A thin uniform rod of length \(L\) and certain mass is kept on a frictionless horizontal table with a massless string of length \(L\) fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point \(\mathrm{O}\). If a horizontal impulse \(P\) is imparted to the rod at a distance \(x=L / n\) from the mid-point of the rod (see figure), then the rod and string revolve together around the point \(\mathrm{O}\), with the rod remaining aligned with the string. In such a case, the value of \(n\) is _______ .

Linear impulse \(\int F d t=\Delta\) momentum
\(\begin{aligned} & =\mathrm{m}\left(\mathrm{V}_{\mathrm{cm}}-0\right) \\ & \mathrm{P}=\mathrm{m}\left(\omega \mathrm{r}_{\mathrm{cm}}\right) \\ & =\mathrm{m} \omega\left(\mathrm{L}+\frac{\mathrm{L}}{2}\right)\end{aligned}\)
\(\mathrm{P}=\mathrm{m} \omega\left(\frac{3 \mathrm{~L}}{2}\right)\) ...(i)
Angular impulse \(\int \tau \mathrm{dt}=\Delta\) angular momentum
\(\begin{aligned} & \int \mathrm{r} \times \mathrm{Fdt}=\Delta \mathrm{L} \\ & \begin{aligned} & \mathrm{r} \times \int \mathrm{Fdt}=\mathrm{I}(\omega-0), \text { and } \mathrm{I} \text { is moment of inertia about axis of rotation. } \\ & \begin{aligned} &\left(\mathrm{L}+\frac{\mathrm{L}}{2}+\mathrm{x}\right) \times \mathrm{P}=\left(\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^2\right) \omega \\ &=\left(\frac{\mathrm{mL}^2}{12}+\mathrm{m}\left(\mathrm{L}+\frac{\mathrm{L}}{2}\right)^2\right) \omega\end{aligned} \\ &\left(\frac{3 \mathrm{~L}}{2}+\mathrm{x}\right) \mathrm{P}=\mathrm{mL}^2\left(\frac{1}{12}+\left(\frac{3}{2}\right)^2\right) \omega\end{aligned}\end{aligned}\)
\(\left(\frac{3 L}{2}+x\right) P=\mathrm{mL}^2\left(\frac{7}{3}\right) \omega\) ...(ii)
Divide eq.-(i) & (ii)
\(\begin{aligned} & \left(\frac{3 L}{2}+x\right)=\frac{L\left(\frac{7}{3}\right)}{\left(\frac{3}{2}\right)} \\ & \frac{3 L}{2}+x=L\left(\frac{14}{9}\right) \\ & x=\frac{L}{18}\end{aligned}\)