Question

The work to be done to blow a soap bubble of radius $3 \times 10^{-3}$ m is nearly
(Surface tension of soap solution = $20 \times 10^{-3}$ N m$^{-1}$)

• A. $4.5 \times 10^{-4}$ J
• B. $4.5 \times 10^{-5}$ J
• C. $4.5 \times 10^{-6}$ J
• D. $4.5 \times 10^{-7}$ J

Hint:

When calculating work done to form a soap bubble, remember to account for both the inner and outer surfaces, doubling the surface area.

Answer 1

The Correct Option is C

The work done to blow a soap bubble is the energy required to increase its surface area. A soap bubble has two surfaces (inner and outer), so the total surface area is \( 2 \times 4\pi r^2 \). Given \( r = 3 \times 10^{-3} \) m, surface area = \( 2 \times 4\pi (3 \times 10^{-3})^2 = 8\pi \times 9 \times 10^{-6} = 72\pi \times 10^{-6} \) m\(^2\).

Surface tension \( \sigma = 20 \times 10^{-3} \) N/m. Work done = surface tension \(\times\) surface area = \( (20 \times 10^{-3}) \times (72\pi \times 10^{-6}) = 20 \times 72\pi \times 10^{-9} = 1440\pi \times 10^{-9} \approx 4523 \times 10^{-9} = 4.523 \times 10^{-6} \) J, which is closest to \( 4.5 \times 10^{-6} \) J.

Thus, the work done is approximately \( 4.5 \times 10^{-6} \) J, suggesting option 3 is correct, but please confirm.