A sealed vertical cylinder of radius R and height h = 0.550 m is initially fille
ID: 2014285 • Letter: A
Question
A sealed vertical cylinder of radius R and height h = 0.550 m is initially filled halfway with water, and the upper half is filled with air. The air is initially at standard atmospheric pressure, p0 = 1.01·10^5 Pa. A small valve at the bottom of the cylinder is opened, and water flows out of the cylinder until the reduced pressure of the air in the upper part of the cylinder prevents any further water from escaping. By what distance is the depth of the water lowered? (Assume that the temperature of water and air do not change and that no air leaks into the cylinder.)Explanation / Answer
Initial volume of air V0 = A * height of air column = A * 0.550 / 2 = 0.275 * A m3 A being area of cross section Let the final pressure and volume be P and V. Water will not leak out if the force (due to reduced paressure ) is same as the weight of water. Let the column of water has lowered by h at this moment. then final volume V = (0.275 + h) * A P * V = P0 * V0 P * (0.275 + h) * A = 0.275 * A * 1.01 * 105 P = 2.78 * 104 / (0.275 + h) Force exerted F = (P0 - P) * A = A * {1.01 * 105 - 2.78 * 104 / (0.275 + h)} Weight of water column m * g = A * remaining height of water column * density * g = A * (0.275 - h) * 1000 * 9.8 = 9800 * A * (0.275 - h) Since F = m * g A * {1.01 * 105 - 2.78 * 104 / (0.275 + h)} = 9800 * A * (0.275 - h) 1.01 * 105 * (0.275 + h) - 2.78 * 104 = 9800 * (0.2752 - h2) 29.425 + 101 * h - 27.8 = 0.741 - 9.8 * h2 9.8 * h2 + 101 * h - 0.884 = 0 solving h = 8.74 * 10-3 m = 8.74 mmRelated Questions
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