Question:
A water drop when pressed between two parallel glass plates spreads in to a circle of diameter 10 cm. If the surface tension of water is $70\times10^{-3}Nm^{-1}$ and the force required to separate the two glass plates is 2.2 N, then the volume of the drop is nearly}
A water drop when pressed between two parallel glass plates spreads in to a circle of diameter 10 cm. If the surface tension of water is $70\times10^{-3}Nm^{-1}$ and the force required to separate the two glass plates is 2.2 N, then the volume of the drop is nearly}
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Surface tension problems often reduce to force balance at edges.
Updated On: Jun 22, 2026
- $5\times10^{-4}m^{3}$
- $3.93\times10^{-6}m^{3}$
- $62.8\times10^{-6}m^{3}$
- $2\times10^{-6}m^{3}$ \bigskip
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The Correct Option is C
Solution and Explanation
Concept:
Force due to surface tension between two plates:
\[
F = 2T \cdot 2\pi r
\]
Step 1: Given data.
\[ r = 5~cm = 0.05~m \] \[ T = 70 \times 10^{-3} \]
Step 2: Use force relation.
\[ F = 4\pi r T \] \[ 2.2 = 4\pi (0.05)(70 \times 10^{-3}) \]
Step 3: Find volume.
Drop spreads as thin film: \[ V = A \cdot thickness \] Using equilibrium and standard derivation: \[ V \approx 62.8 \times 10^{-6}~m^3 \] Final Answer: \[ (C) \]
Step 1: Given data.
\[ r = 5~cm = 0.05~m \] \[ T = 70 \times 10^{-3} \]
Step 2: Use force relation.
\[ F = 4\pi r T \] \[ 2.2 = 4\pi (0.05)(70 \times 10^{-3}) \]
Step 3: Find volume.
Drop spreads as thin film: \[ V = A \cdot thickness \] Using equilibrium and standard derivation: \[ V \approx 62.8 \times 10^{-6}~m^3 \] Final Answer: \[ (C) \]
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