JEE Main 26 Feb 2021 Shift 1 Solved Paper

$({*})$ Let, equivalent capacitance of two capacitors $C_{1}$ and $C_{2}$ connected in parallel be $C_{a}$ and equivalent capacitance of same, when connected in series be $C_{b}$. According to given data, $C_{a} = \;\frac{15}{4} C_{b}$ . . . (i) Since, equivalent capacitance in parallel combination, $C_{\text{eq}} = C_{1} + C_{2}$ $C_{a} = C_{1} + C_{2}$ . . . (ii) and equivalent capacitance in series combination, $\;\frac{1}{C_{\text{eq}}'} \;\; = \;\frac{1}{C_{1}} + \;\frac{1}{C_{2}}$ $\;\frac{1}{C_{b}} \;\; = \;\frac{1}{C_{1}} + \;\frac{1}{C_{2}} = \;\frac{C_{2} + C_{1}}{C_{1} C_{2}}$ $C_{b} \;\; = \;\frac{C_{1} C_{2}}{C_{1} + C_{2}}$. . . (iii) Substituting Eqs. (ii) and (iii) in Eq. (i), we get $C_{1} + C_{2} = \;\frac{15}{4} \;\frac{C_{1} C_{2}}{C_{1}} + C_{2}$ $4(C_{1} + C_{2})^{2} = 15 C_{1} C_{2}$ $\Rightarrow \;\; 4 C_{1}^{2} + 4 C_{2}^{2} + 8 C_{1} C_{2} = 15 C_{1} C_{2}$ $\Rightarrow \;\; 4 C_{1}^{2} + 4 C_{2}^{2} - 7 C_{1} C_{2} = 0$ On dividing both sides by $C_{1}^{2}$, we get $4 + 4 \left( \;\frac{C_{2}}{C_{1}} \right)^{2} - 7 \;\frac{C_{2}}{C_{1}} = 0$ $\;\text{ or } \; \;\; 4 \left( \;\frac{C_{2}}{C_{1}} \right)^{2} - 7 \left( \;\frac{C_{2}}{C_{1}} \right) + 4 = 0$ $\;\text{ If } \; \frac{C_{2}}{C_{1}} = x, \;\text{ then } \; 4 x^{2} - 7 x + 4 = 0$ By using the concept of quadratic equation, $\;x = \;\frac{-b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow x = \;\frac{7 \pm \sqrt{49 - 64}}{8}$ $\Rightarrow \;\; x = \;\frac{C_{2}}{C_{1}} = \;\frac{7 \pm \sqrt{-15}}{8} = \;\frac{7 \pm \sqrt{15} i}{8}$