When an electric current is passed through acidulated water, 112 ml of hydrogen gas at NTP is collected at the cathode in 965 seconds. The current passed in amperes is

$$\begin{gathered} \mathrm{Cathode} \mathrm{reduction} :\hspace{0.17em}2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2( \mathrm{g}) \\ \text{ At N.T.P, }22.4 \mathrm{L}\text{ (or }22400 \mathrm{mL}\text{ ) of }{\mathrm{H}}_2=1\text{ mole of }{\mathrm{H}}_2 \\ 112 \mathrm{mL}\text{ of }{\mathrm{H}}_2=\frac{112}{22400}\times 1=0.005\text{ mole of }{\mathrm{H}}_2 \\ \text{ Moles of }{\mathrm{H}}_2\text{ produced }=\frac{\mathrm{I}\left(\mathrm{A}\right)\times \mathrm{t}\left(\mathrm{s}\right)}{96500\left(\mathrm{C}/\mathrm{mole}\right)}\times \text{mole ratio } \\ \begin{array}{l}0.005 \mathrm{mol}=\frac{\mathrm{I}\left(\mathrm{A}\right)\times 965 \mathrm{s}}{96500\left(\mathrm{C}/\mathrm{mole}\right)}\times \frac{1 {\mathrm{molH}}_2}{2 }\\ \mathrm{I}=1 \mathrm{A}\end{array} \end{gathered}$$

Therefore, the correct option is A