When a block of iron floats in mercury at 0°C, fraction k1 of its volume is submerged, while at the

Q: Why does the density decrease when temperature increases?

As temperature increases, the thermal kinetic energy of the molecules increases, causing them to vibrate more and occupy more space. This increases the volume of the substance. Since mass remains constant, the density (mass per unit volume) must decrease.

Q: Does the expansion of the container holding the mercury affect the submerged fraction?

No. The submerged fraction depends only on the densities of the floating solid (iron) and the liquid (mercury). The expansion of the container changes the total volume of mercury but not its density, so it has no effect on how deep the block sinks.

Q: Why did we use the formula $\rho = \rho_0 / (1 + \gamma \Delta T)$ instead of $\rho = \rho_0(1 - \gamma \Delta T)$?

The formula $\rho = \rho_0 / (1 + \gamma \Delta T)$ is mathematically exact because volume expands as $V = V_0(1 + \gamma \Delta T)$ and density is inversely proportional to volume. The linear approximation $\rho \approx \rho_0(1 - \gamma \Delta T)$ is only an approximation for very small values of $\gamma \Delta T$. Since the options are written in the exact fractional form, we must use the exact relation.

Q: What would happen if the coefficient of volume expansion of mercury was equal to that of iron?

If $\gamma_{\text{Fe}} = \gamma_{\text{Hg}}$, then the ratio $k_1/k_2$ would be exactly $1$. This means the submerged fraction would remain unchanged as the temperature rises, because both densities would decrease at the exact same rate.

Q: Since mercury expands more than iron, does the block sink deeper or float higher at $60^\circ\text{C}$?

Since $\gamma_{\text{Hg}} > \gamma_{\text{Fe}}$, mercury expands more than iron, meaning its density decreases faster than iron's density. As the liquid becomes relatively less dense, the iron block must sink deeper to displace enough mass of mercury to balance its weight. Thus, $k_2 > k_1$.

Question

When a block of iron floats in mercury at $0^\circ	ext{C}$, fraction $k_1$ of its volume is submerged, while at the temperature $60^\circ	ext{C}$, a fraction $k_2$ is seen to be submerged. If the coefficient of volume expansion of iron is $\gamma_{	ext{Fe}}$ and that of mercury is $\gamma_{	ext{Hg}}$, then the ratio $k_1/k_2$ can be expressed as

Options:

(A) $\frac{1 + 60\gamma_{	ext{Fe}}}{1 + 60\gamma_{	ext{Hg}}}$

(B) $\frac{1 - 60\gamma_{	ext{Fe}}}{1 + 60\gamma_{	ext{Hg}}}$

(C) $\frac{1 + 60\gamma_{	ext{Fe}}}{1 - 60\gamma_{	ext{Hg}}}$

(D) $\frac{1 + 60\gamma_{	ext{Hg}}}{1 + 60\gamma_{	ext{Fe}}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine placing a solid block of iron into a pool of liquid mercury. Because mercury is incredibly dense—nearly $13.6	ext{ g/cm}^3$ compared to iron's $7.8	ext{ g/cm}^3$—the iron block floats easily, like an ice cube in water.

At any given temperature, the block settles into a state of **translational equilibrium**. The downward gravitational force acting on the block (its weight) is perfectly balanced by the upward buoyant force exerted by the mercury (the upthrust).

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

Let $V$ be the total volume of the iron block, and $ho_{	ext{Fe}}$ be its density. The weight of the block is:

$$W = V ho_{	ext{Fe}} g$$

If a fraction $k$ of the block's volume is submerged, the volume of mercury displaced is $k V$. The buoyant force is equal to the weight of this displaced mercury:

$$F_B = (k V) ho_{	ext{Hg}} g$$

Equating these two forces gives us our master relation:

$$V ho_{	ext{Fe}} g = k V ho_{	ext{Hg}} g \implies k = \frac{ho_{	ext{Fe}}}{ho_{	ext{Hg}}}$$

This simple, elegant equation tells us that the submerged fraction $k$ is purely determined by the ratio of the density of the floating solid to the density of the liquid.

### The Master Equation at Different Temperatures

Now, let's look at how this system behaves at two different temperatures: $0^\circ	ext{C}$ and $60^\circ	ext{C}$.

At $0^\circ	ext{C}$, the submerged fraction is $k_1$. Using our master relation, we can write:

$$k_1 = \frac{(ho_{	ext{Fe}})_0}{(ho_{	ext{Hg}})_0}$$

At $60^\circ	ext{C}$, the temperature of the system is raised. Both the iron block and the mercury expand, causing their densities to decrease. The new submerged fraction is $k_2$:

$$k_2 = \frac{(ho_{	ext{Fe}})_{60}}{(ho_{	ext{Hg}})_{60}}$$

To find the ratio $k_1/k_2$, we need to understand how density varies with temperature.

### Thermal Expansion of Density

When a material is heated, its molecules vibrate more intensely, pushing each other slightly further apart. This causes the volume of the material to expand. The volume $V_	heta$ at a temperature $	heta$ is related to the initial volume $V_0$ by:

$$V_	heta = V_0(1 + \gamma 	heta)$$

where $\gamma$ is the coefficient of volume expansion.

Since mass $m$ is conserved during heating, the density $ho_	heta$ must decrease as volume increases:

$$ho_	heta = \frac{m}{V_	heta} = \frac{m}{V_0(1 + \gamma 	heta)} = \frac{ho_0}{1 + \gamma 	heta}$$

Applying this physical law to both iron and mercury for a temperature rise of $	heta = 60^\circ	ext{C}$, we get:

$$(ho_{	ext{Fe}})_{60} = \frac{(ho_{	ext{Fe}})_0}{1 + 60\gamma_{	ext{Fe}}}$$

$$(ho_{	ext{Hg}})_{60} = \frac{(ho_{	ext{Hg}})_0}{1 + 60\gamma_{	ext{Hg}}}$$

### Final Calculation

Let's substitute these temperature-dependent densities back into our expression for $k_2$:

$$k_2 = \frac{\frac{(ho_{	ext{Fe}})_0}{1 + 60\gamma_{	ext{Fe}}}}{\frac{(ho_{	ext{Hg}})_0}{1 + 60\gamma_{	ext{Hg}}}} = \frac{(ho_{	ext{Fe}})_0}{(ho_{	ext{Hg}})_0} 	imes \frac{1 + 60\gamma_{	ext{Hg}}}{1 + 60\gamma_{	ext{Fe}}}$$

Notice that the term $\frac{(ho_{	ext{Fe}})_0}{(ho_{	ext{Hg}})_0}$ is exactly equal to $k_1$. Therefore, we can write:

$$k_2 = k_1 	imes \frac{1 + 60\gamma_{	ext{Hg}}}{1 + 60\gamma_{	ext{Fe}}}$$

Rearranging this equation to find the ratio $k_1/k_2$ yields:

$$\frac{k_1}{k_2} = \frac{1 + 60\gamma_{	ext{Fe}}}{1 + 60\gamma_{	ext{Hg}}}$$

This beautifully matches **option (a)**.

Select Answer:

Visualized Solution

Understanding the Floating Condition

Submerged Fraction Formula

Submerged Fraction at $0^{\circ} C$

Submerged Fraction at $6 0^{\circ} C$

Thermal Expansion of Density

Substituting Density Relations into $k_{2}$

Calculating the Ratio $\frac{k _{1}}{k _{2}}$

Matching with Options

Conceptual Extension

The Sigma Insight: Thermal Expansion

Analyzing the Setup

The Master Equation at Different Temperatures

Thermal Expansion of Density

Final Calculation

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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

Select Answer:

Visualized Solution

Understanding the Floating Condition

Submerged Fraction Formula

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

Submerged Fraction at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 60∘C

Thermal Expansion of Density

Substituting Density Relations into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k2​

Calculating the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k2​k1​​

Matching with Options

Conceptual Extension

The Sigma Insight: Thermal Expansion

Analyzing the Setup

The Master Equation at Different Temperatures

Thermal Expansion of Density

Final Calculation

Similar Questions

Submerged Fraction at $0^{\circ} C$

Submerged Fraction at $6 0^{\circ} C$

Substituting Density Relations into $k_{2}$

Calculating the Ratio $\frac{k _{1}}{k _{2}}$