Two solutions, A and B, each of 100 L was made by dissolving 4g of NaOH and 9.8 g of H₂SO₄ in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is

Molarity of solution of NaOH = $\frac{\mathrm{Moles} }{\mathrm{Volume}\left(\mathrm{l}\right)}=\frac{0.1}{100}={10}^{-3}\mathrm{M}$
Molarity of H₂SO₄ solution = $\frac{\mathrm{Moles} }{\mathrm{Volume}}=\frac{0.1}{100}={10}^{-3}\mathrm{M}$
Moles of NaOH in 10L of A = 40X10^-3 
Moles of H₂SO₄ in 10L of B = 10X10^-3
$$\begin{gathered} \left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14} \\ \left[H^{+}\right]=\frac{{10}^{-14}}{4\times {10}^{-4}}=2.5\times {10}^{-11} \\ pH = - \log \left[H^{+}\right] = -\log (2.5\times {10}^{-11})=10.6 \end{gathered}$$

Hence, correct option is: (A) 10.6