A piece of metal floats on mercury. The coefficients of volume expansion of the metal and mercury ar

Q: Why does the density decrease when temperature increases?

When temperature increases, the volume of a substance increases due to thermal expansion while its mass remains constant. Since density is mass divided by volume ($\rho = M/V$), an increase in volume directly leads to a decrease in density.

Q: Can we use linear expansion coefficient instead of volume expansion coefficient?

For liquids like mercury, only volume expansion is defined. For the solid metal, if the linear expansion coefficient $\alpha$ is given, we can approximate the volume expansion coefficient as $\gamma \approx 3\alpha$.

Q: What happens if the container also expands?

The expansion of the container changes its capacity but does not affect the density of the liquid or the solid inside it. Therefore, the floating depth and submerged fraction depend only on the properties of the liquid and the floating body, not the container's expansion.

Q: Why is the formula for density variation $\rho' = \frac{\rho}{1 + \gamma \Delta T}$ and not $\rho' = \rho(1 - \gamma \Delta T)$?

The exact relation is derived from $V' = V(1 + \gamma \Delta T)$. Since $\rho' = M/V'$, we get $\rho' = \frac{M}{V(1 + \gamma \Delta T)} = \frac{\rho}{1 + \gamma \Delta T}$. The linear approximation $\rho' \approx \rho(1 - \gamma \Delta T)$ is only valid for very small values of $\gamma \Delta T$.

Q: What if the metal block completely sinks after heating?

This would happen if the density of mercury becomes less than the density of the metal at the higher temperature ($\rho_2' < \rho_1'$). In our problem, we assume the metal continues to float, which is guaranteed as long as the density of the metal remains significantly lower than that of mercury.

Question

A piece of metal floats on mercury. The coefficients of volume expansion of the metal and mercury are $\gamma_1$ and $\gamma_2$ respectively. If the temperatures of both mercury and the metal are increased by an amount $\Delta T$, the fraction of the volume of the metal submerged in mercury changes by the factor ......

Sigma Percentile · Accepted Answer

### Introduction to Buoyancy and Thermal Expansion Imagine a massive steel ship floating effortlessly on the ocean, or a simple ice cube bobbing in a glass of water. These everyday phenomena are governed by the elegant principles of fluid mechanics—specifically, **Archimedes' Principle**. But what happens when we introduce thermodynamics into the mix? When temperature changes, materials expand, densities shift, and the delicate balance of forces is altered. In this article, we will explore a classic JEE problem that beautifully bridges the worlds of fluid mechanics and thermal physics: a piece of metal floating on mercury under a temperature change. --- ### The Physics of Floatation Let us begin by establishing our physical foundation. When a body floats in a fluid, it is in a state of static equilibrium. This means the net force acting on the body must be zero. There are only two vertical forces acting on our floating metal block: 1. The downward gravitational force, which is the **weight** of the block ($W$). 2. The upward buoyant force, also known as the **upthrust** ($F_B$), exerted by the displaced fluid. Mathematically, we can write: $$W = F_B$$ Let the total volume of the metal block be $V$ and its density at the initial temperature be $\rho_1$. The weight of the block is given by: $$W = V \rho_1 g$$ According to Archimedes' Principle, the buoyant force is equal to the weight of the fluid displaced by the submerged portion of the body. If the submerged volume of the metal block is $V_i$ and the density of mercury is $\rho_2$, then the buoyant force is: $$F_B = V_i \rho_2 g$$ Equating the two forces, we get: $$V \rho_1 g = V_i \rho_2 g$$ We can simplify this by cancelling the acceleration due to gravity ($g$) from both sides: $$V \rho_1 = V_i \rho_2$$ --- ### Defining the Submerged Fraction The problem asks us to track how the "fraction of the volume of the metal submerged in mercury" changes. Let us define this submerged fraction as $x$. By definition, $x$ is the ratio of the submerged volume ($V_i$) to the total volume ($V$): $$x = \frac{V_i}{V}$$ Using our simplified force balance equation, we can express this fraction as a ratio of the densities: $$x = \frac{\rho_1}{\rho_2}$$ This is a remarkably simple and powerful result! It tells us that the fraction of a floating body that is submerged depends solely on the ratio of the density of the floating body to the density of the fluid. For example, if an object is $90\%$ as dense as the liquid it floats in, then $90\%$ of its volume will be submerged. --- ### Enter Thermodynamics: Thermal Expansion Now, let us raise the temperature of both the metal and the mercury by an amount $\Delta T$. As temperature rises, almost all materials undergo thermal expansion. Since the mass of both the metal and the mercury remains constant, an increase in volume must lead to a decrease in density. Let the coefficient of volume expansion of the metal be $\gamma_1$ and that of mercury be $\gamma_2$. The volume of a substance at an elevated temperature is given by: $$V' = V(1 + \gamma \Delta T)$$ Since density is mass divided by volume ($\rho = M/V$), the new density $\rho'$ can be expressed in terms of the initial density $\rho$ as: $$\rho' = \frac{M}{V'} = \frac{M}{V(1 + \gamma \Delta T)} = \frac{\rho}{1 + \gamma \Delta T}$$ Applying this relation to both the metal and the mercury, we find their new densities at temperature $T + \Delta T$: $$\rho_1' = \frac{\rho_1}{1 + \gamma_1 \Delta T}$$ $$\rho_2' = \frac{\rho_2}{1 + \gamma_2 \Delta T}$$ --- ### The Master Derivation With our new densities in hand, we can now calculate the new submerged fraction, which we will call $x'$: $$x' = \frac{\rho_1'}{\rho_2'}$$ Substituting our temperature-dependent density expressions into this equation: $$x' = \frac{\frac{\rho_1}{1 + \gamma_1 \Delta T}}{\frac{\rho_2}{1 + \gamma_2 \Delta T}}$$ To simplify this complex fraction, we can multiply by the reciprocal of the denominator: $$x' = \left(\frac{\rho_1}{\rho_2}\right) \cdot \left(\frac{1 + \gamma_2 \Delta T}{1 + \gamma_1 \Delta T}\right)$$ Recall that the initial submerged fraction was $x = \frac{\rho_1}{\rho_2}$. Substituting this back into our equation yields: $$x' = x \cdot \left(\frac{1 + \gamma_2 \Delta T}{1 + \gamma_1 \Delta T}\right)$$ To find the factor by which the submerged fraction changes, we simply find the ratio of the final fraction to the initial fraction: $$\frac{x'}{x} = \frac{1 + \gamma_2 \Delta T}{1 + \gamma_1 \Delta T}$$ Thus, the submerged fraction changes by the factor **$\frac{1 + \gamma_2 \Delta T}{1 + \gamma_1 \Delta T}$**. --- ### Physical Intuition & Special Cases Let's step back from the algebra and look at the physical story this equation tells us. How does the relative expansion of the two materials affect the floating depth? #### Case 1: $\gamma_2 > \gamma_1$ (Mercury expands more than the metal) If mercury has a higher coefficient of volume expansion, its density will drop faster than that of the metal as temperature rises. Because the liquid is now significantly less dense, the metal block must sink deeper to displace enough mass of mercury to support its weight. Our formula confirms this: if $\gamma_2 > \gamma_1$, then the numerator is larger than the denominator, making the factor greater than $1$, meaning $x' > x$ (the submerged fraction increases). #### Case 2: $\gamma_2 < \gamma_1$ (The metal expands more than mercury) If the metal expands more than the mercury, its density drops faster. Since the floating body is now relatively lighter compared to the liquid, it floats higher in the pool. Our formula reflects this: if $\gamma_2 < \gamma_1$, the factor is less than $1$, meaning $x' < x$ (the submerged fraction decreases). #### Case 3: $\gamma_2 = \gamma_1$ (Equal expansion) If both materials expand at the exact same rate, their relative densities remain perfectly in proportion. In this case, the factor is exactly $1$, and the submerged fraction remains completely unchanged ($x' = x$).

Visualized Solution

Understanding the Floating Condition

The Floatation Equation

Defining the Submerged Fraction

Effect of Temperature on Density

The New Submerged Fraction

Simplifying the Ratio

Finding the Change Factor

Analyzing Special Cases

The Sigma Insight: Thermal Expansion

Introduction to Buoyancy and Thermal Expansion

The Physics of Floatation

Defining the Submerged Fraction

Enter Thermodynamics

The Master Derivation

Physical Intuition & Special Cases

# Case 1: $γ_{2} > γ_{1}$ (Mercury expands more than the metal)

# Case 2: $γ_{2} < γ_{1}$ (The metal expands more than mercury)

# Case 3: $γ_{2} = γ_{1}$ (Equal expansion)

Similar Questions

Two different wires having lengths $L_{1}$ and $L_{2}$ and respective temperature coefficients of linear expansion $α_{1}$ and $α_{2}$ , are joined end-to-end. Then the effective temperature coefficient of linear expansion is

Two rods $A$ and $B$ of identical dimensions are at temperature $3 0^{\circ} C$ . If $A$ is heated upto $18 0^{\circ} C$ and $B$ upto $T^{\circ} C$ , then new lengths are the same. If the ratio of the coefficients of linear expansion of $A$ and $B$ is $4 : 3$ , then the value of $T$ is

Visualized Solution

Understanding the Floating Condition

The Floatation Equation

Defining the Submerged Fraction

Effect of Temperature on Density

The New Submerged Fraction

Simplifying the Ratio

Finding the Change Factor

Analyzing Special Cases

The Sigma Insight: Thermal Expansion

Introduction to Buoyancy and Thermal Expansion

The Physics of Floatation

Defining the Submerged Fraction

Enter Thermodynamics

The Master Derivation

Physical Intuition & Special Cases

# Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } γ2​>γ1​ (Mercury expands more than the metal)

# Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } γ2​<γ1​ (The metal expands more than mercury)

# Case 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } γ2​=γ1​ (Equal expansion)

Similar Questions

# Case 1: $γ_{2} > γ_{1}$ (Mercury expands more than the metal)

# Case 2: $γ_{2} < γ_{1}$ (The metal expands more than mercury)

# Case 3: $γ_{2} = γ_{1}$ (Equal expansion)