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        Find pressure at point A in the given figure.........
one year ago

							Let the adjacent point of sides l2 and l3 be B and that of sides l1 and l2 be D and the free surface of liquid at open end of pipe be E.Pressure at E = atmospheric pressure = Patm.Since fluid is static, pressure difference is due to height of fluid.Pressure at C = Pressure at E + \rhogh1,\rho = density of fluid in pipe and h1 = height of fluid between E and C = l1 sin\theta1.i.e. PC = Patm + \rhogl1 sin\theta1.Pressure at C = Pressure at B + \rhogh2 ,  h2 = l2 sin\theta2.i.e. PC = PB + \rhogl2 sin\theta2   \Rightarrow  PB = PC – \rhogl2 sin\theta2 = Patm + \rhogl1 sin\theta1 – \rhogl2 sin\theta2Similarly, PA = PB + \rhogl3 sin\theta3 = Patm + \rhogl1 sin\theta1 – \rhogl2 sin\theta2 + \rhogl3 sin\theta3PA = \rhog(l1 sin\theta1 – l2 sin\theta2 + l3 sin\theta3) = pressure at A .

11 months ago
							Let the adjacent point of sides l2 and l3 be B and that of sides l1 and l2 be C and the free surface of fluidat open end of pipe be D. Pressure at D = atmospheric pressure = Patm.Since fluid is static, pessure difference is due to height of fluid. Pressure at C = Pressure at D + gh1 . = density of fluid in pipe and h1 = height of fluid between E and C = l1 sin1 .PC = Patm + gl1 sin1 . Pressure at C = Pressure at B + gh2 ,  h2 = l2 sin2 .PC = PB + gl2 sin2      PB = PC – gl2 sin2  =  Patm + gl1 sin1 – gl2 sin2  Similarly, PA = PB + gl2 sin3  =  Patm + gl1 sin1 – gl2 sin2 + gl3 sin3 PA = Patm + g(l1 sin1 – l2 sin2 + l3 sin3) = pressure at A .

11 months ago
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