Question

In a new system of units energy (E) , density (d) and power (P) are taken as fundamental units , then the dimensional formula of universal gravitational constant G will be ?

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Energy}\left(E\right)=\left[M{L}^2{T}^{-2}\right] \\ \mathrm{Density}\left(d\right)=\left[M{L}^{-3}{T}^0\right] \\ \mathrm{Power}\left(P\right)=\left[M{L}^2{T}^{-3}\right] \\ {P}^2={\left[M{L}^2{T}^{-3}\right]}^2=\left[M^2{L}^4{T}^{-6}\right] \\ \left(d\right)(E{)}^2=\left[M{L}^{-3}{T}^0\right]\times {\left[M{L}^2{T}^{-2}\right]}^2=\left[M{L}^{-3}{T}^0\right]\times \left[M^2{L}^4{T}^{-4}\right] \\ \frac{(P{)}^2}{\left(d\right)(E{)}^2}=\frac{\left[M^2{L}^4{T}^{-6}\right]}{\left[M{L}^{-3}{T}^0\right]\times \left[M^2{L}^4{T}^{-4}\right]} \\ \frac{(P{)}^2}{\left(d\right)(E{)}^2}=\left[M^{-1}{L}^3{T}^{-2}\right]=\mathrm{Universal}\mathrm{Gravitational}\mathrm{Constant}(G) \end{gathered}$$
Regards