The work done to increase the radius of a soap bubble is given by W=T×ΔA
Since a soap bubble has two free surfaces, the change in surface area is ΔA=2×4π(r22−r12)
Substituting the given values:
W=3.5×10−2×8π×((2×10−2)2−(1×10−2)2)
W=3.5×10−2×8π×(4×10−4−1×10−4)
W=3.5×10−2×8π×3×10−4
W=84π×10−6
Using π=722:
W=84×722×10−6=264×10−6 J
Comparing with α×10−6 J, we get α=264
Answer: 264