$E$, $m$, $l$ and $G$ denote energy, mass, angular momentum and gravitational constant respectively. The dimen

Question

$E$, $m$, $l$ and $G$ denote energy, mass, angular momentum and gravitational constant respectively. The dimensions of $\frac{El^2}{m^5G^2}$ are the same as that of:

Accepted Answer

Let us determine the dimensional formula of each physical quantity involved:
Energy ($E$) = $[M L^2 T^{-2}]$
Mass ($m$) = $[M]$
Angular momentum ($l$) = $[M L^2 T^{-1}]$
Gravitational constant ($G$) = $[M^{-1} L^3 T^{-2}]$

Substituting these dimensions into the given expression $\frac{E l^2}{m^5 G^2}$:
$$ \frac{[M L^2 T^{-2}] 	imes [M L^2 T^{-1}]^2}{[M]^5 	imes [M^{-1} L^3 T^{-2}]^2} = \frac{[M L^2 T^{-2}] 	imes [M^2 L^4 T^{-2}]}{[M^5] 	imes [M^{-2} L^6 T^{-4}]} $$
$$ = \frac{[M^3 L^6 T^{-4}]}{[M^3 L^6 T^{-4}]} = [M^0 L^0 T^0] $$

The resulting dimension is $[M^0 L^0 T^0]$, which indicates a dimensionless quantity. Among the given options, Angle is defined as the ratio of arc length to radius ($[L]/[L]$) and is a dimensionless quantity.

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