Correct option: A.$\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ}$
a
For weak acid, $\alpha=\frac{\Lambda_m}{\Lambda_0}$
$\mathrm{K}_\alpha=\frac{C \alpha^2}{1-\alpha} \Rightarrow \mathrm{K}_{\mathrm{s}}(1-\alpha)=C \alpha^2$
$\Rightarrow \mathrm{K}_\alpha\left(1-\frac{\Lambda_{\mathrm{m}}}{\Lambda_0}\right)=C\left(\frac{\Lambda_{\mathrm{m}}}{\Lambda_0}\right)^2$
$\Rightarrow \mathrm{K}_2-\frac{\Lambda_{\mathrm{m}} \mathrm{K}_{\mathrm{a}}}{\Lambda_0}=\frac{C \Lambda_{\mathrm{m}}^2}{\left(\Lambda_0\right)^2}$
Diride by ' $\Lambda_{\mathrm{m}}$ '
$\Rightarrow \frac{\mathrm{K}_{\mathrm{a}}}{\Lambda_{\mathrm{m}}}=\frac{\mathrm{C} \Lambda_{\mathrm{m}}}{\left(\Lambda_0\right)^2}+\frac{\mathrm{K}_{\mathrm{a}}}{\Lambda_0}$
$\Rightarrow \frac{1}{\Lambda_{\mathrm{m}}}=\frac{\mathrm{C} \Lambda_{\mathrm{m}}}{\mathrm{K}_{\mathrm{a}}\left(\Lambda_0\right)^2}+\frac{1}{\Lambda_0}$
Plot $\frac{1}{\Lambda_m}$ vi $C \Lambda_m$ has
Slope $=\frac{1}{\mathrm{~K}_{\mathrm{u}}\left(\Lambda_0\right)^2}=\mathrm{S}$
$\mathrm{y} \text {-intercept }=\frac{1}{\Lambda_0}=\mathrm{P}$
Then, $\frac{P}{S}=\frac{\frac{1}{\Lambda_0}}{\frac{1}{K_2\left(\Lambda_0\right)^2}}=\mathrm{K}_2 \Lambda_0$