Given, initial diameter =2cm Initial radius, r1=0.01m Final diameter =4cm Final radius, r2=0.02m Surface Tension, T=3.5×10−2Nm−1 Work done, W=T×∆A where, ∆A is change in surface area ∆A‌=2×4π(r22−r12)‌‌...‌ (ii) ‌ ‌=2×4π[(0.02)2−(0.01)2] ∆A‌=8π(0.0003) ∆A‌=2.4π×10−3m2‌‌...‌ (iii) ‌ Putting value of Eq. (iii) in Eq. (i) Thus, ‌W=3.5×10−2×2.4×π×10−3J ‌W=8.4π×10−5 ‌W=8.4×3.14×10−5=26.3×10−5 W=263×10−6J
