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Assuming that the critical velocity v_{c} of a viscous liquid flowing through a capillary tube depends only upon the radius r of the tube, density ρ and the coefficient of viscosity η of the liquid, find the expression for critical velocity. |
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Answer» udent,◆ Answer -V = k.η/(r.ρ)◆ Explaination -First LET's write down KNOWN dimensions.v = [L1T-1]r = [L1]ρ = [L-3M1]η = [L-1M1T-1]Suppose critical velocity of viscous fluid is GIVEN by -v = k.r^x.ρ^y.η^zIn dimensional form,[v] = [r]^x.[ρ]^y. [η]^z[L1T-1] = [L1]^x.[L-3M1]^y.[L-1M1T-1]^z[L1T-1] = [L^(x-3y-z).M^(y+z).T^(-z)]Comparing two sides,x-3y-z = 1y+z = 0-z = -1Solving we get,x = -1y = -1z = 1Substitute this in initial formula -v = k.r^-1.ρ^-1.η^1v = k.η/(r.ρ)Hope this helps you... |
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