Soap Bubble Pressure vs. Time

Question: A small soap bubble is filled with air from a mechanical pump which increases its volume at a constant rate directly proportional to time \(t\). Then the gauge pressure inside the bubble is proportional to:

(1) \(t^{1/3}\)

(2) \(t^{2/3}\)

(3) \(t^{-2/3}\)

(4) \(t^{-1/3}\)

Detailed Step-by-Step Explanation

Step 1: Volume Increase
The problem states that the volume \(V\) of the soap bubble increases at a constant rate proportional to time: \[ V \propto t. \] We can write this as: \[ V = k\,t, \] where \( k \) is a constant.

Step 2: Relate Volume and Radius
For a spherical bubble, the volume is related to the radius \( R \) by: \[ V = \frac{4}{3}\pi R^3. \] Equating the two expressions for \( V \): \[ \frac{4}{3}\pi R^3 = k\,t. \] Solving for \( R \): \[ R^3 = \frac{3k\,t}{4\pi} \quad \Rightarrow \quad R \propto t^{\frac{1}{3}}. \]

Step 3: Gauge Pressure Inside the Bubble
For a soap bubble, the gauge (excess) pressure \( \Delta P \) is given by: \[ \Delta P \propto \frac{1}{R}. \] Since we found that \( R \propto t^{\frac{1}{3}} \), it follows that: \[ \Delta P \propto \frac{1}{t^{\frac{1}{3}}} = t^{-\frac{1}{3}}. \]

Final Answer: The gauge pressure inside the bubble is proportional to \( t^{-\frac{1}{3}} \), which corresponds to option (4).

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