Explanation

(c)
Given
$H_{2}{SO}_{4}\$- V=100mL, N=0.2 M
NaOH - V=100mL, N=0.2 M
Milliequivalent of $H_{2}SO_{4} = 100\ \times \ 0.2\ \times 2 = 40\ \ \$
($\therefore\ \ It\ is\ dibasic\ acid$)
Milliequivalent of NaOH =$100\ \times \ 0.1\ \times 2 = 20$
$\therefore$ Moilliequivalent Of $H_{2}{SO}_{4}$ left =40-20=20
Total volume = 100mL+100mL=200mL
Normality of $H_{2}{SO}_{4}$ (left0)=$\frac{20}{200}$ =0.1 N