A water drop is divided into 8 equal droplets | vedclass

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(B) Let $R$ be the radius of the big drop and $r$ be the radius of each smaller droplet.Volume of $8$ smaller drops $=$ Volume of the big drop.$8 	im

Vedclass · Accepted Answer

half of that for smaller droplet

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(B) Let $R$ be the radius of the big drop and $r$ be the radius of each smaller droplet.Volume of $8$ smaller drops $=$ Volume of the big drop.$8 	imes \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$$8r^3 = R^3 \Rightarrow R = 2r$ or $r = \frac{R}{2}$.Excess pressure inside a drop is given by $\Delta P = \frac{2T}{r}$,where $T$ is the surface tension.For the big drop,$\Delta P_B = \frac{2T}{R}$.For the smaller droplet,$\Delta P_s = \frac{2T}{r}$.Taking the ratio: $\frac{\Delta P_B}{\Delta P_s} = \frac{r}{R} = \frac{r}{2r} = \frac{1}{2}$.Therefore,$\Delta P_B = \frac{1}{2} \Delta P_s$.

$A$ water drop is divided into $8$ equal droplets. The pressure difference between the inner and outer side of the big drop will be

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