Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is _____. (Nearest integer).

Electrolysis of NaCl is
$\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{aq}\right)\to \mathrm{NaOH}\left(\mathrm{aq}\right)+\frac{1}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)$
Since during electrolysis pH changes to 12
$\text{So, }\left[{\mathrm{OH}}^{\ominus }\right]={10}^{-2}\text{and}\left[{\mathrm{H}}^{+}\right]={10}^{-12}$
So by Faraday law
Gram amount of substance deposited = Amount of electricity passed
$\begin{array}{rl} & {10}^{-2}\times \frac{600}{1000}\times 96500=\mathrm{I}\times \mathrm{t}\\ & \frac{{10}^{-2}\times 600}{1000}\times 96500=\mathrm{I}\times 5\times 60\\ & \mathrm{I}=\frac{{10}^{-2}\times 600\times 96500}{1000\times 5\times 60}\end{array}$
I = 1.93 ampere
So, I = 2 ampere (nearest integer)