Study guides & original practice
Density and buoyancy practice
Density compares mass with volume. Buoyant force equals the weight of displaced fluid, so the displaced volume is central to a floating-object problem.
By ExamCat · Original study lesson · Published September 24, 2026
Understand the method
Use ρ = m/V and, for a uniform fluid, Fb = ρfluid g Vdisplaced. A freely floating object in equilibrium displaces fluid whose weight equals its own. Its entire volume need not be submerged.
Worked example, step by step
A uniform block has density 600 kg/m³ and floats in water of density 1,000 kg/m³. Find the submerged fraction.
- Set buoyant force equal to weight: ρwater g Vsub = ρblock g Vtotal.
- Cancel g and divide by ρwater Vtotal.
- The fraction Vsub/Vtotal is 600/1,000 = 0.60.
- Thus 60% of the block’s volume is submerged in this ideal equilibrium model.
A mistake to watch for
A floating object does not have zero weight. Weight and buoyant force balance. Ignoring the surrounding fluid’s density can reverse a prediction about whether the same object floats.
Try it yourself: three questions
Write an answer before opening the explanation. For calculations, keep your units and intermediate steps; for concepts, state the reason for your choice.
1. A sample has mass 120 g and volume 40 cm³. Find density.
Show answer and explanation
3 g/cm³. Keep both units rather than reporting the bare number three.
2. A floating object weighs 12 N. Find buoyant force at rest.
Show answer and explanation
12 N upward, assuming no other vertical support or applied force.
3. A fully submerged rigid object moves deeper in an incompressible fluid of uniform density. Does buoyancy necessarily increase?
Show answer and explanation
No. If displaced volume and fluid density stay constant, buoyant force stays constant even though pressure is larger at depth.
Build your next practice session
Mark the first step where your reasoning differed from the explanation. Turn that step into a question, answer it without looking, then change one condition in the example and explain what would change. A correct guess is a reason to revisit the method.
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Further reading
OpenStax Physics provides textbook background. The worked scenario and practice questions on this page were written independently for ExamCat.
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