If 'E', 'M', 'L' and 'G' denote energy, mass, angular momentum and constant of gravitaion respectively then \(\left(\frac{\mathrm{EL}^{2}}{\mathrm{G}^{2} \mathrm{M}^{5}}\right)\) has dimensions of

Correct option is
D angle.
\([\mathrm{E}]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right],[\mathrm{m}]\)
\(=[\mathrm{M}][\mathrm{I}]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right],[\mathrm{G}]=\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]\)
\(\therefore\left[\frac{\mathrm{El}^{2}}{\mathrm{~m}^{5} \mathrm{G}^{2}}\right]=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\left[\mathrm{M}^{2} \mathrm{~L}^{4} \mathrm{~T}^{-2}\right]}{\left[\mathrm{M}^{5}\right]\left[\mathrm{M}^{-2} \mathrm{~L}^{6} \mathrm{~T}^{-4}\right]}\)
\(=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]\)
As angle has no dimensions, therefore \(\frac{\mathrm{El}^{5}}{\mathrm{~m}^{5} \mathrm{G}^{2}}\) has the same dimensions as that of angle.