An a.c. source is applied to a series LR circuit with $X_L=3 R$ and power factor is $X_1$. Now a capacitor with $X_C=R$ is added in series to LR circuit and the power factor is $X_2$. The ratio $X_1$ to $X_2$ is

The ratio of power factors $X_1$ to $X_2$ is determined by comparing the impedance characteristics of two circuits.

For the series LR circuit with $X_L = 3R$, the impedance is $Z_1 = \sqrt{R^2 + (3R)^2} = R\sqrt{10}$. The power factor is then $X_1 = \frac{R}{Z_1} = \frac{1}{\sqrt{10}}$.

For the series LCR circuit with $X_L = 3R$ and $X_C = R$, the impedance becomes $Z_2 = \sqrt{R^2 + (3R - R)^2} = R\sqrt{5}$. This yields a power factor of $X_2 = \frac{R}{Z_2} = \frac{1}{\sqrt{5}}$.

The desired ratio is $\frac{X_1}{X_2} = \frac{1/\sqrt{10}}{1/\sqrt{5}} = \frac{\sqrt{5}}{\sqrt{10}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$, corresponding to a ratio of $1 : \sqrt{2}$.