1. Calculate moles of acids mixed

• From $600\text{ mL}$ of $0.01\,\text{M HCl}$: $n(\mathrm{HCl})=0.600\times 0.01=0.006\text{ mol}$ Since HCl is a strong monoprotic acid, it gives: $0.006\text{ mol of }\mathrm{H}^+$

• From $400\text{ mL}$ of $0.01\,\text{M }\mathrm{H_2SO_4}$: $n(\mathrm{H_2SO_4})=0.400\times 0.01=0.004\text{ mol}$ For this level of JEE problem, $\mathrm{H_2SO_4}$ is treated as a strong dibasic acid, so it gives: $\mathrm{H_2SO_4}\rightarrow 2\mathrm{H}^+ + \mathrm{SO_4^{2-}}$ Hence moles of $\mathrm{H}^+$ from sulfuric acid: $2\times 0.004=0.008\text{ mol}$

2. Total moles of $\mathrm{H}^+$

$n(\mathrm{H}^+)=0.006+0.008=0.014\text{ mol}$

3. Total volume after mixing

$V_{\text{total}}=600\text{ mL}+400\text{ mL}=1000\text{ mL}=1.0\text{ L}$

4. Concentration of $\mathrm{H}^+$ in mixture

$[\mathrm{H}^+]=\frac{0.014}{1.0}=0.014\,\text{M}=1.4\times 10^{-2}$

5. Find pH

$\mathrm{pH}=-\log(1.4\times 10^{-2})$ $=2-\log(1.4)$ Now, $1.4=\frac{14}{10}$ so $\log(1.4)=\log 14-1=\log(2\times 7)-1$ Using given values: $\log 2=0.30,\qquad \log 7=0.84$ $\log 14=0.30+0.84=1.14$ Thus, $\log(1.4)=1.14-1=0.14$ Therefore, $\mathrm{pH}=2-0.14=1.86$

6. Match with required format

The question writes pH as: $\underline{\hspace{2cm}}\times 10^{-2}$ Now, $1.86=186\times 10^{-2}$ So the required nearest integer is: $\boxed{186}$

7. Comparison with stored answer

Derived answer $=186$, which matches the stored correct answer.