Hydrostatic Pressure Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB)

Question 1Question →

A cylindrical water storage tank is filled to a depth of $4\text{ meters}$ , creating a hydrostatic pressure of $40,000\text{ Pascals}$ at the tank's flat bottom. If water is drained from the tank until the depth is exactly $1\text{ meter}$ , what is the new hydrostatic pressure at the bottom in Pascals? (Assume the density of water and gravity remain constant.)

Show answer & explanation

Answer: 10000

Answer

10000

Because hydrostatic pressure is directly proportional to depth (

P = \rho g h

), reducing the depth to one-fourth of its original value (

1\text{ meter}

instead of

4\text{ meters}

) also reduces the pressure exactly to one-fourth of its original value (

40,000 / 4 = 10,000

).

Step-by-Step Solution

1

Identify the relationship between hydrostatic pressure and fluid depth.

Hydrostatic pressure is directly proportional to depth (

P \propto h

) when density and gravity are constant.

This fundamental principle allows us to solve the problem using a simple ratio.

2

Calculate the pressure per unit of depth.

\frac{40,000\text{ Pascals}}{4\text{ meters}} = 10,000\text{ Pascals per meter}

.

Finding the rate of pressure change per meter simplifies the calculation for the new depth.

3

Multiply the pressure per meter by the new depth.

10,000\text{ Pascals/meter} \times 1\text{ meter} = 10,000\text{ Pascals}

.

This yields the final pressure at the reduced depth.

Key Concept

Hydrostatic pressure is directly proportional to the depth of the fluid.

Question 2Question →

A U-shaped glass tube is open to the atmosphere at both ends. The left arm of the tube has a cross-sectional area of $10\text{ cm}^2$ , and the right arm has a cross-sectional area of $5\text{ cm}^2$ . The tube initially contains water with a density of $1.0\text{ g/cm}^3$ . A student carefully pours a column of oil measuring $20\text{ cm}$ in length, with a density of $0.8\text{ g/cm}^3$ , into the narrower right arm. Assuming the oil-water interface remains in the vertical section of the right arm, what is the vertical height difference between the top surface of the oil and the top surface of the water once the fluids come to rest?

16\text{ cm}

0\text{ cm}

4\text{ cm}

8\text{ cm}

Show answer & explanation

Answer: $4\text{ cm}$

Answer

4\text{ cm}

Hydrostatic pressure is determined solely by fluid density and depth (

P = \rho g h

), entirely independent of the container's cross-sectional area. Equating the pressures at the fluid interface gives a water column height of

16\text{ cm}

. Subtracting this from the

20\text{ cm}

oil column reveals that the oil surface rests

4\text{ cm}

higher than the water surface.

Step-by-Step Solution

1

Identify the reference point of equal pressure.

The horizontal plane at the oil-water interface in the right arm represents the point where pressure from both sides must be equal.

In a static fluid system, pressure at any given horizontal depth within the same continuous fluid (water) is uniform.

2

Set up the hydrostatic pressure equation.

\rho_{\text{water}} \times g \times h_{\text{water}} = \rho_{\text{oil}} \times g \times h_{\text{oil}}

. The cross-sectional areas (

10\text{ cm}^2

and

5\text{ cm}^2

) are irrelevant and ignored.

Hydrostatic pressure (

P = \rho g h

) depends only on fluid density and depth, not on the container's shape or the total volume of fluid.

3

Calculate the required height of the water column.

1.0\text{ g/cm}^3 \times h_{\text{water}} = 0.8\text{ g/cm}^3 \times 20\text{ cm}

, which simplifies to

h_{\text{water}} = 16\text{ cm}

.

This establishes how high the water surface in the left arm must rise above the interface to balance the downward pressure of the oil.

4

Calculate the difference in surface heights.

20\text{ cm} - 16\text{ cm} = 4\text{ cm}

.

The question asks for the vertical difference between the top of the oil and the top of the water, not just the absolute height of the water column.

Key Concept

Hydrostatic pressure is independent of container shape or cross-sectional area, depending solely on fluid density and vertical column height.

Question 3Question →

Arrange the following points in a fluid system in order of their gauge hydrostatic pressure, from the GREATEST pressure to the LEAST pressure. Assume standard gravity applies equally to all systems.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from greatest to least pressure is: the

15

-meter oil pipe, the layered vat, the

11

-meter lake, the

9

-meter brine tank, and finally the

14

-meter conical vessel.

By calculating the equivalent water depth (Depth

\times

Specific Gravity) for each scenario, we find the values are

12.0

(pipe),

11.5

(layered vat),

11.0

(lake),

10.8

(brine tank), and

9.8

(conical vessel). Ranking these from highest to lowest provides the correct sequence.

Step-by-Step Solution

1

Recall the formula for hydrostatic pressure.

P = \rho \cdot g \cdot h

, where

\rho

is density,

g

is gravity, and

h

is depth.

This establishes that pressure depends solely on fluid density and depth, not on container shape or total volume.

2

Determine a standard unit for comparison across all items.

Use 'equivalent meters of water' by multiplying the depth of each fluid by its Specific Gravity (SG).

This simplifies calculations and provides a single proportional metric to compare.

3

Calculate the equivalent pressure for the single-fluid containers.

Pipe:

15 \times 0.8 = 12.0

. Lake:

11 \times 1.0 = 11.0

. Brine:

9 \times 1.2 = 10.8

. Conical vessel:

14 \times 0.7 = 9.8

.

Applying the formula directly to the given depth and Specific Gravity.

4

Calculate the equivalent pressure for the layered fluids in the vat.

Top layer (water):

5 \times 1.0 = 5.0

. Bottom layer (glycerin):

5 \times 1.3 = 6.5

. Total equivalent depth =

5.0 + 6.5 = 11.5

.

The total pressure at the bottom of layered fluids is the sum of the hydrostatic pressures of each individual layer.

5

Rank the calculated equivalent water depths from greatest to least.

12.0 > 11.5 > 11.0 > 10.8 > 9.8

.

This corresponds to the final required ordering.

Key Concept

Hydrostatic pressure is determined solely by fluid density and depth, completely independent of the container's shape, width, or total fluid volume.

Alternative Method

Instead of calculating equivalent water depth, you can calculate exact pressure if you assign

g = 9.8 \text{ m/s}^2

and density of water as

1000 \text{ kg/m}^3

. For example, the pipe pressure is

0.8 \times 1000 \times 9.8 \times 15 = 117,600 \text{ Pa}

. However, comparing the relative

(h \times \text{SG})

values is much faster for mental arithmetic.

Estimated Time:2m 0s

Question 4Question →

Evaluate the gauge hydrostatic pressure exerted at the specific points in the four situations described below. Rank them in order from the HIGHEST hydrostatic pressure to the LOWEST hydrostatic pressure.

Assume standard gravity ( $g$ ) applies equally to all scenarios.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct ranking from highest to lowest pressure is: the narrow pipe with oil, the scuba diver in the lake, the submersible in the ocean, and finally the municipal water reservoir.

The correct sequence is found by calculating the relative pressure factor (density multiplied by depth) for each situation. The narrow pipe with oil has a factor of 16,000, making it the highest. The scuba diver is next at 15,000. The submersible is third at 12,300, and the municipal reservoir is lowest at 10,000. The shape and total volume of the containers are irrelevant distractors.

Step-by-Step Solution

1

Identify the relevant formula and variables for gauge hydrostatic pressure.

The formula is

P = \rho \cdot g \cdot h

. Since gravity (

g

) is constant across all scenarios, we only need to compare the product of density (

\rho

) and depth (

h

) to determine the relative pressures.

This establishes a direct mathematical way to compare the situations without needing to calculate the exact final pressure in Pascals.

2

Evaluate whether container shape or total fluid volume affects the pressure.

We must ignore the 'extremely narrow' pipe and the 'massive, wide' reservoir descriptions.

According to the principles of fluid mechanics, hydrostatic pressure at a specific depth depends only on depth and density, not on the total volume or the shape of the container (this is known as the hydrostatic paradox).

3

Calculate the pressure factor (

\rho \cdot h

) for each scenario.

Submersible:

1025 \cdot 12 = 12,300

. Narrow pipe:

800 \cdot 20 = 16,000

. Reservoir:

1000 \cdot 10 = 10,000

. Scuba diver:

1000 \cdot 15 = 15,000

.

These proportional values allow us to easily and accurately rank the pressures.

4

Rank the calculated values from highest to lowest.

Highest:

16,000

(Narrow pipe). Second:

15,000

(Scuba diver). Third:

12,300

(Submersible). Lowest:

10,000

(Reservoir).

This provides the final correct ordering required by the question.

Key Concept

Hydrostatic pressure at a given depth is determined solely by the fluid's density and the vertical depth (

P = \rho g h

), and is completely independent of the container's shape or the total volume of fluid.

Question 5Question →

An industrial processing facility uses two large vertical storage tanks that are open to the atmosphere. Tank A contains two distinct layers of non-mixing liquids: a $5\text{-meter}$ top layer of oil (density = $800\text{ kg/m}^3$ ) and a $3\text{-meter}$ bottom layer of water (density = $1000\text{ kg/m}^3$ ). Tank B, which has exactly twice the cross-sectional area of Tank A, contains pure water to a total depth of $8\text{ meters}$ . Which of the following statements correctly compares the gauge hydrostatic pressure acting on the interior bottom surfaces of the two tanks?

The pressure at the bottom of Tank B is greater primarily because its larger cross-sectional area holds a much larger total volume and weight of fluid.

The pressures are exactly equal because the total depth of the fluid column is

8\text{ meters}

in both tanks, regardless of the container's base area.

The pressure at the bottom of Tank B is greater because the entire

8\text{-meter}

fluid column consists of the denser liquid.

The pressure at the bottom of Tank A is greater because the oil layer physically compresses the water layer below it, multiplying the downward force.

Show answer & explanation

Answer: The pressure at the bottom of Tank B is greater because the entire $8\text{-meter}$ fluid column consists of the denser liquid.

Answer

The pressure at the bottom of Tank B is greater because its entire fluid column is composed of the denser water, yielding a higher hydrostatic pressure than Tank A's mixed column.

The correct answer accurately calculates that Tank B has a higher bottom pressure because it contains a full

8\text{-meter}

column of the denser water (

1000\text{ kg/m}^3

), while Tank A replaces

5\text{ meters}

of that column with lighter oil (

800\text{ kg/m}^3

). It also correctly implies that hydrostatic pressure depends only on fluid density and depth, not on the tank's cross-sectional area.

Step-by-Step Solution

1

Identify the formula for gauge hydrostatic pressure.

The hydrostatic pressure

P

at a depth

h

is calculated using

P = \rho gh

, where

\rho

is the fluid density and

g

is the acceleration due to gravity.

This formula is needed to find the pressure at the bottom of each tank.

2

Calculate the relative pressure at the bottom of Tank A.

For multiple layers, sum the individual pressures:

P_A = (\rho_{oil} \times g \times h_{oil}) + (\rho_{water} \times g \times h_{water}) = (800 \times g \times 5) + (1000 \times g \times 3) = 4000g + 3000g = 7000g\text{ Pa}

.

The total bottom pressure in a multi-fluid system is the combined pressure of all stacked fluid layers.

3

Calculate the relative pressure at the bottom of Tank B.

Tank B contains only water:

P_B = \rho_{water} \times g \times h_{total} = 1000 \times g \times 8 = 8000g\text{ Pa}

.

This calculation provides the value needed to compare Tank B against Tank A.

4

Compare the pressures and analyze the effect of tank width.

Since

8000g > 7000g

, Tank B has the higher bottom pressure. The larger cross-sectional area of Tank B is irrelevant to the calculation.

Hydrostatic pressure depends solely on depth and density, meaning a wider container does not artificially increase pressure.

Key Concept

Hydrostatic pressure depends strictly on fluid density, gravity, and depth (

P = \rho gh

), and is independent of the container's shape, base area, or total volume of fluid.

Question 6Question →

A municipal water tower supplies fresh water to a nearby town. Assuming the water is not flowing, which of the following factors primarily determines the hydrostatic pressure of the water at a ground-level faucet in a house?

The vertical distance from the faucet to the top of the water level in the tower

The total mass of the water currently stored inside the tower

The total volume of water the storage tank is designed to hold

The diameter of the main pipe connecting the tower to the house

Show answer & explanation

Answer: The vertical distance from the faucet to the top of the water level in the tower

Answer

The vertical distance from the faucet to the top of the water level in the tower.

According to the principles of fluid mechanics, hydrostatic pressure is determined solely by the fluid's density, gravity, and the vertical depth of the fluid column. Because density and gravity are constant in this scenario, the vertical distance from the water's surface to the faucet is the only factor that dictates the pressure. The total amount of water or the width of the pipes does not change this static pressure.

Step-by-Step Solution

1

Recall the formula for hydrostatic pressure.

P = \rho \cdot g \cdot h

This fundamental equation identifies the variables that affect pressure in a static fluid.

2

Analyze the variables in the context of the water tower.

The density of water (

\rho

) and the acceleration due to gravity (

g

) remain constant.

Since

\rho

and

g

are fixed, the pressure depends entirely on the remaining variable,

h

.

3

Identify what

h

represents in this physical scenario.

The variable

h

represents the vertical height or depth from the fluid's surface down to the measurement point (the faucet).

Hydrostatic pressure is based solely on vertical depth, making the vertical distance from the water level to the faucet the determining factor.

Key Concept

Hydrostatic Pressure Dependency on Depth

Question 7Question →

A sealed industrial holding tank contains water with a pocket of compressed air trapped above the water surface. The total depth of the water in the tank is $8.0\text{ m}$ . A pressure gauge located at the absolute bottom of the tank reads a gauge pressure of $130\text{ kPa}$ . A secondary discharge valve is located on the side of the tank, exactly $5.0\text{ m}$ above the tank's bottom. Assuming the density of water is $1,000\text{ kg/m}^3$ and the acceleration due to gravity is $g = 10\text{ m/s}^2$ , what is the gauge pressure, in kilopascals (kPa), at the secondary discharge valve?

Show answer & explanation

Answer: 80

Answer

80

The pressure at any point in a static fluid is equal to the pressure at a lower point minus the hydrostatic pressure of the fluid column between them. Since the valve is

5.0\text{ m}

above the bottom, the pressure there is less than the bottom pressure by

\rho g \Delta h = 1,000 \times 10 \times 5.0 = 50\text{ kPa}

. Subtracting this from the bottom pressure (

130\text{ kPa}

) gives the correct pressure of

80\text{ kPa}

. Alternatively, calculating the trapped air pressure (

130 - 80 = 50\text{ kPa}

) and adding the

3.0\text{ m}

water column above the valve (

30\text{ kPa}

) yields the same correct result.

Step-by-Step Solution

1

Determine the vertical distance between the pressure gauge at the bottom and the secondary valve.

\Delta h = 5.0\text{ m}

Hydrostatic pressure changes linearly with vertical height within a continuous fluid.

2

Calculate the hydrostatic pressure difference exerted by this

5.0\text{ m}

column of water.

\Delta P = \rho g \Delta h = 1,000 \times 10 \times 5.0 = 50,000\text{ Pa} = 50\text{ kPa}

This represents the amount of pressure lost as you move upward from the bottom to the valve.

3

Subtract the pressure difference from the gauge pressure at the bottom of the tank.

130\text{ kPa} - 50\text{ kPa} = 80\text{ kPa}

The pressure at the valve must be lower than the pressure at the bottom by exactly the weight of the water column between them.

Key Concept

Hydrostatic pressure varies directly with depth. The pressure difference between any two points in a continuous static fluid depends only on the vertical distance between them, regardless of surface air pressure.

Question 8Question →

A municipal water treatment facility utilizes two open-air settling basins with flat bottoms. Basin Alpha is a wide rectangular pool with a base area of $50\text{ m}^2$ , currently holding $150\text{ m}^3$ of water. Basin Beta is a narrow cylindrical well with a base area of $5\text{ m}^2$ , currently holding $30\text{ m}^3$ of water. What is the ratio of the gauge hydrostatic pressure exerted on the bottom of Basin Alpha to the gauge hydrostatic pressure exerted on the bottom of Basin Beta?

1:2

5:1

10:1

2:1

Show answer & explanation

Answer: $1:2$

Answer

The ratio is

1:2

, because gauge hydrostatic pressure depends exclusively on the vertical depth of the fluid, which is

3\text{ m}

in Basin Alpha and

6\text{ m}

in Basin Beta.

Hydrostatic pressure depends exclusively on the depth of the fluid, not the total volume or the base area. The depth of the water in Basin Alpha is

3\text{ m}

(

150\text{ m}^3 / 50\text{ m}^2

). The depth in Basin Beta is

6\text{ m}

(

30\text{ m}^3 / 5\text{ m}^2

). The ratio of these depths is

3:6

, which simplifies to

1:2

.

Step-by-Step Solution

1

Calculate the depth of water in Basin Alpha.

h_{Alpha} = \frac{150\text{ m}^3}{50\text{ m}^2} = 3\text{ m}

Hydrostatic pressure is determined by fluid depth. Depth is found by dividing the total volume by the base area.

2

Calculate the depth of water in Basin Beta.

h_{Beta} = \frac{30\text{ m}^3}{5\text{ m}^2} = 6\text{ m}

The exact same volumetric relationship applies to find the vertical depth in the second basin.

3

Determine the ratio of their hydrostatic pressures.

Ratio =

3\text{ m} : 6\text{ m} = 1:2

Because both basins contain the same fluid (water) under the same gravitational force, the ratio of their bottom pressures is exactly equal to the ratio of their depths (

P = \rho g h

).

Key Concept

Hydrostatic pressure is completely independent of a container's shape, base area, or total fluid volume; it depends exclusively on fluid density, gravity, and vertical depth.

Estimated Time:1m 30s

Question 9Question →

A narrow vertical pipe and a massive, wide cylindrical storage tank are both placed on a flat horizontal surface and filled with fresh water. The water level in both containers reaches an exact depth of 15 feet. Based on the principles of fluid mechanics, how does the hydrostatic pressure at the bottom of the narrow pipe compare to the pressure at the bottom of the wide tank?

The pressure is greater in the tank because it holds a much larger total weight of water.

The pressure is exactly the same at the bottom of both containers.

The pressure is greater in the pipe because the water is concentrated over a smaller bottom area.

The pressure is greater in the tank because the large volume of water compresses the fluid at the bottom.

Show answer & explanation

Answer: The pressure is exactly the same at the bottom of both containers.

Answer

The pressure is exactly the same at the bottom of both containers.

Hydrostatic pressure in an open container is determined entirely by the density of the fluid and its depth. Because both containers hold the same fluid (fresh water) at the exact same depth (15 feet), the pressure exerted on the bottom of each container is identical. The shape of the container and the total volume of the water do not alter the pressure at a given depth.

Step-by-Step Solution

1

Identify the factors that determine hydrostatic pressure in a fluid.

Hydrostatic pressure is determined by the formula

P = \rho g h

, where

\rho

is density,

g

is gravity, and

h

is depth.

This formula establishes that pressure in a resting fluid is independent of the container's shape or total fluid volume.

2

Compare the fluid properties and depth in both containers.

Both containers hold fresh water (same density,

\rho

) and the water reaches exactly 15 feet in both (same depth,

h

).

Comparing the variables in the hydrostatic pressure formula reveals if the pressure differs.

3

Determine the relative pressure at the bottom of each container.

Since both the density and the depth are identical, the pressure calculated by

P = \rho g h

is exactly the same for both containers.

Container width and total water weight do not factor into the calculation for hydrostatic pressure at a specific point.

Key Concept

Hydrostatic pressure depends exclusively on the depth and density of the fluid, not on the shape of the container or the total volume of fluid it holds.

Question 10Question →

A scuba diver records the surrounding water pressure at four different locations in a freshwater lake. Arrange these locations in order from the LOWEST hydrostatic pressure to the HIGHEST hydrostatic pressure experienced by the diver.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest to highest hydrostatic pressure is: Location W (1 meter), Location X (4 meters), Location Y (12 meters), and Location Z (18 meters).

Hydrostatic pressure is determined by the weight of the fluid directly above an object. In a single body of water like a lake, the density of the water is constant. Therefore, the hydrostatic pressure depends entirely on the depth (

h

). Since the question asks for the order from lowest to highest pressure, the locations must be ordered from the shallowest depth to the deepest depth: 1 meter, 4 meters, 12 meters, and finally 18 meters.

Step-by-Step Solution

1

Recall the relationship between hydrostatic pressure and depth.

Hydrostatic pressure is calculated using the formula P = ρgh (where ρ is fluid density, g is gravity, and h is depth). Therefore, pressure increases as depth increases.

Understanding this direct relationship is required to rank the locations.

2

Identify the depth of each location provided in the scenario.

Location W is 1m, Location X is 4m, Location Y is 12m, and Location Z is 18m.

The depth is the only variable changing between the different locations.

3

Order the locations from the smallest depth to the greatest depth.

Smallest depth (1m) is Location W, followed by 4m (Location X), 12m (Location Y), and the greatest depth is 18m (Location Z).

Since pressure increases with depth, ordering from smallest to greatest depth directly corresponds to lowest to highest pressure.

Key Concept

In a fluid of uniform density, hydrostatic pressure increases proportionally with the depth of the fluid.

Question 11Question →

A large dam holds back a deep reservoir of fresh water. Three pressure sensors are installed on the submerged face of the dam at different depths: Sensor X is 10 meters below the water surface, Sensor Y is 2 meters below the water surface, and Sensor Z is 25 meters below the water surface. Arrange the sensors in order from the LOWEST hydrostatic pressure to the HIGHEST hydrostatic pressure.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest to highest hydrostatic pressure is: Sensor Y, Sensor X, and Sensor Z.

Hydrostatic pressure is determined by the density of the fluid, gravity, and the depth of the fluid. Because the fluid (fresh water) is consistent across all sensors, the pressure depends entirely on the depth. The sensor closest to the water's surface (Sensor Y at 2 meters) experiences the least amount of downward force from the water column above it, giving it the lowest pressure. The sensor furthest down (Sensor Z at 25 meters) experiences the most force, resulting in the highest pressure.

Step-by-Step Solution

1

Identify the relationship between depth and hydrostatic pressure.

Hydrostatic pressure increases as the depth of the fluid increases.

The formula for hydrostatic pressure,

P = \rho g h

, demonstrates that pressure (

P

) is directly proportional to depth (

h

).

2

Compare the vertical depths of the three sensors.

Sensor Y is at 2 meters, Sensor X is at 10 meters, and Sensor Z is at 25 meters.

Determining the relative depths is required to compare the relative pressures.

3

Order the sensors based on depth to find the sequence from lowest to highest pressure.

Lowest pressure is at Sensor Y, followed by Sensor X, and the highest is at Sensor Z.

Because the question asks for the order from lowest to highest pressure, the items must be arranged from the shallowest depth to the deepest depth.

Key Concept

Hydrostatic pressure in a fluid increases proportionally with depth.

Question 12Question →

An underwater research drone is deployed in a freshwater lake. It records a hydrostatic pressure reading at a depth of 10 meters. The drone then descends vertically and records a second pressure reading at a depth of 40 meters. Assuming the water density is uniform, how does the second pressure reading compare to the first?

The second reading is four times as large as the first reading.

The second reading is one-fourth the magnitude of the first reading.

Both readings display equivalent values since they were taken in the same body of water.

The pressure difference depends on the weight of the drone rather than the change in depth.

Show answer & explanation

Answer: The second reading is four times as large as the first reading.

Answer

The second reading is four times as large as the first reading.

Hydrostatic pressure is calculated using the equation

P = \rho g h

. Because the density of the water (

\rho

) and the acceleration due to gravity (

g

) remain constant in this scenario, the pressure (

P

) is directly proportional to the depth (

h

). When the drone descends from 10 meters to 40 meters, the depth increases by a factor of 4. Therefore, the hydrostatic pressure also increases by exactly a factor of 4.

Step-by-Step Solution

1

Identify the mathematical formula for hydrostatic pressure.

P = \rho g h

, where

P

is pressure,

\rho

is density,

g

is gravity, and

h

is depth.

This formula establishes the mathematical relationship between fluid pressure and depth.

2

Determine the proportional relationship between the two recorded depths.

The second depth (40 meters) is exactly four times the first depth (10 meters).

Calculating the ratio of the depths allows for a direct comparison of the resulting pressures.

3

Apply the depth ratio to the pressure formula.

Since density and gravity are constant, increasing

h

by a factor of 4 increases

P

by a factor of 4.

Hydrostatic pressure is directly proportional to fluid depth.

Key Concept

Hydrostatic pressure is directly proportional to the depth of the fluid.

Question 13Question →

Two outdoor swimming pools are filled with fresh water to an exact depth of $3\text{ meters}$ . The first is a small backyard wading pool, while the second is a massive Olympic-sized pool holding a much larger total volume of water. How does the hydrostatic pressure at the very bottom of the backyard pool compare to the hydrostatic pressure at the bottom of the Olympic-sized pool?

The pressure is exactly equal in both pools because they share the same fluid density and water depth.

The pressure is significantly greater in the Olympic-sized pool because it contains a larger total mass of water.

The pressure at the bottom of both pools is exactly

3\text{ Pascals}

because the depth is exactly

3\text{ meters}

.

The pressure depends entirely on the buoyant force, which requires a heavy object to be placed in the water.

Show answer & explanation

Answer: The pressure is exactly equal in both pools because they share the same fluid density and water depth.

Answer

The pressure is exactly equal in both pools because they share the same fluid density and water depth.

The correct answer accurately states that hydrostatic pressure relies solely on depth and fluid density. Because both pools contain fresh water and are filled to the exact same depth of

3\text{ meters}

, the pressure at the bottom of both pools is identical, despite their vast difference in total volume.

Step-by-Step Solution

1

Identify the factors that determine hydrostatic pressure in a static fluid.

Hydrostatic pressure is determined by the fluid's density, the acceleration due to gravity, and the depth of the fluid.

To evaluate whether factors like total volume or container shape influence the final pressure reading.

2

Compare the relevant factors between the two distinct swimming pools.

Both pools contain fresh water (same density) and have an exact depth of

3\text{ meters}

.

To determine if there are any differences in the variables that actually calculate pressure.

3

Deduce the relative pressure at the bottom of each pool based on the identified factors.

Since both fluid density and depth are identical, the hydrostatic pressure at the bottom must be identical.

According to the principles of fluid mechanics, hydrostatic pressure is independent of the container's width or the fluid's total mass.

Key Concept

Hydrostatic pressure is dependent only on fluid density and depth, not on the container's shape or the total volume of fluid.

Question 14Question →

A heavy-duty diving bell is being tested in a deep saltwater facility. At a depth of $10\text{ meters}$ , the hydrostatic pressure exerted on its exterior is measured to be $100\text{ kilopascals (kPa)}$ . If the diving bell is lowered further to a depth of $30\text{ meters}$ , what will the new hydrostatic pressure be in $\text{kPa}$ , assuming the fluid density remains constant?

Show answer & explanation

Answer: 300

Answer

The new hydrostatic pressure will be 300 kPa.

Because hydrostatic pressure is directly proportional to depth in a fluid of constant density, tripling the depth from 10 meters to 30 meters results in exactly three times the pressure. Thus, 100 kPa multiplied by 3 equals 300 kPa.

Step-by-Step Solution

1

Identify the relationship between hydrostatic pressure and depth.

Pressure is directly proportional to depth.

This is a fundamental principle of hydrostatic pressure for a fluid with constant density.

2

Calculate the ratio of the new depth to the initial depth.

30\text{ m} / 10\text{ m} = 3

This ratio tells us how much the pressure will increase.

3

Calculate the new pressure.

100\text{ kPa} \times 3 = 300\text{ kPa}

Applying the direct proportion to find the final pressure value.

Key Concept

Hydrostatic pressure increases in direct proportion to the depth of the fluid, assuming fluid density remains constant.

Question 15Question →

A vertical standpipe in a building's fire suppression system is completely filled with fresh water. The top of the water column is exactly $25\text{ meters}$ above a closed test valve on the ground floor. What is the hydrostatic pressure exerted by the water directly on this valve?

(Assume the density of water is $1000\text{ kg/m}^3$ and the acceleration due to gravity is $10\text{ m/s}^2$ . Provide your answer in kilopascals, $\text{kPa}$ .)

Show answer & explanation

Answer: 250

Answer

250

The hydrostatic pressure at the bottom of the water column is found using the equation

P = \rho g h

. Substituting the given values gives

1000 \times 10 \times 25 = 250,000\text{ Pascals}

. To express this in kilopascals, you divide by 1,000, resulting in exactly

250\text{ kPa}

.

Step-by-Step Solution

1

Identify the relevant formula for hydrostatic pressure.

P = \rho g h

Hydrostatic pressure at a given depth depends strictly on the fluid's density, gravity, and the depth of the fluid column.

2

Substitute the given values into the formula.

P = 1000\text{ kg/m}^3 \times 10\text{ m/s}^2 \times 25\text{ m}

These are the precise physical parameters provided in the scenario.

3

Calculate the total pressure in standard SI units (Pascals).

P = 250,000\text{ Pa}

Multiplying the standard SI units yields the pressure in Pascals, since

1\text{ Pa} = 1\text{ kg}/(\text{m}\cdot\text{s}^2)

.

4

Convert the pressure from Pascals to kilopascals.

250,000 / 1000 = 250\text{ kPa}

The question specifically requests the final numerical answer in kilopascals.

Key Concept

Hydrostatic pressure is directly proportional to the density of the fluid, the acceleration due to gravity, and the vertical depth of the fluid column (

P = \rho g h

), regardless of the total fluid volume.

Estimated Time:1m 0s

Question 16Question →

An open-ended U-shaped tube contains two immiscible static liquids, Liquid A and Liquid B, which meet at a distinct separation line. Above this horizontal separation line, the column of Liquid A on the left side is exactly $20\text{ cm}$ tall. The column of Liquid B on the right side is exactly $30\text{ cm}$ tall. If the density of Liquid A is $1.5\text{ g/cm}^3$ , what must be the density of Liquid B to maintain this equilibrium?

1.0\text{ g/cm}^3

2.25\text{ g/cm}^3

0.67\text{ g/cm}^3

1.5\text{ g/cm}^3

Show answer & explanation

Answer: $1.0\text{ g/cm}^3$

Answer

The density of Liquid B is

1.0\text{ g/cm}^3

.

In an open U-tube manometer at equilibrium, the hydrostatic pressure exerted by both fluid columns at their meeting point must be equal. Since pressure equals density times gravity times height (

P = \rho g h

), we can drop the constant gravity and set

\rho_A \times h_A = \rho_B \times h_B

. Multiplying Liquid A's density (

1.5

) by its height (

20

) gives

30

. To balance this, Liquid B must have a density that, when multiplied by its height (

30

), also equals

30

. Therefore, Liquid B's density is

1.0\text{ g/cm}^3

.

Step-by-Step Solution

1

Identify the relevant physics principle for a static fluid system.

The hydrostatic pressure at the horizontal interface must be equal on both sides (

P_A = P_B

).

If the pressures were not equal at the interface, the fluid would move until equilibrium is reached.

2

Set up the pressure equivalence equation.

\rho_A \times g \times h_A = \rho_B \times g \times h_B

Hydrostatic pressure is the product of fluid density, gravitational acceleration, and depth.

3

Cancel out gravity (

g

) and substitute the given values into the equation.

1.5 \times 20 = \rho_B \times 30

Gravity is constant on both sides, and the values provided are density (

1.5\text{ g/cm}^3

) and heights (

20\text{ cm}

and

30\text{ cm}

).

4

Solve for the unknown density (

\rho_B

).

30 = 30 \times \rho_B \Rightarrow \rho_B = 1.0\text{ g/cm}^3

Isolating the variable gives the exact density required to balance the pressure.

Key Concept

Hydrostatic pressure depends on the product of a fluid's density and its vertical depth.

Question 17Question →

A municipal water network is supplied by a central water tower. The free surface of the water inside the tower is maintained at a constant elevation of $50\text{ meters}$ above sea level. The plumbing network is a closed, interconnected system, and the water is completely static.

Arrange the following four locations within this network in order from the LOWEST hydrostatic water pressure to the HIGHEST hydrostatic water pressure:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest pressure to highest pressure is: Hilltop valve, Second-floor residential tap, Street-level fire hydrant, and Basement water main.

In a closed, static fluid system, hydrostatic pressure is determined entirely by the depth of the fluid relative to its highest free surface. By calculating the depth of each location from the 50-meter water tower surface, we find that locations at higher absolute elevations have shallower water columns above them, resulting in lower pressure. Conversely, locations at lower elevations have deeper water columns and thus higher pressure. Therefore, ranking the locations from highest elevation to lowest elevation correctly orders them from lowest hydrostatic pressure to highest hydrostatic pressure.

Step-by-Step Solution

1

Identify the reference point for hydrostatic pressure.

The reference point is the highest free surface of the interconnected water system, which is the water level in the tower at an elevation of 50 meters above sea level.

Hydrostatic pressure in a connected, static fluid system is driven by the depth of the fluid column below its highest free surface.

2

Calculate the depth of each location relative to the free surface.

Hilltop valve depth:

50 - 40 = 10\text{ m}

. Residential tap depth:

50 - 20 = 30\text{ m}

. Fire hydrant depth:

50 - 10 = 40\text{ m}

. Basement main depth:

50 - (-5) = 55\text{ m}

.

Depth (

h

) is found by subtracting the elevation of the specific point from the elevation of the fluid surface.

3

Relate the calculated depths to hydrostatic pressure and order them.

Since pressure increases proportionally with depth, the order from lowest to highest depth (and thus pressure) is

10\text{ m}

,

30\text{ m}

,

40\text{ m}

, and

55\text{ m}

.

Hydrostatic pressure (

P

) is directly proportional to the fluid depth (

h

) according to the formula

P = \rho g h

.

Key Concept

Hydrostatic pressure in communicating vessels depends solely on the depth of the fluid from the highest free surface, independent of the shape of the pipes.

Question 18Question →

A specialized deep-well chemical injection pump is designed to operate safely up to a specific maximum hydrostatic pressure limit. During factory calibration in a test well filled with fresh water (density of $1,000\text{ kg/m}^3$ ), the pump reaches its maximum safe pressure rating at exactly $50\text{ meters}$ of depth. For its field deployment, the pump is lowered into a vertical shaft filled with heavy drilling mud (density of $1,250\text{ kg/m}^3$ ). Assuming the pressure rating remains constant, at what depth will the pump reach its maximum hydrostatic pressure limit in the drilling mud?

40\text{ meters}

62.5\text{ meters}

50\text{ meters}

25\text{ meters}

Show answer & explanation

Answer: $40\text{ meters}$

Answer

The pump will reach its maximum safe pressure limit at

40\text{ meters}

of depth in the drilling mud.

Hydrostatic pressure is determined by the formula

P = \rho g h

. Because the drilling mud is denser (

1,250\text{ kg/m}^3

) than the fresh water (

1,000\text{ kg/m}^3

), it exerts more pressure for every meter of depth. Since the mud is

1.25

times as dense as the water (

1,250 / 1,000 = 1.25

), the pump will reach the maximum pressure limit at

1/1.25

(or

4/5

) of the original depth. Multiplying

50\text{ meters}

by

0.8

yields exactly

40\text{ meters}

.

Step-by-Step Solution

1

Identify the relationship between hydrostatic pressure, fluid density, and depth.

P = \rho \cdot g \cdot h

Hydrostatic pressure is directly proportional to both the fluid's density and the depth.

2

Set up an equation equating the pressure in both fluids.

\rho_{\text{water}} \cdot h_{\text{water}} = \rho_{\text{mud}} \cdot h_{\text{mud}}

Since the maximum pressure (

P

) and gravity (

g

) are identical in both scenarios, the product of density and depth must remain constant.

3

Substitute the known values and solve for the target depth.

1,000 \cdot 50 = 1,250 \cdot h_{\text{mud}}

, which simplifies to

50,000 = 1,250 \cdot h_{\text{mud}}

, resulting in

h_{\text{mud}} = 40

.

Dividing the constant pressure factor by the new, higher density yields the shallower depth required.

Key Concept

Hydrostatic pressure at a given depth is directly proportional to the density of the fluid. A denser fluid exerts more pressure per meter, requiring less depth to reach the same total pressure.

Question 19Question →

An open industrial mixing vat contains a liquid chemical with a specific gravity of $1.25$ . The liquid in the vat reaches a depth of $4\text{ meters}$ . Assuming the density of fresh water is $1,000\text{ kg/m}^3$ and the acceleration due to gravity is $10\text{ m/s}^2$ , what is the hydrostatic pressure exerted by the liquid at the bottom of the vat? (Enter your answer in kilopascals)

Show answer & explanation

Answer: 50

Answer

50

The hydrostatic pressure at the bottom of a fluid column is determined by the formula

P = \rho g h

. First, the specific gravity of

1.25

indicates the fluid is

1.25

times as dense as water, giving a density of

1,250\text{ kg/m}^3

. Multiplying this density by the acceleration due to gravity (

10\text{ m/s}^2

) and the depth (

4\text{ meters}

) yields

50,000\text{ Pascals}

. Since

1,000\text{ Pa}

equals

1\text{ kPa}

, this is equivalent to

50\text{ kPa}

.

Step-by-Step Solution

1

Determine the density of the fluid.

Density = 1,250 kg/m³

Specific gravity is the ratio of the fluid's density to the density of water. Multiplying 1.25 by 1,000 kg/m³ gives the actual density.

2

Calculate the hydrostatic pressure in Pascals.

Pressure = 50,000 Pa

Hydrostatic pressure is the product of density, gravity, and depth (1,250 × 10 × 4 = 50,000).

3

Convert the pressure to kilopascals (kPa).

50 kPa

The question asks for the answer in kilopascals, so divide the Pascal value by 1,000.

Key Concept

Hydrostatic Pressure

Question 20Question →

A remote-controlled submersible is conducting an inspection in a freshwater lake. It initially hovers at a static depth of $10\text{ meters}$ below the surface. Later, the operator commands the submersible to descend to a depth of $3,000\text{ centimeters}$ . Assuming the water density remains constant, how does the hydrostatic pressure on the submersible at the new depth compare to the pressure at the initial depth?

The hydrostatic pressure is exactly

3\text{ times}

as great.

The hydrostatic pressure is exactly

300\text{ times}

as great.

The hydrostatic pressure remains the same because the volume of the submersible displacing the water has not changed.

The hydrostatic pressure increases exponentially because the deeper water is highly compressed to a greater density.

Show answer & explanation

Answer: The hydrostatic pressure is exactly $3\text{ times}$ as great.

Answer

The hydrostatic pressure is exactly 3 times as great.

Hydrostatic pressure is calculated using the formula

P = \rho \cdot g \cdot h

. Because the fluid (freshwater) and gravity remain constant, the pressure is directly proportional to the depth (

h

). To accurately compare the depths, they must be in the same unit. Converting the new depth to meters (

3,000\text{ cm} = 30\text{ m}

) reveals that the submersible has gone from

10\text{ meters}

to

30\text{ meters}

. Since the depth has increased by a factor of 3, the hydrostatic pressure also increases by a factor of 3.

Step-by-Step Solution

1

Identify the formula for hydrostatic pressure.

P = \rho \cdot g \cdot h

, meaning pressure is directly proportional to depth (

h

) when density (

\rho

) and gravity (

g

) are constant.

Establishing the mathematical relationship is necessary to compare the two states.

2

Convert the second depth into meters to match the initial depth's units.

3,000\text{ cm} \div 100\text{ cm/m} = 30\text{ meters}

.

Physical quantities must be in the same units before they can be accurately compared or divided.

3

Compare the new depth to the initial depth.

30\text{ meters} \div 10\text{ meters} = 3

.

Because the depth has tripled, the directly proportional hydrostatic pressure also triples.

Key Concept

Hydrostatic pressure in an incompressible fluid is directly proportional to its depth.

Estimated Time:1m 0s

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