Question:medium
If the excess pressure inside a spherical mercury drop is \(1240\,Nm^{-2}\), then the radius of the mercury drop is:
If the excess pressure inside a spherical mercury drop is \(1240\,Nm^{-2}\), then the radius of the mercury drop is:
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For a liquid drop:
\[
\Delta P=\frac{2T}{R}
\]
whereas for a soap bubble:
\[
\Delta P=\frac{4T}{R}.
\]
Updated On: Jun 18, 2026
- \(0.75\,mm\)
- \(1.5\,mm\)
- \(0.375\,mm\)
- \(2.25\,mm\)
Show Solution
The Correct Option is A
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