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JEE Main 2026 · 2 Apr · Shift 2, Physics question 33

The surface tension of a soap bubble is 0.03 N/m. The work done in increasing the diameter of bubble from 2 cm to 6 cm is απ×10−4\alpha\pi\times10^{-4} J. The value of α\alpha is ________. (Take π=3.14\pi=3.14)

Choose one option
Answer and worked solution

Official answer: C

A bubble has two surfaces. W=2T ΔA=2T⋅4π(r22−r12)W=2T\,\Delta A=2T\cdot4\pi(r_2^2-r_1^2).

Values. r1=1r_1=1 cm, r2=3r_2=3 cm, so 4π(9−1)=32π4\pi(9-1)=32\pi cm² =32π×10−4=32\pi\times10^{-4} m².

Work. W=2(0.03)(32π×10−4)=1.92π×10−4W=2(0.03)(32\pi\times10^{-4})=1.92\pi\times10^{-4} J, so α=1.92\alpha=1.92.

(C)

Topics:Surface tension
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