Solution: (2)
$
\mathrm{V}_{\mathrm{c}}=\eta^{\mathrm{x}} \rho^{\mathrm{y}} \mathrm{r}^{\mathrm{z}}
$
Critical velocity is given by $V_{c}=\frac{R \eta}{2 \rho r}$
So, $x=1$
$
\begin{aligned}
y=-1 \\
z=-1
\end{aligned}
$
If dimensions of critical velocity $v_{c}$ of a liquid flowing through a tube are expressed as $\left[\eta^{x} \rho^{y} r^{2}\right]$, 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: (1) $\quad 1,1,1$ (2) $1,-1,-1$ (3) $-1,-1,1$ (4) $\quad-1,-1,-1$
If dimensions of critical velocity $v_{c}$ of a liquid flowing through a tube are expressed as $\left[\eta^{x} \rho^{y} r^{2}\right]$, 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: (1) $\quad 1,1,1$ (2) $1,-1,-1$ (3) $-1,-1,1$ (4) $\quad-1,-1,-1$