Question

Work done in increasing the size of a soap bubble from radius of 3 cm to 5 cm in millijoule is nearly (surface tension of soap solution = $0.03\ \text{Nm}^{-1}$)

• A. $0.4\pi$
• B. $0.2\pi$
• C. $4\pi$
• D. $2\pi$

Hint:

Always remember the multiplier factor of 2 for soap bubbles due to their double-sided film structure. For a solid liquid droplet, there is only one surface layer, which would require exactly half as much work ($4\pi T \Delta (r^2)$).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks for the mechanical work required to expand a spherical soap bubble from an initial radius of $3\ \text{cm}$ to a final radius of $5\ \text{cm}$, given the surface tension ($T$) of the fluid. The final value must be expressed in millijoules (mJ).

Step 2: Key Formula or Approach:
A soap bubble is a hollow shell that contains two active liquid-air interfaces (an inner surface and an outer surface). The work done ($W$) to expand its surface area is given by:
$$W = T \cdot \Delta A$$ where $\Delta A$ is the total change across both surfaces. For a sphere with area $4\pi r^2$:
$$\Delta A = 2 \times \left(4\pi r_2^2 - 4\pi r_1^2\right) = 8\pi\left(r_2^2 - r_1^2\right)$$

Step 3: Detailed Explanation:
Let's write down the given data in standard SI units:
Surface Tension, $T = 0.03\ \text{Nm}^{-1}$ Initial radius, $r_1 = 3\ \text{cm} = 0.03\ \text{m}$ Final radius, $r_2 = 5\ \text{cm} = 0.05\ \text{m}$ Now, substitute these parameters into the work formula:
$$W = 8\pi T \left(r_2^2 - r_1^2\right)$$ $$W = 8\pi \times 0.03 \times \left[(0.05)^2 - (0.03)^2\right]$$ Calculate the square values inside the bracket:
$$W = 0.24\pi \times \left[0.0025 - 0.0009\right]$$ $$W = 0.24\pi \times 0.0016$$ $$W = 0.000384\pi\ \text{Joules}$$ To convert the work from Joules to millijoules (mJ), multiply the result by $10^3$:
$$W = 0.000384\pi \times 1000\ \text{mJ} = 0.384\pi\ \text{mJ}$$ Rounding this numerical value to one decimal place gives approximately $0.4\pi\ \text{mJ}$.

Step 4: Final Answer:
The total work done is approximately $0.4\pi\ \text{mJ}$, which corresponds to option (A).