Initial radius r1=22=1 cm =10−2 m
Final radius r2=26=3 cm =3×10−2 m
Surface tension T=0.03 N/m
A soap bubble has two free surfaces. The change in total surface area is given by:
ΔA=2×4π(r22−r12)
ΔA=8π((3×10−2)2−(10−2)2)
ΔA=8π(9×10−4−10−4)=64π×10−4 m2
Work done W=TΔA
W=0.03×64π×10−4
W=1.92π×10−4 J
Comparing with απ×10−4 J, we get α=1.92.
Answer: 1.92