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(B) Let $R$ be the radius of the big drop and $r$ be the radius of each smaller droplet.Since the volume remains constant:$8 	imes \frac{4}{3} \pi r^

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$\Delta P_{B}=\frac{1}{2} \Delta P_{S}$

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(B) Let $R$ be the radius of the big drop and $r$ be the radius of each smaller droplet.Since the volume remains constant:$8 	imes \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$$8r^3 = R^3$$2r = R \implies r = \frac{R}{2}$Excess pressure inside a drop is given by $\Delta P = \frac{2T}{r}$,where $T$ is surface tension.For the smaller droplet: $\Delta P_{S} = \frac{2T}{r}$For the bigger drop: $\Delta P_{B} = \frac{2T}{R}$Taking the ratio:$\frac{\Delta P_{B}}{\Delta P_{S}} = \frac{2T/R}{2T/r} = \frac{r}{R}$Substituting $r = \frac{R}{2}$:$\frac{\Delta P_{B}}{\Delta P_{S}} = \frac{R/2}{R} = \frac{1}{2}$Therefore,$\Delta P_{B} = \frac{1}{2} \Delta P_{S}$.

$A$ big water drop is divided into $8$ equal droplets. $\Delta P_{S}$ and $\Delta P_{B}$ are the excess pressure inside a smaller and bigger drop respectively. The relation between $\Delta P_{S}$ and $\Delta P_{B}$ is

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