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Question:

If dimensions of critical velocity vc of a liquid flowing through a tube are expressed as [ηxρyrz], where η, ρ and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then values of x, y and z are given by:

1, -1, -1

-1, -1, -1

1,1,1

-1, -1, 1

Solution:

[v]=[LT-1] η=FA(dv/dx) ⇒[η]=[MLT-2][L2][T-1]=[M1L-1T-1] [ρ]=[ML-3] ⇒[LT-1]=[M1L-1T-1]x[ML-3]y[L]z Equating the exponents of M, L and T on both LHS and RHS ⇒M0=M(x+y) ⇒y=-x For T,-1=-x x = 1 y=-x=-1 For L 1=-x-3y+z ⇒-1+3+z 1=2+z ⇒z=-1 ⇒z=-1