Q 11-09-080JEE MainJEE Main 2023 (29 Jan, Shift 1)Medium
Surface tension of a soap bubble is $2.0\times10^{-2}\ \text{N m}^{-1}$. Work done to increase the radius of soap bubble from $3.5\ \text{cm}$ to $7\ \text{cm}$ will be [Take $\pi=\frac{22}{7}$]
Answer: (C) $18.48\times10^{-4}\ \text{J}$
A soap bubble has two surfaces, so $W=T\times2\times4\pi(r_2^2-r_1^2)$.
$$W=2\times10^{-2}\times8\times\frac{22}{7}\times(0.07^2-0.035^2)=2\times10^{-2}\times\frac{176}{7}\times3.675\times10^{-3}$$
$$W=18.48\times10^{-4}\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics