Question:medium

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:

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For a soap bubble: \[ U = 8\pi T R^2 \] because a soap bubble has two liquid surfaces.
Updated On: May 29, 2026
  • $\dfrac{5}{3}W$
  • $\dfrac{4}{3}W$
  • $\dfrac{7}{3}W$
  • $W$
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
Work done in expanding a bubble is equivalent to the change in its surface potential energy.
Surface energy is defined as the product of Surface Tension (\(T\)) and the total surface area of the film.
A soap bubble is unique because it consists of a thin film of liquid with air on both the inside and the outside.
Consequently, it has **two** free surfaces.
Total Surface Area of a soap bubble \(A = 2 \times (4\pi r^2) = 8\pi r^2\).
Key Formula or Approach:
Work Done \(W = T \times \Delta A = T \times 8\pi(r_{final}^2 - r_{initial}^2)\).
Since \(T\) and \(8\pi\) are constant for a given bubble, we can say \(W \propto (r_{final}^2 - r_{initial}^2)\).
Step 2: Detailed Explanation:
1. First Expansion (from \(R\) to \(2R\)):
Work done is given as \(W\).
\[ W = 8\pi T [(2R)^2 - R^2] \]
\[ W = 8\pi T [4R^2 - R^2] \]
\[ W = 8\pi T (3R^2) \]
This gives us a base value: \(8\pi T R^2 = \frac{W}{3}\).

2. Second Expansion (from \(2R\) to \(3R\)):
Let the required work be \(W'\).
\[ W' = 8\pi T [(3R)^2 - (2R)^2] \]
\[ W' = 8\pi T [9R^2 - 4R^2] \]
\[ W' = 8\pi T (5R^2) \]

3. Comparing \(W'\) and \(W\):
We substitute the value of \(8\pi T R^2\) from our first step:
\[ W' = 5 \times (8\pi T R^2) = 5 \times \frac{W}{3} = \frac{5}{3} W \]

The work increases because even though the radius increases by the same increment \(R\), the total area added is much larger when starting from a larger initial radius.
Step 3: Final Answer:
The work done for the further increase is \(\frac{5}{3} W\).
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