If E,M,L,and G denote energy,mass,angular mom | vedclass

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(A) The dimensions of the given physical quantities are:$[E] = [M L^{2} T^{-2}]$$[M] = [M]$$[L] = [M L^{2} T^{-1}]$$[G] = [M^{-1} L^{3} T^{-2}]$Now,su

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(A) The dimensions of the given physical quantities are:$[E] = [M L^{2} T^{-2}]$$[M] = [M]$$[L] = [M L^{2} T^{-1}]$$[G] = [M^{-1} L^{3} T^{-2}]$Now,substitute these into the expression $\left(\frac{E L^{2}}{G^{2} M^{5}}\right)$:Dimensions $= \frac{[M L^{2} T^{-2}] \cdot [M L^{2} T^{-1}]^{2}}{[M^{-1} L^{3} T^{-2}]^{2} \cdot [M]^{5}}$$= \frac{[M L^{2} T^{-2}] \cdot [M^{2} L^{4} T^{-2}]}{[M^{-2} L^{6} T^{-4}] \cdot [M^{5}]}$$= \frac{[M^{3} L^{6} T^{-4}]}{[M^{3} L^{6} T^{-4}]}$$= [M^{0} L^{0} T^{0}]$Since the resulting dimension is dimensionless,it corresponds to the dimensions of an angle (which is a dimensionless quantity).

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

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