Question

The surface tension of a soap bubble is \(2 \times 10^{-2} \, \text{N/m}\). Work done to increase the radius of the bubble from 3.5 cm to 7 cm will be:

• A. \(4.072 \times 10^{-3} \, \text{J}\)
• B. \(5.76 \times 10^{-3} \, \text{J}\)
• C. \(9.24 \times 10^{-3} \, \text{J}\)
• D. \(1.848 \times 10^{-3} \, \text{J}\)

Hint:

For work done on bubbles:
• Account for both inner and outer surfaces when calculating surface area.
• Use the formula W = T ×∆A consistently with units.

Answer 1

The Correct Option is D

The problem at hand involves calculating the work done to increase the radius of a soap bubble, given the surface tension. To solve this, we need to understand the relationship between work done and surface tension for a bubble.

The formula to calculate the work done due to surface tension is given by:

W = \Delta A \cdot T

where:

• W is the work done,
• \Delta A is the change in surface area,
• T is the surface tension.

For a soap bubble, there are two surfaces (inner and outer), so the total change in surface area is twice the increase in the area of one sphere.

The surface area of a sphere is given by:

A = 4 \pi r^2

Therefore, the change in surface area when the radius changes from r_1 to r_2 is:

\Delta A = 2 (4 \pi r_2^2 - 4 \pi r_1^2)

\Delta A = 8 \pi (r_2^2 - r_1^2)

Substituting the given values, r_1 = 3.5 \, \text{cm} = 0.035 \, \text{m} and r_2 = 7 \, \text{cm} = 0.07 \, \text{m}, we get:

r_1^2 = (0.035)^2 = 0.001225 \, \text{m}^2

r_2^2 = (0.07)^2 = 0.0049 \, \text{m}^2

Thus,

\Delta A = 8 \pi (0.0049 - 0.001225)

\Delta A = 8 \pi \times 0.003675 \, \text{m}^2

\Delta A = 0.09228 \, \text{m}^2

The given surface tension is T = 2 \times 10^{-2} \, \text{N/m}.

Therefore, the work done is:

W = 0.09228 \times 2 \times 10^{-2} = 1.8456 \times 10^{-3} \, \text{J}

So, rounding to significant figures, the work done is approximately:

W \approx 1.848 \times 10^{-3} \, \text{J}

Therefore, the correct answer is \(1.848 \times 10^{-3} \, \text{J}\).