A cube of coefficient of linear expansion alpha_s is floating in a bath containing a liquid of coeff

Q: Why does the weight of the cube remain constant when heated?

Heating an object changes its volume and density, but its mass remains strictly conserved. Since weight is mass times gravity ($W = mg$), the weight of the cube does not change.

Q: Why do we use $2\alpha_s$ for the expansion of the cube's area?

The coefficient of linear expansion is $\alpha_s$. For a two-dimensional surface like the base area of the cube, the coefficient of superficial (areal) expansion is $\beta = 2\alpha_s$.

Q: Why does the density of the liquid decrease?

Density is mass divided by volume ($\rho = M/V$). When the liquid is heated, its volume increases due to thermal expansion. Since its mass is constant, the density must decrease.

Q: Does the total volume of the liquid in the bath matter?

No, Archimedes' principle only depends on the density of the liquid and the volume of the liquid displaced by the submerged part of the object. The total volume of the bath is irrelevant.

Q: What would happen if $\gamma_l$ was greater than $2\alpha_s$?

If $\gamma_l > 2\alpha_s$, the liquid's density would decrease faster than the cube's area expands. To displace enough weight of liquid to stay afloat, the cube would have to sink deeper, meaning $h_i$ would increase.

Question

A cube of coefficient of linear expansion $\alpha_s$ is floating in a bath containing a liquid of coefficient of volume expansion $\gamma_l$. When the temperature is raised by $\Delta T$, the depth upto which the cube is submerged in the liquid remains the same. Find the relation between $\alpha_s$ and $\gamma_l$ showing all the steps.

Sigma Percentile · Accepted Answer

The problem of a floating cube undergoing thermal expansion is a beautiful intersection of Archimedes' principle and thermodynamics. It challenges us to think about how multiple physical properties—volume, area, and density—change simultaneously when heat is applied.

### The Principle of Flotation

Imagine a solid cube floating peacefully in a liquid bath. For it to remain in equilibrium, the upward buoyant force, commonly known as **upthrust**, must perfectly balance the downward pull of gravity, which is its weight.

According to Archimedes' principle, the upthrust is equal to the weight of the liquid displaced by the submerged portion of the cube. If we let the cube's cross-sectional area be $A$ and its submerged depth be $h_i$, the immersed volume is $V_i = A h_i$. Therefore, the initial upthrust $F_B$ can be written as:

$$F_B = A h_i \rho_L g$$

where $\rho_L$ is the initial density of the liquid and $g$ is the acceleration due to gravity.

### Turning Up the Heat

Now, we introduce thermal energy into the system, raising the temperature by $\Delta T$. This causes both the solid cube and the liquid to expand. However, the problem provides a fascinating constraint: the submerged depth $h_i$ remains exactly the same.

Even though the cube has expanded, its mass is conserved, meaning its weight remains constant. Consequently, the new upthrust $F_B'$ must still equal the original weight to maintain flotation:

$$F_B' = F_B$$

### The Master Equation

Let's determine the new parameters after heating. The cube's cross-sectional area expands according to the laws of superficial expansion. Since the coefficient of linear expansion is $\alpha_s$, the areal expansion coefficient is $2\alpha_s$. The new area $A'$ is:

$$A' = A(1 + 2\alpha_s \Delta T)$$

Simultaneously, the liquid expands volumetrically. As its volume increases, its density decreases. The new density $\rho_L'$ is given by:

$$\rho_L' = \frac{\rho_L}{1 + \gamma_l \Delta T}$$

where $\gamma_l$ is the coefficient of volume expansion of the liquid. The new upthrust is the product of the new immersed volume ($A' h_i$), the new density, and gravity:

$$F_B' = A(1 + 2\alpha_s \Delta T) h_i \left( \frac{\rho_L}{1 + \gamma_l \Delta T} \right) g$$

### The Elegant Conclusion

Equating the initial and final upthrusts, we get our master equation:

$$A h_i \rho_L g = A(1 + 2\alpha_s \Delta T) h_i \left( \frac{\rho_L}{1 + \gamma_l \Delta T} \right) g$$

Notice how the physical constants and initial dimensions—$A$, $h_i$, $\rho_L$, and $g$—appear on both sides. Canceling them out simplifies the equation dramatically:

$$1 = \frac{1 + 2\alpha_s \Delta T}{1 + \gamma_l \Delta T}$$

Cross-multiplying yields:

$$1 + \gamma_l \Delta T = 1 + 2\alpha_s \Delta T$$

Subtracting 1 from both sides and dividing by $\Delta T$, we arrive at the final, elegant relation:

$$\gamma_l = 2\alpha_s$$

This result tells us a profound physical truth: for the submerged depth to remain constant, the volumetric expansion of the liquid must perfectly match the areal expansion of the solid's base!

Visualized Solution

\text{Condition for Flotation}

\text{Initial Upthrust}

\text{Heating the System}

\text{Final Upthrust}

\text{Thermal Expansion Formulas}

\text{Equating Upthrusts}

\text{Simplification}

\text{Final Relation}

The Sigma Insight: Thermal Expansion

The Principle of Flotation

Turning Up the Heat

The Master Equation

The Elegant Conclusion

Similar Questions

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