TS EAMCET 9 May 2024 Shift 1 Solved Paper

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