Gram equivalent of H2SO4 = normality × volume (L)
=15×10010001=1005000
Gram equivalent of NaOH =110×5001000
=0.05
nature of the solution will be basic as gram equivalent of NaOH are more
so after uses of 0.02 eq H2SO4 , eq left = 0.05 - 0.02
= 0.03(600ml/0.62)
so normality of excess react ant left is = gramequivalentvolume(L)
so option c is correct = 0.030.6 = 0.05 N