60 ml of $0.1\mathrm{M}\mathrm{HCl}$ is mixed with 40ml of $0.05 \mathrm{M}{\mathrm{H}}_2{\mathrm{SO}}_4.$ The pH value of the resulting solution is

60ml of 0.1M HCL
No.of moles = molarity x volume (litres)
Final normality = $\frac{N_1{V}_1+N_2{V}_2}{V_1+V_2}$  
$\frac{60x0.1+40x0.05x2}{100}(\therefore H_{2}S{O}_{4}nfactoris2)$ 
$\frac{6+4}{100}=0.1$ 
${ }^{\left[H^{+}\right]}=0.1$ 
$P^H=-\log \left(0.1\right)=1$