A soap bubble is blown to a diameter of 7 cm. 36960 erg of work is done in blowing it further. If surface tension of soap solution is 40 dyne/cm then the new radius is ________ cm Take (π=22/7).

Solution

Physics sir · 4 steps

A soap bubble has two surfaces, so the work equals T times twice the increase in 4π r².

1. Two surfaces. A soap film has an inner and an outer surface. When the radius grows from r₁ to r₂, the total new surface area is 2 × 4π(r₂² - r₁²). The work done is surface tension times this new area.

   W = 8π T(r₂² - r₁²)

2. Put in numbers. The diameter is 7 cm, so r₁ = 3.5 cm and r₁² = 12.25 cm². Also 8π T = 8 × 22/7 × 40 = 7040/7 erg cm⁻².
3. Solve for r2. r₂² - 12.25 = (36960 × 7)/7040 = 36.75, so r₂² = 49 cm².
4. So. r₂ = √49 = 7 cm. The new radius is 7.

The common slip: 14. Reporting the new diameter instead of the radius gives 14, not 7. The question asks for the radius.

The idea to keep. For a soap bubble, always count both surfaces: work = 2T × Δ(4π r²). For a liquid drop there is only one surface. Check whether you were given a radius or a diameter.

Answer from the NTA final answer key. Solution by PrepWiser.