JEE Main 2026 — 24 January, Evening Shift — Question 35

For $\mathrm{C}_{\mathrm{A}}$ : Let $\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=\mathrm{C}$ $\therefore \mathrm{C}_{\mathrm{A}}=\frac{\mathrm{K}_{1} \mathrm{C}}{2}+\frac{\mathrm{K}_{1} \mathrm{~K}_{2} \mathrm{C}}{\mathrm{K}_{1}+\mathrm{K}_{2}}$ $=\mathrm{K}_{1} \mathrm{C}\left[\frac{\mathrm{K}_{1}+2 \mathrm{~K}_{2}}{2\left(\mathrm{~K}_{1}+\mathrm{K}_{2}\right)}\right]$

For $\mathrm{C}_{\mathrm{B}}$ :

$\begin{aligned} \mathrm{C}_{\mathrm{B}} & =\frac{\mathrm{K}_{2} \mathrm{C}}{2}+\frac{\mathrm{K}_{1} \mathrm{~K}_{2} \mathrm{C}}{\mathrm{~K}_{1}+\mathrm{K}_{2}} & =\mathrm{K}_{2} \mathrm{C}\left[\frac{\mathrm{~K}_{1}+2 \mathrm{~K}_{2}}{2\left(\mathrm{~K}_{1}+\mathrm{K}_{2}\right)}\right] \end{aligned}$ For $\mathrm{C}_{\mathrm{C}}$ : $\mathrm{C}_{\mathrm{C}}=\frac{2 \mathrm{~K}_{1} \mathrm{~K}_{2} \mathrm{C}}{\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right)}$ $\mathrm{C}_{\mathrm{A}}>\mathrm{C}_{\mathrm{C}}>\mathrm{C}_{\mathrm{B}}$