An empty plastic mug floats in a bucket of water. When a solid iron ball is kept in the mug it doesn't sink. When the ball is put in the water it sinks. Compared to when the ball is in water, the water level in the bucket when the ball is in the mug is

Question

Prepp Team · Accepted Answer

Understanding the Physics Principles

This problem involves understanding the concepts of buoyancy and Archimedes' principle. Buoyancy is the upward force exerted by a fluid that opposes the weight of an immersed object. Archimedes' principle states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

Key points to consider:

A floating object displaces a volume of fluid whose weight is equal to the object's own weight.
A submerged object displaces a volume of fluid equal to its own volume.
The water level in a container rises based on the volume of water displaced.

Scenario Analysis

Let's analyze the two situations described in the question:

Situation 1: Iron ball inside the floating mug

When the empty mug floats, it displaces a volume of water equivalent to its own weight. When the iron ball is placed inside the mug, the mug (with the ball) continues to float. This means the combined weight of the mug and the iron ball is supported by the buoyant force acting on the mug. According to Archimedes' principle, the mug displaces a volume of water whose weight is equal to the total weight of the mug plus the ball.

Let $W_{mug}$ be the weight of the mug and $W_{ball}$ be the weight of the iron ball. Let $V_{displaced1}$ be the volume of water displaced in this situation.

The buoyant force $F_{B1}$ equals the total weight:

$F_{B1} = W_{mug} + W_{ball}$

This buoyant force is also equal to the weight of the displaced water:

$F_{B1} = \rho_{water} \times V_{displaced1} \times g$

Where $\rho_{water}$ is the density of water and $g$ is the acceleration due to gravity.

Equating these, we get:

$\rho_{water} \times V_{displaced1} \times g = W_{mug} + W_{ball}$

Therefore, the volume of water displaced is:

$V_{displaced1} = \frac{W_{mug} + W_{ball}}{\rho_{water} \times g}$

Situation 2: Iron ball sinks in water, mug floats empty

In this case, the mug is floating empty. It displaces a volume of water, $V_{mug\_displaced}$, such that its weight equals the mug's weight:

$\rho_{water} \times V_{mug\_displaced} \times g = W_{mug}$ $V_{mug\_displaced} = \frac{W_{mug}}{\rho_{water} \times g}$

The iron ball sinks. A sinking object displaces a volume of fluid equal to its own volume. Let $V_{ball}$ be the volume of the iron ball.

The total volume of water displaced in this situation, $V_{displaced2}$, is the sum of the volume displaced by the floating mug and the volume displaced by the submerged ball:

$V_{displaced2} = V_{mug\_displaced} + V_{ball}$ $V_{displaced2} = \frac{W_{mug}}{\rho_{water} \times g} + V_{ball}$

Comparing Displaced Volumes

Now we compare $V_{displaced1}$ and $V_{displaced2}$.

We know that the weight of the ball $W_{ball}$ can be expressed as $W_{ball} = m_{ball} \times g = \rho_{ball} \times V_{ball} \times g$, where $\rho_{ball}$ is the density of the iron ball.

Substitute this into the expression for $V_{displaced1}$:

$V_{displaced1} = \frac{W_{mug}}{\rho_{water} \times g} + \frac{\rho_{ball} \times V_{ball} \times g}{\rho_{water} \times g}$

Simplifying this:

$V_{displaced1} = \frac{W_{mug}}{\rho_{water} \times g} + \frac{\rho_{ball}}{\rho_{water}} \times V_{ball}$

Recall that $V_{mug\_displaced} = \frac{W_{mug}}{\rho_{water} \times g}$. So,

$V_{displaced1} = V_{mug\_displaced} + \frac{\rho_{ball}}{\rho_{water}} \times V_{ball}$

Comparing this with $V_{displaced2} = V_{mug\_displaced} + V_{ball}$, we need to compare $\frac{\rho_{ball}}{\rho_{water}} \times V_{ball}$ with $V_{ball}$.

Since iron is much denser than water, the density of the iron ball ($\rho_{ball}$) is significantly greater than the density of water ($\rho_{water}$). This means the ratio $\frac{\rho_{ball}}{\rho_{water}} > 1$.

Therefore:

$\frac{\rho_{ball}}{\rho_{water}} \times V_{ball} > V_{ball}$

This implies that $V_{displaced1} > V_{displaced2}$.

Conclusion on Water Level

The volume of water displaced in Situation 1 (ball in the mug) is greater than the volume of water displaced in Situation 2 (ball sunk in water). Since the water level rises proportionally to the volume of water displaced, the water level will be higher when the ball is in the mug.