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(C) The excess pressure $p$ inside a spherical drop of radius $r$ is given by $p = \frac{2T}{r}$,where $T$ is the surface tension.Given that the exces

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$1: 9$

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(C) The excess pressure $p$ inside a spherical drop of radius $r$ is given by $p = \frac{2T}{r}$,where $T$ is the surface tension.Given that the excess pressure of the first drop is three times that of the second drop,we have $p_1 = 3p_2$.Substituting the formula for excess pressure: $\frac{2T}{r_1} = 3 	imes \frac{2T}{r_2}$.This simplifies to $\frac{1}{r_1} = \frac{3}{r_2}$,which implies $r_2 = 3r_1$ or $\frac{r_1}{r_2} = \frac{1}{3}$.The surface area $A$ of a spherical drop is given by $A = 4\pi r^2$.The ratio of their surface areas is $\frac{A_1}{A_2} = \frac{4\pi r_1^2}{4\pi r_2^2} = \left(\frac{r_1}{r_2}ight)^2$.Substituting the ratio of radii: $\frac{A_1}{A_2} = \left(\frac{1}{3}ight)^2 = \frac{1}{9}$.

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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