Physics - Capacitor Question with Solution | TestHub

When connected in parallel

$C_{eq}=C_1+C_2$

When in series

$C_{\mathrm{eq}}^{\text{'}}=\frac{C_1{C}_2}{C_1+C_2}$

$C_1+C_2=10\left(\frac{C_1{C}_2}{C_1+C_2}\right)$

${\left(C_1+C_2\right)}^2=10C_1{C}_2$

$C_1^2+C_2^2-8C_1{C}_2=0$

dividing by $C_1^2$

${\left(\frac{C_2}{C_1}\right)}^2-\frac{8C_2}{C_1}+1=0$

Let $\frac{C_2}{C_1}=x$

$x^2-8x+1=0$

$x=\frac{8\pm \sqrt{64-4}}{2}=4+\sqrt{15}$

This question was Bonus in JEE Main 2021 exam because of incorrect data in original question.