Question

If $E,M,l$ and $G$ denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of $P$ in the formula $P=E{l}^2{M}^{-5}{G}^{-2}$ are

• A. $\left[M^1{L}^1{T}^{-2}\right]$
• B. $\left[M^0{L}^0{T}^0\right]$
• C. $\left[M^{-1}{L}^{-1}{T}^2\right]$
• D. $\left[M^0{L}^1{T}^0\right]$

Answer 1

The correct option is B $\left[M^0{L}^0{T}^0\right]$

From the question, following are the dimensions,

Energy, $\left[E\right]=\left[M{L}^2{T}^{-2}\right]$,
Mass $\left[m\right]=\left[M\right]$,
Angular momentum $\left[l\right]=\left[M{L}^2{T}^{-1}\right]$,
Gravitational constant $\left[G\right]=\left[M^{-1}{L}^3{T}^{-2}\right]$

So, for $P=E{l}^2{M}^{-5}{G}^{-2}$

We have,

$\left[P\right]=\left[M{L}^2{T}^{-2}\right]\times {\left[M{L}^2{T}^{-1}\right]}^2\times {\left[M\right]}^{-5}\times {\left[M^{-1}{L}^3{T}^{-2}\right]}^{-2}$

Combining like terms,

$\left[P\right]=\left[M^{1+2-5+2}{L}^{2+4-6}{T}^{-2-2+4}\right]$

$\left[P\right]=\left[M^0{L}^0{T}^0\right]$

It shows that, $P$ is a dimensionless quantity.