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Question
Eight droplets of water each of radius 0.2 mm coalesce into a single drop. Find the decrease in the surface area.
Solution
r = radius of droplet
= 0.2 mm = 2 × 10-4 m
R = radius of single drop
Volume of 8 droplets = Volume of a single drop
`8 xx 4/3 pi "r"^3 = 4/3 pi"R"^3`
∴ R3 = 8r3
∴ R = 2r
Decrease in the surface area
ΔA = A1 - A2
= 8 × 4πr2 - 4πR2
= 32πr2 - 4π(2r)2
= 32πr2 - 16πr2
= 16πr2
= 16 × 3.142 × (2 × 10-4)2
ΔA = 2.011 × 10-6 m2
∴ Decrease in the surface area is 2.011 × 10-6 m2.
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