Understanding Weight and Buoyancy

This problem involves understanding how buoyancy affects the weight of objects when submerged in a fluid, like water. The weight of an object measured in air is called its true weight, while the weight measured when submerged in a fluid is called its apparent weight.

The difference between the true weight and the apparent weight is the buoyant force exerted by the fluid on the object. This is described by Archimedes' Principle.

• True Weight (\(W_{air}\)) = Mass (\(m\)) \(\times\) Acceleration due to gravity (\(g\))
• Mass (\(m\)) = Density (\(\rho\)) \(\times\) Volume (\(V\))
• So, \(W_{air} = \rho_{object} V_{object} g\)
• Buoyant Force (\(F_B\)) = Weight of the fluid displaced by the object
• \(F_B = \rho_{fluid} V_{submerged} g\)
• Since the objects are completely immersed, \(V_{submerged} = V_{object}\).
• \(F_B = \rho_{fluid} V_{object} g\)
• Apparent Weight (\(W_{app}\)) = True Weight (\(W_{air}\)) - Buoyant Force (\(F_B\))
• \(W_{app} = W_{air} - \rho_{fluid} V_{object} g\)

Analyzing the Given Information

We are given two solid pieces, one of gold and one of silver. They are identical, which in this context implies they have the same shape, and likely the same volume if made of the same material. However, they are made of different materials with different densities. A key piece of information is that when completely immersed in water, they exhibit equal apparent weights.

• Let \(W_{app, Au}\) be the apparent weight of the gold piece in water.
• Let \(W_{app, Ag}\) be the apparent weight of the silver piece in water.
• We are given: \(W_{app, Au} = W_{app, Ag}\).

Using the formula for apparent weight:

• \(W_{air, Au} - F_{B, Au} = W_{air, Ag} - F_{B, Ag}\)

Now, let's express the true weights (\(W_{air}\)) and buoyant forces (\(F_B\)) in terms of density (\(\rho\)), volume (\(V\)), and gravity (\(g\)). Let \(\rho_{Au}\) be the density of gold, \(\rho_{Ag}\) be the density of silver, \(\rho_{water}\) be the density of water, \(V_{Au}\) be the volume of the gold piece, and \(V_{Ag}\) be the volume of the silver piece.

• \(W_{air, Au} = \rho_{Au} V_{Au} g\)
• \(F_{B, Au} = \rho_{water} V_{Au} g\)
• \(W_{air, Ag} = \rho_{Ag} V_{Ag} g\)
• \(F_{B, Ag} = \rho_{water} V_{Ag} g\)

Substituting these into the equality of apparent weights:

\(\rho_{Au} V_{Au} g - \rho_{water} V_{Au} g = \rho_{Ag} V_{Ag} g - \rho_{water} V_{Ag} g\)

We can divide both sides by \(g\) (since \(g\) is not zero):

\(\rho_{Au} V_{Au} - \rho_{water} V_{Au} = \rho_{Ag} V_{Ag} - \rho_{water} V_{Ag}\)

Factoring out the volumes:

\((\rho_{Au} - \rho_{water}) V_{Au} = (\rho_{Ag} - \rho_{water}) V_{Ag}\)

Comparing Volumes Based on Densities

We are given that the density of gold is greater than the density of silver: \(\rho_{Au} > \rho_{Ag}\). Both gold and silver are denser than water (\(\rho_{Au} > \rho_{water}\) and \(\rho_{Ag} > \rho_{water}\)) since they sink when placed in water.

From the equation \((\rho_{Au} - \rho_{water}) V_{Au} = (\rho_{Ag} - \rho_{water}) V_{Ag}\), let's compare the terms in the parentheses:

• Since \(\rho_{Au} > \rho_{Ag}\), and \(\rho_{water}\) is subtracted from both, it follows that \((\rho_{Au} - \rho_{water}) > (\rho_{Ag} - \rho_{water})\).

For the product on the left side to equal the product on the right side, if the first factor on the left is larger than the first factor on the right, the second factor on the left (\(V_{Au}\)) must be smaller than the second factor on the right (\(V_{Ag}\)).

Therefore, \(V_{Au} < V_{Ag}\). The gold piece has a smaller volume than the silver piece, given that they have equal apparent weights in water.

Comparing Weights in Air

Now we want to compare their weights in air:

• \(W_{air, Au} = \rho_{Au} V_{Au} g\)
• \(W_{air, Ag} = \rho_{Ag} V_{Ag} g\)

We know \(\rho_{Au} > \rho_{Ag}\) and \(V_{Au} < V_{Ag}\). This doesn't immediately tell us which product is larger. However, let's go back to the apparent weight equation:

\(W_{app, Au} = W_{air, Au} - F_{B, Au}\)

\(W_{app, Ag} = W_{air, Ag} - F_{B, Ag}\)

Since \(W_{app, Au} = W_{app, Ag}\), let's call this value \(W_{app}\).

• \(W_{air, Au} = W_{app} + F_{B, Au}\)
• \(W_{air, Ag} = W_{app} + F_{B, Ag}\)

The buoyant force depends on the volume:

• \(F_{B, Au} = \rho_{water} V_{Au} g\)
• \(F_{B, Ag} = \rho_{water} V_{Ag} g\)

Since we found \(V_{Au} < V_{Ag}\), and \(\rho_{water}\) and \(g\) are the same for both, it follows that \(F_{B, Au} < F_{B, Ag}\).

Now substitute this back into the equations for weight in air:

• \(W_{air, Au} = W_{app} + F_{B, Au}\)
• \(W_{air, Ag} = W_{app} + F_{B, Ag}\)

Since \(W_{app}\) is the same for both, and \(F_{B, Au}\) is less than \(F_{B, Ag}\), adding a smaller value (\(F_{B, Au}\)) to \(W_{app}\) will result in a smaller sum compared to adding a larger value (\(F_{B, Ag}\)) to \(W_{app}\).

Therefore, \(W_{air, Au} < W_{air, Ag}\).

The silver piece weighs more in air than the gold piece.

Summary Table

Based on this analysis, the silver piece will weigh more in air.

Revision Table: Key Concepts Review

Additional Information on Buoyancy and Density

Archimedes' Principle: This 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. This force acts upwards, opposing gravity.

Why does buoyancy matter? When an object is in a fluid, the pressure at the bottom surface is greater than the pressure at the top surface due to the depth difference. This pressure difference results in a net upward force, which is the buoyant force. For an object to float, the buoyant force must be equal to or greater than its weight. If the weight is greater, the object sinks.

Density and Sinking/Floating: An object sinks in a fluid if its average density is greater than the density of the fluid. It floats if its average density is less than or equal to the density of the fluid. In this problem, both gold and silver sink in water, confirming their densities are greater than the density of water.

Relating Volume and Buoyancy: The buoyant force is directly proportional to the volume of the object submerged. If two objects are in the same fluid, the one with the larger volume will experience a larger buoyant force.

In this specific problem, the fact that the less dense silver piece has the same apparent weight in water as the denser gold piece implies that the silver piece must have a larger volume to compensate for its lower density and experience a larger buoyant force. This larger buoyant force acting on the silver is necessary for its apparent weight to be reduced down to the same value as the gold piece, which experiences a smaller buoyant force due to its smaller volume.