A vessel contains oil (density = 0.8 g/cm3) over mercury (density = 13.6 g/cm3). A homogeneous spher

Q: Why do we add the buoyant forces of both liquids instead of just taking the denser one?

Buoyancy is caused by the pressure difference across the submerged surface of an object. Since the sphere is partially submerged in both fluids, both fluids exert upward pressure forces on their respective contact areas. Therefore, both fluids contribute to the total upward buoyant force.

Q: Does the shape of the object matter in this problem?

No, the shape does not matter as long as exactly half of the total volume is in one liquid and the other half is in the second liquid. The volume terms cancel out regardless of the geometry.

Q: What would happen if the sphere was 1/3 in oil and 2/3 in mercury?

The equilibrium equation would change to reflect the new volume fractions: $\rho = \frac{1}{3}\rho_{oil} + \frac{2}{3}\rho_{Hg}$. The density of the sphere would be a weighted average of the two liquid densities.

Q: Why does the volume of the sphere cancel out in the final calculation?

Because both the weight of the sphere and the buoyant forces are directly proportional to the volume of the sphere. This scaling means the equilibrium position depends only on the relative densities, not the size of the object.

Q: Is it possible for a sphere to float at the interface if its density is greater than mercury?

No. If the sphere's density is greater than mercury ($\rho > 13.6\text{ g/cm}^3$), it will sink to the bottom of the mercury layer because the maximum possible buoyant force from both liquids would still be less than its weight.

Question

A vessel contains oil (density $= 0.8	ext{ g/cm}^3$) over mercury (density $= 13.6	ext{ g/cm}^3$). A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of the sphere in $	ext{g/cm}^3$ is

Options:

(A) 3.3

(B) 6.4

(C) 7.2

(D) 12.8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a container holding two immiscible liquids of vastly different densities. At the bottom, we have heavy, metallic **mercury** with a density of $ho_{Hg} = 13.6	ext{ g/cm}^3$. Floating directly on top of it is a layer of **oil** with a density of $ho_{oil} = 0.8	ext{ g/cm}^3$.

Now, a homogeneous solid sphere is gently lowered into this vessel. It doesn't sink to the very bottom, nor does it float entirely on top of the oil. Instead, it finds its perfect equilibrium position right at the interface of the two liquids, with exactly **half of its volume** submerged in the mercury and the other half in the oil.

Our goal is to determine the density of this sphere, $ho$.

### The Physics of Multi-Liquid Buoyancy

For any object floating in static equilibrium, the net force acting on it must be zero. In the vertical direction, two competing forces are at play:

1. The downward force of **gravity** (the weight of the sphere, $W$).
2. The upward force of **buoyancy** ($F_B$).

Since the sphere is in contact with two different fluids, both fluids contribute to the total buoyant force. According to **Archimedes' Principle**, the buoyant force exerted by a fluid is equal to the weight of the fluid displaced by the object.

Therefore, the total buoyant force is the sum of the buoyant force from the oil layer ($F_{B1}$) and the buoyant force from the mercury layer ($F_{B2}$):

$$F_B = F_{B1} + F_{B2}$$

### Formulating the Equations

Let the total volume of the sphere be $V$ and its density be $ho$. The weight of the sphere is:

$$W = V ho g$$

Since half of the sphere's volume is in the oil, the volume of oil displaced is $V_1 = \frac{V}{2}$. The buoyant force from the oil is:

$$F_{B1} = \left(\frac{V}{2}ight) ho_{oil} g$$

Similarly, the other half of the sphere is in the mercury, so the volume of mercury displaced is $V_2 = \frac{V}{2}$. The buoyant force from the mercury is:

$$F_{B2} = \left(\frac{V}{2}ight) ho_{Hg} g$$

### Solving for Density

At equilibrium, we equate the downward force to the total upward force:

$$W = F_{B1} + F_{B2}$$

Substituting our expressions into this balance equation yields:

$$V ho g = \frac{V}{2} ho_{oil} g + \frac{V}{2} ho_{Hg} g$$

Notice how beautifully the math simplifies here. The total volume $V$ and the acceleration due to gravity $g$ are present in every single term. We can divide the entire equation by $V g$:

$$ho = \frac{ho_{oil} + ho_{Hg}}{2}$$

This elegant result shows that when an object floats half-submerged in two liquids, its density is simply the **arithmetic mean** (the simple average) of the densities of the two liquids!

Now, we substitute the given values:

$$ho = \frac{0.8 + 13.6}{2}$$

$$ho = \frac{14.4}{2} = 7.2	ext{ g/cm}^3$$

Thus, the density of the material of the sphere is **$7.2	ext{ g/cm}^3$**, which corresponds to option **(c)**.

Select Answer:

Visualized Solution

Visualizing the Interface

Identifying the Forces

Archimedes' Principle

Formulating the Weight

Buoyancy from the Oil Layer

Buoyancy from the Mercury Layer

Setting up the Equilibrium Equation

Simplifying the Equation

Substituting the Values

Calculating the Final Density

The Sigma Insight: Buoyancy and Archimedes' Principle

Analyzing the Setup

The Physics of Multi-Liquid Buoyancy

Formulating the Equations

Solving for Density

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