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Question
A capillary tube of radius 0.50 mm is dipped vertically in a pot of water. Find the difference between the pressure of the water in the tube 5.0 cm below the surface and the atmospheric pressure. Surface tension of water = 0.075 N m−1.
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Solution
Given:
Radius of capillary tube r = 0.5 mm = 5 × 10−4 m
Depth (where pressure is to be found) h = 5.0 cm = 5 × 10−2 m
Surface tension of water T = 0.075 N/m
Excess pressure at 5 cm before the surface:
P = ρhg = 1000 × (5 × 10−2) × 9.8 = 490 N/m2
Excess pressure at the surface is given by:
\[P_0 = \frac{2T}{r} = \frac{2 \times \left( 0 . 75 \right)}{\left( 5 \times {10}^{- 4} \right)}\]
\[ = 300 \text{ N/ m}^2\]
Difference in pressure: P0 − P
\[= 490 - 300 = 190 \text{ N/ m}^2\]
Hence, the required difference in pressure is 190 N/m2.
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