Chapter 8, Problem 1P

a)

To determine

The given ratio is dimensionless or not.

Explanation of Solution

Given:

$\frac{\mu V}{L^3\rho }$

Calculation:

If the ratio of two or more quantities is cancelled to give the final result of 1, it is known as dimensionless ratio.

Refer Table $\text{8-1}$, write the required dimensions for the quantities or variables as in below Table.

Quantity	Symbol	$M\text{-}L\text{-}T$
Density	$\rho$	$M{L}^{-3}$
velocity	$V$	$L{T}^{-1}$
Pressure	$p$	$M{L}^{-1}{T}^{-2}$
Surface tension	$\sigma$	$M{T}^{-2}$
gravity	$g$	$L{T}^{-2}$
Dynamic viscosity	$\mu$	$M{L}^{-1}{T}^{-1}$

Write the ratio of quantities or variables.

  $\begin{array}{c}\frac{\mu V}{L^3\rho }=\frac{\left(\frac{M}{LT}\right)\left(\frac{L}{T}\right)}{L^3\left(\frac{M}{L^3}\right)}\\ =\frac{1}{T^2}\end{array}$

Thus, the given ratio is $\underset{\_}{\text{not}}$ a dimensionless ratio.

b)

To determine

The given ratio is dimensionless or not.

Explanation of Solution

Given:

  $\frac{\mu }{\rho VL}$

Calculation:

Write the ratio of quantities or variables.

  $\begin{array}{c}\frac{\mu }{\rho VL}=\frac{\frac{M}{LT}}{\left(\frac{M}{L^3}\right)\left(\frac{L}{T}\right)L}\\ =1\end{array}$

Thus, the given ratio is a dimensionless ratio.

c)

To determine

The given ratio is dimensionless or not.

Explanation of Solution

Given:

  $\frac{\sqrt{gL}}{V}$

Calculation:

Write the ratio of quantities.

  $\begin{array}{c}\frac{\sqrt{gL}}{V}=\frac{{\left[\left(\frac{L}{T^2}\right)(L)\right]}^{1/2}}{\frac{L}{T}}\\ =1\end{array}$

Thus, the given ratio is a dimensionless ratio.

d)

To determine

The given ratio is dimensionless or not.

Explanation of Solution

Given:

  $\frac{\sigma }{\rho L}$

Calculation:

Write the ratio of quantities.

  $\begin{array}{c}\frac{\sigma }{\rho L}=\frac{\left(\frac{M}{T^2}\right)}{\left(\frac{M}{L^3}\right)(L)}\\ =\frac{L^2}{T^2}\end{array}$

Thus, the given ratio is not a dimensionless ratio.