Chapter 9, Problem 1ICA

To determine

Fill the appropriate dimensions for given quantity and complete the table.

Answer to Problem 1ICA

The completed table with appropriate dimension for given quantity is,

			Dimensions
	Quality	SI Units	M	L	T	$\Theta$	N	J	I
Example	Acoustic impedance	$\left(\text{Pa}\cdot \text{s}\right)/\text{m}$	1	$-2$	$-1$	0	0	0	0
(a)	Circuit resistance	$\text{V}/\text{A}$	$1$	$2$	$-3$	0	0	0	$-2$
(b)	Luminous efficiency	$\text{cd}/\text{W}$	$-1$	$-2$	3	0	0	1	0
(c)	Molar concentration	$\text{mol}/\text{L}$	0	$-3$	0	0	1	0	0
(d)	Thermal conductivity	$\text{cal}/\left(\text{cm}\cdot \text{s }°\text{C}\right)$	1	1	$-3$	$-1$	0	0	0

Explanation of Solution

Given data:

The SI unit of circuit resistance is $\text{V}/\text{A}$.

The SI unit of luminous efficacy is $\text{cd}/\text{W}$.

The SI unit of Molar concentration is $\text{mol}/\text{L}$.

The SI unit of thermal conductivity is $\text{cal}/\left(\text{cm}\cdot \text{s }\cdot °\text{C}\right)$.

Calculation:

Refer to the Table 9-1 in the textbook for Fundamental dimensions and base units.

Refer to the Table 8-1 in the textbook for Common derived units in the SI system to find the fundamental dimension of voltage.

$1\text{V}=1\frac{\text{kg}\cdot {\text{m}}^2}{{\text{s}}^3\cdot \text{A}}$

Substitute M for kg, L for $\text{m}$, I for A, and T for s to find dimension of voltage.

$\begin{array}{c}1\text{V}=1\frac{\text{M}\cdot {\text{L}}^2}{{\text{T}}^3\cdot \text{I}}\\ ={\text{ML}}^2{\text{T}}^{-3}{\text{I}}^{-1}\end{array}$

Therefore, the dimensions of the voltage is ${\text{ML}}^2{\text{T}}^{-3}{\text{I}}^{-1}$.

Find the fundamental dimension of resistance as follows.

$\text{Circuit resistance}=\text{V}/\text{A}$

Substitute ${\text{ML}}^2{\text{T}}^{-3}{\text{I}}^{-1}$ for V and I for A to find the dimension of resistance.

$\begin{array}{c}\text{Circuit resistance}=\frac{{\text{ML}}^2{\text{T}}^{-3}{\text{I}}^{-1}}{\text{I}}\\ ={\text{ML}}^2{\text{T}}^{-3}{\text{I}}^{-2}\end{array}$

Therefore, the dimensions of the circuit resistance is ${\text{ML}}^2{\text{T}}^{-3}{\text{I}}^{-2}$.

Find the fundamental dimension of luminous efficiency as follows.

Luminous efficacy is the measure of visible light produced by the source and is the ratio of luminous flux to power.

$\text{luminous efficacy}=\frac{\text{luminous flux output}}{\text{Electrical input power}}$

$\text{luminous efficacy}=\text{cd}/\text{W}$ (1)

Fundamental dimension of luminous flux output (cd) is J.

Find the fundamental dimension of power.

Refer to the Table 8-14 in the textbook for Summary of electrical properties.

$\text{W}=\text{VA}$

Substitute ${\text{ML}}^2{\text{T}}^{-3}{\text{I}}^{-1}$ for V and I for A to find the dimension of power.

$\begin{array}{c}\text{W}=\left({\text{ML}}^2{\text{T}}^{-3}{\text{I}}^{-1}\right)\left(\text{I}\right)\\ ={\text{ML}}^2{\text{T}}^{-3}\end{array}$

Substitute J for cd and ${\text{ML}}^2{\text{T}}^{-3}$ for W in equation (1) to obtain the dimension of power.

$\begin{array}{c}\text{luminous efficacy}=\frac{\text{J}}{{\text{ML}}^2{\text{T}}^{-3}}\\ ={\text{M}}^{-1}{\text{L}}^{-2}{\text{T}}^3\text{J}\end{array}$

Therefore, the dimensional formula of the luminous efficacy is ${\text{M}}^{-1}{\text{L}}^{-2}{\text{T}}^3\text{J}$.

Consider the conversion factor to convert Liter to ${\text{m}}^3$.

$1\text{L}=0.001{\text{m}}^{\text{3}}$

$\text{Molar concentration}=\frac{\text{mol}}{\text{L}}$ (2)

Substitute $0.001{\text{m}}^{\text{3}}$ for L in equation (2).

$\text{Molar concentration}=\frac{\text{1}\text{mol}}{0.001{\text{m}}^{\text{3}}}$

$\text{Molar concentration}={10}^3\frac{\text{mol}}{{\text{m}}^{\text{3}}}$

As the value ${10}^3$ is dimensionless, it is not considered to find the dimensional formula of molar concentration.

Substitute N for mol and ${\text{L}}^{\text{3}}$ for m in equation (2).

$\begin{array}{c}\text{Molar concentration}=\frac{\text{N}}{{\text{L}}^{\text{3}}}\\ ={\text{NL}}^{-3}\end{array}$

Therefore, the dimensional formula of the molar concentration is ${\text{NL}}^{-3}$.

Find the fundamental dimension of thermal conductivity.

$\text{Thermal conductivity}=\text{cal}/\left(\text{cm}\cdot \text{s }\cdot °\text{C}\right)$ (3)

Calories can be measured in units of Joules.

Refer to the Table 8-11 in the textbook for Dimension of energy to represent the dimensions of required parameters.

Substitute J for cal in equation (3) to find the thermal conductivity in terms of Joule.

$\text{thermal conductivity}=\text{J}/\left(\text{cm}\cdot \text{s}\cdot °\text{C}\right)$

Substitute ${\text{ML}}^2{\text{T}}^{-2}$ for J, L for cm, T for s and $\Theta$ for $°\text{C}$ to find the fundamental dimension of thermal conductivity.

$\begin{array}{c}\text{thermal conductivity}=\frac{{\text{ML}}^2{\text{T}}^{-2}}{\left(\text{L}\right)\left(\text{T}\right)\left(\Theta \right)}\\ ={\text{ML}}^{2-1}{\text{T}}^{-2-1}{\Theta }^{-1}\\ ={\text{ML}}^1{\text{T}}^{-3}{\Theta }^{-1}\end{array}$

Obtained dimensions for circuit resistance, luminous efficacy and thermal conductivity is tabulated in Table 1.

Table 1

			Dimensions
	Quality	SI Units	M	L	T	$\Theta$	N	J	I
Example	Acoustic impedance	$\left(\text{Pa}\cdot \text{s}\right)/\text{m}$	1	$-2$	$-1$	0	0	0	0
(a)	Circuit resistance	$\text{V}/\text{A}$	$1$	$2$	$-3$	0	0	0	$-2$
(b)	Luminous efficiency	$\text{cd}/\text{W}$	$-1$	$-2$	3	0	0	1	0
(c)	Molar concentration	$\text{mol}/\text{L}$	0	$-3$	0	0	1	0	0
(d)	Thermal conductivity	$\text{cal}/\left(\text{cm}\cdot \text{s }°\text{C}\right)$	1	1	$-3$	$-1$	0	0	0

Conclusion:

Thus, the completed table of appropriate dimension for given quantity is shown in Table 1.