The surface tension of a soap solution is 3.5×10⁻² N·m⁻¹. The radius of a soap bubble is increased from 1 cm to 2 cm. Taking π = 22 / 7, if the work required is α×10⁻⁶ J, find α.

Question

G1 Physics · Accepted Answer

A soap bubble has two surfaces, so its surface energy is U = 2T(4πR²) = 8πTR². Therefore the work required to increase the radius from R₁ to R₂ is W = 8πT(R₂²-R₁²). Substitute: T = 3.5×10⁻² N·m⁻¹, R₁ = 1×10⁻² m, R₂ = 2×10⁻² m. Thus, W = 8×(22 ÷ 7)×3.5×10⁻²×[(2×10⁻²)²-(1×10⁻²)²] = 264×10⁻⁶ J. Hence, α = 264.