Question

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

Answer 1

For this given problem, the dimensional unit of the velocity of a liquid is equated with the density of the liquid, the viscosity of the liquid, and the radius of the tube. By using the known dimensional units then the values of $x$, $y$, and $z$ can be calculated.

Complete step by step solution:
The dimension of critical velocity of the liquid is expressed as,
${v_c} = K{\eta ^x}{\rho ^y}{r^z}\,.............\left( 1 \right)$
Where, ${v_c}$ is the critical velocity of the liquid, $\eta$ is the coefficient of the liquid, $\rho$ is the density of the liquid, $r$ is the radius of the tube and $K$ is constant.
By the method of dimensional analysis,
1. Dimension for velocity
${v_c} = m{s^{ - 1}}$ (unit)
$\Rightarrow {v_c} = \left[ {{M^0}L{T^{ - 1}}} \right]$
$\Rightarrow {v_c} = \left[ {L{T^{ - 1}}} \right]$
2. Dimension for coefficient of viscosity
$\eta = \dfrac{F}{{A\left( {\dfrac{{dv}}{{dx}}} \right)}}$
Where, $F$ is the force, $A$ is the area and $\left( {\dfrac{{dv}}{{dx}}} \right)$ is the velocity gradient.
$F = m \times a$, Where, $m$ is the mass and $a$ is the acceleration.
Dimension for force,
$F = kgm{s^{ - 2}}$ (unit)
$\Rightarrow F = \left[ M \right] \times \left[ {L{T^{ - 2}}} \right]$
$\Rightarrow F = \left[ {ML{T^{ - 2}}} \right]$
Dimension for area,
$A = {m^2}$ (unit)
$\Rightarrow A = \left[ {{L^2}} \right]$
Dimension for velocity gradient,
$\dfrac{{dv}}{{dx}} = {s^{ - 1}}$
$\Rightarrow \dfrac{{dv}}{{dx}} = \left[ {{T^{ - 1}}} \right]$
Then,
$\eta = \dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{L^2}{T^{ - 1}}} \right]}} \\\ \Rightarrow \eta = \left[ {M{L^{ - 1}}{T^{ - 1}}} \right] \\\$
3. Dimension for density
$\rho = kg.{m^{ - 3}}$ (unit)
$\Rightarrow \rho = \left[ {M{L^{ - 3}}} \right]$
4. Dimension for radius
$r = m$ (unit)
$\Rightarrow r = \left[ L \right]$
Substituting the dimensions for each term in equation (1)
${M^0}L{T^{ - 1}} = {\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]^x}{\left[ {M{L^{ - 3}}} \right]^y}{\left[ L \right]^z}$
Now equating the power of $M$,
$0 = x + y\,.............\left( 2 \right)$
Now equating the of $L$,
$\Rightarrow 1 = - x - 3y + z\,.............\left( 3 \right)$
Now equating the power of $T$,
$\Rightarrow - 1 = - x$
By cancelling the negative sign on both sides, so the value of $x$ is
$x = 1\,............\left( 4 \right)$
Substituting the equation (4) in equation (2),
$\Rightarrow 0 = 1 + y \\\ \Rightarrow y = - 1 \\\$
So, the value of $y = - 1$
Now substitute the values of $x$ and $y$ in equation (3),
$\Rightarrow 1 = - \left( 1 \right) - 3\left( { - 1} \right) + z \\\ \Rightarrow 1 = - 1 + 3 + z \\\$
By arranging the above equations,
$\Rightarrow 1 + 1 - 3 = z \\\ \Rightarrow - 1 = z \\\$
$\therefore z = - 1$

Thus, the value of $x$, $y$ and $z$ are $1$, $- 1$ and $- 1$ respectively.

Note:
In the velocity gradient step, normally the unit of velocity is $m{s^{ - 1}}$ but in velocity gradient the mass value gets cancelled because the mass value is in the denominator. So, the unit for velocity gradient is not $m{s^{ - 1}}$, the unit is ${s^{ - 1}}$ only. Also, while solving the force unit, care should be taken. It has a unit Newton $\left( N \right)$ also expressed as $kgm{s^{ - 1}}$.