A series LCR circuit has 120$\mathrm{\Omega }$ resistance. When the angular frequency of the source is 4 x 10⁵ rad s^-1 the voltage across resistance , inductance and capacitance are 60V , 40V, and 40V respectively. At n x 10⁵ rad/s angular frequency of the source the current in the circuit will lag behind the source voltage by $\frac{\mathrm{\pi }}{4}$ then n=?

With $\mathrm{\omega }$₀ = 4 x 10⁵ rad s^-1, the circuit is in resonance $\left(\because {\mathrm{V}}_{\mathrm{L}}={\mathrm{V}}_{\mathrm{C}}\right)$
$\therefore$Z=R=12$\mathrm{\Omega }$
$\therefore$RMS current $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{60}{120}=0.5\mathrm{A}$
[Note that source voltage =VR in resonance]
Now $1{\mathrm{X}}_{\mathrm{L}}={\mathrm{V}}_{\mathrm{L}}\Rightarrow 0.5\left(\mathrm{\omega L}\right)=40$
$\Rightarrow \mathrm{L}=\frac{40}{0.5\times 4\times {10}^5}=0.2\mathrm{mH}$
and 1X_C = V_C
$\Rightarrow 0.5\frac{1}{\mathrm{\omega c}}=40\Rightarrow \mathrm{C}=\frac{0.5}{4\times {10}^5\times 40}=31.25\mathrm{nF}$
If current lags behind the voltage by $\frac{\mathrm{\pi }}{4}$ we must have
$\begin{array}{l}\tan \left(\frac{\mathrm{\pi }}{4}\right)=\frac{{\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{c}}}{\mathrm{R}}\\ \mathrm{\omega L}-\frac{1}{\mathrm{\omega C}}=1\times 120\\ 0.2\times {10}^{-3}\mathrm{\omega }-\frac{1}{\mathrm{\omega }\times 31.25\times {10}^{-9}}=120\\ 0.2\times 31.25\times {10}^{-12}{\mathrm{\omega }}^2-120\times 31.25\times {10}^{-9}\mathrm{\omega }-1=0\\ \Rightarrow 6.25{\mathrm{\omega }}^2-3.75\times {10}^6\mathrm{\omega }-{10}^{12}=0\end{array}$
Solving this quadratic equation gives
$\mathrm{\omega }$ = 8 x 10⁵ rad s^-1