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(C) $A \rightarrow (iii)$: According to Gauss's law for electricity,the total electric flux through a closed surface is $\oint E \cdot dA = \frac{Q}{\

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$A-(iii), B-(i), C-(ii), D-(iv)$

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(C) $A \rightarrow (iii)$: According to Gauss's law for electricity,the total electric flux through a closed surface is $\oint E \cdot dA = \frac{Q}{\varepsilon_0}$.$B \rightarrow (i)$: According to Gauss's law for magnetism,the net magnetic flux through any closed surface is zero,i.e.,$\oint B \cdot dA = 0$.$C \rightarrow (ii)$: According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is the line integral of the electric field,$\oint E \cdot dl = -\frac{d\phi_B}{dt}$.$D \rightarrow (iv)$: According to the Ampere-Maxwell law,the line integral of the magnetic field is $\oint B \cdot dl = \mu_0(i_c + i_d)$,where $i_c$ is conduction current and $i_d$ is displacement current.

$A$. $\oint E \cdot dA$	$(i)$ $0$
$B$. $\oint B \cdot dA$	$(ii)$ $-\frac{d\phi_B}{dt}$
$C$. $\oint E \cdot dl$	$(iii)$ $\frac{Q}{\varepsilon_0}$
$D$. $\oint B \cdot dl$	$(iv)$ $\mu_0(i_c + i_d)$

List-$I$	List-$II$
$(a)$ Gauss’ law for electrostatics	$(i)$ $\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$
$(b)$ Gauss’ law for magnetism	(ii) $\oint \vec{B} \cdot d\vec{l} = \mu_0 [i_c + \epsilon_0 \frac{d\phi_E}{dt}]$
$(c)$ Faraday’s law	(iii) $\oint \vec{B} \cdot d\vec{A} = 0$
$(d)$ Ampere-Maxwell’s law	(iv) $\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$

Match the following List-$I$ with List-$II$.
$A$. $\oint E \cdot dA$ $(i)$ $0$
$B$. $\oint B \cdot dA$ $(ii)$ $-\frac{d\phi_B}{dt}$
$C$. $\oint E \cdot dl$ $(iii)$ $\frac{Q}{\varepsilon_0}$
$D$. $\oint B \cdot dl$ $(iv)$ $\mu_0(i_c + i_d)$

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Match the following List-$I$ with List-$II$.$A$. $\oint E \cdot dA$$(i)$ $0$$B$. $\oint B \cdot dA$$(ii)$ $-\frac{d\phi_B}{dt}$$C$. $\oint E \cdot dl$$(iii)$ $\frac{Q}{\varepsilon_0}$$D$. $\oint B \cdot dl$$(iv)$ $\mu_0(i_c + i_d)$