Match List-I with List-II.List-IList-II(a) Ca | vedclass

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(A) $1$. Capacitance $(C)$: From $q = CV$,we have $[C] = [q/V] = [q^2 / (Work)] = [A^2 T^2 / (M L^2 T^{-2})] = M^{-1} L^{-2} T^4 A^2$. Thus,$(a) \righ

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$(a) \rightarrow (iii), (b) \rightarrow (ii), (c) \rightarrow (iv), (d) \rightarrow (i)$

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(A) $1$. Capacitance $(C)$: From $q = CV$,we have $[C] = [q/V] = [q^2 / (Work)] = [A^2 T^2 / (M L^2 T^{-2})] = M^{-1} L^{-2} T^4 A^2$. Thus,$(a) \rightarrow (iii)$.$2$. Permittivity of free space $(\varepsilon_0)$: From Coulomb's law $F = (q_1 q_2) / (4 \pi \varepsilon_0 r^2)$,we have $[\varepsilon_0] = [q^2 / (F L^2)] = [A^2 T^2 / (M L T^{-2} L^2)] = M^{-1} L^{-3} T^4 A^2$. Thus,$(b) \rightarrow (ii)$.$3$. Permeability of free space $(\mu_0)$: Using $c = 1 / \sqrt{\mu_0 \varepsilon_0}$,we get $\mu_0 = 1 / (\varepsilon_0 c^2)$. Substituting dimensions,$[\mu_0] = [1 / (M^{-1} L^{-3} T^4 A^2 \cdot L^2 T^{-2})] = M^1 L^1 T^{-2} A^{-2}$. Thus,$(c) \rightarrow (iv)$.$4$. Electric field $(E)$: From $F = qE$,we have $[E] = [F/q] = [M L T^{-2} / (A T)] = M^1 L^1 T^{-3} A^{-1}$. Thus,$(d) \rightarrow (i)$.The correct matching is $(a) \rightarrow (iii), (b) \rightarrow (ii), (c) \rightarrow (iv), (d) \rightarrow (i)$.

List-$I$	List-$II$
$(a)$ Capacitance,$C$	$(i)$ $M^{1} L^{1} T^{-3} A^{-1}$
$(b)$ Permittivity of free space,$\varepsilon_{0}$	$(ii)$ $M^{-1} L^{-3} T^{4} A^{2}$
$(c)$ Permeability of free space,$\mu_{0}$	$(iii)$ $M^{-1} L^{-2} T^{4} A^{2}$
$(d)$ Electric field,$E$	$(iv)$ $M^{1} L^{1} T^{-2} A^{-2}$

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