In an LC circuit shown in figure, C = 1F, L = 4H. At time t = 0, charge in the capacitor is 4C and it is decreasing at the rate of $\sqrt{5}{\mathrm{Cs}}^{-1}$. Choose the correct statement.

$\begin{array}{l}Given : \frac{dq}{dt}=i=\sqrt{5}\mathrm{A}\\ At some time t,\\ \Rightarrow \frac{q_m^2}{2C}=\frac{q^2}{2C}+\frac{1}{2}L{i}^2\\ \Rightarrow \frac{q_m^2}{2}=\frac{16}{2}+\frac{1}{2}\left(4\right)\left(5\right)=18\\ \Rightarrow q_{max}=6\mathrm{C}\end{array}$

$q=Q\cos (\omega t+\delta ) at t = 0, q = 4$

$$\begin{gathered} \Rightarrow 4=6\cos \delta \\ \Rightarrow \cos \delta =\frac{2}{3} \\ \Rightarrow \delta ={\cos }^{-1}\left(\frac{2}{3}\right) \end{gathered}$$

Charge on the capacitor is maximum at t = t₁ for this $\omega t_1+\delta =\pi$

$\Rightarrow t_1=\frac{1}{\omega }(\pi -\delta )=\sqrt{LC}(\pi -\delta )=2\left[\pi -{\cos }^{-1}\left(\frac{2}{3}\right)\right]$