Mensuration-55

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A spherical drop of soap water of radius 1/2 mm is blown into a bubble of outer radius 20 cm.
Find approximately the thickness of the bubble.

Solution: Radius of the spherical drop = 1/2 mm = 1/20 cm

Volume of the spherical drop = 4/3 Π r3 = 4/3 Π (1/20)3 = Π/6000 cm3

Outer radius of the bubble = 20 cm

Therefore, Outer surface area of the bubble = 4 Π r2 = 4 Π (20)2 sq cm = 1600 Π sq cm

Volume of drop = Outer surface of the bubble x thickness of the bubble

Therefore, Thickness of the bubble = Volume of the drop / Outer surface of the bubble = Π / 6000 / 1600 Π cm

= 1 / (6000 x 1600) cm = 10 / (96 x 105) mm = 1/96 x 104 mm = 10-4 / 96 mm

= 10-4 mm (nearly)

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