Pressure, Density and Buoyancy: Lesson Test

Answer: A

Correct answer: 2,000 kg/m³. Density is mass divided by volume:

/m³.

• A (correct): Dividing mass by volume gives 6 ÷ 0.003 = 2,000 kg/m³.
• B: This slips a decimal place when dividing 6 by 0.003.
• C: This adds an extra zero; 6 ÷ 0.003 is 2,000, not 20,000.
• D: This divides volume by mass, which inverts the density formula.

Answer: B

Correct answer: 4,000 Pa. Use

2,400 N ÷ 0.6 m² = 4,000 Pa.

• A: This drops a zero; 2,400 divided by 0.6 is 4,000.
• B (correct): Pressure is force over area: 2,400 ÷ 0.6 = 4,000 Pa.
• C: This multiplies the force by the area instead of dividing.
• D: This divides the area by the force, which inverts the pressure formula.

Answer: D

Correct answer: 340 kPa. A tire gauge shows pressure above the atmosphere. Absolute pressure is

• A: This is the gauge reading, which leaves out the surrounding atmosphere.
• B: This is only the atmospheric pressure, not the pressure inside the tire.
• C: This subtracts the atmosphere instead of adding it.
• D (correct): Absolute pressure adds the atmosphere to the gauge reading: 240 + 100 = 340 kPa.

Answer: D

Correct answer: 60 kPa.

Pa = 60 kPa. Gauge pressure leaves out the atmosphere.

• A: This adds atmospheric pressure, giving absolute rather than gauge pressure.
• B: This adds an extra factor of ten to the depth calculation.
• C: This drops a factor of ten; each meter of fresh water adds about 10 kPa, not 1 kPa.
• D (correct): Water pressure is 1,000 × 10 × 6 = 60,000 Pa, which is 60 kPa.

Answer: A

Correct answer: 300 N. Archimedes' principle:

• A (correct): The displaced water weighs 1,000 × 10 × 0.03 = 300 N.
• B: This slips a decimal place in the volume, using 0.003 m³ instead of 0.03 m³.
• C: This adds an extra factor of ten to the displaced water's weight.
• D: This uses a water density of 10 kg/m³ instead of 1,000 kg/m³.

Answer: A

Correct answer: 0.92; it floats on the water.

Because 0.92 is less than 1, the oil is less dense than water and floats on top.

• A (correct): Specific gravity is 920 ÷ 1,000 = 0.92, and a value below 1 means it floats on fresh water.
• B: This divides water's density by the oil's, which inverts the ratio.
• C: The ratio is right, but a specific gravity below 1 means the liquid floats, not sinks.
• D: This forgets to divide by water's density; specific gravity has no units.

Answer: B

Correct answer: 0.04 m³. Since 1 m³ = 1,000 L, divide by 1,000:

• A: This uses 1 L = 0.0001 m³, which is ten times too small.
• B (correct): One liter is 0.001 m³, so 40 L is 40 × 0.001 = 0.04 m³.
• C: This uses 1 L = 0.1 m³, which is a hundred times too large.
• D: This uses 1 L = 0.01 m³, which is ten times too large.

Answer: A

Correct answer: It is the same at all three bottoms. The pressure from a still liquid is $\rho gh$, which contains no width or volume. With the same liquid and the same 2 m depth, the bottom pressure is identical in all three.

• A (correct): Liquid pressure depends only on density, g, and depth, and the depth is 2 m in every container.
• B: Slanted or flared walls do not change the pressure at a given depth.
• C: The tank holds more water, but the extra water rests on a larger floor, so pressure does not increase.
• D: Concentrating the water in a thin tube does not raise the pressure; only depth matters.

Answer: A

Correct answer: It sinks to the bottom. An object floats only if it can push aside its own weight of fluid before it is fully submerged. When fully under, the buoyant force equals the weight of the displaced water, and since 1,000 kg/m³ is less than 1,200 kg/m³, that force is always smaller than the block's weight. The trap is computing a percentage underwater; that ratio applies only when the object is less dense than the fluid.

• A (correct): The block is denser than fresh water (1,200 versus 1,000 kg/m³), so its weight exceeds the largest buoyant force it can get.
• B: This divides water density by block density; a block denser than water cannot float.
• C: Hovering happens only when the densities are equal, and these are not.
• D: This treats the extra 200 kg/m³ as a floating margin; only objects less dense than water float.

Answer: C

Correct answer: 40 kg. Mass measures the matter in an object and stays the same in air, water, or on the Moon. The lead still has a mass of 40 kg; only its apparent weight on a scale is reduced.

• A: Being supported by water does not make the object massless; its mass is unchanged.
• B: The water reduces the scale reading through buoyancy, but it does not remove any matter.
• C (correct): Mass is the amount of matter and does not change when the object is placed in water.
• D: This multiplies by g, which turns mass into weight in newtons rather than giving mass.

Answer: B

Correct answer: 600 kg/m³. The fraction submerged equals the density ratio. Submerged fraction =

/m³.

• A: This adds the 40% to water's density; a floating object must be less dense than water.
• B (correct): With 40% above, 60% is submerged, so the density is 0.60 × 1,000 = 600 kg/m³.
• C: This divides water's density by 0.4, which inverts the ratio and would make the block sink.
• D: This uses the fraction above the water instead of the fraction below it.

Answer: D

Correct answer: 14.4 kg. Rearrange

to

• A: This divides density by volume instead of multiplying.
• B: This misplaces the decimal; 720 × 0.02 equals 14.4.
• C: This shifts the decimal the other way; 720 × 0.02 equals 14.4.
• D (correct): Mass is density times volume: 720 × 0.02 = 14.4 kg.

Answer: C

Correct answer: 2,500 kg/m³. Since 1 m³ = 1,000,000 cm³, 600 cm³ = 0.0006 m³.

• A: This divides by the volume in liters (0.6 L) and labels the result kg/m³; 2.5 kg/L is actually 2,500 kg/m³.
• B: This converts 600 cm³ to 0.006 m³, which is ten times too large.
• C (correct): Converting 600 cm³ to 0.0006 m³ gives 1.5 ÷ 0.0006 = 2,500 kg/m³.
• D: This divides by 600 without converting cubic centimeters to cubic meters.

Answer: A

Correct answer: 80 kPa. Add the contact areas:

Pa = 80 kPa.

• A (correct): Total contact area is 4 × 0.15 = 0.6 m², so pressure is 48,000 ÷ 0.6 = 80,000 Pa, or 80 kPa.
• B: This multiplies the weight by one tire's area instead of dividing by the total area.
• C: This divides by the total area and then divides by four again, counting the tires twice.
• D: This divides the full weight by the area of just one tire.

Answer: C

Correct answer: 250 kPa. Water pressure:

Pa = 150 kPa.

Absolute pressure =

• A: This subtracts the atmosphere from the water pressure instead of adding it.
• B: This is only the gauge pressure from the water; absolute pressure also includes the atmosphere.
• C (correct): Water adds 1,000 × 10 × 15 = 150 kPa, and adding the 100 kPa atmosphere gives 250 kPa.
• D: This adds the depth in meters directly to the atmosphere instead of computing the water pressure.

Answer: A

Correct answer: 120 kPa. Liquid pressure grows in direct proportion to depth. Going from 4 m to 10 m multiplies the depth by 2.5, so the gauge pressure becomes

• A (correct): Gauge pressure is proportional to depth, so 48 × (10 ÷ 4) = 120 kPa.
• B: This adds 100 kPa of atmosphere, but the question asks for gauge pressure.
• C: This adds the 6 m depth change as if each meter added only 1 kPa.
• D: This uses the depth ratio upside down, giving a smaller pressure deeper down.

Answer: D

Correct answer: 2,050 N.

The fluid here is seawater, so its density must be used.

• A: This adds an extra factor of ten to the product.
• B: This uses fresh water's density of 1,000 kg/m³ instead of seawater's.
• C: This drops a factor of ten from the product.
• D (correct): Buoyant force is 1,025 × 10 × 0.2 = 2,050 N.

Answer: D

Correct answer: 136 N. Weight:

Buoyant force:

Apparent weight =

• A: This divides the correct reading by g, giving a mass-like number instead of the force the scale shows.
• B: This is the bracket's true weight, ignoring the water's upward push.
• C: This adds the buoyant force to the weight, but buoyancy pushes up and lowers the reading.
• D (correct): Weight is 156 N and the buoyant force is 20 N, so the scale reads 156 − 20 = 136 N.

Answer: A

Correct answer: 0.26 m³. A floating object displaces its total weight of water. Total weight =

• A (correct): The total weight is 2,600 N, and 2,600 ÷ (1,000 × 10) = 0.26 m³.
• B: This divides by 1,000 instead of by 10,000 N per cubic meter.
• C: This counts only the empty raft and leaves out the campers.
• D: This counts only the campers and leaves out the raft's own weight.

Answer: B

Correct answer: 9 cm. Extra displaced volume:

Depth change =

m = 9 cm.

• A: This multiplies the extra volume by the waterline area instead of dividing.
• B (correct): The extra displaced volume is 720 ÷ 10,000 = 0.072 m³, and 0.072 ÷ 0.8 = 0.09 m, or 9 cm.
• C: This slips a decimal place when converting 0.09 m to centimeters.
• D: This finds the extra volume, 0.072 m³, but reads it as a depth without dividing by the waterline area.

Answer: D

Correct answer: 4,500 N. Pascal's principle carries the same pressure to both pistons.

• A: This multiplies the input force by the large area without first finding the pressure.
• B: This divides by the area ratio instead of multiplying, so the force shrinks instead of grows.
• C: This is the fluid pressure in pascals, not the force on the large piston.
• D (correct): The pressure is 150 ÷ 0.02 = 7,500 Pa, and 7,500 × 0.6 = 4,500 N.

Answer: B

Correct answer: 6,800 Pa. Convert 5 cm to 0.05 m, then

• A: This adds the atmosphere, but the height difference gives gauge pressure.
• B (correct): Gauge pressure is 13,600 × 10 × 0.05 = 6,800 Pa.
• C: This uses water's density instead of mercury's.
• D: This uses 0.5 m for 5 cm, a factor of ten too large.

Answer: C

Correct answer: It rises, because the buoyant force grows while its weight stays the same. At neutral buoyancy, weight equals buoyant force. Adding salt raises the water's density, so ${\rho }_{\text{fluid}}gV$ grows while the weight does not, giving a net upward force.

• A: Pressure rises on the top and bottom alike, and the stronger push on the deeper bottom face adds lift.
• B: The displaced volume is the same, but that volume of salt water now weighs more.
• C (correct): Denser water makes the displaced water heavier, so the buoyant force now exceeds the unchanged weight.
• D: The bottle's mass and volume are unchanged, so its density stays the same; the water's density changed.

Answer: C

Correct answer: It stays the same. Buoyant force equals the weight of displaced water. A rigid case displaces the same volume at any depth, so the buoyant force is unchanged even though the pressure is higher.

• A: The case does not shrink, so it keeps displacing the same volume and gets the same lift.
• B: The pressure grows tenfold in the water term, but the top and bottom pressures rise together, so their difference does not.
• C (correct): For a rigid object, the displaced volume and the water's density do not change, so the buoyant force is constant.
• D: Greater pressure does not add lift to a rigid object; only the difference between top and bottom pressure matters.

Answer: B

Correct answer: 650 kg/m³; it floats. Average density = total mass ÷ outside volume =

/m³. Since 650 is less than 1,000, the drum floats.

• A: This is the empty drum's average density, before the gear was added.
• B (correct): Total mass 52 kg divided by 0.08 m³ gives 650 kg/m³, which is less than water's 1,000 kg/m³.
• C: This misplaces the decimal; 52 ÷ 0.08 is 650, not 6,500.
• D: This uses only the 40 kg of gear and leaves out the drum's own 12 kg.

Answer: A

Correct answer: 1.83 m³. For a floating object,

So the submerged volume is 0.917 × 2 ≈ 1.83 m³.

• A (correct): The submerged fraction is 917 ÷ 1,000 = 0.917, and 0.917 × 2 = 1.834 m³.
• B: This divides water density by ice density, giving more than the block's whole volume.
• C: This uses seawater's density of 1,025 kg/m³, but the pond is fresh water.
• D: This is the volume above the waterline, not below it.

Answer: C

Correct answer: 30 kPa. Since 1 m² = 10,000 cm², 30 cm² = 0.003 m². Then

Pa = 30 kPa.

• A: This converts 30 cm² to 0.03 m², which is ten times too large an area.
• B: This divides by 30 without converting square centimeters to square meters.
• C (correct): Converting 30 cm² to 0.003 m² gives 90 ÷ 0.003 = 30,000 Pa, or 30 kPa.
• D: This converts 30 cm² to 0.0003 m², which is ten times too small an area.

Answer: B

Correct answer: 30,000 N. Pressure:

Force:

• A: This is the pressure in pascals, not the total force on the floor.
• B (correct): The water pressure is 15,000 Pa, and multiplying by 2 m² gives 30,000 N.
• C: This divides the pressure by the area instead of multiplying.
• D: This multiplies density, depth, and area but leaves out g.

Answer: C

Correct answer: 2,000 N. At 5 m,

The force is

• A: This is the difference between the two hatch forces, not the force on the deeper one.
• B: This is the force on the shallower hatch at 2 m.
• C (correct): Pressure at 5 m is 50,000 Pa, and 50,000 × 0.04 = 2,000 N.
• D: This is the pressure in pascals; it still has to be multiplied by the hatch area.

Answer: B

Correct answer: 36 kPa.

Pa = 36 kPa. Atmospheric pressure acts on both sensors and cancels.

• A: This adds the atmosphere, which cancels because both sensors feel it equally.
• B (correct): The depth difference is 4 m, so 900 × 10 × 4 = 36,000 Pa, or 36 kPa.
• C: This uses the density of water instead of the oil's 900 kg/m³.
• D: This adds the two depths to get 8 m instead of subtracting them.

Answer: B

Correct answer: 7.1 m. Gauge pressure at the lower sensor =

With g = 10, fresh water adds 10 kPa per meter, so the depth is

• A: This is the depth of the upper sensor, not the lower one.
• B (correct): Subtracting the atmosphere leaves 71 kPa of water pressure, and at 10 kPa per meter that is 7.1 m.
• C: This is the spacing between the two sensors, not the depth of the lower one.
• D: This converts the full 171 kPa to depth without first removing the 100 kPa atmosphere.

Answer: D

Correct answer: 40%. The block's weight equals the weight of oil plus water displaced. With fraction x in water:

Solving, 200x = 80, so x = 0.40, meaning 40% in water and 60% in oil.

• A: This is the fraction in the oil layer, not the water layer.
• B: This uses the single-liquid rule with water alone, ignoring the lift from the oil.
• C: This divides the block's excess density over oil, 80, by the oil's density instead of by the 200 difference between the liquids.
• D (correct): Setting 1,000x + 800(1 − x) = 880 gives 200x = 80, so x = 0.40.

Answer: C

Correct answer: 4,000 N. Maximum buoyant force:

Available for load:

• A: This adds the raft's weight to the buoyant force instead of subtracting it.
• B: This is the maximum buoyant force, which must also hold up the raft itself.
• C (correct): Fully submerged, the raft gets 5,000 N of buoyancy; subtracting its own 1,000 N leaves 4,000 N for the load.
• D: This leaves out g and treats 500 kg of displaced water as 500 N.

Answer: A

Correct answer: 320 N. Upward:

Downward: weight 80 N plus tension T.

Equilibrium:

so T = 320 N.

• A (correct): Buoyancy is 400 N and weight is 80 N, so the rope must pull down 400 − 80 = 320 N.
• B: This adds the weight to the buoyant force, but both the weight and the rope act downward against the buoyancy.
• C: This is the buoyant force alone, ignoring the buoy's own weight that also pulls down.
• D: This is only the buoy's weight, not the force needed to hold it under.

Answer: D

Correct answer: 4,000 kg/m³. Buoyant force =

so the displaced volume is

Mass =

and density =

/m³.

• A: This divides the underwater reading by the weight in air, giving a density that would float.
• B: This divides the underwater reading, not the true weight, by the buoyant force.
• C: This divides the weight in air by the reading in water instead of by the buoyant force.
• D (correct): Buoyancy is 15 N, so the volume is 0.0015 m³; the mass is 6 kg, and 6 ÷ 0.0015 = 4,000 kg/m³.

Answer: C

Correct answer: 400 m³. The barge's weight is unchanged:

In seawater,

so the barge rides slightly higher.

• A: This assumes the displaced volume stays fixed, but denser water needs less volume to give the same lift.
• B: This is the change in displaced volume, not the new total.
• C (correct): Its weight is 1,000 × 10 × 410 = 4,100,000 N, and 4,100,000 ÷ (1,025 × 10) = 400 m³.
• D: This multiplies by 1.025 instead of dividing, which would make the barge ride deeper in denser water.

Answer: C

Correct answer: 12 cm. Volume:

Average density:

/m³, so the submerged fraction is 0.24.

Depth below surface: 0.24 × 0.5 m = 0.12 m = 12 cm.

• A: This is the height of the cube above the waterline, not below it.
• B: This divides the 0.03 m³ submerged volume by the 0.5 m edge instead of by the 0.25 m² face area.
• C (correct): Its density is 30 ÷ 0.125 = 240 kg/m³, so 24% of the 0.5 m height, or 0.12 m, is underwater.
• D: This reads the 0.24 submerged fraction as a depth in meters without multiplying by the 0.5 m height.

Answer: B

Correct answer: 180 N. Weight:

Buoyancy:

Upward forces must balance the weight:

so the bottom pushes up with 180 N.

• A: This is the buoyant force, not the support from the pond bottom.
• B (correct): The rock weighs 300 N and gets 120 N of buoyancy, so the bottom supplies the remaining 180 N.
• C: This ignores the buoyant force, which still acts on a rock resting on the bottom.
• D: This adds the buoyant force to the weight, but buoyancy helps hold the rock up.

Answer: D

Correct answer: 200 N; the load rises 0.5 cm. Area ratio:

Equal pressure means the input force is

The fluid volume moved is the same on both sides, so the large piston moves 20 cm ÷ 40 = 0.5 cm.

• A: This multiplies the load by the area ratio instead of dividing it.
• B: This assumes both pistons move the same distance, which would create energy.
• C: The force is right, but this multiplies the distance by 40; the large piston moves less, not more.
• D (correct): The area ratio is 40, so the force is 8,000 ÷ 40 = 200 N and the load moves 20 ÷ 40 = 0.5 cm.

Answer: D

Correct answer: It sinks, and keeps sinking, because it is squeezed smaller and displaces less water. A balloon is compressible. Deeper water squeezes it, so it displaces less water and its buoyant force falls below its weight. Sinking deeper squeezes it more, so it keeps sinking faster.

• A: Neutral buoyancy held only at 5 m; at 6 m the balloon is compressed and no longer balanced.
• B: Buoyancy depends on displaced volume, which shrinks as the balloon is compressed.
• C: Greater depth adds pressure on all sides; for a squeezable balloon it reduces lift rather than adding it.
• D (correct): Higher pressure compresses the balloon, cutting its displaced volume and buoyancy, so it sinks further and is squeezed even more.

Answer: D

Correct answer: 18 kg. Water has a density of 1,000 kg/m³, which is the same as 1 kg per liter.

Mass = density × volume = 1 kg/L × 18 L = 18 kg.

• A: This slips a decimal place; each liter of water is about 1 kg, not 0.1 kg.
• B: This multiplies by 10 as if converting to weight in newtons; 180 would be the weight in N, not the mass in kg.
• C: This converts liters to cubic meters but then forgets to multiply by the density of 1,000 kg/m³.
• D (correct): Fresh water has about 1 kg of mass in every liter, so 18 L has a mass of about 18 kg.

Answer: C

Correct answer: 2,400,000 Pa. Convert the area first: 1 m² = 10,000 cm², so 0.5 cm² =

That is 2.4 MPa, which is why a sharp edge cuts so easily.

• A: This divides 120 by 0.5 without converting square centimeters to square meters.
• B: This uses 1 m² = 100 cm², as if converting a length instead of an area.
• C (correct): Converting 0.5 cm² to 0.00005 m² and dividing 120 N by it gives 2,400,000 Pa, or 2.4 MPa.
• D: This uses 1 m² = 1,000 cm²; an area conversion needs 10,000 cm² per square meter.

Answer: C

Correct answer: 205 kPa. Gauge pressure from a liquid is

That is 205 kPa. Gauge pressure leaves out the atmosphere, so nothing is added.

• A: This uses the density of fresh water, 1,000 kg/m³, instead of seawater.
• B: This adds about 100 kPa of atmosphere, which gives absolute pressure, not gauge pressure.
• C (correct): The water pressure is 1,025 × 10 × 20 = 205,000 Pa, which is 205 kPa of gauge pressure.
• D: This slips a decimal place when converting 205,000 Pa to kilopascals.

Answer: D

Correct answer: Both blocks receive the same buoyant force. By Archimedes' principle, the buoyant force equals the weight of the fluid displaced,

Both blocks are fully under and have the same volume, so they displace the same amount of water and get the same buoyant force. Their materials only decide what happens when they are let go.

• A: This confuses the object's density with the fluid's; a heavier material does not displace more water.
• B: Pine floats when released, but while both are held fully under they displace the same water and get equal lift.
• C: Any object in water displaces some of it and feels an upward buoyant force.
• D (correct): Equal outside volumes displace equal volumes of water, and the buoyant force depends only on the displaced water.

Answer: A

Correct answer: 60 N. Weight:

Buoyant force:

The diver must supply the apparent weight:

• A (correct): The stone weighs 90 N and the water pushes up with 30 N, so the diver supplies the remaining 60 N.
• B: This is the stone's full weight in air; it ignores the buoyant force the water provides.
• C: This is the buoyant force itself, not the extra force the diver must add.
• D: This adds the buoyant force to the weight, but buoyancy pushes up and helps the diver.

Answer: C

Correct answer: 75%. In fresh water the submerged fraction equals the density ratio, so the float's density is

/m³.

In the oil:

So 75% of the float is below the surface. A less dense liquid gives less lift per cubic meter, so the float sinks deeper.

• A: This multiplies 0.6 by 0.8; in a less dense liquid the float must sink deeper, not ride higher.
• B: This assumes the fraction stays the same, but the submerged fraction depends on the liquid's density.
• C (correct): The float's density is 600 kg/m³, and 600 ÷ 800 = 0.75, so 75% sits below the oil surface.
• D: This is the fraction that stays above the oil, not the fraction below it.

Answer: B

Correct answer: 31.5 kPa. Liquid pressure $\rho gh$ is proportional to density when the depth stays the same.

The density ratio is

so the new reading is

• A: This treats the pressure as depending only on depth; the liquid's density matters too.
• B (correct): At the same depth, liquid pressure scales with density: 25 kPa × 1.26 = 31.5 kPa.
• C: This divides by 1.26 instead of multiplying; a denser liquid gives more pressure.
• D: This adds about 100 kPa of atmosphere, but the gauge reads only the pressure above atmospheric.

Answer: B

Correct answer: 2,000 N. Buoyant force from the cool outside air:

Weight of the warm air inside:

Lift left over:

the most payload the balloon can carry.

• A: This subtracts the warm air's weight but forgets to subtract the 3,000 N envelope and basket.
• B (correct): Buoyancy of 24,000 N minus 19,000 N of warm air and 3,000 N of envelope and basket leaves 2,000 N for payload.
• C: This forgets that the heated air inside also has weight that must be lifted.
• D: This uses the inside air's density to find the buoyant force; buoyancy comes from the outside air displaced.

Answer: A

Correct answer: 2,000 Pa. Pressure is force divided by area:

Lying flat:

Standing would give

eight times as much, so lying flat is far safer on thin ice.

• A (correct): Dividing the 720 N weight by the 0.36 m² contact area gives 2,000 Pa.
• B: This is the pressure while standing, found with the 0.045 m² boot area.
• C: This multiplies the force by the area instead of dividing.
• D: This slips a decimal place, dividing 720 by 3.6 instead of 0.36.

Answer: D

Correct answer: 60%. For a floating object, the submerged fraction equals object density divided by fluid density.

So 60% of the float is under the seawater and 40% sits above it.

• A: This uses fresh water at 1,000 kg/m³; seawater is denser, so the float rides a little higher.
• B: This is the fraction above the surface, not the fraction below.
• C: This multiplies the two densities instead of dividing the float's density by the seawater's.
• D (correct): The submerged fraction is the density ratio: 615 ÷ 1,025 = 0.60.