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Question
Two large glass plates are placed vertically and parallel to each other inside a tank of water with separation between the plates equal to 1 mm. Find the rise of water in the space between the plates. Surface tension of water = 0.075 Nm−1.
Solution
Given:
Surface tension of water T = 0.075 N/m
Separation between the glass plates d = 1 mm = 10−3 m
Density of water ρ = 103 kg/m3
Applying law of conservation of energy:
T (2L) = [1 × (10−3) × h] ρg
\[\Rightarrow \text{ h} = \frac{2 \times \left( 0 . 075 \right)}{{10}^{- 3} \times {10}^3 \times 10}\]
\[ = 0 . 015 \text{ m = 1 . 5 cm }\]
Therefore, the rise of water in the space between the plates is 1.5 cm.
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