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Figure (14-Q1) shows a capillary tube of radius r dipped into water. If the atmospheric pressure is P0, the pressure at point A is(a) P0(b) P0 + 2S/r(c) P0 - 2S/r(d) P0 - 4S/r. |
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Answer» (c) Po -2S/r. Explanation: Let pressure at A = P. Net pressure =P₀-P. Net resultant pressure on the hemispherical depression =(P₀-P)*πr² along the axis of the tube. The surface tension along the circular edge of the depression =2πrS, along the axis of the tube. Equating we get, (P₀-P)*πr² = 2πrS → (P₀-P) = 2S/r →P = P₀ - 2S/r |
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