If dimensions of critical velocity $v_c$ of a liquid flowing through a tube are expressed as $[\eta^x \rho^y r^z]$, where $\eta$, $\rho$ and $r$ are the coefficient of viscosity of the liquid, the density of liquid and radius of the tube respectively, then the values of $x$, $y$ and $z$ are given by

Let the dimensional formula of critical velocity $v_c$ be related to coefficient of viscosity $\eta$, density $\rho$, and radius $r$ as: $[v_c] = [\eta]^x [\rho]^y [r]^z$ Substituting the dimensions of each quantity: Dimension of velocity, $v_c = [L T^{-1}]$ Dimension of coefficient of viscosity, $\eta = [M L^{-1} T^{-1}]$ Dimension of density, $\rho = [M L^{-3}]$ Dimension of radius, $r = [L]$

$[M^0 L^1 T^{-1}] = [M L^{-1} T^{-1}]^x [M L^{-3}]^y [L]^z$ $[M^0 L^1 T^{-1}] = [M^{x+y} L^{-x-3y+z} T^{-x}]$

Equating the powers of $M$, $L$, and $T$ on both sides, we get: For $M$: $x + y = 0 \implies y = -x$ For $T$: $-x = -1 \implies x = 1$ Since $x = 1$, we have $y = -1$. For $L$: $-x - 3y + z = 1 \implies -(1) - 3(-1) + z = 1 \implies -1 + 3 + z = 1 \implies 2 + z = 1 \implies z = -1$

Therefore, the values of $x, y$ and $z$ are $1, -1, -1$ respectively.