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.