A leak proof cylinder of length 1 m, made of a metal which has very low coefficient of expansion is

Q: Why do we ignore the thermal expansion of the cylinder?

The problem explicitly states that the metal has a 'very low coefficient of expansion'. This means its volume and cross-sectional area remain practically constant between $0^\circ\text{C}$ and $4^\circ\text{C}$.

Q: Why does the cylinder rise when the temperature increases to $4^\circ\text{C}$?

Water exhibits anomalous expansion, meaning its density increases as it is heated from $0^\circ\text{C}$ to $4^\circ\text{C}$. Since the water becomes denser, a smaller submerged volume is required to displace the same weight of water.

Q: Can we use the total volume of the cylinder in the buoyant force formula?

No, Archimedes' principle states that the buoyant force depends only on the *submerged* volume of the object, which is the volume of the displaced fluid.

Q: What happens to the buoyant force when the temperature changes?

The buoyant force remains exactly the same! It must always perfectly balance the constant weight of the cylinder for it to remain floating in equilibrium.

Q: Why is the density ratio $\frac{\rho_4}{\rho_0}$ greater than 1?

Because water is denser at $4^\circ\text{C}$ than at $0^\circ\text{C}$. The maximum density of water occurs at $4^\circ\text{C}$, making the ratio slightly greater than 1.

Question

A leak proof cylinder of length $1	ext{ m}$, made of a metal which has very low coefficient of expansion is floating vertically in water at $0^{\circ}	ext{C}$ such that its height above the water surface is $20	ext{ cm}$. When the temperature of water is increased to $4^{\circ}	ext{C}$, the height of the cylinder above the water surface becomes $21	ext{ cm}$. The density of water at $T = 4^{\circ}	ext{C}$, relative to the density at $T = 0^{\circ}	ext{C}$ is close to

Options:

(A) 1.26

(B) 1.03

(C) 1.01

(D) 1.04

Sigma Percentile · Accepted Answer

### The Mystery of the Floating Cylinder

Imagine you are observing a leak-proof metal cylinder floating peacefully in a beaker of water. The water is at exactly $0^\circ	ext{C}$. The total length of this cylinder is $1	ext{ m}$, which is $100	ext{ cm}$. You notice that $20	ext{ cm}$ of the cylinder is peeking out above the water surface. This immediately tells us a crucial piece of information: the submerged length of the cylinder is $100	ext{ cm} - 20	ext{ cm} = 80	ext{ cm}$.

Now, the environment changes. The water is slowly heated to $4^\circ	ext{C}$. You might expect the cylinder to sink a bit due to the thermal expansion of the water, but instead, it rises! The height of the cylinder above the water surface becomes $21	ext{ cm}$. This means the submerged length has decreased to $100	ext{ cm} - 21	ext{ cm} = 79	ext{ cm}$. The question asks us to find the ratio of the density of water at $4^\circ	ext{C}$ to its density at $0^\circ	ext{C}$. Let's dive into the physics behind this fascinating phenomenon.

### Archimedes to the Rescue

To solve this, we need to invoke one of the most elegant principles in fluid mechanics: **Archimedes' Principle**. It states that any object floating in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces.

Because the cylinder is floating in equilibrium, the downward force of gravity (its weight, $W$) is perfectly balanced by the upward buoyant force ($F_B$).

$$W = F_B$$

The buoyant force can be mathematically expressed as the volume of the displaced fluid multiplied by the density of the fluid and the acceleration due to gravity:

$$F_B = V_{	ext{sub}} \cdot ho_{	ext{liquid}} \cdot g$$

Since the cylinder has a uniform cross-sectional area $A$, the submerged volume is simply $V_{	ext{sub}} = A \cdot h_{	ext{sub}}$.

### The First State: Freezing Point

Let's apply this to our first scenario at $0^\circ	ext{C}$. The submerged height is $80	ext{ cm}$. Let the density of water at this temperature be $ho_0$. The weight of the cylinder is balanced by the buoyant force:

$$W = (A \cdot 80) \cdot ho_0 \cdot g$$

This is our first master equation. It perfectly describes the physical state of the cylinder at the freezing point of water.

### The Second State: The Anomaly of Water

Now, let's look at the second scenario at $4^\circ	ext{C}$. The problem explicitly states that the metal has a "very low coefficient of expansion." This is a critical hint! It means we can safely assume that the volume and cross-sectional area $A$ of the cylinder remain constant despite the temperature change. Its weight $W$ also remains constant.

However, the water has changed. The submerged height is now $79	ext{ cm}$. Let the density of water at $4^\circ	ext{C}$ be $ho_4$. We can write the new equilibrium equation:

$$W = (A \cdot 79) \cdot ho_4 \cdot g$$

### The Master Equation

Since the weight $W$ of the cylinder is the same in both cases, we can equate the two buoyant forces. This is the logical bridge that connects the two states:

$$(A \cdot 80) \cdot ho_0 \cdot g = (A \cdot 79) \cdot ho_4 \cdot g$$

This equation is beautiful because it allows us to eliminate the unknowns. The cross-sectional area $A$ and the acceleration due to gravity $g$ are present on both sides. We can simply cancel them out:

$$80 \cdot ho_0 = 79 \cdot ho_4$$

### The Final Calculation

We are asked to find the relative density of water at $4^\circ	ext{C}$ with respect to $0^\circ	ext{C}$, which is the ratio $\frac{ho_4}{ho_0}$. Rearranging our simplified equation, we get:

$$\frac{ho_4}{ho_0} = \frac{80}{79}$$

Now, it's just a matter of simple division.

$$\frac{80}{79} \approx 1.0126$$

Looking at our options, the closest value is **1.01**.

This result is not just a number; it represents a profound physical reality known as the **anomalous expansion of water**. Unlike most liquids that expand and become less dense as they are heated, water actually contracts and becomes *more* dense as it is heated from $0^\circ	ext{C}$ to $4^\circ	ext{C}$. It reaches its maximum density at exactly $4^\circ	ext{C}$. Because the water at $4^\circ	ext{C}$ is denser, a smaller volume of it needs to be displaced to support the same weight of the cylinder, which is exactly why the cylinder rises!

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Visualized Solution

$Initial State at 0^{\circ} C$

$Archimedes’ Principle$

$Buoyancy at 0^{\circ} C$

$Final State at 4^{\circ} C$

$Buoyancy at 4^{\circ} C$

$Equating the Weights$

$Simplifying the Equation$

$Calculating the Ratio$

$Anomalous Expansion of Water$

The Sigma Insight: Buoyancy and Archimedes' Principle

The Mystery of the Floating Cylinder

Archimedes to the Rescue

The First State

The Second State

The Master Equation

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial State at 0∘C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Archimedes’ Principle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Buoyancy at 0∘C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final State at 4∘C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Buoyancy at 4∘C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating the Weights

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Anomalous Expansion of Water