AP Physics 1 Quiz: Pressure

What this quiz covers

This quiz focuses on Pressure, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Point A, because it is closer to the surface
2. Point B, because pressure increases with depth (correct answer)
3. They are equal because the container is sealed
4. They are equal because pressure depends only on container shape

Explanation: This question assesses understanding of hydrostatic pressure in fluids, which increases with depth. In a fluid at rest, pressure at a depth h is P = ρgh + P_atm, where ρ is density, g is gravity, and h is depth, showing pressure grows linearly with depth. Point A at 0.20 m has less pressure than Point B at 0.50 m due to the greater overlying fluid weight at B. Thus, Point B experiences greater water pressure. Distractor C suggests equal pressure because the container is sealed, but sealing does not affect the depth dependence. A useful strategy is to compare depths directly, as pressure differences depend solely on Δh in the same fluid.

Question 2

1. $1.0\times10^5\,\text{Pa}$ (correct answer)
2. $4.0\times10^5\,\text{Pa}$
3. $1.0\times10^2\,\text{Pa}$
4. $4.0\times10^2\,\text{Pa}$

Explanation: This question tests pressure due to contact force, defined as force per unit area. Pressure is P = F/A, where F is the perpendicular force on the surface. The 400 N crate is supported by four feet, so each foot bears 100 N, and with area 1.0 × 10^{-3} m², the pressure per foot is 100 / 0.001 = 1.0 × 10^5 Pa. This calculation assumes even weight distribution across the feet. Distractor B (4.0 × 10^5 Pa) might come from using the total force instead of per foot. When analyzing multi-support contacts, divide the total force equally among supports before applying P = F/A.

Question 3

1. Less at D, because deeper water has less room to expand
2. Equal, because both points are in the same lake
3. Greater at D, because pressure increases with depth (correct answer)
4. Equal, because pressure depends on the lake’s shape, not depth

Explanation: This question tests hydrostatic pressure, emphasizing its increase with fluid depth. Hydrostatic pressure is P = ρgh + P0, directly proportional to depth h below the surface. Point C at 1.0 m has lower pressure than Point D at 3.0 m, where the pressure is three times greater due to triple the depth. Therefore, pressure is greater at D because of the increased depth. Choice B is a distractor claiming equal pressure since both are in the same lake, ignoring depth's role. For such problems, always focus on the depth from the surface as the determining factor for pressure comparisons.

Question 4

1. It doubles because pressure is proportional to area.
2. It stays the same because the force is unchanged.
3. It is cut in half because the same force is distributed over twice the area. (correct answer)
4. It becomes zero because the force is spread out.

Explanation: This question tests understanding of the inverse relationship between pressure and area. Pressure is defined as P = F/A, so if force remains constant at 200 N while area doubles, pressure becomes half of its original value. Mathematically, if initial pressure is P₁ = F/A, then new pressure is P₂ = F/(2A) = (F/A)/2 = P₁/2. Choice B incorrectly suggests pressure remains constant, ignoring the role of area in the pressure equation. To solve pressure problems with changing conditions, apply P = F/A systematically and note which variables change.

Question 5

1. $p_M=p_N$ because the water’s density is the same at both points.
2. $p_N>p_M$ because pressure decreases as you go down.
3. $p_M>p_N$ because deeper points have more fluid above them. (correct answer)
4. The pressures depend on the container’s shape, not depth.

Explanation: This question tests understanding of pressure variation with depth in fluids. In any static fluid, pressure increases with depth following P = P₀ + ρgh. Since point M is 0.20 m deeper than point N, M experiences greater pressure due to the additional weight of water above it. The pressure difference equals ρg(0.20 m), where ρ is water density and g is gravitational acceleration. Choice B incorrectly suggests pressure decreases with depth, which violates basic fluid statics principles. When analyzing fluid pressure, remember that pressure always increases as you go deeper, regardless of the fluid type or container.

Question 6

1. $P_A>P_B$, because A is directly under the surface
2. $P_A<P_B$, because B is farther from the container wall
3. $P_A=P_B$, because pressure depends only on depth in a static fluid (correct answer)
4. Cannot be determined without the container’s shape

Explanation: This question tests understanding that pressure depends only on depth in a static fluid. In any static fluid, pressure at a given depth is P = P₀ + ρgh, which depends only on the vertical depth h below the surface, not on horizontal position. Since both points A and B are at the same 0.30 m depth, they experience identical pressure regardless of their horizontal separation or position relative to container walls. This principle allows pressure to be transmitted equally throughout a fluid at the same depth. Choice A incorrectly suggests that being "directly under" matters, confusing vertical depth with horizontal position. When analyzing fluid pressure, only the vertical depth below the surface determines the pressure value.

Question 7

1. Point R, because it is closer to the surface
2. Point S, because pressure increases with depth (correct answer)
3. They are equal, because the container could be wide or narrow
4. They are equal, because oil is less dense than water

Explanation: This question tests understanding of pressure increasing with fluid depth. Pressure in a static fluid follows P = P₀ + ρgh, where ρ is the oil's density and h is depth. Point S at 0.40 m depth has four times the gauge pressure (ρgh) compared to Point R at 0.10 m depth, regardless of the oil's specific density. The absolute pressure includes atmospheric pressure P₀ at both points, but the pressure difference depends only on the depth difference. Choice D incorrectly implies that oil's lower density compared to water affects which point has higher pressure, when depth alone determines the ranking. For any fluid, deeper points have higher pressure following the same P = P₀ + ρgh relationship.

Question 8

1. $p_X<p_Y$ because the higher point has more water above it.
2. $p_X>p_Y$ because the deeper point has greater pressure. (correct answer)
3. $p_X=p_Y$ because water transmits pressure equally.
4. The pressure depends on the pool’s width at each depth.

Explanation: This question tests understanding of pressure variation with depth in fluids. In static fluids, pressure increases with depth according to P = P₀ + ρgh. Point X at 1.5 m depth has greater pressure than point Y at 0.5 m depth because X has more water weight above it. The pressure difference is ρg(1.5 - 0.5) = ρg(1.0 m). Choice A incorrectly reverses this relationship, perhaps misunderstanding which point is deeper. When comparing fluid pressures, always identify which point is deeper below the surface, as deeper points have higher pressure.

Question 9

1. Case 2, because larger area means larger pressure
2. Case 1, because the same force acts on a smaller area (correct answer)
3. Same in both cases because the block’s weight is $50\,\text{N}$
4. Cannot be determined without knowing the block’s shape

Explanation: This question tests understanding of pressure as force divided by contact area. The block's weight (50 N) creates the same downward force in both cases. Case 1: pressure = 50 N / 0.005 m² = 10,000 Pa. Case 2: pressure = 50 N / 0.010 m² = 5,000 Pa. Case 1 has twice the pressure because it has half the contact area. Choice A reverses the relationship, incorrectly thinking larger area means larger pressure when actually pressure and area are inversely related for constant force. To find pressure from contact forces, always divide the perpendicular force by the contact area.

Question 10

1. four times larger, because there are four feet
2. the same, because the total weight is unchanged
3. four times smaller, because the force per foot is smaller while area per foot is the same (correct answer)
4. dependent on the spacing between the feet, not on force and area

Explanation: This question tests understanding of pressure distribution across multiple contact points. When the 300 N machine rests on four identical feet, each foot supports 300 N / 4 = 75 N. The pressure at each foot is P = 75 N / (2.0×10⁻³ m²) = 37,500 Pa. If one foot supported the full weight, pressure would be 300 N / (2.0×10⁻³ m²) = 150,000 Pa, which is four times larger. The four-foot configuration reduces pressure to one-fourth because each foot carries one-fourth the force while maintaining the same contact area. Choice A incorrectly multiplies by the number of feet. When weight is distributed across multiple supports, divide the total force by the number of supports to find force per support.

Question 11

1. $P_E>P_F$ because the wall adds extra pressure
2. $P_E<P_F$ because pressure is greater near the wall
3. $P_E=P_F$ because they are at the same depth in the same fluid (correct answer)
4. $P_E=P_F$ only if the pool has vertical sides

Explanation: This question tests understanding of pressure in fluids at equal depths. In a static fluid, pressure depends only on the vertical depth below the surface, not on horizontal position or proximity to walls. Both points E and F are 1.5 m below the surface, so they experience the same pressure: P = P_atmospheric + ρgh, where h = 1.5 m for both points. The wall's presence doesn't create additional pressure at point F because fluids transmit pressure equally in all directions at a given depth. Choice A incorrectly assumes walls add pressure, but walls only provide reaction forces. For fluid pressure problems, focus on vertical depth and ignore horizontal variations.

Question 12

1. Same pressure in both orientations because the force is the same
2. Greater pressure on the $0.10\,\text{m}^2$ face (correct answer)
3. Greater pressure on the $0.20\,\text{m}^2$ face because it has more contact area
4. Greater pressure depends on the crate’s shape, not the contact area

Explanation: This question tests understanding of pressure as force per unit area. Pressure is defined as P = F/A, where F is the perpendicular force and A is the contact area. Since the crate's weight (600 N) remains constant regardless of orientation, the force on the floor is always 600 N. When the crate sits on the 0.10 m² face, pressure = 600 N / 0.10 m² = 6000 Pa, while on the 0.20 m² face, pressure = 600 N / 0.20 m² = 3000 Pa. Choice C incorrectly assumes larger area means greater pressure, but pressure is inversely proportional to area when force is constant. To solve pressure problems involving contact forces, identify the perpendicular force and divide by the contact area.

Question 13

1. $P_G>P_H$ because the beaker is narrower near the bottom
2. $P_G=P_H$ because both points are in the same beaker
3. $P_H>P_G$ because pressure increases with depth (correct answer)
4. $P_H=P_G$ because water exerts the same force everywhere

Explanation: This question tests understanding of pressure increasing with depth in fluids. In static water, pressure follows P = P_surface + ρgh, where h is depth below surface. Point G at 0.40 m depth has pressure P_G = P_0 + ρg(0.40), while point H at 0.80 m depth has pressure P_H = P_0 + ρg(0.80). Since 0.80 > 0.40, point H experiences greater pressure from the taller column of water above it. The pressure difference is ΔP = ρg(0.80 - 0.40) = 0.40ρg. Choice D incorrectly claims equal pressure, confusing force with pressure. Remember that in any fluid at rest, pressure always increases linearly with depth.

Question 14

1. $P_J>P_K$ because the narrower section concentrates pressure
2. $P_J<P_K$ because the wider section has more fluid above it
3. $P_J=P_K$ because pressure depends on depth, not container shape (correct answer)
4. $P_J=P_K$ only if the tank is open to the atmosphere

Explanation: This question tests understanding that fluid pressure depends only on depth, not container shape. In a static fluid, pressure at any point is P = P_surface + ρgh, where h is the vertical depth below the surface. Points J and K are both at 2.0 m depth, so they experience identical pressure regardless of the tank's width at their locations. The wider section at K doesn't affect pressure because pressure depends on the weight of the vertical column of fluid above, not the total volume. Choice A incorrectly assumes narrow sections concentrate pressure, confusing this with flow situations. For static fluids, remember that pressure at a given depth is the same everywhere in connected fluid.

Question 15

1. Point $G$
2. Point $H$ (correct answer)
3. Same, because the container is sealed.
4. Same, because pressure is uniform in a closed container.

Explanation: This question assesses hydrostatic pressure in AP Physics 1, due to fluid depth. In a static fluid, pressure increases with depth because of the overlying fluid's weight, P = ρgh. Greater depth results in higher pressure from the fluid. Point H is three times deeper than G, so it has greater pressure from the oil. Choice D is a distractor, mistakenly claiming uniform pressure in closed containers, but pressure varies with depth. A transferable strategy is to measure depth from the fluid surface when evaluating pressure in sealed or open containers.

Question 16

1. Point G, because it is closer to the shore
2. Point H, because pressure increases with horizontal distance
3. They are equal, because both points are at the same depth (correct answer)
4. They are unequal, because pressure depends on the container’s shape

Explanation: This question assesses hydrostatic pressure, focusing on its dependence on vertical depth rather than horizontal position. In fluids, pressure increases with depth h as P = ρgh + P0, but is independent of horizontal distance. Points G and H are both at 2.0 m depth, so they have equal pressure despite the 10 m horizontal separation. The pressures are the same because only vertical depth matters. Choice B is a distractor that incorrectly claims pressure increases with horizontal distance, which is not true in static fluids. A transferable strategy is to ignore horizontal positions and base comparisons solely on vertical depths from the surface.

Question 17

1. Pressure is greater for $A_1$ because the book is more stable.
2. Pressure is greater for $A_2$ because the same force acts on smaller area. (correct answer)
3. Pressures are equal because the weight is the same.
4. Pressure depends on the table material, so it cannot be compared.

Explanation: This question tests understanding of pressure as force per unit area. With the same 40 N weight but different contact areas, pressure varies inversely with area according to P = F/A. For area A₁ = 0.020 m², pressure is P₁ = 40/0.020 = 2000 Pa. For smaller area A₂ = 0.010 m², pressure is P₂ = 40/0.010 = 4000 Pa, which is greater. Choice C incorrectly assumes equal pressures because weight is constant, missing that pressure depends on both force and area. To compare pressures with constant force, remember that smaller contact area always produces greater pressure.

Question 18

1. $p_Q>p_P$ because pressure increases with depth in a fluid at rest. (correct answer)
2. $p_Q=p_P$ because water pushes equally in all directions.
3. $p_P>p_Q$ because the water above $P$ has less area to push on.
4. The deeper point has lower pressure because its volume is smaller.

Explanation: This question tests understanding of pressure variation with depth in fluids. In a static fluid, pressure increases linearly with depth according to P = P₀ + ρgh, where ρ is fluid density, g is gravitational acceleration, and h is depth below the surface. Since point Q is 2.0 m deeper than point P, the pressure at Q must be greater than at P due to the additional weight of water above it. Choice B incorrectly suggests equal pressures, confusing the fact that pressure acts in all directions with pressure magnitude. When comparing pressures in fluids, always consider the depth difference and remember that deeper points have higher pressure.

Question 19

1. Case A, because the area is larger
2. Case B, because the same normal force is distributed over less area (correct answer)
3. They are equal, because the crate’s weight is unchanged
4. Case A, because friction increases pressure

Explanation: This question tests understanding of pressure as force per unit area. Pressure is calculated as P = F/A, where F is specifically the normal force (perpendicular to the surface) and A is the contact area. Since the normal force is the same in both cases but Case B has half the contact area (0.25 m²) compared to Case A (0.50 m²), Case B produces twice the pressure. The horizontal friction force from pulling does not contribute to vertical pressure. Choice D incorrectly suggests friction affects pressure, when only the perpendicular normal force matters. For pressure calculations, always identify which force component is perpendicular to the surface.

Question 20

1. Trial 1, because the area is larger
2. Trial 2, because the same force acts on a smaller area (correct answer)
3. They are equal, because the force is the same
4. Cannot be determined without the block’s mass

Explanation: This question tests understanding of pressure as force per unit area. Pressure is defined as P = F/A, where F is the perpendicular force and A is the contact area. Since the block's weight (force) remains constant but the contact area changes between trials, the pressure will be different. In Trial 2, the same force acts on a smaller area (0.010 m²) compared to Trial 1 (0.020 m²), resulting in pressure that is twice as large. Choice C incorrectly assumes that equal force means equal pressure, ignoring the crucial role of area. When solving pressure problems involving contact forces, always identify both the force and the area, then apply P = F/A to compare pressures.