Question

If dimensions of critical velocity vc of a liquid flowing through a tube are expressed as [ n^x p^y r^z], where n,p and r are the coefficients of viscosity of a liquid, density of liquid and radius of the tube respectively, then values of x,y and z are given by:
a) -1, -1, 1 b) -1,-1,-1 c) 1,1,1 d) 1,-1,-1

Answer 1

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 $\bf{\propto}$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

Answer 2

Answer:

See attachment

please