AP Physics 1: Algebra-Based Practice Quiz: Pressure

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 15 questions to check your progress.

Question 1 of 15

According to the definition provided, pressure is the magnitude of which component of force exerted per unit area?

A)The parallel componentB)The perpendicular componentC)The total force, regardless of directionD)The gravitational component

All Questions (15)

According to the definition provided, pressure is the magnitude of which component of force exerted per unit area?

A) The parallel component

B) The perpendicular component

C) The total force, regardless of direction

D) The gravitational component

Correct Answer: B

The content explicitly states that pressure is defined as 'the magnitude of the perpendicular force component exerted per unit area'.

What is the fundamental cause of the pressure exerted by a fluid on a surface?

A) The weight of the entire fluid in the container.

B) The chemical properties of the fluid's constituent particles.

C) The entirety of the interactions between the fluid’s particles and the surface.

D) The external atmospheric pressure acting on the fluid.

Correct Answer: C

The text states that 'The pressure exerted by a fluid is the result of the entirety of the interactions between the fluid’s constituent particles and the surface with which those particles interact.'

Absolute pressure is defined as the sum of a reference pressure, $P_0$, and what other quantity?

A) Atmospheric pressure, $P_{atm}$

B) Gauge pressure, $P_{gauge}$

C) Fluid density, $\rho$

D) Hydrostatic pressure, $\rho gh$

Correct Answer: B

The content specifies that 'The absolute pressure of a fluid at a given point is equal to the sum of a reference pressure $P_0$ ... and the gauge pressure $P_{gauge}$.' While hydrostatic pressure is often the gauge pressure, 'gauge pressure' is the correct general term.

A diver swims in a freshwater lake (density $\rho$) at a depth $h$. If the diver descends to a new depth of $4h$, how does the new gauge pressure compare to the original gauge pressure?

A) It is the same.

B) It is doubled.

C) It is quadrupled.

D) It is sixteen times greater.

Correct Answer: C

Gauge pressure in a fluid is given by $P_{gauge} = \rho gh$. Since pressure is directly proportional to the depth $h$, increasing the depth by a factor of 4 will increase the gauge pressure by a factor of 4.

Which of the following expressions correctly represents the absolute pressure, $P$, at a depth $h$ in a fluid of density $\rho$, open to the atmosphere where the pressure is $P_{atm}$?

A) $P = \rho gh$

B) $P = P_{atm} - \rho gh$

C) $P = P_{atm} + \rho gh$

D) $P = P_{atm} / (\rho gh)$

Correct Answer: C

Absolute pressure is the sum of the reference pressure (in this case, $P_{atm}$) and the gauge pressure ($P_{gauge} = \rho gh$). Therefore, $P = P_{atm} + \rho gh$.

A person is standing on the ground. To decrease the pressure they exert on the ground without changing their weight, the person should:

A) Stand on one foot instead of two.

B) Lie down on the ground.

C) Jump up and down.

D) Wear shoes with smaller soles.

Correct Answer: B

Pressure is defined as force per unit area ($P = F/A$). To decrease pressure while keeping the force (weight) constant, one must increase the area over which the force is distributed. Lying down significantly increases this area.

In the equation for gauge pressure, $P_{gauge} = \rho gh$, what does the variable $\rho$ represent?

A) The pressure at the surface.

B) The depth of the fluid column.

C) The acceleration due to gravity.

D) The density of the fluid.

Correct Answer: D

In the formula $P_{gauge} = \rho gh$, $\rho$ is the symbol for the density of the fluid, $g$ is the acceleration due to gravity, and $h$ is the height or depth of the fluid column.

The absolute pressure at the bottom of a container of liquid is measured to be $P_{abs}$. The atmospheric pressure at the surface is $P_{atm}$. Which expression represents the gauge pressure at the bottom of the container?

A) $P_{abs} + P_{atm}$

B) $P_{abs} - P_{atm}$

C) $P_{atm} - P_{abs}$

D) $P_{abs}$

Correct Answer: B

The relationship between absolute, atmospheric, and gauge pressure is $P_{abs} = P_{atm} + P_{gauge}$. Rearranging this equation to solve for gauge pressure gives $P_{gauge} = P_{abs} - P_{atm}$.

Two open containers are filled to the same height $h$. Container 1 is filled with oil (density $\rho_{oil}$) and Container 2 is filled with water (density $\rho_{water}$). Given that $\rho_{water} > \rho_{oil}$, how does the gauge pressure at the bottom of Container 1 ($P_1$) compare to the gauge pressure at the bottom of Container 2 ($P_2$)?

A) $P_1 > P_2$

B) $P_1 < P_2$

C) $P_1 = P_2$

D) The relationship cannot be determined without knowing the container shape.

Correct Answer: B

Gauge pressure is calculated as $P_{gauge} = \rho gh$. Since $g$ and $h$ are the same for both containers, the pressure is directly proportional to the density $\rho$. Because water has a higher density than oil, the gauge pressure at the bottom of the water container will be greater.

A submarine is at a depth where the gauge pressure is equal to the atmospheric pressure at the surface, $P_{atm}$. What is the absolute pressure at this depth?

A) $0.5 P_{atm}$

B) $P_{atm}$

C) $2 P_{atm}$

D) $3 P_{atm}$

Correct Answer: C

Absolute pressure is the sum of the reference (atmospheric) pressure and the gauge pressure: $P_{abs} = P_{atm} + P_{gauge}$. If $P_{gauge} = P_{atm}$, then $P_{abs} = P_{atm} + P_{atm} = 2 P_{atm}$.

A constant force is applied to two different surfaces. Surface A has an area of $A_1$ and Surface B has an area of $2A_1$. How does the pressure on Surface A ($P_A$) compare to the pressure on Surface B ($P_B$)?

A) $P_A = 0.5 P_B$

B) $P_A = P_B$

C) $P_A = 2 P_B$

D) $P_A = 4 P_B$

Correct Answer: C

Pressure is force per unit area ($P = F/A$). Since pressure is inversely proportional to area, the surface with the smaller area ($A_1$) will experience twice the pressure as the surface with the larger area ($2A_1$) for the same applied force.

An astronaut takes an open container of a fluid with density $\rho$ and height $h$ from Earth to the Moon, where the acceleration due to gravity is approximately 1/6th that of Earth's. How does the gauge pressure at the bottom of the container on the Moon compare to the gauge pressure on Earth?

A) It is 6 times greater.

B) It is the same.

C) It is 1/6th as large.

D) It is 1/36th as large.

Correct Answer: C

Gauge pressure is calculated by $P_{gauge} = \rho gh$. Since $\rho$ and $h$ remain constant, the gauge pressure is directly proportional to the acceleration due to gravity, $g$. On the Moon, $g$ is 1/6th of its value on Earth, so the gauge pressure will also be 1/6th as large.

Which of the following is the correct equation for the gauge pressure of a vertical column of fluid with density $\rho$ and height $h$?

A) $P_{gauge} = \rho g / h$

B) $P_{gauge} = g h / \rho$

C) $P_{gauge} = \rho h / g$

D) $P_{gauge} = \rho g h$

Correct Answer: D

The provided content explicitly gives the equation for the gauge pressure of a vertical column of fluid as $P_{gauge} = \rho gh$.

A pressure gauge on a deep-sea submersible reads $P_{gauge}$. The submersible then ascends to a new depth where the gauge pressure is $P_{gauge}/2$. Assuming the fluid density is constant, what is the new depth in relation to the original depth, $h_{original}$?

A) $2 h_{original}$

B) $4 h_{original}$

C) $h_{original} / 4$

D) $h_{original} / 2$

Correct Answer: D

Gauge pressure is directly proportional to depth ($P_{gauge} = \rho gh$). If the gauge pressure is halved, and $\rho$ and $g$ are constant, the depth $h$ must also be halved. Therefore, the new depth is $h_{original} / 2$.

The absolute pressure at a depth $h$ in a liquid is $P_{abs}$. If the reference pressure at the surface, $P_0$, were to double, what would be the new absolute pressure at the same depth $h$?

A) $P_{abs} + P_0$

B) $2 P_{abs}$

C) $P_{abs} - P_0$

D) $P_{abs}$

Correct Answer: A

The original absolute pressure is $P_{abs} = P_0 + P_{gauge}$. The gauge pressure, $\rho gh$, depends only on the fluid and depth, so it remains unchanged. If the reference pressure becomes $2P_0$, the new absolute pressure will be $P_{new} = (2P_0) + P_{gauge}$. We can rewrite this as $P_{new} = P_0 + (P_0 + P_{gauge}) = P_0 + P_{abs}$.