Consider the circuit diagram shown.

Column - I		Column - II
(i)	Potential difference across A and D in steady state is	p.	2 V
(ii)	Potential difference across capacitor in steady state is	q.	1.8 V
(iii)	Value of Q for which no energy is stored across capacitor is	r.	0.2 V
(iv)	Potential difference across A and B in steady state is	s.	$\frac{14}{3}\Omega$

Now, match the given columns and select the correct option from the codes given below

As $3\Omega$ and $7\Omega$ are in series across 6 v and in series potential divides in proportion to resistance.

So ${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{D}}=\frac{\mathrm{R}\times \mathrm{X}}{\left(\mathrm{R}+\mathrm{S}\right)}=\frac{3\times 6}{3+7}=\frac{18}{10}=1.8\mathrm{V}$

Hence (i) $\to$ (q)

Similarly, ${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=\frac{\mathrm{P}\times \mathrm{V}}{\left(\mathrm{P}+\mathrm{Q}\right)}=\frac{2\times 6}{2+4}=2\mathrm{V}$

$\left({\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}\right)-\left({\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{D}}\right)={\mathrm{V}}_{\mathrm{D}}-{\mathrm{V}}_{\mathrm{B}}=2-1.8=0.2\mathrm{V}$

Hence (ii) $\to$ (r)

Energy stored in the capacitor will be zero if V_B = V_D

i.e., $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}}\Rightarrow \frac{2}{\mathrm{Q}}=\frac{3}{7}\mathrm{or}\mathrm{Q}=\frac{14}{3}\mathrm{\Omega }$

So (iii) $\to$ (s)

Hence, [i-q], [ii-r], [iii-s] and [iv-p]