Animated Solution for Physics - Properties of Solids and Liquids: Work done in increasing the size of a soap bubble from radius of $3\text{ cm}$ to $5\text{ cm}$ is nearly (surface tension of soap solution = $0.03{\text{ Nm}}^{-1}$)

Imagine you are gently blowing air into a soap bubble. As you blow, the bubble expands, its surface stretches, and it grows larger. But this expansion doesn't happen for free! The liquid film of the bubble has a property called surface tension, which acts like a stretched elastic sheet trying to pull the bubble back to a smaller size. To make the bubble grow, you have to do work against this inward pull.

According to the Work-Energy Theorem, the work you do in expanding the bubble is stored within the bubble's surface as potential energy. Therefore, the work done is exactly equal to the change in the surface energy of the bubble.

Now, how do we calculate this surface energy? The surface energy $U$ is simply the product of the surface tension $T$ and the total surface area $A$.

But here is where many students make a classic mistake! A soap bubble is not a solid drop of liquid. It is a thin, hollow film of soapy water with air on the inside and air on the outside. Because of this, the liquid film has two free surfaces—an inner surface and an outer surface.

When the bubble expands, both of these surfaces stretch. Therefore, the effective surface area is twice the surface area of a standard sphere. Instead of $4\pi R^2$, we must use $8\pi R^2$.

With our effective area figured out, we can write the master equation for the work done. The work done is the final surface energy minus the initial surface energy.

This elegant equation tells us exactly how much energy is required to stretch the bubble from an initial radius $R_1$ to a final radius $R_2$.

Let's plug in the numbers from our problem. The surface tension $T$ is given as $0.03{\text{ Nm}}^{-1}$. The initial radius $R_1$ is $3\text{ cm}$ and the final radius $R_2$ is $5\text{ cm}$.

Before we substitute, we must be extremely careful with units! We need to convert the radii from centimeters to meters by multiplying by $10^{-2}$.

Now, we carefully square the terms inside the bracket. Five squared is $25$, three squared is $9$, and squaring $10^{-2}$ gives us $10^{-4}$.

Subtracting the terms inside the bracket gives us $16\times 10^{-4}$.

Finally, we multiply the numbers together. $8\times 0.03\times 16=3.84$.

The options are given in millijoules ($\text{mJ}$). To convert Joules to millijoules, we multiply by $10^3$.

Looking at our options, $0.384\pi$ is nearly $0.4\pi$. Thus, the work done is approximately $0.4\pi \text{ mJ}$.