Question

Using dimension show that \[1joule = {10^7}erg\].

Answer 1

Hint: Joule is the standard units of work and energy, where erg is its CGS unit. Dimensions of a physical quantity are the power to which that fundamental unit is raised to obtain one unit of that quantity. There are seven primary dimensions which are independent and also known as fundamental dimensions from which other dimensions can be obtained. They are mass, length, time, temperature, electric current, amount of light and amount of matter.

Complete step by step solution:
Joule is the SI unit of energy which is the measure of the capacity to do work or generate work. 1 Joule equals the work done by Force of one newton (N) acting over a distance of one meter (m).
Dimensions for all form of energy will be the same; hence we can say the dimension of energy is the same as the dimension of work done, where (work is done = force x displacement)
where Force= mass x acceleration
\[
  Force = mass \times acceleration \\
   = \left( {{M^1}} \right)\left( {{L^1}{T^{ - 2}}} \right) \\
   = \left[ {{M^1}{L^1}{T^{ - 2}}} \right] \\
 \]
Hence energy will be
\[
  E = Force \times displacement \\
   = \left[ {{M^1}{L^1}{T^{ - 2}}} \right]\left[ {{L^1}} \right] \\
   = \left[ {{M^1}{L^2}{T^{ - 2}}} \right] \\
 \]
Since both the units Joule and erg are the units of energy in SI and CGS units, where the dimension of the energy is \[\left[ {{M^1}{L^2}{T^{ - 2}}} \right]\]
Now let us assume that 1 joule = X erg
\[
  1\left[ {M_1^1L_1^2T_1^{ - 2}} \right] = X\left[ {M_2^1L_2^2T_2^{ - 2}} \right] \\
  X = \dfrac{{\left[ {M_1^1L_1^2T_1^{ - 2}} \right]}}{{\left[ {M_2^1L_2^2T_2^{ - 2}} \right]}} \\
  X = \left[ {\dfrac{{{M_1}}}{{{M_2}}}} \right]{\left[ {\dfrac{{{L_1}}}{{{L_2}}}} \right]^2}{\left[ {\dfrac{{{T_1}}}{{{T_2}}}} \right]^2} \\
  X = \left[ {\dfrac{{Kg}}{{gm}}} \right]{\left[ {\dfrac{m}{{cm}}} \right]^2}{\left[ {\dfrac{s}{s}} \right]^{ - 2}} \\
   = \left[ {\dfrac{{1000gm}}{{1gm}}} \right]{\left[ {\dfrac{{100cm}}{{1cm}}} \right]^2}{\left( 1 \right)^{ - 2}} \\
   = {\left( {10} \right)^3}{\left( {10} \right)^4} \\
   = {\left( {10} \right)^7} \\
 \]

Hence we can say \[1joule = {10^7}erg\].

Note: Please note that while writing the dimensional formula only SI units of the measuring quantities should be used and should be bifurcated further.