Question:
A soap bubble of radius $1$ cm has surface tension $0.03$ N/m. Excess pressure inside is:
A soap bubble of radius $1$ cm has surface tension $0.03$ N/m. Excess pressure inside is:
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Always remember the difference between a liquid drop ($\frac{2T}{R}$) and a soap bubble ($\frac{4T}{R}$). The "double surface" of the soap bubble doubles the excess pressure compared to a drop of the same size and surface tension.
Updated On: Jun 3, 2026
- $3$ Pa
- $6$ Pa
- $12$ Pa
- $24$ Pa
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The Correct Option is C
Solution and Explanation
Concept:
The excess pressure inside a soap bubble is due to surface tension. Unlike a liquid drop which has only one free surface, a soap bubble has two free surfaces (inner and outer). Therefore, the formula for excess pressure ($\Delta P$) is:
\[ \Delta P = \frac{4T}{R} \]
Where:
• $T$ is the surface tension of the soap solution.
• $R$ is the radius of the bubble.
Step 1: Identify and convert the given values.
From the question:
• Surface tension ($T$) = $0.03$ N/m
• Radius ($R$) = $1$ cm = $0.01$ m (converting to SI units)
Step 2: Apply the excess pressure formula.
Substitute the values into the equation: \[ \Delta P = \frac{4 \times 0.03}{0.01} \] \[ \Delta P = \frac{0.12}{0.01} \]
Step 3: Calculate the final result.
\[ \Delta P = 12 \text{ Pa} \]
• $T$ is the surface tension of the soap solution.
• $R$ is the radius of the bubble.
Step 1: Identify and convert the given values.
From the question:
• Surface tension ($T$) = $0.03$ N/m
• Radius ($R$) = $1$ cm = $0.01$ m (converting to SI units)
Step 2: Apply the excess pressure formula.
Substitute the values into the equation: \[ \Delta P = \frac{4 \times 0.03}{0.01} \] \[ \Delta P = \frac{0.12}{0.01} \]
Step 3: Calculate the final result.
\[ \Delta P = 12 \text{ Pa} \]
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