Answer:
Given that
[tex]V=\dfrac{\Delta Pd^2}{32\mu L}[/tex]
LHS of above given equation have dimension [tex][M^oL^{1}T^{-1}][/tex].
Now find the dimension of RHS
Dimension of P = [tex][ML^{-1}T^{-2}][/tex].
Dimension of d= [tex][M^{0}L^{1}T^{0}][/tex].
Dimension of μ = [tex][ML^{-1}T^{-1}][/tex].
Dimension of L= [tex][M^{0}L^{1}T^{0}][/tex].
So
[tex]\dfrac{\Delta Pd^2}{32\mu L}=\dfrac{[ML^{-1}T^{-2}].[M^{0}L^{1}T^{0}]^2}{[ML^{-1}T^{-1}].[M^{0}L^{1}T^{0}]}[/tex]
[tex]\dfrac{\Delta Pd^2}{32\mu L}=[M^0L^{1}T^{-1}][/tex]
It means that both sides have same dimensions.
