JEE Main 2026 Physics question
The surface tension of a soap solution is 3.5 × 10⁻² N/m. The work required to increase the radius of a soap bubble from 1 cm to 2 cm is α × 10⁻⁶ J. The value of α is . (π = 22/7)
Answer: 264
Solution
W = 8πT(r₂²−r₁²) = 8(22/7)(3.5×10⁻²)(0.0004−0.0001) = 264×10⁻⁶ J.