Question

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

• A. mass
• B. length
• C. time
• D. angle.

Answer 1

The correct option is D angle.

$\left[E\right]=\left[M{L}^2{T}^{-2}\right],\left[m\right]$

$=\left[M\right]\left[l\right]=\left[M{L}^2{T}^{-1}\right],\left[G\right]=\left[M^{-1}{L}^3{T}^{-2}\right]$

$\therefore \left[\frac{E{l}^2}{m^5{G}^2}\right]=\frac{\left[M{L}^2{T}^{-2}\right]\left[M^2{L}^4{T}^{-2}\right]}{\left[M^5\right]\left[M^{-2}{L}^6{T}^{-4}\right]}$

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