Soap bubble has two surfaces. Work done: W=ΔA×σ
Initial radius: ri=3.5 cm = 0.035 m, Final radius: rf=7 cm = 0.07 m
Change in surface area (both surfaces): ΔA=8π(rf2−ri2)=8π(0.072−0.0352)
ΔA=8π(0.0049−0.001225)=8π×0.003675=0.0294π m²
W=0.0294π×0.04=0.001176π J = 0.001176×722=0.003696 J = 3696 μJ
Therefore: x=15000−3696=11304