A liquid column of height 0.04 cm balances excess pressure of a soap bubble of certain radius. If density of liquid is 8 × 10³ kg m⁻³ and surface tension of soap solution is 0.28 Nm⁻¹, then diameter of the soap bubble is __________ cm. (if g=10 m s⁻² )

Solution

Physics sir · 4 steps

A soap bubble has two surfaces, so use 4T/r and set it equal to the pressure of the column, hρg.

1. Two surfaces. A soap bubble is a thin film with an inner and an outer surface. Each one adds 2T/r, so the excess pressure inside is 4T/r.

   4T/r = hρ g

2. Column pressure. Convert the height first: h = 0.04 cm = 4 × 10⁻⁴ m. Then hρ g = 4 × 10⁻⁴ × 8 × 10³ × 10 = 32 Pa.
3. Solve for r. r = 4T/(hρ g) = (4 × 0.28)/32 = 1.12/32 = 0.035 m = 3.5 cm.
4. So. The question asks for the diameter: 2r = 2 × 3.5 = 7 cm.

The common slip: 3.5. Using 2T/r as for a drop, or stopping at the radius, gives 3.5 instead of the diameter 7 cm.

The idea to keep. Count the surfaces before you write the formula. A drop or an air bubble inside a liquid has one surface, so use 2T/r. A soap bubble has two, so use 4T/r.

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