1. Reaction at the cathode

In electrolysis of acidified water, hydrogen is produced at the cathode:

$2H^+ + 2e^- \rightarrow H_2$

So, 1 mole of $H_2$ requires 2 moles of electrons.

2. Moles of hydrogen collected

At N.T.P., $1$ mole of gas occupies $22.4\,\text{L} = 22400\,\text{mL}$.

Given volume of hydrogen:

$112\,\text{mL}$

Hence,

$n(H_2)=\frac{112}{22400}=\frac{1}{200}=0.005\,\text{mol}$

3. Moles of electrons required

From the cathode reaction:

$1\,\text{mol } H_2 \text{ needs } 2\,\text{mol } e^-$

Therefore,

$n(e^-)=2 \times 0.005=0.01\,\text{mol}$

4. Total charge passed

Using Faraday constant,

$1\,\text{mol } e^- = 96500\,\text{C}$

So charge passed:

$Q=0.01 \times 96500=965\,\text{C}$

5. Calculate current

Given time:

$t=965\,\text{s}$

Using

$I=\frac{Q}{t}$

$I=\frac{965}{965}=1.0\,\text{A}$

6. Match with options

The correct option is:

$\boxed{\text{A: }1.0\,\text{A}}$