Question

The work done in increasing the diameter of a soap bubble from 2 cm to 4 cm is (Surface tension of soap solution \(=3.5\times10^{-2}Nm^{-1}\))

• A. \(528\times10^{-6}\) J
• B. \(264\times10^{-6}\) J
• C. \(132\times10^{-6}\) J
• D. \(178\times10^{-6}\) J

Hint:

Remember that a soap bubble has two surfaces (inner and outer), so the surface area is twice the area of a sphere.

Answer 1

The Correct Option is B

The work done in increasing the surface area of a soap bubble is given by: \[ W = T \Delta A \] where \( T \) is the surface tension and \( \Delta A \) is the change in surface area. The surface area of a soap bubble is given by \( A = 2 \times 4\pi r^2 \) since it has two surfaces. Given: \begin{itemize} \item Initial diameter \( d_1 = 2 \) cm, so initial radius \( r_1 = 1 \) cm \(= 0.01 \) m \item Final diameter \( d_2 = 4 \) cm, so final radius \( r_2 = 2 \) cm \(= 0.02 \) m \item Surface tension \( T = 3.5 \times 10^{-2} \) Nm\(^{-1}\) \end{itemize} Initial surface area: \[ A_1 = 2 \times 4\pi r_1^2 = 8\pi (0.01)^2 = 8\pi \times 10^{-4} \text{ m}^2 \] Final surface area: \[ A_2 = 2 \times 4\pi r_2^2 = 8\pi (0.02)^2 = 8\pi \times 4 \times 10^{-4} = 32\pi \times 10^{-4} \text{ m}^2 \] Change in surface area: \[ \Delta A = A_2 - A_1 = 32\pi \times 10^{-4} - 8\pi \times 10^{-4} = 24\pi \times 10^{-4} \text{ m}^2 \] Work done: \[ W = T \Delta A = 3.5 \times 10^{-2} \times 24\pi \times 10^{-4} \] \[ W = 84\pi \times 10^{-6} \] \[ W \approx 84 \times 3.14 \times 10^{-6} \] \[ W \approx 263.76 \times 10^{-6} \text{ J} \] This is approximately \( 264 \times 10^{-6} \) J. Final Answer: (2) \(264\times10^{-6}\) J