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 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 O, with the rod remaining aligned with the string. In such a case, the value of n is _____.

 To solve the problem, we need to determine the value of 

$n$ such that when a horizontal impulse $P$ is applied to a thin uniform rod at a distance $x=\frac{L}{n}$ from its midpoint, the rod and string revolve together around the pivot point $O$ with the rod remaining aligned with the string.

Key Steps:

1. Moment of Inertia Calculation:

   • The rod rotates about point $O$, which is $L$ away from one end of the rod.

   • The moment of inertia $I$ of the rod about point $O$ is calculated as:

     $$I={\int }_L^{2L}\frac{m}{L}{x}^{2}dx=\frac{7}{3}m{L}^2$$

2. Angular Impulse and Momentum:

   • The angular impulse imparted by the impulse $P$ is given by $P\cdot d$, where $d$ is the distance from $O$ to the point of application of the impulse.

   • The angular momentum after the impulse is $I\omega$.

3. Relating Linear and Angular Quantities:

   • The linear impulse $P$ equals the change in linear momentum: $P=m{v}_{\text{cm}}$, where $v_{\text{cm}}=1.5L\omega$.

   • Substituting $v_{\text{cm}}$ into the angular momentum equation:

     $$P\cdot d=\frac{7}{3}m{L}^2\omega$$

   • Solving for $d$:

     $$d=\frac{14}{9}L$$

4. Determining $n$:

   • The distance from $O$ to the point of application is $1.5L+\frac{L}{n}$.

   • Setting this equal to $\frac{14}{9}L$:

     $$1.5L+\frac{L}{n}=\frac{14}{9}L$$

   • Solving for $n$:

     $$\frac{3}{2}+\frac{1}{n}=\frac{14}{9}⟹\frac{1}{n}=\frac{1}{18}⟹n=18$$

Final Answer

The value of $n$ is $\boxed{{\displaystyle 18}}$.