Q 11-09-033JEE MainJEE Main 2026 (4 Apr, Shift 1)Easy
The surface tension of a soap solution is $3.5 \times 10^{-2}$ N/m. The work required to increase the radius of a soap bubble from $1$ cm to $2$ cm is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is ______. ($\pi = 22/7$)
Numerical value type. Enter your answer.
Answer: 264
A soap bubble has two surfaces: $W = T \times 2 \times 4\pi(r_2^2 - r_1^2)$.
$$W = 3.5 \times 10^{-2} \times 8 \times \frac{22}{7} \times 3 \times 10^{-4} = 264 \times 10^{-6}\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics