Fluid Mechanics Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 3

Question 41Question →

When a steady flow of fluid passes through a horizontal pipe that suddenly narrows in diameter, the fluid's velocity increases in the constricted section, resulting in a higher static pressure against the pipe walls.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because Bernoulli's principle dictates that as a fluid's velocity increases in a constricted horizontal pipe, its static pressure decreases, creating a low-pressure zone rather than a high-pressure one.

Step-by-Step Solution

1

Analyze the change in fluid velocity through the constriction.

Based on the continuity equation (

A_1V_1 = A_2V_2

), the fluid's velocity increases as it enters the narrower section of the pipe.

The same volume of fluid must pass through a smaller area in the same amount of time, requiring it to speed up.

2

Determine the effect of increased velocity on static pressure.

Static pressure decreases.

Bernoulli's principle states that for a horizontal flow, an increase in the speed of the fluid occurs simultaneously with a decrease in static pressure.

3

Evaluate the statement's claim against the established physical principles.

The statement claims static pressure increases, which contradicts the actual outcome (a decrease in static pressure).

The inverse relationship between fluid velocity and static pressure confirms the statement is false.

Key Concept

Bernoulli's principle and the continuity equation for incompressible fluids.

Estimated Time:45s

Question 42Question →

A mechanic uses a hydraulic floor jack to lift the front axle of a delivery truck. The jack's output piston has a cross-sectional area of $25 \text{ in}^2$ and must support a load of $5,000 \text{ lbs}$ . If the input piston has a cross-sectional area of $0.5 \text{ in}^2$ , what is the minimum input force the mechanic must apply to lift the truck? (Assume a frictionless, ideal system)

100 \text{ lbs}

200 \text{ lbs}

10,000 \text{ lbs}

250,000 \text{ lbs}

Show answer & explanation

Answer: $100 \text{ lbs}$

Answer

100 \text{ lbs}

In a closed hydraulic system, pressure is uniform. The pressure created by the load on the output piston is its force divided by its area (

5,000 / 25 = 200 \text{ psi}

). To produce this same pressure at the input piston, you multiply the pressure by the input piston's area (

200 \times 0.5

), which equals an input force of

100 \text{ lbs}

.

Step-by-Step Solution

1

Calculate the fluid pressure required at the output piston using the formula P = F/A.

5,000 \text{ lbs} / 25 \text{ in}^2 = 200 \text{ psi}

Pascal's Principle states that pressure applied to a confined fluid is transmitted equally in all directions.

2

Apply the calculated pressure to the input piston to find the required input force using F = P × A.

200 \text{ psi} \times 0.5 \text{ in}^2 = 100 \text{ lbs}

The input piston must generate the identical system pressure to balance and lift the load.

Key Concept

Pascal's Principle (Hydraulic Mechanical Advantage)

Question 43Question →

A small metal anchor weighs $120 \text{ N}$ in the air. When the anchor is completely submerged in water, the tension on its lifting cable is measured to be $85 \text{ N}$ . What is the magnitude of the buoyant force acting on the submerged anchor?

Show answer & explanation

Answer: 35

Answer

35

According to Archimedes' principle, the upward buoyant force exerted on a submerged object reduces its apparent weight. The buoyant force is equal to the difference between its true weight in the air and its apparent weight in the fluid. Subtracting the apparent weight (85 N) from the true weight (120 N) gives a buoyant force of 35 N.

Step-by-Step Solution

1

Identify the actual weight and the apparent weight.

Actual weight = 120 N, Apparent weight = 85 N

These values represent the downward force of gravity and the net downward force when submerged, respectively.

2

Calculate the buoyant force.

120 N - 85 N = 35 N

The buoyant force is the upward force that accounts for the reduction in weight. It is found by subtracting the apparent weight from the actual weight.

Key Concept

Apparent Weight and Buoyant Force

Question 44Question →

An industrial cardboard baler utilizes a closed hydraulic system to compress recyclable materials. The system features an input pump piston with a cross-sectional area of $3 \text{ in}^2$ and a large output driving piston with a cross-sectional area of $45 \text{ in}^2$ . If the input piston is pushed downward for a single stroke distance of $15 \text{ inches}$ , how many inches will the output piston move? (Assume no fluid is lost and the fluid is incompressible.)

Show answer & explanation

Answer: 1

Answer

1

In a hydraulic system, the volume of fluid displaced must be conserved. The output piston has 15 times the cross-sectional area of the input piston (

45 \div 3 = 15

). Because it is 15 times larger, it will only travel 1/15th of the distance to displace the identical volume of fluid. Therefore, an input distance of 15 inches results in an output distance of exactly 1 inch.

Step-by-Step Solution

1

Identify the relationship for fluid displacement in a closed hydraulic system.

A_1 \times d_1 = A_2 \times d_2

Because liquids are incompressible, the volume of fluid pushed out by the input piston (

A_1 \times d_1

) must equal the volume of fluid pushing up the output piston (

A_2 \times d_2

).

2

Substitute the given values for the input piston (

A_1 = 3

,

d_1 = 15

) and output piston (

A_2 = 45

).

3 \times 15 = 45 \times d_2

This sets up the equation with all known variables to solve for the unknown distance (

d_2

).

3

Calculate the total volume displaced by the input piston.

45 = 45 \times d_2

Multiplying the input area by its travel distance gives a total displaced volume of 45 cubic inches.

4

Solve for the output distance (

d_2

) by dividing the total volume by the output piston's area.

d_2 = 1

Dividing 45 cubic inches by 45 square inches yields a travel distance of 1 inch.

Key Concept

Conservation of fluid volume and the distance-force tradeoff in hydraulics

Question 45Question →

An industrial pneumatic press system utilizes trapped gas to generate force. The gas is initially at an absolute pressure of $30 \text{ psi}$ and occupies a volume of $200 \text{ in}^3$ . During the operation cycle, a mechanical ram pushes down, reducing the gas volume to $50 \text{ in}^3$ while the temperature remains constant. If the face of the piston has a cross-sectional area of $5 \text{ in}^2$ , what is the total upward force, in pounds, exerted by the compressed gas on the piston at this new volume?

Show answer & explanation

Answer: 600

Answer

600

Since the temperature is constant, Boyle's Law (

P_1V_1 = P_2V_2

) applies. The volume is reduced from

200 \text{ in}^3

to

50 \text{ in}^3

, which is one-quarter of its original size. Because pressure and volume are inversely proportional for a fixed amount of gas at a constant temperature, the absolute pressure must increase by a factor of four (

30 \text{ psi} \times 4 = 120 \text{ psi}

). The total force exerted by a pressurized fluid on a surface is the product of its pressure and the surface area (

F = P \times A

). Multiplying the new pressure of

120 \text{ psi}

by the piston area of

5 \text{ in}^2

gives a total upward force of

600 \text{ lbs}

.

Step-by-Step Solution

1

Calculate the final pressure of the gas after compression.

P_2 = 120 \text{ psi}

According to Boyle's Law, when temperature is constant, the product of absolute pressure and volume is constant (

P_1V_1 = P_2V_2

).

2

Calculate the total upward force exerted by the gas.

F = 600 \text{ lbs}

Force is equal to pressure multiplied by the area over which it is applied (

F = P \times A

).

Key Concept

Combining Boyle's Law for isothermal compression with the basic relationship between pressure, force, and area.

Question 46Question →

A hydraulic dump truck bed lift system operates using a small input pump cylinder and a larger output lifting cylinder. The pump cylinder has a cross-sectional area of $2 \text{ in}^2$ and the lifting cylinder has a cross-sectional area of $20 \text{ in}^2$ . To raise the truck bed, the lifting cylinder must extend a distance of $8 \text{ inches}$ . Assuming an ideal closed hydraulic system, what is the total distance the pump cylinder must travel to achieve this lift?

80 \text{ inches}

0.8 \text{ inches}

160 \text{ inches}

10 \text{ inches}

Show answer & explanation

Answer: $80 \text{ inches}$

Answer

The input pump cylinder must travel a total distance of

80 \text{ inches}

.

Because liquids are essentially incompressible, the volume of fluid pushed by the input cylinder must equal the volume received by the output cylinder. Volume is calculated as Area multiplied by Distance (

V = A \times d

). Setting the volumes equal gives

2 \times d_1 = 20 \times 8

. Solving this equation shows that the input cylinder must travel

80 \text{ inches}

to move the larger cylinder

8 \text{ inches}

.

Step-by-Step Solution

1

Identify the given values for the hydraulic system.

A_1 = 2 \text{ in}^2

,

A_2 = 20 \text{ in}^2

, and

d_2 = 8 \text{ inches}

.

These values are needed to apply the principle of volume conservation in an incompressible fluid system.

2

Apply the volume conservation formula for hydraulic systems.

A_1 \times d_1 = A_2 \times d_2

In a closed hydraulic system, the volume of fluid displaced by the input piston must equal the volume of fluid displacing the output piston.

3

Substitute the known values into the equation and solve for the input distance (

d_1

).

2 \times d_1 = 20 \times 8 \Rightarrow 2 \times d_1 = 160 \Rightarrow d_1 = 80 \text{ inches}

.

Isolating

d_1

provides the total travel distance required by the input pump to move the lifting cylinder by the required amount.

Key Concept

In a hydraulic system, work and fluid volume are conserved. A smaller input piston must travel a proportionally greater distance to move a larger output piston a shorter distance.

Question 47Question →

A heavy-duty hydraulic press is designed to operate using hydraulic oil with a specific gravity of 0.85. During maintenance, a large volume of air is accidentally trapped in the sealed main pressure line. When the pump applies 2,000 psi of pressure to the system, which of the following best describes how the properties of the contents will affect the machine's operation?

The trapped air will compress significantly before the oil can transmit full force, causing a delayed and spongy cylinder response.

The oil and the trapped air will compress equally, generating standard force but causing the cylinder to actuate at a much faster speed.

The oil will compress into a smaller volume due to its specific gravity being less than 1.0, while the air will resist compression and block fluid flow.

The trapped air will maintain a constant volume but increase in density, acting as a solid barrier that prevents the hydraulic oil from transmitting any pressure.

Show answer & explanation

Answer: The trapped air will compress significantly before the oil can transmit full force, causing a delayed and spongy cylinder response.

Answer

The trapped air will compress significantly before the oil can transmit full force, causing a delayed and spongy cylinder response.

Hydraulic systems rely on the incompressibility of liquids to transmit force instantaneously and efficiently. When air (a gas) enters the system, it acts like a weak mechanical spring because gases are highly compressible. The pump's applied pressure will first crush the air pocket into a smaller volume before enough pressure builds up to move the heavy cylinder, resulting in a delayed, inefficient, and spongy mechanical action.

Step-by-Step Solution

1

Identify the two different fluids in the closed hydraulic system.

The system contains a liquid (hydraulic oil) and a gas (air).

Different phases of matter respond completely differently to applied pressure.

2

Evaluate the compressibility of each fluid.

Liquids are nearly incompressible, meaning their volume remains virtually constant under high pressure. Gases are highly compressible, meaning their volume decreases significantly when pressure is applied.

Understanding compressibility is essential to predicting the physical behavior of the system.

3

Determine the mechanical effect on the hydraulic press.

When the 2,000 psi is applied, the work energy initially goes into compressing the trapped air pocket rather than moving the cylinder. This creates a delay and a soft, 'spongy' feel until the air is fully compressed.

Force cannot be efficiently transmitted through a fluid medium until all highly compressible elements in the line are reduced in volume.

Key Concept

Fluid Compressibility

Estimated Time:1m 0s

Question 48Question →

A deep-sea recovery team locates a lost concrete mooring block resting on the ocean floor. The solid concrete block has a physical volume of $2 \text{ m}^3$ and a total downward weight of $45,000 \text{ N}$ . The surrounding seawater has a weight density of $10,000 \text{ N/m}^3$ . According to Archimedes' principle, what is the magnitude of the upward buoyant force acting on the completely submerged block?

45,000 \text{ N}

20,000 \text{ N}

25,000 \text{ N}

65,000 \text{ N}

Show answer & explanation

Answer: $20,000 \text{ N}$

Answer

The correct answer is

20,000 \text{ N}

, which is found by multiplying the block's volume by the weight density of the seawater.

According to Archimedes' Principle, the buoyant force exerted on an object is exactly equal to the weight of the fluid that the object displaces. Because the block is completely submerged, it displaces a volume of water equal to its own volume (

2 \text{ m}^3

). Multiplying this volume by the water's weight density (

10,000 \text{ N/m}^3

) yields an upward buoyant force of

20,000 \text{ N}

. The actual weight of the concrete block is irrelevant to calculating the buoyant force.

Step-by-Step Solution

1

Identify the formula for buoyant force based on Archimedes' Principle.

Buoyant Force = Volume of displaced fluid × Weight density of the fluid.

To establish that buoyancy depends exclusively on fluid properties and displaced volume, rather than the object's own weight.

2

Extract the necessary physical values from the problem description.

The displaced volume is equal to the block's volume (

2 \text{ m}^3

), and the fluid's weight density is

10,000 \text{ N/m}^3

.

To separate the relevant variables needed for the calculation from the distractor information (the block's

45,000 \text{ N}

weight).

3

Calculate the magnitude of the upward buoyant force.

2 \text{ m}^3 \times 10,000 \text{ N/m}^3 = 20,000 \text{ N}

.

To determine the final upward thrust exerted by the seawater.

Key Concept

Archimedes' Principle states that the buoyant force on a completely submerged object equals the weight of the fluid it displaces, independent of the object's own downward weight.

Question 49Question →

A heavy-duty pneumatic shock absorber relies on a sealed chamber of air to dampen impacts. Before an impact, the air inside the chamber has an absolute pressure of $90 \text{ psi}$ , occupies a volume of $60 \text{ in}^3$ , and is at a temperature of $300 \text{ K}$ . During a severe impact, the piston rapidly compresses the air volume to $20 \text{ in}^3$ , causing the internal temperature to spike to $400 \text{ K}$ . Assuming no air escapes, what is the new absolute pressure inside the shock absorber?

40 \text{ psi}

120 \text{ psi}

270 \text{ psi}

360 \text{ psi}

Show answer & explanation

Answer: $360 \text{ psi}$

Answer

360 \text{ psi}

The correct answer correctly applies the Combined Gas Law. The volume is reduced to one-third of its original size, which alone would triple the pressure to

270 \text{ psi}

. Simultaneously, the absolute temperature increases by a factor of

4/3

, which further scales the pressure to

270 \times \frac{4}{3} = 360 \text{ psi}

.

Step-by-Step Solution

1

Identify the initial state (

P_1, V_1, T_1

) and final state (

V_2, T_2

) variables from the problem.

P_1 = 90 \text{ psi}

,

V_1 = 60 \text{ in}^3

,

T_1 = 300 \text{ K}

;

V_2 = 20 \text{ in}^3

,

T_2 = 400 \text{ K}

.

Establishing known parameters is required to use the Combined Gas Law.

2

Set up the Combined Gas Law equation:

\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}

.

\frac{90 \times 60}{300} = \frac{P_2 \times 20}{400}

Since both volume and temperature are changing simultaneously, the combined relationship must be used.

3

Rearrange the formula to isolate the final pressure (

P_2

).

P_2 = P_1 \times \left(\frac{V_1}{V_2}\right) \times \left(\frac{T_2}{T_1}\right)

This setup groups the variables into easy-to-calculate ratios.

4

Substitute the values and calculate the final result.

P_2 = 90 \times \left(\frac{60}{20}\right) \times \left(\frac{400}{300}\right) = 90 \times 3 \times \frac{4}{3} = 360 \text{ psi}

.

Mathematical evaluation yields the final absolute pressure inside the shock absorber.

Key Concept

Combined Gas Law and Compressibility

Question 50Question →

Four different objects are completely submerged in a tank of fresh water. The mass of each object and the volume of water it displaces are recorded below. Arrange the objects in order of the buoyant force exerted on them by the water, from LEAST buoyant force to GREATEST buoyant force.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from least to greatest buoyant force is: Iron weight, Lead brick, Aluminum gear, Wooden block.

According to Archimedes' principle, the buoyant force acting on an object is equal to the weight of the fluid it displaces. Because all the objects are in the same fresh water, the buoyant force is directly proportional to the displaced volume. By ordering the displaced volumes from smallest to largest (

0.2 \text{ L}

,

0.5 \text{ L}

,

1.0 \text{ L}

,

2.0 \text{ L}

), we accurately order the buoyant forces from least to greatest.

Step-by-Step Solution

1

Recall the definition of Archimedes' principle.

The buoyant force acting on a submerged object is equal to the weight of the fluid that the object displaces.

This principle establishes that when objects are submerged in the same fluid, the buoyant force depends entirely on the volume of fluid displaced, not the mass of the objects themselves.

2

Extract the relevant data for each object.

Iron =

0.2 \text{ L}

, Lead =

0.5 \text{ L}

, Aluminum =

1.0 \text{ L}

, Wood =

2.0 \text{ L}

.

The mass values provided (

0.5 \text{ kg}

to

3.0 \text{ kg}

) are distractor information and should be ignored for calculating buoyant force.

3

Rank the objects based on their displaced volumes from least to greatest.

Iron (

0.2 \text{ L}

) < Lead (

0.5 \text{ L}

) < Aluminum (

1.0 \text{ L}

) < Wood (

2.0 \text{ L}

).

Since greater displaced volume equals greater buoyant force, this is the final correct order.

Key Concept

According to Archimedes' Principle, the buoyant force on a submerged object depends only on the density of the fluid and the volume of the fluid displaced, completely independent of the object's mass or weight.

Estimated Time:45s

Question 51Question →

Match each closed hydraulic system scenario to its calculated unknown value. Assume the fluid is incompressible, there is no friction, and conditions are ideal.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches are based on applying Pascal's Principle and the conservation of work. Output Force is 250 lbs, Output Area is 15 in², Input Force is 50 lbs, and Output Distance is 1 in.

Each matching pair applies either Pascal's Principle (

P = \frac{F_1}{A_1} = \frac{F_2}{A_2}

) or the conservation of work (

F_1 \cdot d_1 = F_2 \cdot d_2

) to solve for the missing variable in the hydraulic system.

Step-by-Step Solution

1

Calculate the Output Force for the first scenario.

250 lbs

Using Pascal's Principle (

F_1/A_1 = F_2/A_2

), the system pressure is 50 lbs / 2 in² = 25 psi. The output force is 25 psi * 10 in² = 250 lbs.

2

Calculate the Output Area for the second scenario.

15 in²

The system pressure is 30 lbs / 3 in² = 10 psi. Rearranging the formula for area (

A = F/P

), the output area is 150 lbs / 10 psi = 15 in².

3

Calculate the Input Force for the third scenario.

50 lbs

System pressure is determined by the output side: 100 lbs / 20 in² = 5 psi. The required input force is 5 psi * 10 in² = 50 lbs.

4

Calculate the Output Distance for the fourth scenario.

1 in

By conservation of work (

F_1 \times d_1 = F_2 \times d_2

), the input work is 40 lbs * 5 in = 200 in-lbs. Thus, the output distance is 200 in-lbs / 200 lbs = 1 in.

Key Concept

Pascal's Principle states that pressure applied to an enclosed fluid is transmitted undiminished. The work done on the input piston equals the work done by the output piston.

Question 52Question →

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 53Question →

A hydraulic lifting mechanism on a heavy-duty forklift contains an input cylinder and an output cylinder. The output cylinder has $5$ times the cross-sectional area of the input cylinder. If the operator applies an input force of $50 \text{ lbs}$ to move the input piston down by $15 \text{ inches}$ , what is the resulting lifting force and how far does the output piston move?

250 \text{ lbs}

of force and

3 \text{ inches}

of movement

250 \text{ lbs}

of force and

75 \text{ inches}

of movement

10 \text{ lbs}

of force and

3 \text{ inches}

of movement

10 \text{ lbs}

of force and

75 \text{ inches}

of movement

Show answer & explanation

Answer: $250 \text{ lbs}$ of force and $3 \text{ inches}$ of movement

Answer

The output cylinder produces

250 \text{ lbs}

of lifting force and moves

3 \text{ inches}

.

The hydraulic system provides a mechanical advantage equal to the ratio of the piston areas. Since the output cylinder has

5

times the area, the output force is

5

times the input force (

50 \times 5 = 250 \text{ lbs}

). However, to conserve energy and fluid volume, the output piston only moves

1/5

of the distance of the input piston (

15 / 5 = 3 \text{ inches}

).

Step-by-Step Solution

1

Determine the mechanical advantage of the hydraulic system.

Mechanical Advantage =

5

.

The output cylinder has

5

times the cross-sectional area of the input cylinder.

2

Calculate the output lifting force.

F_2 = 50 \times 5 = 250 \text{ lbs}

.

According to Pascal's Principle, force is directly proportional to the area. The larger piston exerts

5

times the force.

3

Calculate the distance the output piston moves.

d_2 = 15 / 5 = 3 \text{ inches}

.

Because the volume of incompressible fluid displaced remains constant, the distance moved is inversely proportional to the area.

Key Concept

Pascal's Principle and Conservation of Energy in Hydraulics

Estimated Time:1m 0s

Question 54Question →

A solid steel engine block with a true weight of $2,000 \text{ N}$ is completely submerged in a freshwater lake. The block displaces a volume of water that weighs $250 \text{ N}$ . According to Archimedes' principle, what is the magnitude of the upward buoyant force acting on the engine block?

2,000 \text{ N}

1,750 \text{ N}

2,250 \text{ N}

250 \text{ N}

Show answer & explanation

Answer: $250 \text{ N}$

Answer

250 \text{ N}

According to Archimedes' principle, the upward buoyant force on an object submerged in a fluid is exactly equal to the weight of the fluid it displaces. Since the engine block displaces water weighing

250 \text{ N}

, the buoyant force acting upon it is exactly

250 \text{ N}

.

Step-by-Step Solution

1

Identify the values provided in the scenario.

The true weight of the block is

2,000 \text{ N}

and the weight of the displaced water is

250 \text{ N}

.

Extracting given values clarifies which numbers represent which physical properties.

2

Apply Archimedes' principle to determine the buoyant force.

Archimedes' principle states that the upward buoyant force is equal to the weight of the displaced fluid.

This principle directly links the volume of fluid displaced to the upward force exerted by the fluid.

3

Select the value that represents the weight of the displaced fluid.

The weight of the displaced water is

250 \text{ N}

, therefore the buoyant force is

250 \text{ N}

.

The object's own weight does not determine the buoyant force; only the displaced fluid's weight matters.

Key Concept

Archimedes' Principle and Buoyant Force

Question 55Question →

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 56Question →

A barber chair operates using a closed hydraulic lifting system to raise a customer. The input pump piston has a cross-sectional area of $4 \text{ in}^2$ , and the output lifting piston has a cross-sectional area of $60 \text{ in}^2$ . If the barber pushes the input piston down a distance of $7.5 \text{ inches}$ , how many inches does the output piston move upward?

Show answer & explanation

Answer: 0.5

Answer

0.5

In a hydraulic system, the mechanical advantage that multiplies the output force must correspondingly divide the output distance. The output piston has an area 15 times larger than the input piston (

60 / 4 = 15

). Therefore, while it can lift 15 times the force, it will only travel

1/15\text{th}

of the input distance (

7.5 / 15 = 0.5 \text{ inches}

).

Step-by-Step Solution

1

Identify the relevant principle for hydraulic piston travel distances.

The volume of fluid displaced by the input piston equals the volume gained by the output piston:

A_1 \times d_1 = A_2 \times d_2

.

Because hydraulic fluid is essentially incompressible, the total volume in the closed system remains constant.

2

Set up the equation using the given values.

4 \times 7.5 = 60 \times d_2

Input area (

A_1

) is 4, input distance (

d_1

) is 7.5, and output area (

A_2

) is 60.

3

Calculate the total fluid volume displaced.

4 \times 7.5 = 30 \text{ in}^3

Multiplying area by distance gives the volume of fluid moved.

4

Solve for the output piston's upward distance (

d_2

).

30 / 60 = 0.5 \text{ inches}

Dividing the displaced volume by the output piston's larger area yields the smaller travel distance.

Key Concept

Conservation of volume in hydraulic systems dictates that distance and area are inversely proportional (Mechanical Advantage trades distance for force).

Question 57Question →

Evaluate whether the following statement is true or false:

In a mechanical carburetor, outside air flows through a narrow central constriction. As the air accelerates and its velocity increases through this narrow section, the static pressure of the air increases, forcing fuel to mix into the airstream.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because Bernoulli's principle establishes an inverse relationship between a fluid's velocity and its static pressure. As the air accelerates through the carburetor's narrow constriction, its static pressure drops, creating a low-pressure area (a partial vacuum) that siphons fuel into the airstream, rather than an area of increased pressure.

Step-by-Step Solution

1

Recall Bernoulli's principle regarding fluid velocity and static pressure.

Bernoulli's principle states that as the velocity of a fluid increases, its static pressure decreases.

Establishing the fundamental physical law governing fluids moving through a constriction.

2

Evaluate the claim made in the statement.

The statement incorrectly claims that an increase in air velocity causes the static pressure to increase.

Comparing the scenario's claim directly against the established physics principle to determine its validity.

Key Concept

Inverse relationship between fluid velocity and static pressure (Bernoulli's Principle)

Question 58Question →

In an aircraft's hydraulic braking system, the master cylinder (input) connects to a brake caliper cylinder (output). The cross-sectional area of the caliper cylinder is $5$ times the area of the master cylinder. If the master cylinder piston is pushed with $40 \text{ lbs}$ of force and travels $5 \text{ inches}$ , what is the resulting force and travel distance of the caliper cylinder piston?

200 \text{ lbs}

of force and

1 \text{ inch}

of travel

200 \text{ lbs}

of force and

25 \text{ inches}

of travel

8 \text{ lbs}

of force and

25 \text{ inches}

of travel

8 \text{ lbs}

of force and

1 \text{ inch}

of travel

Show answer & explanation

Answer: $200 \text{ lbs}$ of force and $1 \text{ inch}$ of travel

Answer

The caliper cylinder piston will output

200 \text{ lbs}

of force and move

1 \text{ inch}

.

Based on Pascal's principle and the conservation of energy, the larger output cylinder multiplies the input force by the area ratio (

40 \times 5 = 200 \text{ lbs}

). Correspondingly, the output distance must be divided by that same ratio (

5 \div 5 = 1 \text{ inch}

) to ensure the work input equals the work output.

Step-by-Step Solution

1

Determine the area ratio between the input and output cylinders.

Area Ratio = 5.

The problem states the caliper cylinder has an area 5 times larger than the master cylinder.

2

Calculate the resulting force at the output cylinder.

Output Force = Input Force x Area Ratio =

40 \text{ lbs} \times 5 = 200 \text{ lbs}

.

According to Pascal's principle, pressure is transmitted equally, so the force is multiplied by the ratio of the piston areas.

3

Calculate the travel distance of the output cylinder.

Output Distance = Input Distance / Area Ratio =

5 \text{ inches} / 5 = 1 \text{ inch}

.

Because fluid is incompressible, the volume displaced by the input piston equals the volume displaced by the output piston. A larger area requires proportionally less travel distance.

Key Concept

In a closed hydraulic system, moving from a smaller to a larger cylinder multiplies the force by the area ratio but decreases the travel distance by that same ratio, conserving the total work done.

Question 59Question →

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 60Question →

During flight, an aircraft wing (airfoil) is designed so that air travels significantly faster over its curved upper surface compared to its flatter lower surface. According to Bernoulli's principle, what is the result of this difference in air velocity?

The static pressure above the wing becomes lower than the pressure below it, generating upward lift.

The faster-moving air above the wing generates higher static pressure, which pulls the aircraft upward.

The difference in air speed creates a buoyant force based on the displaced weight of the air, lifting the wing.

The faster air over the top of the wing compresses tightly, increasing its density and pushing the wing upward.

Show answer & explanation

Answer: The static pressure above the wing becomes lower than the pressure below it, generating upward lift.

Answer

The static pressure above the wing becomes lower than the pressure below it, generating upward lift.

According to Bernoulli's principle, as the velocity of a fluid (such as air) increases, its static pressure decreases. Because the wing is shaped to make the air travel faster over its top surface, a region of low pressure is created above the wing. The slower-moving air beneath the wing maintains a relatively higher pressure. This difference in pressure—high on the bottom, low on the top—results in a net upward force that lifts the aircraft.

Step-by-Step Solution

1

Recall the fundamental rule of Bernoulli's principle regarding fluid flow.

In a flowing fluid, an increase in velocity is accompanied by a decrease in static pressure.

This inverse relationship is the core concept of Bernoulli's principle.

2

Apply the principle to the specific scenario of the aircraft wing described in the question.

The air moving over the top of the wing has a higher velocity than the air moving underneath. Therefore, the static pressure on the top surface must be lower than the static pressure on the bottom surface.

Mapping the physical conditions (faster top, slower bottom) to the principle reveals the pressure distribution.

3

Determine the physical consequence of this pressure difference.

The relatively higher pressure beneath the wing pushes upward toward the area of lower pressure above the wing, creating a net upward force known as lift.

Objects are naturally pushed from areas of higher pressure toward areas of lower pressure.

Key Concept

According to Bernoulli's principle, an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure. This pressure differential across surfaces like airfoils creates mechanical forces such as lift.

Estimated Time:45s

Fluid Mechanics

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