Animated Solution for Physics - Oscillations: A tank contains two immiscible liquids of densities $6\rho$ and $2\rho$. The higher density liquid is filled up to a height $L/2$ 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, 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:

A Dance of Fluids and Gravity

Imagine a serene tank filled with two distinct, immiscible liquids. The heavier liquid, dense and resolute at $6\rho$, settles at the bottom, claiming exactly half the height, $L/2$. Above it floats the lighter liquid, with a density of $2\rho$.

Now, introduce the protagonist of our physics play: a thin, uniform rod of length $L$ and density $\rho$. It is hinged perfectly at the bottom of the tank, standing tall and fully submerged. But what happens when we disrupt this perfect equilibrium? If we give the rod a slight nudge, displacing it by a tiny angle $\theta$, it doesn't just fall over. It begins to oscillate, caught in a beautiful tug-of-war between gravity pulling it down and buoyancy pushing it back up. Our mission is to decode the rhythm of this dance and find its time period of oscillation.

Buoyancy vs. Weight

To understand the motion, we must first unmask the invisible forces acting on the rod.

First, there is the undeniable pull of gravity. The weight of the rod, $W=mg$, acts straight down from its center of mass. Since the rod is uniform and has a length $L$, its center of mass is exactly at the midpoint, a distance of $L/2$ from the hinge.

But the liquids are not passive observers; they fight back with buoyancy! Because the rod spans across two different liquids, it experiences two distinct buoyant forces. - The lower half of the rod is submerged in the denser liquid ($6\rho$). This liquid exerts an upward buoyant force, $F_{B2}$, acting at the center of this lower submerged section, which is at a distance of $L/4$ from the hinge. - The upper half of the rod is bathed in the lighter liquid ($2\rho$). This liquid exerts its own upward buoyant force, $F_{B1}$, acting at the center of the upper section. This point is located at $L/2+L/4=3L/4$ from the hinge.

Balancing the Torques

With our forces identified, we can now set up the master equation for rotational motion. The rod is hinged, so we must calculate the torque ($\tau$) about this pivot point.

When the rod is tilted by a small angle $\theta$, the perpendicular distance for each force becomes the distance along the rod multiplied by $\sin \theta$. The buoyant forces try to restore the rod to its vertical position (positive torque), while gravity tries to pull it further down (negative torque).

The net restoring torque is:

For small oscillations, we can use the small-angle approximation, $\sin \theta \approx \theta$. Now, let's substitute the actual values for the forces. The mass of the rod is $m=\rho AL$. The buoyant forces are equal to the weight of the displaced liquids:

Plugging these into our torque equation:

Factoring out $\rho Ag{L}^2$, we get a beautifully simple expression:

Finding the Time Period

We have our restoring torque, and now we bring in Newton's Second Law for rotation: $\tau =-I\alpha$, where $I$ is the moment of inertia and $\alpha$ is the angular acceleration.

For a uniform rod hinged at one end, the moment of inertia is $I=\frac{m{L}^2}{3}$. Substituting $m=\rho AL$, we get $I=\frac{\rho A{L}^3}{3}$.

Equating our torques:

Solving for $\alpha$, we find the defining equation of Simple Harmonic Motion:

This matches the standard SHM form $\alpha =-{\omega }^2\theta$, revealing that the square of the angular frequency is ${\omega }^2=\frac{3g}{L}$.

The time period $T$ is simply $2\pi$ divided by $\omega$:

Unlocking the Value of $n$

We are at the finish line. The problem states that the time period is given by the expression $T=\frac{2\pi }{n}\sqrt{\frac{L}{g}}$.

By comparing our derived formula with the given expression:

It is immediately clear that $n=\sqrt{3}$.

Calculating the numerical value, we know that $\sqrt{3}\approx 1.732$. Rounding to two decimal places as required for numerical answer type questions, we arrive at our final, triumphant answer: 1.73.