Question

A physical quantity of the dimension of length that can be formed out of c, G and $\dfrac{{{e^2}}}{{4\pi { \in _0}}}$ is
[c is the velocity of light, G is universal gravitational constant and e is charge]

Answer 1

Hint: A physical quantity of desired dimensions can be constructed from given quantities by using their dimensional formulas. The dimensional formulae of given quantities can be used to obtain relation between interdependent quantities.

Complete Step-by-Step solution:
Dimensional formula: A dimensional formula of a physical quantity is an expression which describes the dependence of that quantity on the fundamental quantities.
All physical quantities can be expressed in terms of certain fundamental quantities. Following table contains the fundamental quantities and their units and dimensional notation respectively.

No.	Quantities	unit	Dimensional formula
1.	Length	metre (m)	$\left[{{M^0}{L^1}{T^0}}\right]$
2.	Mass	kilogram (g)	$\left[{{M^1}{L^0}{T^0}}\right]$
3.	Time	second (s)	$\left[{{M^0}{L^0}{T^1}}\right]$
4.	Electric current	ampere (A)	$\left[{{M^0}{L^0}{T^0}{A^1}}\right]$
5.	Temperature	kelvin [K]	$\left[{{M^0}{L^0}{T^0}{K^1}}\right]$
6.	Amount of substance	mole [mol]	$\left[{{M^0}{L^0}{T^0}mo{l^1}}\right]$
7.	Luminous intensity	candela [cd]	$\left[{{M^0}{L^0}{T^0}C{d^1}}\right]$

We are given three quantities: c = velocity of light, G=universal gravitational constant and
$\dfrac{{{e^2}}}{{4\pi { \in _0}}}$
 Their dimensional formulae are following calculations from basic laws relating them with fundamental quantities given in the table. The dimensional formula for
$\dfrac{{{e^2}}}{{4\pi { \in _0}}}$
can be calculated as follows:
$\begin{gathered}
  \dfrac{{{e^2}}}{{4\pi { \in _0}}} = \dfrac{{\left[ {AT} \right]\left[ {AT} \right]}}{{\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]}} = \dfrac{{\left[ {{A^2}{T^2}} \right]}}{{\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]}} = \left[ {{M^1}{L^3}{T^{2 - 4}}{A^{2 - 2}}} \right] \\
   \Rightarrow \dfrac{{{e^2}}}{{4\pi { \in _0}}} = \left[ {{M^1}{L^3}{T^{ - 2}}} \right] = B(let) \\
\end{gathered} $
Similarly, c and G have following dimensional formulae.
$\begin{gathered}
  c = \left[ {{M^0}{L^1}{T^{ - 1}}} \right] \\
  G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right] \\
\end{gathered} $
Now, we need a dimensional formula of length [L] from these three quantities which can be obtained as follows in terms of the three quantities.
$[L] = {c^x}{G^y}{B^z}$
Where we need to find numerical values of x, y and z. Using the respective dimensional formulae, we get
$\left[ L \right] = {\left[ {{M^0}{L^1}{T^{ - 1}}} \right]^x}{\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]^y}{\left[ {{M^1}{L^3}{T^{ - 2}}} \right]^z}$
Simplifying this expression, we get
$\left[ {{M^0}{L^1}{T^0}} \right] = \left[ {{M^{ - y + z}}{L^{x + 3y + 3z}}{T^{ - x - 2y - 2z}}} \right]$
Equating the powers on both sides, we get the following equations.
$\begin{gathered}
   - y + z = 0 \Rightarrow y = z{\text{ }}...{\text{(i)}} \\
  x + 3y + 3z = 1{\text{ }}...{\text{(ii)}} \\
   - x - 2y - 2z = 0{\text{ or }}x + 4y = 0{\text{ }}...{\text{(iii)}} \\
\end{gathered} $
Solving these equations simultaneously, we get
$x = - 2,y = \dfrac{1}{2}{\text{ and }}z = \dfrac{1}{2}$
Putting the values of x, y and z, we get the required quantity which is:
${c^{ - 2}}{G^{\dfrac{1}{2}}}{\left( {\dfrac{{{e^2}}}{{4\pi { \in _0}}}} \right)^{\dfrac{1}{2}}} = \dfrac{1}{{{c^2}}}\sqrt {\dfrac{{G{e^2}}}{{4\pi { \in _0}}}} $

Note: Mass, length and time are most commonly encountered fundamental quantities so they must be specified in all dimensional formulas. The square bracket notation is used only for dimensional formulas.