Question

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

Answer 1

Hint: \[1\]\[joule\] is the amount of work done when a force of \[1\] Newton displaces a body through a distance of \[1\]\[m\]in the direction of the force applied. \[joule\] is the standard unit of work and energy. As we know that \[erg\] is the unit of energy in the CGS system whereas \[joule\] is the unit of energy in the MKS system.

Complete answer:
Given that \[1\]\[joule\]=\[{10^7}erg\].
\[erg\] is the unit of energy in the CGS system whereas \[joule\] is the unit of energy in the MKS system. CGS system stands for centimeter-gram-second and MKS system stands for meter-kilogram-seconds. Let us consider \[1\]\[joule\]= \[Xerg\]

As we know that dimension of\[joule\] = \[\left[ {{M_1}L_1^2T_1^{ - 2}} \right]\] and dimension of \[erg\]= \[\left[ {{M_2}L_2^2T_2^{ - 2}} \right]\]
Substitute these dimensions
\[1\] \[\left[ {{M_1}L_1^2T_1^{ - 2}} \right]\] =\[X\]\[\left[ {{M_2}L_2^2T_2^{ - 2}} \right]\]
Now take all the dimensions values on one side,
\[X\]= \[\dfrac{{\left[ {{M_1}L_1^2T_1^{ - 2}} \right]}}{{\left[ {{M_2}L_2^2T_2^{ - 2}} \right]}}\]
\[\Rightarrow X\]= \[\left[ {\dfrac{{{M_1}}}{{{M_2}}}} \right]\]\[{\left[ {\dfrac{{{L_1}}}{{{L_2}}}} \right]^{^2}}\]\[{\left[ {\dfrac{{{T_1}}}{{{T_2}}}} \right]^2}\]
Now substitute the units of mass length and time in this resultant equation, we get-
\[X\]= \[\left[ {\dfrac{{Kg}}{{gm}}} \right]\]\[{\left[ {\dfrac{m}{{cm}}} \right]^2}\]\[{\left[ {\dfrac{s}{s}} \right]^{ - 2}}\]
\[\Rightarrow X\]= \[\left[ {\dfrac{{1000gm}}{{gm}}} \right]\]\[{\left[ {\dfrac{{100cm}}{{cm}}} \right]^2}\]\[{\left( 1 \right)^{ - 2}}\]
\[\Rightarrow X\]= \[{\left( {10} \right)^3}\]\[\left( {{{10}^4}} \right)\]
\[\therefore X\]= \[{\left( {10} \right)^7}\]

Hence, \[1\]\[joule\]=\[{10^7}erg\].

Note: \[joule\] is the measurement of work and energy. Measurements of physical quantities are expressed in terms of units, which are standardized values. The word energy we often use in our daily life. In the same way to measure the energy its standard value is \[joule\]. We need a standard unit for measurement to make calculation easy and accurate.
By using the dimensional analysis, we can change or do the conversion of units. But remember to take the correct dimensions for mass, length and time.