Question

Radius of a soap bubble is increased from \(1\,\text{cm}\) to \(2\,\text{cm}\). Work done in the process is ( \(S\) is surface tension ):

• A. \( \pi S \times 10^{-2} \, \text{J} \)
• B. \( 1.2 \pi S \, \text{J} \)
• C. \( 2.4 \pi S \times 10^{-3} \, \text{J} \)
• D. \( \pi S \times 10^{-3} \, \text{J} \)

Hint:

For a \textbf{soap bubble}, two liquid surfaces are present. Hence total surface energy \(E = 2S \times A = 8\pi S r^2\). Work done in expansion \(= 8\pi S (r_2^2 - r_1^2)\).

Answer 1

The Correct Option is C

Concept: A soap bubble has two liquid surfaces (inner and outer). The surface energy of a liquid surface is: \[ E = S \times \text{Area} \] Since a soap bubble has two surfaces, \[ E = 2S \times A \] For a spherical bubble: \[ A = 4\pi r^2 \] Thus total surface energy becomes \[ E = 2S(4\pi r^2) = 8\pi S r^2 \] The work done in expanding the bubble equals the increase in surface energy. Step 1: Write the expression for work done.} \[ W = \Delta E = 8\pi S (r_2^2 - r_1^2) \] Step 2: Substitute the given radii.} \[ r_1 = 1\,\text{cm} = 10^{-2}\,\text{m}, \qquad r_2 = 2\,\text{cm} = 2 \times 10^{-2}\,\text{m} \] \[ r_2^2 - r_1^2 = (2\times10^{-2})^2 - (10^{-2})^2 \] \[ = 4\times10^{-4} - 1\times10^{-4} \] \[ = 3\times10^{-4} \] Step 3: Calculate the work done.} \[ W = 8\pi S (3\times10^{-4}) \] \[ W = 24\pi S \times 10^{-4} \] \[ W = 2.4\pi S \times 10^{-3}\ \text{J} \]