Work Done Calculation for Soap Bubble

The work done ($W$) to increase the surface area of a liquid surface is given by the product of the surface tension ($S$) and the change in surface area ($\Delta A$). For a soap bubble, which has two surfaces (inner and outer), the formula is:

$W = S \times \Delta A_{total}$

where $\Delta A_{total}$ is the change in the total surface area.

Surface Area Calculations

The surface area of a sphere is $4\pi r^2$. Since a soap bubble has two surfaces, the total surface area is $A_{total} = 2 \times 4\pi r^2 = 8\pi r^2$. The radius ($r$) is half the diameter ($d$).

• Initial diameter $d_1 = 2 \text{ cm} = 0.02 \text{ m}$, so initial radius $r_1 = \frac{0.02}{2} = 0.01 \text{ m}$.
• Final diameter $d_2 = 6 \text{ cm} = 0.06 \text{ m}$, so final radius $r_2 = \frac{0.06}{2} = 0.03 \text{ m}$.

Calculate the initial and final total surface areas:

$A_1 = 8\pi r_1^2 = 8\pi (0.01)^2 = 8\pi (0.0001) \text{ m}^2$

$A_2 = 8\pi r_2^2 = 8\pi (0.03)^2 = 8\pi (0.0009) \text{ m}^2$

The change in total surface area is:

$\Delta A_{total} = A_2 - A_1 = 8\pi (0.0009) - 8\pi (0.0001) = 8\pi (0.0009 - 0.0001) = 8\pi (0.0008) \text{ m}^2$

Work Done Calculation

Now, calculate the work done using the surface tension ($S = 0.03 \text{ N/m}$):

$W = S \times \Delta A_{total} = (0.03 \text{ N/m}) \times (8\pi \times 0.0008 \text{ m}^2)$

$W = 0.03 \times 8 \times 0.0008 \times \pi \text{ J}$

$W = 0.24 \times 0.0008 \times \pi \text{ J}$

$W = 0.000192 \times \pi \text{ J}$

$W = 1.92 \times 10^{-4} \times \pi \text{ J}$

Determining the Value of $\alpha$

The problem states the work done is $\alpha\pi \times 10^{-4} \text{ J}$. By comparing this with the calculated work done:

$\alpha\pi \times 10^{-4} \text{ J} = 1.92 \times \pi \times 10^{-4} \text{ J}$

Therefore, the value of $\alpha$ is $1.92$.