JEE Main 202329 Jan, Morning shiftPhysicsEasyProperties of Solids and Liquids184× since 2002Surface tension of a soap bubble is 2.0×10−2Nm−12.0 \times 10^{-2} \mathrm{Nm}^{-1}2.0×10−2Nm−1. Work done to increase the radius of soap bubble from 3.5 cm3.5 \mathrm{~cm}3.5 cm to 7 cm7 \mathrm{~cm}7 cm will be: Take [π=227]\left[\pi=\frac{22}{7}\right][π=722]AOption A. 18.48×10−4 J18 .48 \times 10^{-4} \mathrm{~J}18.48×10−4 JBOption B. 5.76×10−4 J5.76 \times 10^{-4} \mathrm{~J}5.76×10−4 JCOption C. 0.72×10−4 J0.72 \times 10^{-4} \mathrm{~J}0.72×10−4 JDOption D. 9.24×10−4 J9.24 \times 10^{-4} \mathrm{~J}9.24×10−4 JCheckSkip · show solution
