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

To determine the work done in increasing the radius of a soap bubble from 1 cm to 2 cm, we need to consider the change in the surface area of the bubble and the surface tension involved. The work done, \(W\), in increasing the surface area against surface tension is given by the formula:

\(W = S \cdot \Delta A\)

where \(S\) is the surface tension and \(\Delta A\) is the change in the surface area of the bubble.

For a soap bubble, which has two surfaces (inner and outer), the total surface area \(A\) is:

\(A = 2 \times 4 \pi r^2 = 8 \pi r^2\)

where \(r\) is the radius of the bubble.

Initially, when the radius \(r_1\) is 1 cm:

\(A_1 = 8 \pi (1)^2 = 8 \pi \, \text{cm}^2\)

Finally, when the radius \(r_2\) is 2 cm:

\(A_2 = 8 \pi (2)^2 = 32 \pi \, \text{cm}^2\)

Thus, the change in surface area, \(\Delta A\), is:

\(\Delta A = A_2 - A_1 = 32 \pi - 8 \pi = 24 \pi \, \text{cm}^2\)

Therefore, the work done is:

\(W = S \cdot 24 \pi \, \text{cm}^2\)

Converting the surface area from square centimeters to square meters (since 1 cm\( ^2 \) = 10\( ^{-4} \) m\( ^2 \)):

\(W = S \cdot 24 \pi \times 10^{-4} \, \text{m}^2\)

Finally, simplifying gives:

\(W = 2.4 \pi S \times 10^{-3} \, \text{J}\)

Therefore, the correct answer is \( 2.4 \pi S \times 10^{-3} \, \text{J} \), as indicated.