Question 9

To find the time period of small oscillations, we first determine the restoring torque acting on the rod when it is displaced by a small angle $\theta$ from the vertical.

1. Properties of the Rod: Let the cross-sectional area of the rod be $A$. Mass of the rod, $M = \rho A L$

Moment of inertia about the hinge at the bottom, $I = \frac{1}{3} M L^2 = \frac{1}{3} (\rho A L) L^2 = \frac{1}{3} \rho A L^3$

2. Torque due to Gravity: Gravity acts at the center of mass ($L/2$ from the bottom).

$\tau_g = Mg \frac{L}{2} \sin \theta \approx (\rho A L) g \frac{L}{2} \theta = \frac{1}{2} \rho A L^2 g \theta$

This torque is destabilizing (tending to increase $\theta$).

3. Torque due to Buoyancy: The rod is in two liquids. The bottom half ($0$ to $L/2$) is in liquid of density $6\rho$. The top half ($L/2$ to $L$) is in liquid of density $2\rho$.

Buoyant force on lower part: $F_{b1} = (6\rho) (A \frac{L}{2}) g = 3\rho A L g$, acting at a distance $L/4$ from the hinge.

Torque $\tau_{b1} = F_{b1} \frac{L}{4} \theta = (3\rho A L g) \frac{L}{4} \theta = \frac{3}{4} \rho A L^2 g \theta$

Buoyant force on upper part: $F_{b2} = (2\rho) (A \frac{L}{2}) g = \rho A L g$, acting at a distance $(\frac{L}{2} + \frac{L}{4}) = \frac{3L}{4}$ from the hinge.

Torque $\tau_{b2} = F_{b2} \frac{3L}{4} \theta = (\rho A L g) \frac{3L}{4} \theta = \frac{3}{4} \rho A L^2 g \theta$

Total restoring torque from buoyancy: $\tau_b = \tau_{b1} + \tau_{b2} = \frac{3}{4} \rho A L^2 g \theta + \frac{3}{4} \rho A L^2 g \theta = \frac{3}{2} \rho A L^2 g \theta$

4. Net Restoring Torque:

$\tau_{net} = \tau_b - \tau_g = (\frac{3}{2} - \frac{1}{2}) \rho A L^2 g \theta = \rho A L^2 g \theta$

5. Equation of Motion: $I \frac{d^2\theta}{dt^2} = -\tau_{net}$

$\frac{1}{3} \rho A L^3 \frac{d^2\theta}{dt^2} = -\rho A L^2 g \theta$

$\frac{d^2\theta}{dt^2} = -\frac{3g}{L} \theta$

This is SHM with $\omega^2 = \frac{3g}{L} \implies \omega = \sqrt{\frac{3g}{L}}$.

The time period is: $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{L}{3g}} = \frac{2\pi}{\sqrt{3}} \sqrt{\frac{L}{g}}$

Comparing with the given form $T = \frac{2\pi}{n} \sqrt{\frac{L}{g}}$, we find:

$n = \sqrt{3} \approx 1.732$

Rounding off to two decimal places, $n = 1.73$.