Question

The work done in increasing the radius of a soap bubble from \(R\) to \(2R\) is \(W\). The work done in further increasing its radius from \(2R\) to \(3R\) will be:

• A. \(\dfrac{5}{3}W\)
• B. \(\dfrac{4}{3}W\)
• C. \(\dfrac{7}{3}W\)
• D. \(W\)

Hint:

For a soap bubble: \[ E=8\pi TR^2 \] because a soap bubble has two surfaces. For a liquid drop: \[ E=4\pi TR^2 \] because it has only one surface.

Answer 1

The Correct Option is A

Concept: A soap bubble has two surfaces. Surface energy: \[ E=2T(4\pi R^2) \] \[ E=8\pi TR^2 \] where: \[ T=\text{surface tension} \] Work done equals increase in surface energy.

Step 1: Finding work done from \(R\) to \(2R\).
Initial energy: \[ E_1=8\pi TR^2 \] Final energy: \[ E_2 = 8\pi T(2R)^2 = 32\pi TR^2 \] Thus, \[ W=E_2-E_1 \] \[ W = 32\pi TR^2-8\pi TR^2 \] \[ W=24\pi TR^2 \]

Step 2: Finding work done from \(2R\) to \(3R\).
Energy at radius \(3R\): \[ E_3 = 8\pi T(3R)^2 = 72\pi TR^2 \] Required work: \[ W' = E_3-E_2 \] \[ W' = 72\pi TR^2-32\pi TR^2 \] \[ W' = 40\pi TR^2 \]

Step 3: Relating \(W'\) with \(W\).
Since: \[ W=24\pi TR^2 \] therefore, \[ \frac{W'}{W} = \frac{40}{24} = \frac{5}{3} \] Hence, \[ W'=\frac53 W \] Therefore, the correct answer is: \[ \boxed{\frac53 W} \]