The pH of 0.01 M NaOH(a q) solution will be

$\mathrm{NaOH}$ is a strong base, thus \begin{aligned} {\left[\mathrm{OH}^{-}\right] } & =0.0 \mathrm{M}=10^{-2} \mathrm{M} \\ \mathrm{pOH} & =-\log \left[\mathrm{OH}^{-}\right]=-\log \left(10^{-2}\right)=2 \end{aligned} We know that, $\mathrm{pH}+\mathrm{pOH}=14$ $\therefore \quad \mathrm{pH}=14-2=12$ Thus, option (3) is correct.