Buoyancy Principles: Weight Loss in Water

This question delves into the principles of buoyancy. It asks what characteristic must be identical between two objects if they both exhibit the same reduction in weight when submerged in water.

Archimedes' Principle Explained

To solve this, we rely on Archimedes' principle. This principle states that a buoyant force acts on a body immersed in a fluid, and this force is equal to the weight of the fluid that the body displaces. Essentially, the upward buoyant force counteracts the object's weight.

When an object is weighed first in air and then in a fluid (like water), the difference represents the loss of weight. This loss is exactly equal to the buoyant force exerted by the fluid.

Mathematical Basis

Let's denote:

• The weight of an object in air as $W_{air}$.
• The apparent weight of the object when submerged in water as $W_{water}$.
• The buoyant force as $F_B$.

The loss of weight is calculated as:

Loss of Weight = $W_{air} - W_{water}$

According to Archimedes' principle, this loss of weight is equal to the buoyant force:

Loss of Weight = $F_B$

The buoyant force ($F_B$) is determined by the weight of the displaced fluid. The formula for buoyant force is:

$F_B = \rho_{fluid} \times V_{displaced} \times g$

Where:

• $\rho_{fluid}$ is the density of the fluid (water, in this case).
• $V_{displaced}$ is the volume of the fluid displaced by the object.
• $g$ is the acceleration due to gravity.

The problem states that two objects (let's call them Object 1 and Object 2) lose the same weight in water. This implies that the buoyant force acting on both objects is the same:

$F_{B1} = F_{B2}$

Since both objects are submerged in the same fluid (water), $\rho_{fluid}$ is constant. Gravity ($g$) is also constant.

Therefore, we can write:

$\rho_{fluid} \times g \times V_{displaced1} = \rho_{fluid} \times g \times V_{displaced2}$

By canceling out the constants ($\rho_{fluid}$ and $g$), we find:

$V_{displaced1} = V_{displaced2}$

This crucial result means that both objects displace the same volume of water. When an object is fully submerged, the volume of the displaced fluid is equal to the volume of the object itself. Assuming the objects are fully submerged for weighing, their volumes must be identical.

$V_{object1} = V_{object2}$

Option Analysis: Weight Loss Implications

Let's examine why the other options aren't necessarily correct:

• Specific gravities: Specific gravity is the ratio of an object's density to the fluid's density ($\frac{\rho_{object}}{\rho_{water}}$). If specific gravities were identical, densities would be identical. While identical densities and identical volumes lead to the same buoyant force, the problem only requires the *volume* to be identical, not necessarily the density.
• Weights in air: The weight in air ($W_{air}$) depends on the object's mass (density $\times$ volume). Even if the volumes are the same, the weights in air could differ if the objects have different densities. The question concerns the *loss* of weight, not the initial weight.
• Densities: If densities were identical but volumes were different, the volume of displaced water would differ, leading to different buoyant forces and different weight losses. For the weight loss to be the same, the volume of displaced water must be equal, which doesn't require densities to match.
• Volumes: As shown through the application of Archimedes' principle, the condition of equal weight loss directly implies that the volumes of water displaced are equal. Assuming full submersion, this means the objects themselves must have identical volumes.

Volumes Identical: Conclusion from Buoyancy

The key takeaway is that the buoyant force, which causes the apparent loss of weight, depends directly on the volume of fluid displaced. When two objects experience the same loss of weight in water, they must be displacing equal volumes of water. Because the volume of displaced fluid equals the object's volume (when fully submerged), the two objects must have identical volumes.