A capacitor is discharging through a resistor $R$. Consider in time $t_1$, the energy stored in the capacitor reduces to half of its initial value and in time $t_2$, the charge stored reduces to one eighth of its initial value. The ratio $\frac{t_1}{t_2}$ will be

We know that in the discharging circuit the charge is given by,

$q=q_0{e}^{\frac{-t}{RC}}$  $\Rightarrow \frac{t}{RC}=\ln \left(\frac{q_0}{q}\right)$

In time $t_1$ the energy stored reduces to half. Hence, $\frac{{q_1}^2}{2C}=\frac{1}{2}\times \frac{{q_0}^2}{2C}\Rightarrow \frac{q_0}{q_1}=\sqrt{2}$. Therefore,

$\frac{t_1}{RC}=\ln \left(\sqrt{2}\right)=\frac{1}{2}\ln 2$.

In time $t_2$ the charge stored reduces to ${\frac{1}{8}}^{\mathrm{th}}$ of initial value. Hence, $\frac{q_2}{q_0}=\frac{1}{8}$. Therefore,

$\frac{t_2}{RC}=\ln \left(8\right)=3\ln 2$.

Hence,

$\frac{t_1}{t_2}=\frac{{\displaystyle \frac{1}{2}}\ln 2}{3\ln 2}=\frac{1}{6}$