A tank contains two immiscible liquids of densities 6 and 2 . The...

Let $A$ be the cross-sectional area of the rod.

Mass of the rod, $M = \rho A L$.

Weight of the rod, $W = \rho A L g$, acting at a distance $L/2$ from the hinge.

When the rod is displaced by a small angle $\theta$, the restoring torque is provided by the buoyant forces from the two liquids.

Buoyant force due to the lower liquid, $F_{B1} = (6\rho) A (L/2) g = 3\rho A L g$.

This force acts at the center of the immersed part in the lower liquid, which is at a distance $L/4$ from the hinge.

Torque due to $F_{B1}$ is $\tau_{B1} = F_{B1} (L/4) \theta = (3\rho A L g) (L/4) \theta = \dfrac{3}{4} \rho A L^2 g \theta$.

Buoyant force due to the upper liquid, $F_{B2} = (2\rho) A (L/2) g = \rho A L g$.

This force acts at the center of the immersed part in the upper liquid, which is at a distance $L/2 + L/4 = 3L/4$ from the hinge.

Torque due to $F_{B2}$ is $\tau_{B2} = F_{B2} (3L/4) \theta = (\rho A L g) (3L/4) \theta = \dfrac{3}{4} \rho A L^2 g \theta$.

Destabilizing torque due to gravity is $\tau_g = W (L/2) \theta = (\rho A L g) (L/2) \theta = \dfrac{1}{2} \rho A L^2 g \theta$.

Net restoring torque, $\tau_{net} = \tau_{B1} + \tau_{B2} - \tau_g = \left(\dfrac{3}{4} + \dfrac{3}{4} - \dfrac{1}{2}\right) \rho A L^2 g \theta = \rho A L^2 g \theta$.

Moment of inertia of the rod about the hinge, $I = \dfrac{M L^2}{3} = \dfrac{\rho A L^3}{3}$.

The equation of motion is $I \alpha = -\tau_{net}$

$\dfrac{\rho A L^3}{3} \alpha = -\rho A L^2 g \theta$

$\alpha = -\dfrac{3g}{L} \theta$

This represents simple harmonic motion with $\omega^2 = \dfrac{3g}{L}$.

Time period, $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{\sqrt{3}} \sqrt{\dfrac{L}{g}}$.

Answer: $\sqrt{3} = 1.73$