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

## Answers : (2)

Samyak Jain
325 Points
							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 .

14 days ago
Samyak Jain
325 Points
							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 + $\dpi{100} \rho$gh1 .$\dpi{100} \rho$ = density of fluid in pipe and h1 = height of fluid between E and C = l1 sin$\dpi{100} \theta$1 .PC = Patm + $\dpi{100} \rho$gl1 sin$\dpi{100} \theta$1 . Pressure at C = Pressure at B + $\dpi{100} \rho$gh2 ,  h2 = l2 sin$\dpi{100} \theta$2 .PC = PB + $\dpi{100} \rho$gl2 sin$\dpi{100} \theta$2   $\dpi{100} \Rightarrow$   PB = PC – $\dpi{100} \rho$gl2 sin$\dpi{100} \theta$2  =  Patm + $\dpi{100} \rho$gl1 sin$\dpi{100} \theta$1 – $\dpi{100} \rho$gl2 sin$\dpi{100} \theta$2  Similarly, PA = PB + $\dpi{100} \rho$gl2 sin$\dpi{100} \theta$3  =  Patm + $\dpi{100} \rho$gl1 sin$\dpi{100} \theta$1 – $\dpi{100} \rho$gl2 sin$\dpi{100} \theta$2 + $\dpi{100} \rho$gl3 sin$\dpi{100} \theta$3 PA = Patm + $\dpi{100} \rho$g(l1 sin$\dpi{100} \theta$1 – l2 sin$\dpi{100} \theta$2 + l3 sin$\dpi{100} \theta$3) = pressure at A .

14 days ago
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