# Mensuration-55

### From Homeworkwiki

**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 r^{3} = 4/3 Π (1/20)^{3} = Π/6000 cm^{3}

Outer radius of the bubble = 20 cm

Therefore, Outer surface area of the bubble = 4 Π r^{2} = 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 10^{5}) mm = 1/96 x 10^{4} mm = 10^{-4} / 96 mm

= 10^{-4} mm (nearly)