Question

A soap bubble of radius $r$ is blown up to form a bubble of radius $2r$ under isothermal conditions. If $\sigma$ is the surface tension of soap solution, the energy spent in doing so is

• A. $6\pi\sigma r^2$
• B. $3\pi\sigma r^2$
• C. $24\pi\sigma r^2$
• D. $12\pi\sigma r^2$
• E. $9\pi\sigma r^2$

Hint:

Always remember the factor of 2 for soap bubbles due to their two surfaces. If it were a liquid drop, you would only consider one surface ($4\pi r^2$).

Answer 1

The Correct Option is C

Concept:
A soap bubble has two free surfaces (inner and outer). The surface energy $U$ is given by $U = \text{Surface Tension} \times \text{Total Surface Area}$. [itemsep=8pt]
• Total Surface Area of a soap bubble $= 2 \times (4\pi r^2) = 8\pi r^2$.
• Energy spent (Work done) $= \text{Final Surface Energy} - \text{Initial Surface Energy}$.

Step 1: Calculate initial and final surface energies.
Initial Surface Energy ($U_i$): \[ U_i = \sigma \times (8\pi r^2) = 8\pi\sigma r^2 \] Final Surface Energy ($U_f$) for radius $2r$: \[ U_f = \sigma \times 2 \times [4\pi (2r)^2] = \sigma \times 2 \times [16\pi r^2] = 32\pi\sigma r^2 \]

Step 2: Calculate the energy spent.
\[ \text{Work Done} = U_f - U_i = 32\pi\sigma r^2 - 8\pi\sigma r^2 = 24\pi\sigma r^2 \]