Question:
Due to surface tension, the excess pressure inside a smaller drop is \(9\) units. If \(27\) smaller drops combine then excess pressure inside the bigger drop is
Due to surface tension, the excess pressure inside a smaller drop is \(9\) units. If \(27\) smaller drops combine then excess pressure inside the bigger drop is
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Volume conservation gives R = 3r; excess pressure is inversely proportional to radius.
Updated On: Oct 1, 2026
- \(3\) units
- \(9\) units
- \(18\) units
- \(6\) units
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
The excess pressure inside a drop is \(P=\dfrac{2T}{r}\), so it is inversely proportional to the radius.
Step 2: Find the radius of the big drop:
\(27\times\dfrac43\pi r^3=\dfrac43\pi R^3\), so \(R=3r\).
Step 3: Find the new pressure:
\(P_{big}=\dfrac{2T}{R}=\dfrac{2T}{3r}=\dfrac13P_{small}=\dfrac93=3\) units. Option A.
Step 4: Why the other options are wrong.
9 units ignores the change of radius. 18 units and 6 units come from multiplying instead of dividing, or from using 3 as the factor on the wrong side.
Final Answer:
The excess pressure is 3 units. \[ \boxed{\text{(A) }3\ \text{units}} \]
The excess pressure inside a drop is \(P=\dfrac{2T}{r}\), so it is inversely proportional to the radius.
Step 2: Find the radius of the big drop:
\(27\times\dfrac43\pi r^3=\dfrac43\pi R^3\), so \(R=3r\).
Step 3: Find the new pressure:
\(P_{big}=\dfrac{2T}{R}=\dfrac{2T}{3r}=\dfrac13P_{small}=\dfrac93=3\) units. Option A.
Step 4: Why the other options are wrong.
9 units ignores the change of radius. 18 units and 6 units come from multiplying instead of dividing, or from using 3 as the factor on the wrong side.
Final Answer:
The excess pressure is 3 units. \[ \boxed{\text{(A) }3\ \text{units}} \]
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