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 $t_{1} / t_{2}$ will be ................
JEE MAIN 2022, Diffcult
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In $t_{1}$ time energy becomes half so charge will become $\frac{1}{\sqrt{2}}$ time
$q = Q _{0} e ^{\frac{ T _{1}}{ RC }}=\frac{ Q _{0}}{\sqrt{2}}$
and $q = Q _{0} e ^{\frac{ t _{1}}{ RC }}=\frac{ Q _{0}}{8}=\left(\frac{ Q _{0}}{\sqrt{2}}\right)^{6}$
$t _{2}=6 t _{1}$
$\frac{ t _{1}}{ t _{2}}=\frac{1}{6}$
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