Question

If dimensions of critical velocity (Vc) of liquid flowing through a tube is expressed as (Nx Py Rz) where N,P and R the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively then the value of the x,y,z is given by

Answer 1

Dear student, Dimension of coefficient of viscosity , η = [MLT⁻²]/[L][LT⁻¹] = [M¹L⁻¹T⁻¹] [Because viscous force = 6πηrv ] dimension of density of liquid , ρ = [ML⁻³] dimension of radius of tube , r = [L] Now, terminal velocity η^x ρ^y r^z [LT⁻¹] = [M⁻¹L⁻¹T⁻¹]^x [ML⁻³]^y [L]^z [LT⁻¹] = [M]^(x + y) [L]^(-x-3y+z) [T]^(-x) compare both sides, x + y = 0 ⇒x = -y -x - 3y + z = 1 ⇒z = -1 -x = -1 ⇒x = 1 And y = -1 Hence, x = 1 , y = -1 and z = -1 Regards