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

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$- 1, - 1, 1$

$- 1, - 1, - 1$

$1, 1, 1$

$1, - 1, - 1$

answer is D.

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Detailed Solution

$\begin{array}{l} Applying dimensional method: \\ v_{c} = η^{x} ρ^{y} r^{Z} \\ [M^{0} {LT}^{- 1}] = {[{ML}^{- 1} T^{- 1}]}^{x} {[{ML}^{- 3} T^{0}]}^{y} {[M^{0} {LT}^{0}]}^{2} \\ Equating powers both sides \\ x + y = 0; - x = - 1 ∴ x = 1 \\ 1 + y = 0 ∴ y = - 1 \\ -x-3 y+z=1 \\ - 1 - 3 (- 1) + z = 1 \\ -1+3+z=1 \\ ∴ Z = - 1 \end{array}$