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Step-by-Step Solution
Step 1: Identify the Relevant Concept
The work done in inflating (or increasing the size of) a soap bubble involves surface tension. A soap bubble has two surfaces (inner and outer), so the effective surface tension contribution must be doubled.
Step 2: Write Down the Formula for Work Done
The general expression for the work done in changing the radius of a soap bubble from $r_1$ to $r_2$ is:
$ W = 2T \times \Delta A $
where:
$ T $ is the surface tension of the soap solution.
$ \Delta A $ is the change in the total surface area of the bubble.
The factor of 2 appears because a soap bubble has two surfaces.
Step 3: Calculate the Change in Surface Area
The area of one spherical surface of radius $r$ is $ 4\pi r^2 $. For a bubble having two surfaces, the total surface area is $ 2 \times 4\pi r^2 = 8\pi r^2 $. Therefore,
$ \Delta A = 8 \pi \left( r_2^2 - r_1^2 \right).
$
However, this can also be written directly as $ 2 \times 4\pi (r_2^2 - r_1^2) $, which is equivalent.
Step 4: Insert the Given Numerical Values
From the problem statement:
Initial radius: $ r_1 = 3 \text{ cm} = 3 \times 10^{-2} \text{ m} $
Final radius: $ r_2 = 5 \text{ cm} = 5 \times 10^{-2} \text{ m} $
Surface tension of soap solution: $ T = 0.03 \text{ N/m} $
In the solution provided, the change in area is directly computed using the radius values in centimeters, and then a factor of $10^{-4}$ is multiplied in the end to convert $ \text{cm}^2 $ to $ \text{m}^2 $ (since $1 \text{ cm}^2 = 10^{-4} \text{ m}^2$).
Step 5: Compute the Work Done
Using the formula (with a direct approach in cm followed by appropriate unit conversion):
$
W = 2 \times T \times 4\pi \left[ (5)^2 - (3)^2 \right] \times 10^{-4} \text{ J}
$
Substitute $ T = 0.03 \text{ N/m}: $
$
W = 2 \times 0.03 \times 4\pi \left[ 25 - 9 \right] \times 10^{-4} \text{ J}
$
$
W = 2 \times 0.03 \times 4\pi \times 16 \times 10^{-4} \text{ J}
$
$
W = (2 \times 0.03 \times 64\pi) \times 10^{-4} \text{ J}
$
$
W = 3.84 \pi \times 10^{-4} \text{ J}
$
$
W = 0.384 \pi \times 10^{-3} \text{ J}
$
Rounded suitably, this becomes approximately:
$
W = 0.4 \pi \times 10^{-3} \text{ J} = 0.4\pi \, \text{mJ}.
$
Step 6: Conclude the Answer
Hence, the work done in increasing the radius of the soap bubble from $3 \text{ cm}$ to $5 \text{ cm}$ is approximately $0.4\pi \,\text{mJ}$. This matches the given correct answer.
