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

Using dimensions, show that 1$$$$joule=107erg{10^7}erg.

Explanation

Solution

1$$$$joule is the amount of work done when a force of 11 Newton displaces a body through a distance of 1$$$$min the direction of the force applied. joulejoule is the standard unit of work and energy. As we know that ergerg is the unit of energy in the CGS system whereas joulejoule is the unit of energy in the MKS system.

Complete answer:
Given that 1$$$$joule=107erg{10^7}erg.
ergerg is the unit of energy in the CGS system whereas joulejoule is the unit of energy in the MKS system. CGS system stands for centimeter-gram-second and MKS system stands for meter-kilogram-seconds. Let us consider 1$$$$joule= XergXerg

As we know that dimension ofjoulejoule = [M1L12T1−2]\left[ {{M_1}L_1^2T_1^{ - 2}} \right] and dimension of ergerg= [M2L22T2−2]\left[ {{M_2}L_2^2T_2^{ - 2}} \right]
Substitute these dimensions
11 [M1L12T1−2]\left[ {{M_1}L_1^2T_1^{ - 2}} \right] =X$$$$\left[ {{M_2}L_2^2T_2^{ - 2}} \right]
Now take all the dimensions values on one side,
XX= [M1L12T1−2][M2L22T2−2]\dfrac{{\left[ {{M_1}L_1^2T_1^{ - 2}} \right]}}{{\left[ {{M_2}L_2^2T_2^{ - 2}} \right]}}
⇒X\Rightarrow X= \left[ {\dfrac{{{M_1}}}{{{M_2}}}} \right]$$$${\left[ {\dfrac{{{L_1}}}{{{L_2}}}} \right]^{^2}}$$$${\left[ {\dfrac{{{T_1}}}{{{T_2}}}} \right]^2}
Now substitute the units of mass length and time in this resultant equation, we get-
XX= \left[ {\dfrac{{Kg}}{{gm}}} \right]$$$${\left[ {\dfrac{m}{{cm}}} \right]^2}$$$${\left[ {\dfrac{s}{s}} \right]^{ - 2}}
⇒X\Rightarrow X= \left[ {\dfrac{{1000gm}}{{gm}}} \right]$$$${\left[ {\dfrac{{100cm}}{{cm}}} \right]^2}$$$${\left( 1 \right)^{ - 2}}
⇒X\Rightarrow X= {\left( {10} \right)^3}$$$$\left( {{{10}^4}} \right)
∴X\therefore X= (10)7{\left( {10} \right)^7}

Hence, 1$$$$joule=107erg{10^7}erg.

Note: joulejoule 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 joulejoule. 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.