Question

If $E,M,L$ and $G$ denote energy, mass, angular momentum and gravitational constant respectively, then the quantity $\left(E^2{L}^2/M^5{G}^2\right)$ has the dimensions of

• A. angle
• B. length
• C. mass
• D. None of these

Answer 1

Answer: (A)

$[E]=\left[M{L}^2{T}^{-2}\right],[m]$
$=[M][l]=\left[M{L}^2{T}^{-1}\right],$
$[G]=\left[M^{-1}{L}^3{T}^{-2}\right]$
$\therefore \left[\frac{E{l}^2}{m^5{G}^2}\right]=\frac{\left[M{L}^2{T}^{-2}\right]\left[M^2{L}^4{T}^{-2}\right]}{\left[M^5\right]\left[M^{-2}{L}^6{T}^{-4}\right]}$
$=\left[M^0{L}^0{T}^0\right]$
As angle has no dimensions,
$\therefore \frac{E{l}^5}{m^5{G}^2}$ has the same dimensions as that of angle.