Select the correct statement(s).

For strong electrolyte, variation in graph is uniform, hence ${\wedge }_m$  can be extrapolated to $\sqrt{C}\to 0$, to obtain  ${\wedge }_m^{\infty }$
In case of weak electrolyte, variation of ${\wedge }_m$  with $\sqrt{\text{C}}$ is rapid in lower concentration range (higher dilution) hence graph is not uniform and  ${\wedge }_m^{\infty }$ cannot be obtained by extrapolation to  $\sqrt{C}\to 0$. Thus (a) is incorrect 

${\text{Al}}_2\left(S{O}_4\right)=2A{l}^{3+}+3S{O}_4^{2-}$
${\wedge }_m^{\infty }=2{\lambda }_{A{l}^{+}}^{\infty }+3{\lambda }_{S{O}_4^{2-}}^{\infty }$

Thus (a) is true
C) Also  ${\wedge }_{BaS{O}_4}^{\infty }={\wedge }_{BaC{l}_2}^{\infty }+{\wedge }_{H_{2}S{O}_4}^{\infty }-2{\wedge }_{HCl}^{\infty }$
Thus, it is also correct based on Kohlrausch’s Law
d)  ${\wedge }_{AgCl}^{\infty }={\wedge }_{AgN{O}_3}^{\infty }+{\wedge }_{HCl}^{\infty }-{\wedge }_{HN{O}_3}^{\infty }={\wedge }_{A{g}^{+}}^{\infty }+{\wedge }_{c{l}^{-}}^{\infty }$
Thus, correct based on Kohlrausch’s law