17. A homogeneous solid cylinder of length L and cross-sectional area A/5 is immersed such that it f

Q: Why do we add the buoyant forces from both liquids?

Because the cylinder is partially submerged in both liquids, both liquids are being displaced and therefore both exert an upward buoyant force on the cylinder according to Archimedes' Principle.

Q: Does the atmospheric pressure $p_0$ affect the equilibrium of the cylinder?

No, because the atmospheric pressure acts equally on the top surface of the upper liquid and is transmitted throughout, resulting in no net vertical force on the cylinder due to atmospheric pressure alone.

Q: What is the condition for an object to float at the interface of two liquids?

The density of the object must be strictly between the densities of the two liquids, i.e., $d_{\text{upper}} < D < d_{\text{lower}}$.

Q: Why did the cross-sectional area $A/5$ cancel out in the final calculation?

Both the weight and the buoyant forces are directly proportional to the cross-sectional area of the cylinder, so the area does not affect the floating depth or the density ratio.

Q: What would happen if we pushed the cylinder down slightly and released it?

It would experience a net restoring force proportional to the displacement, causing it to undergo Simple Harmonic Motion (SHM) about its equilibrium position.

Question

A homogeneous solid cylinder of length $L$ and cross-sectional area $A/5$ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length $L/4$ in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure $p_0$. Then, density $D$ of solid is given by

Options:

(A) $\frac{5}{4}d$

(B) $\frac{4}{5}d$

(C) $4d$

(D) $\frac{d}{5}$

Sigma Percentile · Accepted Answer

### Introduction: The Magic of Floating Interfaces

Imagine standing at the boundary of two completely different worlds—one light and airy, the other dense and deep.

In the world of fluid mechanics, this boundary is the interface between two immiscible liquids.

When a solid object is placed at this interface, it experiences a unique tug-of-war of buoyant forces from both sides.

This classic JEE Advanced problem from 1995 invites us to explore this delicate balance of forces and find the exact density of a floating cylinder. Let's dive in!

---

### Analyzing the Setup

Let's first visualize the physical scenario. We have a container holding two immiscible liquids:
1. An **upper liquid** of lower density $d$.
2. A **lower liquid** of higher density $2d$.

A homogeneous solid cylinder of total length $L$, cross-sectional area $A/5$, and density $D$ is floating vertically at the interface.

The problem tells us that a length of $L/4$ of the cylinder is submerged in the denser lower liquid.

Since the total length of the cylinder is $L$, the remaining length floating in the lighter upper liquid must be:

$$L - \frac{L}{4} = \frac{3L}{4}$$

This simple geometric breakdown is our first crucial step!

---

### The Master Equation: Translational Equilibrium

For any object to float stably without sinking or rising, it must be in a state of **translational equilibrium**.

This means the net vertical force acting on the cylinder must be exactly zero.

Let's identify the forces acting on our cylinder:
- A downward gravitational force, which is the **weight** ($W$) of the cylinder.
- An upward buoyant force, or **upthrust** ($F_B$), exerted by the displaced liquids.

At equilibrium, these forces must balance perfectly:

$$	ext{Weight} = 	ext{Total Upthrust}$$

$$W = F_{B1} + F_{B2}$$

where $F_{B1}$ is the upthrust from the upper liquid and $F_{B2}$ is the upthrust from the lower liquid.

---

### Calculating the Forces

Let's express each of these forces mathematically.

#### 1. Weight of the Cylinder ($W$)
The weight is given by the mass of the cylinder times gravity ($Mg$). Since mass is volume times density:

$$W = 	ext{Volume} 	imes 	ext{Density} 	imes g$$

The volume of the cylinder is its cross-sectional area ($A/5$) multiplied by its total length ($L$):

$$W = \left( \frac{A}{5} 	imes L ight) 	imes D 	imes g$$

#### 2. Upthrust from the Upper Liquid ($F_{B1}$)
According to Archimedes' Principle, the upthrust is equal to the weight of the liquid displaced.

The volume of the cylinder in the upper liquid is $\frac{A}{5} 	imes \frac{3L}{4}$. Since the density of this liquid is $d$:

$$F_{B1} = \left( \frac{A}{5} 	imes \frac{3L}{4} ight) 	imes d 	imes g$$

#### 3. Upthrust from the Lower Liquid ($F_{B2}$)
Similarly, the volume of the cylinder in the lower liquid is $\frac{A}{5} 	imes \frac{L}{4}$. Since the density of this liquid is $2d$:

$$F_{B2} = \left( \frac{A}{5} 	imes \frac{L}{4} ight) 	imes (2d) 	imes g$$

---

### Solving for the Density $D$

Now, let's substitute these expressions back into our force balance equation:

$$\left( \frac{A}{5} 	imes L ight) D g = \left( \frac{A}{5} 	imes \frac{3L}{4} ight) d g + \left( \frac{A}{5} 	imes \frac{L}{4} ight) (2d) g$$

Notice how beautifully the common terms cancel out! The factor $\left( \frac{A}{5} ight) L g$ is present in every single term. Dividing both sides by this factor, we get:

$$D = \frac{3}{4} d + \frac{1}{4} (2d)$$

Simplifying the right-hand side:

$$D = \frac{3}{4} d + \frac{2}{4} d$$

$$D = \frac{5}{4} d$$

Thus, the density of the solid cylinder is **$\frac{5}{4}d$** (or $1.25d$).

---

### Deep Physical Insights

Let's look at our final result: $D = 1.25d$.

Notice that $d < 1.25d < 2d$. This is a beautiful sanity check!

For any object to float at the interface of two liquids, its density **must** be greater than the upper liquid's density but less than the lower liquid's density.

If $D$ were less than $d$, the cylinder would float entirely in the upper liquid (or even on top of it). If $D$ were greater than $2d$, it would sink to the very bottom of the container.

This elegant balance is what keeps the cylinder suspended perfectly at the interface!

Select Answer:

Visualized Solution

Understanding the Physical Setup

Condition for Translational Equilibrium

Calculating the Weight of the Cylinder

Understanding Archimedes' Principle

Upthrust from the Lighter Liquid

Upthrust from the Denser Liquid

Setting up the Master Equation

Canceling Common Terms

Solving for Solid Density $D$

Pedagogical Takeaway & Extensions

The Sigma Insight: Buoyancy and Archimedes' Principle

Introduction

Analyzing the Setup

The Master Equation

Calculating the Forces

# 1

# 2

# 3

Solving for the Density $D$

Deep Physical Insights

Similar Questions

A vessel contains oil (density $= 0.8 g/cm^{3}$ ) over mercury (density $= 13.6 g/cm^{3}$ ). A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of the sphere in $g/cm^{3}$ is

Consider a solid sphere of radius $R$ and mass density $ρ (r) = ρ_{0} (1 - \frac{r ^{2}}{R ^{2}})$ , $0 < r \leq R$ . The min. density of a liquid in which it will float is

A thin uniform cylindrical shell, closed at both ends, is partially filled with water. It is floating vertically in water in half-submerged state. If $ρ_{c}$ is the relative density of the material of the shell with respect to water, then the correct statement is that the shell is

A body floats in a liquid contained in a beaker. The whole system as shown in figure falls freely under gravity. The upthrust on the body is