop is given by $m = V \rho = \left( \frac{4}{3} \pi r^3 \right) \rho$.Since both are water drops,the density $\rho$ is the same.Therefore,the ratio of masses is $\frac{m_A}{m_B} = \frac{r_A^3}{r_B^3} = \left( \frac{r_A}{r_B} \right)^3$.Substituting the ratio of radii,$\frac{m_A}{m_B} = \left( \frac{1}{4} \right)^3 = \frac{1}{64}$.

The excess pressure inside a spherical drop of water $A$ is four times that of another drop $B$. Then the ratio of mass of drop $A$ to that of drop $B$ is

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