Joule: SI system , erg: CGS system
Work = Force `xx` Distance = Mass `xx` Acceleration `xx` Length
= Mass `xx ( "length")/("Time")^(2) xx "Length"`
Dimensions of work =` [W] = [ M^(1) L^(2) T^(2) ]`
`:. a = 1, b = 2 , c = -2`
Now
`|{:(SI" system",,CGS" system"),(M_(1)=1kg,,M_(2)=1g),(L_(1)=1m,,L_(2)=1cm),(T_(1)=1s,,T_(2)=1s):}|`
Here `N_(1) = 1 , N_(2) = ?`
`:.` Using `N_(2) = N_(1) [(M_(1))/(M_(2))]^(a) [[(L_(1))/(L_(2))]^(b)[(T_(1))/(T_(2))]^(c)`
`= 1[( 1 kg)/( 1 g)]^(1) [( 1 m)/( 1cm)]^(1) [( 1 s)/( 1 s)]^(-2) = 10^(5)`
`1 [ (1000 g)/( 1 g)]^(1) [ (100 cm)/(1 cm)]^(2) = 10^(7)`
` :.N_(2) = 10^(7)`
So `1 J = 10^(7) erg`
