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A liquid of density rho flows along a horizontal pipe of uniform cross - section A with a velocity v through a righ angled bend as shown in fig. What force has to be exerted at the bend to hold the pipe in equilibrium ? |
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Answer» Solution :Change in momentum of mass `Delta m` of liquid as it passes through the bend `Delta p = P_(f)-P_(i)` `= Delta MV sqrt(2)` `F = (Delta P)/(Delta t)=(sqrt(2))v(Delta m)/(Delta t) , [as Delta m = RHO A Delta L]` `F = sqrt(2)v ((rho.A Delta L))/(Delta t), [as Delta L//Delta t=v]` `F = sqrt(2)rho A v^(2)` `therefore` So the force to be applied at the bend to hold it in position is `sqrt(2)rho Av^(2)` in the DIRECTION as shown in fig. (Otherwise pipe will move in opposite direction.). 
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