According to 2nd law of Faraday’s electrolysis the correct one is 
I) $\frac{\text{ wt.of }{\mathrm{H}}_2\text{ liberated }}{\text{ wt.of }{\mathrm{Cl}}_2\text{ liberated }}=\frac{\text{ eq.wt.of }{\mathrm{H}}_2}{\text{ eq.wt.of }{\mathrm{Cl}}_2}$   ii) $\frac{{\mathrm{m}}_{\mathrm{Ag}}}{{\mathrm{m}}_{\mathrm{Cu}}}=\frac{{\mathrm{E}}_{\mathrm{Ag}}}{{\mathrm{E}}_{\mathrm{Cu}}}$  iii) $\frac{{\mathrm{m}}_{\mathrm{Ag}}}{{\mathrm{m}}_{\mathrm{Cu}}}=\frac{{\mathrm{E}}_{\mathrm{Cu}}}{{\mathrm{E}}_{\mathrm{Ag}}}$  iv) $\frac{{\mathrm{m}}_{{\mathrm{H}}_2}}{{\mathrm{m}}_{\mathrm{Cu}}}=\frac{{\mathrm{E}}_{{\mathrm{H}}_2}}{{\mathrm{E}}_{\mathrm{Cu}}}$  The correct combination is

Faraday's second law: 

Faraday's Second Law of Electrolysis states that the mass of a substance deposited or liberated at any electrode on passing a certain amount of charge is directly proportional to its chemical equivalent weight.

$W \alpha E$

$\frac{W_1}{E_1}=\frac{W_2}{E_2}$

W₁ and W₂ are masses mass of the substances deposited. 

E₁ and E₂ are the equivalent weights of the substances.