Two solid pieces, one made of gold and the other of silver, are chosen so that when they are completely immersed in water they show equal apparent weights. Given that the density of gold is greater than the density of silver, what can be said about their actual weights when measured in air?

Introduction / Context:
This question combines ideas from fluid statics, buoyancy and density. When an object is immersed in a fluid such as water, it appears to weigh less than in air because the fluid exerts an upward buoyant force equal to the weight of the displaced fluid. The apparent weight is the true weight minus the buoyant force. Here, two different materials gold and silver are compared under the condition that their apparent weights in water are equal. Since gold is denser than silver, the volumes required to give the same apparent weight will differ, and this affects their true weights in air. The question asks which piece will weigh more in air, given equal apparent weights in water.

Given Data / Assumptions:
• One solid piece is made of gold, the other of silver. • When completely immersed in water, both show equal apparent weights. • Density of gold is greater than density of silver. • Water has the same density for both experiments and is at rest. • Shapes of the pieces do not matter; only volumes and densities are relevant.

Concept / Approach:
The apparent weight W_app of an immersed object is given by W_app = W_air minus B, where W_air is its true weight in air and B is the buoyant force. The buoyant force equals the weight of displaced water, B = rho_water * V * g, where V is the volume of the object. The true weight in air is W_air = rho_object * V * g. The condition of equal apparent weights means that for gold and silver, W_air gold minus B_gold equals W_air silver minus B_silver. Using densities, this becomes (rho_gold minus rho_water) * V_gold * g equals (rho_silver minus rho_water) * V_silver * g. Because rho_gold is greater than rho_silver, the volume of the gold piece must be smaller than the volume of the silver piece to keep the products equal. When we then compute the true weight in air, W_air = rho_object * V * g, the piece with larger volume of less dense material can end up heavier in air.

Step-by-Step Solution:
Step 1: Write apparent weight expressions: W_app gold = (rho_gold * V_gold * g) minus (rho_water * V_gold * g) and W_app silver = (rho_silver * V_silver * g) minus (rho_water * V_silver * g). Step 2: Factor common terms to get W_app gold = (rho_gold minus rho_water) * V_gold * g and W_app silver = (rho_silver minus rho_water) * V_silver * g. Step 3: Given that apparent weights are equal, set these equal: (rho_gold minus rho_water) * V_gold = (rho_silver minus rho_water) * V_silver. Step 4: Because rho_gold is greater than rho_silver, the factor (rho_gold minus rho_water) is greater than (rho_silver minus rho_water). Step 5: For the products to be equal, V_gold must be smaller than V_silver, since the larger density difference on the gold side is multiplied by a smaller volume. Step 6: Compute true weights in air: W_air gold = rho_gold * V_gold * g and W_air silver = rho_silver * V_silver * g, and compare them using the relationships between densities and volumes.

Verification / Alternative check:
From the equality condition, V_gold divided by V_silver equals (rho_silver minus rho_water) divided by (rho_gold minus rho_water), which is less than one because rho_gold is greater than rho_silver. Thus V_gold is less than V_silver. Now compare true weights: W_air gold minus W_air silver equals rho_water * g times (V_gold minus V_silver), obtained by manipulating the expressions and the equality of apparent weights. Since V_gold is less than V_silver, V_gold minus V_silver is negative, so W_air gold minus W_air silver is negative. This means W_air silver is greater than W_air gold, so the silver piece weighs more in air, even though both appear equal in water. This algebraic check confirms the qualitative reasoning.

Why Other Options Are Wrong:
Option A, stating that gold weighs more in air, might seem intuitive because gold is denser and often associated with heaviness, but it neglects the fact that the volumes have been adjusted to equalise apparent weight in water. Option C, that both weigh equal in air, would be true only if their volumes and densities happened to produce equal true weights, which conflicts with the equal apparent weight condition and the density difference. Option D, claiming the weighing depends only on mass and cannot be predicted, ignores the precise relationship between mass, volume, density and buoyant force; with density information given, a definite prediction is possible. Only option B matches the correct physical analysis.

Common Pitfalls:
A common mistake is to think only in terms of density and forget that equal apparent weights in water imply different volumes and masses. Many students automatically assume that the denser material must always be heavier, even when the volumes are chosen specifically to adjust buoyant effects. Another pitfall is to ignore buoyant force when comparing immersed weights, treating all objects as if they were weighed in air. Remember that apparent weight depends on both object density and displaced fluid weight; when these are carefully balanced, surprising comparisons like a less dense material weighing more in air can arise. Carefully using formulas for buoyancy and apparent weight helps avoid these intuitive traps.

Final Answer:
The correct choice is The silver piece will weigh more in air, because to have equal apparent weights in water, the less dense silver piece must have a larger volume, and when true weights in air are compared, this larger volume of silver leads to a greater actual weight than the smaller, denser gold piece.