Eight droplets of water each of radius 0.2 mm coalesce into a sin…

Question

Eight droplets of water each of radius $0.2 mm$ coalesce into a single drop. Fine the decrease in the surface area

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Answer

$r=$ radius of droplet
$=0.2 mm=2 \times 10^{-4} m$
$R =$ radius of single drop
Volume of 8 droplets = Volume of a single drop
$8 \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3$
$\therefore R^3=8 r^3$
$\therefore R=2 r$
Decrease in the surface area
$\Delta A=A_1-A_2$
$=8 \times 4 \pi r^2-4 \pi R^2$
$=32 \pi r^2-4 \pi(2 r)^2$
$=32 \pi r^2-16 \pi r^2$
$=16 \pi r^2$
$=16 \times 3.142 \times\left(2 \times 10^{-4}\right)^2$
$\Delta A=2.011 \times 10^{-6} m^2$
$\therefore$ Decrease in the surface area is $2.011 \times 10-6 m 2$.

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