A bakelite beaker has volume capacity of 500 cc at 30°C. When it is partially filled with V_m volume

Q: Why do we use the volume expansion coefficient ($\gamma$) instead of linear ($\alpha$)?

Because we are dealing with liquids (mercury) and the capacity of a container (beaker), both of which are 3D volumetric quantities. Liquids don't have a linear expansion coefficient since they take the shape of their container.

Q: Does the shape of the beaker matter for this problem?

Not at all! The volume expansion formula $\Delta V = V \gamma \Delta T$ depends only on the initial total volume, not the geometric shape. Whether it's a cylinder, a sphere, or an irregular flask, the volumetric expansion remains the same.

Q: Why did we cancel out $\Delta T$? What if the temperature change is very large?

We cancelled $\Delta T$ because it appears on both sides of the equation ($\Delta V_b = \Delta V_m$). As long as the temperature change is uniform across both the beaker and the mercury, it factors out. However, for extremely large $\Delta T$, the approximation $\Delta V = V \gamma \Delta T$ might lose accuracy as $\gamma$ itself can vary slightly with temperature.

Q: What would happen if $\gamma_{\text{beaker}}$ was greater than $\gamma_{\text{mercury}}$?

If the beaker expanded more rapidly than the mercury, the total capacity of the beaker would increase faster than the mercury could fill it. As a result, the unfilled volume (the empty space) would actually increase as the temperature rises.

Q: Why is the volume of mercury so small compared to the beaker?

Look at the expansion coefficients! Mercury's $\gamma$ ($1.5 \times 10^{-4}$) is 25 times larger than the beaker's $\gamma$ ($6 \times 10^{-6}$). Because mercury expands so much more aggressively, you only need a tiny amount of it (20 cc) to match the total expansion of the massive 500 cc beaker.

Question

A bakelite beaker has volume capacity of $500 	ext{ cc}$ at $30^\circ	ext{C}$. When it is partially filled with $V_m$ volume (at $30^\circ	ext{C}$) of mercury, it is found that the unfilled volume of the beaker remains constant as temperature is varied. If $\gamma_{	ext{(beaker)}} = 6 	imes 10^{-6} \, ^\circ	ext{C}^{-1}$ and $\gamma_{	ext{(mercury)}} = 1.5 	imes 10^{-4} \, ^\circ	ext{C}^{-1}$, where $\gamma$ is the coefficient of volume expansion, then $V_m$ (in cc) is close to ........... .

Sigma Percentile · Accepted Answer

### The Physics of the Empty Space

Imagine you are holding a sturdy bakelite beaker in your hands. It has a total volume capacity of $500 	ext{ cc}$. Now, you pour a small amount of liquid mercury into it. The mercury settles at the bottom, occupying a volume $V_m$.

The space that remains empty above the mercury is what we call the **unfilled volume**. Mathematically, this is simply the difference between the total capacity of the beaker and the volume of the mercury inside it.

$$V_{	ext{unfilled}} = V_b - V_m$$

This equation is our visual anchor. It perfectly describes the physical reality of the setup at any given temperature.

### The Balancing Act of Thermal Expansion

Now, the problem introduces a fascinating constraint: as we heat the system and the temperature varies, this unfilled volume remains absolutely constant.

Let's think about what happens when we increase the temperature. The bakelite beaker will undergo thermal expansion, increasing its total capacity. If the mercury didn't expand, the empty space would simply get larger. However, the mercury also expands, and it does so quite aggressively!

For the empty space to remain exactly the same, the extra volume created by the expanding beaker must be perfectly filled by the expanding mercury. This is a beautiful balancing act of thermal physics.

$$\Delta V_{	ext{unfilled}} = 0$$

$$\Delta V_b - \Delta V_m = 0 \implies \Delta V_b = \Delta V_m$$

This tells us that the **absolute change in volume** for both the beaker and the mercury must be identical.

### The Master Equation

To proceed, we need to bring in the fundamental law of volumetric thermal expansion. The change in volume $\Delta V$ of any substance is proportional to its initial volume $V$, its coefficient of volume expansion $\gamma$, and the change in temperature $\Delta T$.

$$\Delta V = V \gamma \Delta T$$

We can substitute this powerful tool into our balancing equation for both the beaker and the mercury.

$$V_b \gamma_b \Delta T = V_m \gamma_m \Delta T$$

Notice how the temperature change $\Delta T$ appears on both sides of the equation. Because the beaker and the mercury are in thermal equilibrium and experience the same temperature change, we can elegantly cancel $\Delta T$ out.

$$V_b \gamma_b = V_m \gamma_m$$

### Crunching the Numbers

We have now isolated the core relationship. Our goal is to find the initial volume of the mercury, $V_m$. Let's rearrange the equation to solve for it.

$$V_m = \frac{V_b \gamma_b}{\gamma_m}$$

Now, we carefully substitute the numerical values provided in the problem. The volume of the beaker $V_b$ is $500 	ext{ cc}$. The expansion coefficient of the bakelite beaker $\gamma_b$ is $6 	imes 10^{-6} \, ^\circ	ext{C}^{-1}$, and for the mercury $\gamma_m$ is $1.5 	imes 10^{-4} \, ^\circ	ext{C}^{-1}$.

$$V_m = \frac{500 	imes 6 	imes 10^{-6}}{1.5 	imes 10^{-4}}$$

Let's simplify the numerator first. Multiplying $500$ by $6$ gives $3000$, which can be written as $3 	imes 10^3$.

$$V_m = \frac{3 	imes 10^3 	imes 10^{-6}}{1.5 	imes 10^{-4}} = \frac{3 	imes 10^{-3}}{1.5 	imes 10^{-4}}$$

Finally, we divide the coefficients and the powers of ten. $3$ divided by $1.5$ is exactly $2$. And $10^{-3}$ divided by $10^{-4}$ leaves us with $10^1$, or $10$.

$$V_m = 2 	imes 10 = 20 	ext{ cc}$$

### The Real-World Application

The required volume of mercury is exactly **$20 	ext{ cc}$**.

Take a moment to appreciate this result. The mercury's volume is only $4\%$ of the beaker's total volume. Yet, because mercury's expansion coefficient is $25$ times larger than that of bakelite, this tiny puddle of liquid expands just enough to perfectly match the expansion of the entire massive beaker!

This exact principle—balancing different rates of thermal expansion—is the foundational engineering concept behind designing accurate liquid-in-glass thermometers and compensating pendulums in grandfather clocks.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Condition for Constant Unfilled Volume

Equating Volume Changes

Formula for Thermal Expansion

Substituting Expansion Formulas

Isolating $V_{m}$

Substituting Given Values

Simplifying the Expression

Final Answer

The Way Forward

The Sigma Insight: Thermal Expansion

The Physics of the Empty Space

The Balancing Act of Thermal Expansion

The Master Equation

Crunching the Numbers

The Real-World Application

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