JEE Main 2026 Physics question
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 J. The value of α is . (Take π = 3.14)
Answer: (C)
Solution
W = 8πT(r₂² − r₁²) (two surfaces) = 8π(0.03)(0.03² − 0.01²) = 8π(0.03)(0.0008) = 1.92π×10⁻⁴ J.