If we use permittivity varepsilon,resistance | vedclass

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(D) The dimensions of the given quantities are:$\varepsilon = L^{-3} M^{-1} T^4 I^2 = L^{-3} M^{-1} T^2 Q^2$$R = L^2 M T^{-3} I^{-2} = L^2 M T^{-1} Q^

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all of the above

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(D) The dimensions of the given quantities are:$\varepsilon = L^{-3} M^{-1} T^4 I^2 = L^{-3} M^{-1} T^2 Q^2$$R = L^2 M T^{-3} I^{-2} = L^2 M T^{-1} Q^{-2}$$V = L^2 M T^{-3} I^{-1} = L^2 M T^{-2} Q^{-1}$$G = L^3 M^{-1} T^{-2}$$1$. Angular displacement $	heta$ is a dimensionless quantity,so its dimension is $M^0 L^0 T^0 Q^0$. Thus,$[	heta] = \varepsilon^0 R^0 G^0 V^0$ is correct.$2$. Velocity $[v] = L T^{-1}$.Checking option $B$: $\varepsilon^{-1} R^{-1} = (L^3 M T^{-2} Q^{-2}) (L^{-2} M^{-1} T^1 Q^2) = L^1 T^{-1} = [v]$. This is correct.$3$. Force $[F] = M L T^{-2}$.Checking option $C$: $\varepsilon^1 V^2 = (L^{-3} M^{-1} T^2 Q^2) (L^4 M^2 T^{-4} Q^{-2}) = L^1 M^1 T^{-2} = [F]$. This is correct.Since all statements are correct,the answer is $D$.

If we use permittivity $\varepsilon$,resistance $R$,gravitational constant $G$,and voltage $V$ as fundamental physical quantities,then:

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