How will you convert 10 joule into erg using dimensional analysis ?

Explanation

Joule is the unit of work in SI system. Dimensional formula of work `=(ML^(2)T^(-2)]`. In this question, we have to convert SI units of work into CGS units of work . Here,

We know that `n_(2)=n_(1)[(M_(1))/(M_(2))]^(a)[(L_(1))/(L_(2))]^(b)[T_(1)/(T_(2))]^(c )`
Here `a=1`, `b=2` and `c=-2` therefore
`n_(2)=10[(1kg)/(1g)]^(1)[(1m)/(1cm)]^(2)[(1s)/(1s)]^(-2)=10[(1000g)/(1g)][(100cm)/(1cm)]^(2)[1]^(-2)=10^(8)`
Using `n_(1)u_(1)=n_(2)u_(2)`, we get `10` joule `=10^(8)erg`.