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Physics Question on Dimensional Analysis

If EE and GG respectively denote energy and gravitational constant, then EG\frac{ E }{ G } has the dimensions of:

A

[M2][L−1][T0]\left[ M ^{2}\right]\left[ L ^{-1}\right]\left[ T ^{0}\right]

B

[M][L−1][T−1][ M ]\left[ L ^{-1}\right]\left[ T ^{-1}\right]

C

[M][L0][T0][ M ]\left[ L ^{0}\right]\left[ T ^{0}\right]

D

[M2][L−2][T−1]\left[ M ^{2}\right]\left[ L ^{-2}\right]\left[ T ^{-1}\right]

Answer

[M2][L−1][T0]\left[ M ^{2}\right]\left[ L ^{-1}\right]\left[ T ^{0}\right]

Explanation

Solution

[E]=ML2T−2[ E ]= ML ^{2} T ^{-2}
[G]=[Fr2][m1m2]=[MLT−2L2M2]=M−1L3T−2[ G ]=\frac{\left[ Fr ^{2}\right]}{\left[ m _{1} m _{2}\right]}=\left[\frac{ ML T ^{-2} L ^{2}}{ M ^{2}}\right]= M ^{-1} L ^{3} T ^{-2}
∴[EG]=ML2T−2M−1L3T−2=[M2L−1T0]\therefore\left[\frac{ E }{ G }\right]=\frac{ ML ^{2} T ^{-2}}{ M ^{-1} L ^{3} T ^{-2}}=\left[ M ^{2} L ^{-1} T ^{0}\right]