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If dimensions of critical velocity $v_{c}$ of a liquid flowing through a tube are expressed as $[\eta^{x}\rho^{y}r^{x}]$, where $\eta$, $\rho$ and 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:
Explanation
Applying dimensional method: $v_{c} = \eta^{x}\rho^{y}r^{z}$ $[M^{0}LT^{-1}] = [ML^{-1}T^{-1}]^{x}[ML^{-3}T^{0}]^{y}[M^{0}LT^{0}]^{z}$ Equating powers both sides $x + y = 0$; $-x = -1 \therefore x = 1$. $1 + y = 0 \therefore y = -1$. $-x - 3y + z = 1$. $-1 - 3(-1) + z = 1$. $-1 + 3 + z = 1 \therefore z = -1$