Explanation

(c)
Given, $conc.of\ HCl = 0.04\ N$
$Volume\ of\ HCl = 100\ mL$
$conc.of\ NaOH = 0.02\ N$
$Volume\ of\ NaOH = 100\ mL$
$pH = ?$
Milliequivalents of $HCl = 0.04 \times 100 = 4$
Milliequivalents of $NaOH = 0.02 \times 100 = 2$
$HCl + NaOH \rightarrow NaCl + H_{2}O$
M. eq. before reaction 4 2 0 0
M. eq. after reaction 2 0 2 2
Total volume of solution = 100 mL+100 mL=200 mL
$\therefore$ 2 milliequivalents of $HCl$ are present in 200 mL solution.
$\therefore$ 10 milliequivalents of $HCl$ are present in = 1000 mL solution.
$\therefore$ concentration of $HCl = \frac{10}{1000} = 0.01\ N$
$HCl \rightarrow H^{+} + \ Cl^{-}$
$\therefore$ $\left\lbrack H^{+} \right\rbrack in\ solution = 0.01$
$N\ HCl = 0.01\ M\ HCl$
($\therefore HCl$ is monobasic acid)
$pH = - \log{\lbrack H^{+}\rbrack}$
$= - \log{(0.01)}$
$= 2$