A metal ball immersed in alcohol weighs w1 at 0°C and w2 at 50°C. The coefficient of cubical expansi

Q: Why does the density of the liquid decrease when heated?

When a liquid is heated, its molecules gain kinetic energy and move further apart, increasing the total volume. Since mass remains constant, the increase in volume results in a decrease in density according to the relation $\rho = m/V$.

Q: Does the mass of the metal ball change with temperature?

No, the mass of the metal ball remains strictly constant. Only its volume and density change due to thermal expansion.

Q: What would happen if the metal ball expanded more than the alcohol?

If $\gamma_s > \gamma_L$, the numerator in the upthrust ratio would be larger than the denominator, meaning the buoyant force would increase with temperature ($F' > F$). Consequently, the apparent weight would decrease, making $w_1 > w_2$.

Q: Why is the density of the metal assumed to be large compared to alcohol?

This assumption ensures that the metal ball remains completely submerged and does not float, allowing us to analyze it as a fully immersed body where the submerged volume equals the total volume of the solid.

Q: Does gravity change with temperature?

No, the acceleration due to gravity $g$ is a constant environmental factor and does not depend on the temperature of the system.

Question

A barometer made of a very narrow tube (see figure) is placed at normal temperature and pressure. The coefficient of volume expansion of mercury is $0.00018/^{\circ}	ext{C}$ and that of the tube is negligible. The temperature of mercury in the barometer is now raised by $1^{\circ}	ext{C}$ but the temperature of the atmosphere does not change. Then, the mercury height in the tube remains unchanged.

Options:

(A) $w_1 > w_2$

(B) $w_1 = w_2$

(C) $w_1 < w_2$

(D) All of these

Sigma Percentile · Accepted Answer

### Introduction to Buoyancy and Thermal Expansion

Imagine holding a heavy metal ball in your hand, and then lowering it into a pool of liquid.

Suddenly, the ball feels lighter!

This magical reduction in weight is not magic at all—it is the beautiful physics of **buoyancy**, first formulated by Archimedes over two thousand years ago.

But what happens when we heat the entire system?

Both the solid metal ball and the surrounding liquid will expand, changing their volumes and densities.

This problem challenges us to analyze the delicate competition between the thermal expansion of a solid and a liquid, and how it ultimately dictates the apparent weight of the submerged object.

---

### Analyzing the Forces

Let's first establish the force balance on the submerged metal ball.

When the ball is completely immersed in alcohol, it experiences two primary forces:

1. The downward gravitational force, which is its **true weight**:
   $$W_{	ext{actual}} = m g$$

2. The upward buoyant force, or **upthrust**:
   $$F = V_s ho_L g$$
   where $V_s$ is the volume of the submerged solid, and $ho_L$ is the density of the liquid.

The apparent weight $w$ measured by a scale or spring balance is the net downward force:
$$w = W_{	ext{actual}} - F$$

At $0^{\circ}	ext{C}$, the apparent weight is:
$$w_1 = W_{	ext{actual}} - F$$

At $50^{\circ}	ext{C}$, the apparent weight is:
$$w_2 = W_{	ext{actual}} - F'$$
where $F'$ is the upthrust at the elevated temperature.

---

### The Battle of Expansions

As the temperature rises by $\Delta T = 50^{\circ}	ext{C}$, both the metal ball and the alcohol expand.

Let's write down how their physical properties change:

- The volume of the solid metal ball increases to:
  $$V'_s = V_s (1 + \gamma_s \Delta T)$$

- The density of the alcohol decreases to:
  $$ho'_L = \frac{ho_L}{1 + \gamma_L \Delta T}$$

Now, let's look at the new upthrust $F'$ at $50^{\circ}	ext{C}$:
$$F' = V'_s ho'_L g = V_s (1 + \gamma_s \Delta T) \left( \frac{ho_L}{1 + \gamma_L \Delta T} ight) g$$

By comparing this to the initial upthrust $F = V_s ho_L g$, we can write the ratio:
$$\frac{F'}{F} = \frac{1 + \gamma_s \Delta T}{1 + \gamma_L \Delta T}$$

---

### Resolving the Inequality

We are given a crucial piece of information:
$$\gamma_s < \gamma_L$$

The coefficient of volume expansion of the metal is strictly less than that of the alcohol.

Since the temperature change $\Delta T$ is positive ($50^{\circ}	ext{C}$):
$$\gamma_s \Delta T < \gamma_L \Delta T$$

Adding $1$ to both sides preserves the inequality:
$$1 + \gamma_s \Delta T < 1 + \gamma_L \Delta T$$

This means the numerator of our ratio is smaller than the denominator!

Therefore:
$$\frac{F'}{F} < 1 \implies F' < F$$

Even though the metal ball expanded and displaced more volume, the alcohol expanded much more rapidly, causing its density to drop significantly.

As a result, the overall upward buoyant force **decreased**.

---

### Final Conclusion

Now, let's substitute this back into our apparent weight equations:
$$w_1 = W_{	ext{actual}} - F$$
$$w_2 = W_{	ext{actual}} - F'$$

Since $F' < F$, we are subtracting a smaller upward force at $50^{\circ}	ext{C}$ than at $0^{\circ}	ext{C}$.

Therefore, the apparent weight at the higher temperature must be larger:
$$w_2 > w_1 \implies w_1 < w_2$$

This beautifully confirms that the ball feels heavier at $50^{\circ}	ext{C}$ than at $0^{\circ}	ext{C}$ because the liquid's ability to support it via buoyancy has diminished.

Thus, the correct option is **(c)**.

Select Answer:

Visualized Solution

Understanding the Physical Setup

The Apparent Weight Equation

Effect of Temperature on Volume and Density

Setting up the Ratio of Upthrusts

Simplifying the Upthrust Ratio

Comparing the Coefficients of Expansion

Deducing the Change in Upthrust

Comparing Apparent Weights

Exploring Alternative Scenarios

The Sigma Insight: Buoyancy and Archimedes' Principle

Introduction to Buoyancy and Thermal Expansion

Analyzing the Forces

The Battle of Expansions

Resolving the Inequality

Final Conclusion

Similar Questions

Comprehension Passage

The value of $X$ is _______.

The value of $Y$ is _______.

Comprehension Passage

As the bubble moves upwards, besides the buoyancy force the following forces are acting on it.

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Select Answer:

Visualized Solution

Understanding the Physical Setup

The Apparent Weight Equation

Effect of Temperature on Volume and Density

Setting up the Ratio of Upthrusts

Simplifying the Upthrust Ratio

Comparing the Coefficients of Expansion

Deducing the Change in Upthrust

Comparing Apparent Weights

Exploring Alternative Scenarios

The Sigma Insight: Buoyancy and Archimedes' Principle

Introduction to Buoyancy and Thermal Expansion

Analyzing the Forces

The Battle of Expansions

Resolving the Inequality

Final Conclusion

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of X is _______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of Y is _______.

Comprehension Passage

As the bubble moves upwards, besides the buoyancy force the following forces are acting on it.

When the gas bubble is at a height y from the bottom, its temperature is

The buoyancy force acting on the gas bubble is (Assume R is the universal gas constant)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A wooden block, with a coin placed on its top, floats in water as shown in figure. The distance l and h are shown there. After sometime the coin falls into the water. Then

The value of $X$ is _______.

The value of $Y$ is _______.

A wooden block, with a coin placed on its top, floats in water as shown in figure. The distance $l$ and $h$ are shown there. After sometime the coin falls into the water. Then