(B) The dimensional formula for energy $E$ is $[M L^{2} T^{-2}]$.The dimensional formula for the gravitational constant $G$ is $[M^{-1} L^{3} T^{-2}]$

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(B) The dimensional formula for energy $E$ is $[M L^{2} T^{-2}]$.The dimensional formula for the gravitational constant $G$ is $[M^{-1} L^{3} T^{-2}]$.Therefore,the dimensions of $\frac{E}{G}$ are given by:$\frac{[E]}{[G]} = \frac{[M L^{2} T^{-2}]}{[M^{-1} L^{3} T^{-2}]}$$= [M^{1 - (-1)}] [L^{2 - 3}] [T^{-2 - (-2)}]$$= [M^{2}] [L^{-1}] [T^{0}]$

Class 11 Physics Units, Dimensions and Measurement Dimensions and Dimensional Formula

Medium

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

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A
$[M][L^{-1}][T^{-1}]$
B
$[M^{2}][L^{-1}][T^{0}]$
C
$[M][L^{0}][T^{0}]$
D
$[M^{2}][L^{-2}][T^{-1}]$

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Match the following two columns:

Column-$I$ Column-$II$

$(A)$ Electrical resistance $(p)$ $M L^3 T^{-3} A^{-2}$

$(B)$ Electrical potential $(q)$ $M L^2 T^{-3} A^{-2}$

$(C)$ Specific resistance $(r)$ $M L^2 T^{-3} A^{-1}$

$(D)$ Specific conductance $(s)$ None of these

Column-$I$	Column-$II$
$(A)$ Electrical resistance	$(p)$ $M L^3 T^{-3} A^{-2}$
$(B)$ Electrical potential	$(q)$ $M L^2 T^{-3} A^{-2}$
$(C)$ Specific resistance	$(r)$ $M L^2 T^{-3} A^{-1}$
$(D)$ Specific conductance	$(s)$ None of these