A soap bubble of surface tension 0.04 N/m is blown to a diameter of 7 cm. If (15000 - x) μJ of work is done in blowing it further to make its diameter 14 cm, then the value of x i $s ____.$
(π = 22/7)

Question

A soap bubble of surface tension 0.04 N/m is blown to a diameter of 7 cm. If (15000 - x) μJ of work is done in blowing it further to make its diameter 14 cm, then the value of x i s ____. (π = 22/7)

G1 Physics · Accepted Answer

A soap bubble has two surfaces. Therefore its surface energy is U = 2S(4πr²) = 8πSr². Initial diameter: d₁ = 7 cm, so r₁ = 3.5 cm = 0.035 m. Final diameter: d₂ = 14 cm, so r₂ = 7 cm = 0.070 m. Work done is the increase in surface energy: W = 8πS(r₂²-r₁²). Substitute S = 0.04 N/m, π = 22/7. Thus, W = 8×(22/7)×0.04 [(0.070)²-(0.035)²] ≈ 3.696×10⁻³ J = 3696 μJ. Given, W = (15000-x) μJ. Therefore, 15000-x = 3696. Hence, x = 15000-3696 = 11304. Therefore the answer is 11304.