Question

The work done in blowing a soap bubble of volume $V$ is $W$. The work required to blow a soap bubble of volume $2V$ is [$T$ = surface tension of soap solution]

• A. $2^{2/3}W$
• B. $2W$
• C. $W$
• D. $2^{1/3}W$

Hint:

For any three-dimensional spherical object, surface area scales with volume via the geometric exponent factor $\frac{2}{3}$ (i.e., $A \propto V^{2/3}$). Since work done against surface tension directly tracks the surface area, doubling the volume ($2 \times$) will always increase the work by a factor of $2^{2/3}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the work required ($W'$) to blow a soap bubble to a volume of $2V$, given that the work needed to blow a bubble of volume $V$ is $W$.
A soap bubble has two free surfaces (inner and outer) exposed to air, and the work done goes into increasing the surface energy of these boundaries.

Step 2: Key Formula or Approach:
The work done in blowing a soap bubble of radius $r$ is equal to its surface energy:
$$W = T \cdot \Delta A = T \cdot (2 \times 4\pi r^2) = 8\pi r^2 T$$ The volume of a spherical bubble is given by:
$$V = \frac{4}{3}\pi r^3 \implies r = \left(\frac{3V}{4\pi}\right)^{1/3}$$ Thus, the radius scales with volume as $r \propto V^{1/3}$. Substituting this into the work formula shows how work scales with volume:
$$W \propto r^2 \propto (V^{1/3})^2 \implies W \propto V^{2/3}$$

Step 3: Detailed Explanation:
Let's establish the proportional relationship between the two states:
$$\frac{W'}{W} = \left(\frac{V'}{V}\right)^{2/3}$$ The problem states that the final volume is doubled, so $V' = 2V$:
$$\frac{W'}{W} = \left(\frac{2V}{V}\right)^{2/3} = 2^{2/3}$$ Isolating the new work term $W'$ yields:
$$W' = 2^{2/3}W$$

Step 4: Final Answer:
The work required to blow a soap bubble of volume $2V$ is $2^{2/3}W$, matching option (A).