A tank contains two immiscible liquids of densities $6\rho$ and $2\rho$. The higher density liquid is filled up to a height $\frac{L}{2}$ from the bottom. A thin rod of density $\rho$ and length $L$ is fully immersed and hinged at the bottom so that it can oscillate freely, as shown in the figure. If the rod is slightly disturbed from its equilibrium position, the time period of small oscillations is $\frac{2\pi }{n}\sqrt{\frac{L}{g}}$ where $g$ is the acceleration due to gravity. The value of $n$ is _______. 

The correct answer is 3.8.

Let us analyze the forces acting on the thin rod when it is displaced by a small angle $\theta$ from its vertical equilibrium position.

1. Parameters of the System:
- Total length of the rod = $L$
- Cross-sectional area of the rod = $A$
- Mass of the rod: $m=\rho AL$
- Density of lower liquid layer (height $\frac{L}{2}$) = $6\rho$
- Density of upper liquid layer (height $\frac{L}{2}$) = $2\rho$
- Density of the rod = $\rho$

2. Moment of Inertia:
The moment of inertia of the rod about the hinge at the bottom is given by:

$I=\frac{1}{3}m{L}^2=\frac{1}{3}\rho A{L}^3$

3. Restoration Torque Calculation:
When the rod is tilted by a small angle $\theta$:
- Weight of the rod ($W$): Acts downwards at its center of mass ($\frac{L}{2}$ from the hinge).
Torque due to gravity about the hinge (destabilizing/deflecting torque):

${\mathrm{\tau ;}}_g=mg\left(\frac{L}{2}\right)sin\theta \approx \left(\rho AL\right)g\left(\frac{L}{2}\right)\theta =\frac{1}{2}\rho Ag{L}^2\theta$

- Buoyant Force from lower liquid ($F_{\mathrm{b;}1}$):
Length immersed in lower liquid = $\frac{L}{2}$
Magnitude: $F_{\mathrm{b;}1}=\left(6\rho \right)\left(A\frac{L}{2}\right)g=3\rho ALg$
This force acts vertically upwards at the midpoint of the lower section, i.e., at a distance of $\frac{L}{4}$ from the hinge.
Restoring torque due to lower buoyant force:

${\mathrm{\tau ;}}_{\mathrm{b;}1}=F_{\mathrm{b;}1}\left(\frac{L}{4}\right)\theta =\left(3\rho ALg\right)\left(\frac{L}{4}\right)\theta =\frac{3}{4}\rho Ag{L}^2\theta$

- Buoyant Force from upper liquid ($F_{\mathrm{b;}2}$):
Length immersed in upper liquid = $\frac{L}{2}$
Magnitude: $F_{\mathrm{b;}2}=\left(2\rho \right)\left(A\frac{L}{2}\right)g=\rho ALg$
This force acts vertically upwards at the center of the upper segment, i.e., at a distance of $\frac{L}{2}+\frac{L}{4}=\frac{3L}{4}$ from the hinge.
Restoring torque due to upper buoyant force:

${\mathrm{\tau ;}}_{\mathrm{b;}2}=F_{\mathrm{b;}2}\left(\frac{3L}{4}\right)\theta =\left(\rho ALg\right)\left(\frac{3L}{4}\right)\theta =\frac{3}{4}\rho Ag{L}^2\theta$

4. Net Restoring Torque:

${\mathrm{\tau ;}}_{\text{net}}={\mathrm{\tau ;}}_{\mathrm{b;}1}+{\mathrm{\tau ;}}_{\mathrm{b;}2}-{\mathrm{\tau ;}}_g$

${\mathrm{\tau ;}}_{\text{net}}=\left(\frac{3}{4}+\frac{3}{4}-\frac{1}{2}\right)\rho Ag{L}^2\theta =\left(\frac{6}{4}-\frac{2}{4}\right)\rho Ag{L}^2\theta =\rho Ag{L}^2\theta$

5. Time Period of Small Oscillations:
Using equation of motion for angular oscillations