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Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (surface tension of soap solution =0.03 N m⁻¹)

Asked in AIEEE 2011 · Work done on a soap bubble

Answer: (4) 0.4π mJ

Step-by-step solution

Given: T=0.03 N m⁻¹, r₁=0.03 m, r₂=0.05 m.

Two surfaces on a soap bubble, so W=8π T(r₂²-r₁²).

r₂²-r₁²=25×10⁻⁴-9×10⁻⁴=16×10⁻⁴ m².

W=8π(0.03)(16×10⁻⁴)=3.84×10⁻⁴π J=0.384π mJ.

That is nearly 0.4π mJ.

Why the other options are wrong

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