A student observes that the absolute pressure inside a soap bubble of radius 2.0 mm in the atmosphere is exactly equal to the absolute pressure at a depth of 1.6 cm inside an oil tank. If the density of the oil is 800 kg m ⁻³ and the acceleration due to gravity is 10 m s ⁻² , what is the surface tension of the soap solution?

Step-by-step solution

The absolute pressure inside a soap bubble of radius R and surface tension T is given by: P_ bubble = P₀ + 4T R The absolute pressure at a depth h in a liquid of density is: P_ depth = P₀ + g h Equating the two pressures: P₀ + 4T R = P₀ + g h 4T R = g h T = g h R 4 Given values: R = 2.0 mm = 2 10⁻³ m h = 1.6 cm = 1.6 10⁻² m = 800 kg m ⁻³ g = 10 m s ⁻² Substituting the values: T = 800 10 1.6 10⁻² 2 10⁻³ 4 T = 128 2 10⁻³ 4 T = 64 10⁻³ N m ⁻¹ = 0.064 N m ⁻¹ Answer: 0.064 N m ⁻¹