Drag force fD=f(D,V,P,$\mu)$

fD=$k-D^{a}.V^{b}.P^{c}.\mu^{d}$

substituting the dimensions on both sides

$MLT^{2}=k(L)^{a}.(LT^{-1})^{b}.(ML^{-2})^{c}.(ML^{-1}T^{-1})^{d}$

on equating the dimensions on both sides

Power of M:1=c+d

Power of L:=1=a+b-3c-d

Power of T=-2=-b-d

We have only three equation whereas we have unknowns therefore the values of the power abcd cannot be determined out of these any three variables can be expressed in terms of fourth chosen variable which has the important rate to play.Tn the present problem the drag force depends mainly on the viscosity of the fluid therefore we can express the powers a,b,, and c in terms of d

from Equation (ii):c=1-d

from Equation (iv) b=2-d

from Equation (iii) 1=a+b=3c-d

1=a+(2-d)-3(1-d)-d=a-1 & d

a=2-d

on substituting of values a,b,c in equation i, we get

$F_{D}=k.D^{2-d}.v^{2-d}.p^{1-d}.\mu^{d}$

FD=$k.D^{2}.v^{2}.p.(D^{-d}.v^{-d}.p^{-d}.\mu^{d})$

FD=$kD^{2}.v^{2}.p.(\frac{D.v.p}{\mu})^{-d}=k.D^{2}.v^{2}.p.(\frac{\mu}{D.v.p})^{-d}$

FD=$k.D^{2}.v^{2}.p\Phi(\frac{\mu}{D.v.p})$