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Animated Solution for Physics - Properties of Solids and Liquids: A ball is made of a material of density $ρ$ , where $ρ_{oil} < ρ < ρ_{water}$ with $ρ_{oil}$ and $ρ_{water}$ representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position?

Q: Why doesn't the ball just stay in the middle of the oil?

Because its density is strictly greater than the oil's density ($\rho > \rho_{\text{oil}}$). An object will only suspend in a fluid if its density exactly matches the fluid's density.

Q: Could the ball float partially above the oil?

No, for an object to float on the surface of a fluid, its density must be less than that fluid. Here, the ball is denser than the oil.

Q: What does 'immiscible' mean and why is it important here?

Immiscible means the liquids do not mix (like oil and water). This ensures there is a distinct, sharp boundary (interface) between them where the ball can rest.

Q: Does the size or shape of the ball matter?

Not for determining its equilibrium position! Whether it's a tiny marble or a giant sphere, as long as the density condition $\rho_{\text{oil}} < \rho < \rho_{\text{water}}$ holds, it will rest at the interface.

Q: How do we calculate the exact fraction of the ball in each liquid?

You would use Archimedes' principle! The total weight of the ball ($V \rho g$) must equal the sum of the buoyant forces from the displaced oil ($V_{\text{oil}} \rho_{\text{oil}} g$) and displaced water ($V_{\text{water}} \rho_{\text{water}} g$).

Select Answer:

Visualized Solution

$ρ_{oil} < ρ_{water}$

$ρ_{oil} < ρ_{water}$
Oil is less dense than water, so it forms the top layer.

$ρ > ρ_{oil}$

$ρ > ρ_{oil}$
The ball is denser than oil, so it will sink through the oil layer.

$ρ < ρ_{water}$

$ρ < ρ_{water}$
The ball is less dense than water, so it cannot sink into the water layer.

$ρ_{oil} < ρ < ρ_{water}$

$ρ_{oil} < ρ < ρ_{water}$
The ball rests exactly at the interface of the two immiscible liquids.

$W = F_{B, oil} + F_{B, water}$

$W = F_{B, oil} + F_{B, water}$
$ρ V g = ρ_{oil} V_{oil} g + ρ_{water} V_{water} g$

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The Sigma Insight: Buoyancy and Archimedes' Principle

Solution Diagram

Analyzing the Setup

Imagine you are holding a solid ball over a container filled with two distinct liquids: oil and water. Because oil and water are immiscible—meaning they refuse to mix—they form two separate layers.

We are given a crucial piece of information:

ρ_{oil} < ρ_{water}

. This tells us that the oil is less dense than the water. Naturally, the lighter oil will float on top, forming the upper layer, while the heavier water settles at the bottom.

The Descent Through Oil

Now, we drop the ball into the container. What happens first? The ball enters the oil layer. To determine its fate, we must compare the density of the ball,

ρ

, with the density of the oil.

The problem states that

ρ > ρ_{oil}

. Because the ball is denser than the oil, the buoyant force exerted by the oil is not strong enough to support the ball's weight. Consequently, the ball will sink right through the oil layer.

The Water Barrier

As the ball continues its descent, it eventually reaches the boundary between the oil and the water. Will it keep sinking all the way to the bottom of the container?

Let's look at the second part of our density inequality:

ρ < ρ_{water}

. The ball is strictly less dense than the water. This means that if the ball were completely submerged in water, the buoyant force would exceed its weight, pushing it back up. Therefore, the ball cannot sink into the water layer.

The Final Equilibrium

So, we have a fascinating situation: the ball is too heavy to float in the oil, but too light to sink in the water. It is effectively trapped!

The ball will come to rest exactly at the boundary between the two liquids. This is its stable equilibrium position, resting perfectly at the oil-water interface. A portion of its volume will be submerged in the water, and the remaining portion will be surrounded by the oil, such that the combined buoyant forces perfectly balance its weight.

Common Traps & FAQs

Why doesn't the ball just stay in the middle of the oil?

Because its density is strictly greater than the oil's density (

ρ > ρ_{oil}

). An object will only suspend in a fluid if its density exactly matches the fluid's density.

Could the ball float partially above the oil?

No, for an object to float on the surface of a fluid, its density must be less than that fluid. Here, the ball is denser than the oil.

What does 'immiscible' mean and why is it important here?

Immiscible means the liquids do not mix (like oil and water). This ensures there is a distinct, sharp boundary (interface) between them where the ball can rest.

Does the size or shape of the ball matter?

Not for determining its equilibrium position! Whether it's a tiny marble or a giant sphere, as long as the density condition

ρ_{oil} < ρ < ρ_{water}

holds, it will rest at the interface.

How do we calculate the exact fraction of the ball in each liquid?

You would use Archimedes' principle! The total weight of the ball (

V ρ g

) must equal the sum of the buoyant forces from the displaced oil (

V_{oil} ρ_{oil} g

) and displaced water (

V_{water} ρ_{water} g

Select Answer:

Visualized Solution

$ρ_{oil} < ρ_{water}$

$ρ > ρ_{oil}$

$ρ < ρ_{water}$

$ρ_{oil} < ρ < ρ_{water}$

$W = F_{B, oil} + F_{B, water}$

The Sigma Insight: Buoyancy and Archimedes' Principle

Analyzing the Setup

The Descent Through Oil

The Water Barrier

The Final Equilibrium

Similar Questions

A vessel contains oil (density $= 0.8 g/cm^{3}$ ) over mercury (density $= 13.6 g/cm^{3}$ ). 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 $g/cm^{3}$ is

Consider a solid sphere of radius $R$ and mass density $ρ (r) = ρ_{0} (1 - \frac{r ^{2}}{R ^{2}})$ , $0 < r \leq R$ . The min. density of a liquid in which it will float is

A thin uniform cylindrical shell, closed at both ends, is partially filled with water. It is floating vertically in water in half-submerged state. If $ρ_{c}$ is the relative density of the material of the shell with respect to water, then the correct statement is that the shell is

A body floats in a liquid contained in a beaker. The whole system as shown in figure falls freely under gravity. The upthrust on the body is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ρoil​<ρwater​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ρ>ρoil​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ρ<ρwater​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ρoil​<ρ<ρwater​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W=FB,oil​+FB,water​

The Sigma Insight: Buoyancy and Archimedes' Principle

Analyzing the Setup

The Descent Through Oil

The Water Barrier

The Final Equilibrium

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider a solid sphere of radius R and mass density ρ(r)=ρ0​(1−R2r2​), 0<r≤R. The min. density of a liquid in which it will float is

A body floats in a liquid contained in a beaker. The whole system as shown in figure falls freely under gravity. The upthrust on the body is

$ρ_{oil} < ρ_{water}$

$ρ > ρ_{oil}$

$ρ < ρ_{water}$

$ρ_{oil} < ρ < ρ_{water}$

$W = F_{B, oil} + F_{B, water}$

Consider a solid sphere of radius $R$ and mass density $ρ (r) = ρ_{0} (1 - \frac{r ^{2}}{R ^{2}})$ , $0 < r \leq R$ . The min. density of a liquid in which it will float is