The pH of the resulting solution when equal volumes of 0.1M NaOH and 0.02 M HCl are mixed is (Take log 2 = 0.30)

One mole of NaOH is completely neutralized by one mole of HCl
Hence, 0.02 mole of NaOH will be completely neutralised by 0.02 mole of HCl
$\Rightarrow$NaOH left unneutralised = 0.1 – 0.02 = 0.08 mol

As equal volumes of two solutions are mixed 

${\left[OH\right]}^{-}=\frac{0.08}{2}=0.04M\Rightarrow pOH=-\log \left(0.04\right)=1.4$

$\therefore pH=14-1.40=12.60\cong 13$