Fluids

Question 1

What is a fluid?

Answer 1

It is a phase of matter that is capable of yielding to pressure and changing its shape to fit a container.

On the MCAT, fluids come in two forms: liquids and gases.

Question 2

What are the properties of an ideal fluid?

Answer 2

• incompressible
• non-viscous (no friction)
• changes shape to fit its container
• flows without turbulence

On the MCAT, all fluids are assumed to be ideal unless otherwise noted.

Question 3

What term is given to fluid flow that lacks turbulence?

Answer 3

Fluid flow that is not turbulent is known as laminar flow. Ideal fluids are assumed to be laminar.

When a fluid molecule is moving in a laminar pattern, its motion is orderly and occurs in the same direction as neighboring molecules.

Question 4

Water is flowing through a level pipe of uniform diameter. What can be said about the nature of its flow?

Answer 4

Since water can be assumed to be an ideal fluid, the flow will be frictionless and non-turbulent.

The speed of the flow will be constant at all points along the pipe, and the pressure on the walls of the pipe will be constant at all points.

Question 5

Define:

density (ρ)

Answer 5

A substance’s density (ρ) is the mass per unit volume of that substance. Put in other words, density equals mass over volume.

The SI units for density are kg/m³. Other units may appear on the MCAT, such as g/cm³ or g/mL.

Question 6

What units of density are most commonly used?

Answer 6

• g/cm³
• g/mL, kg/L
• kg/m³

The first three of the above values are equivalent in magnitude. The last value, kg/m³, differs by a factor of 1000.

Question 7

What is the density of water in g/cm³ and g/mL, respectively?

Answer 7

1 g/cm³ and 1 g/mL

Since 1 cm³ approximately equals 1 mL, the density of water is the same for both sets of units.

The density of water in various units should be memorized for the MCAT.

Question 8

What is the density of water in kg/m³ and kg/L, respectively?

Answer 8

1000 kg/m³ and 1 kg/L.

The SI units for density are kg/m³. The density of water in various units should be memorized for the MCAT.

Question 9

Define:

specific gravity

Answer 9

A substance’s specific gravity is that substance’s density in comparison to the density of water.

The magnitude of a substance’s specific gravity is identical to the substance’s density as expressed in g/mL. However, specific gravity is a unitless quantity.

Question 10

Define:

buoyancy

Please provide the definition of buoyancy, not the formula for buoyant force (that’ll come later)!

Answer 10

It is the tendency of an object to weigh less when partially or fully submerged in a fluid.

Buoyancy is generated by the fluid displaced by the object. The fluid pushes up on the object with a force equal to the weight of the fluid displaced.

Question 11

How is the buoyant force of a submerged object calculated?

Answer 11

The buoyant force of a submerged object is simply the weight of the fluid displaced by the portion of the object that is submerged.

F_B = (ρ_fluid)(V_sub)(g)

Where:

• F_B = buoyant force pushing up on the object (N)
• ρ_fluid = density of the liquid (kg/m³)
• V_sub = volume of the object submerged in the liquid (m3)
• g = 10 m/s²

Question 12

An object with a volume of 1.5 L is fully submerged in water. What is the buoyant force on the object?

Answer 12

15 N

Using the definition of buoyant force:

F_B = (ρ_fluid)(V_sub)(g)
F_B = (1 kg/L)(1.5 L)(10 m/s²)
F_B = 15 N

Question 13

Equivalent objects are submerged in ethyl alcohol (ρ = 0.79 kg/L) and mercury (ρ = 5.43 kg/L). Which object experiences the larger buoyant force?

Answer 13

The object submerged in mercury will experience the larger buoyant force.

The buoyant force is equal to (ρ_fluid)(V_sub)(g). Since the objects are identical, the submerged volume is equal in the two cases, as is the value of g. The fluid with the larger density must then produce the higher buoyant force.

Question 14

If a solid object is dropped into a liquid, under what conditions will it float?

Answer 14

The object will float if its density is less than or equal to the density of the surrounding liquid.

Otherwise, the object will sink.

Question 15

What information can be discerned from the effective weight of a submerged object?

Answer 15

It is equal to the object’s weight on dry land minus the buoyant force.

If the weight on land and the effective weight are known, buoyant force can be calculated.

W_eff = m_objg - (ρ_fluid)(V_obj)(g)

Where:

• m_obj = mass of object (kg)
• ρ_fluid = density of fluid (kg/m³)
• V_obj = volume of object (m³)
• g = 10 m/s²

Question 16

An aluminum ball with a density of 3 g/cm³ has a mass of 9 kg. What is its effective weight when it is submerged in water?

Answer 16

60 N

With a mass of 9 kg and a density of 3 g/cm³, the ball’s volume must be 3000 cm³, or 3 L. It will then displace 3 L of water when submerged. The buoyant force is simply the weight of 3 L of water:

F_B = (ρ_liq)(V_sub)(g)
(1 kg/L) (3 L) (10 m/s²) = 30 N

With a mass of 9 kg, the ball’s weight on land is 90 N. Subtracting the buoyant force yields the final answer, 60 N.

Question 17

A plastic ball with a density of 2 g/cm³ has a mass of 6 kg. What shortcut can help calculate the effective weight of the ball when it is submerged in water?

Answer 17

The proportion of the object’s weight cancelled by the buoyant force is exactly equal to the ratio of the fluid’s density to the object’s density.

In this case, water has half the density of the object. Thus, the proportion of the object’s weight cancelled by buoyancy is one-half of the object’s dry land weight.

With a mass of 6 kg, the object’s dry land weight is 60 N. Half of that amount (30 N) is cancelled by buoyancy, so the remaining effective weight is 60 - 30 = 30 N.

Question 18

Define:

Pascal’s Law

Answer 18

This states that a fluid will carry pressure undiminished. In other words, pressure exerted on any part of a fluid will be carried equally to all walls of the container.

Pascal’s Law is most commonly tested on the MCAT using hydraulic lifts.

Question 19

What does Pascal’s Law predict about the pressure on Plates 1 and 2 in the diagram below?

Answer 19

pressures will be equal

The force (F₁) will exert a pressure (P₁) on the fluid behind Plate 1. The fluid will exert that pressure, undiminished, on all the walls that it contacts, including Plate 2.

Question 20

In the hydraulic lift pictured below, how does the force F₂ change if Plate 2 doubles in area?

Answer 20

The force F₂ doubles.

According to Pascal’s Law, the pressures on Plates 1 and 2 must be equivalent. From the definition of pressure:

P₁ = F₁/A₁ = F₂/A₂ = P₂

So, force and area are directly proportional, and doubling A₂ will double F₂ as well.

Question 21

In the hydraulic lift pictured below, how does the distance D₂ compare to the distance D₁ if Plate 2 is double the area of Plate 1?

Answer 21

D₂ = ½D₁

Since the liquid between the plates in incompressible, the volume displaced by Plate 1 must be equal to the volume displaced by Plate 2:

V₁ = A₁D₁ = A₂D₂ = V₂

So, distance and area are inversely proportional, and a plate with double the area will move half of the distance.

Question 22

Define:

hydrostatic pressure

Answer 22

It is the pressure exerted on a submerged object by the fluid in which the object is submerged.

Hydrostatic pressure increases proportionately with the distance beneath the surface of the liquid.

Question 23

What pressure is exerted on an object submerged a distance z below the surface of a liquid?

Answer 23

Hydrostatic pressure, often called gauge pressure, is calculated as:

P_H = (ρ_fluid)(g)(z)

Where:

P_H = hydrostatic pressure (Pa)
ρ_fluid = density of the fluid (kg/m³)
g = 10 m/s²
z = distance beneath the liquid’s surface (m)

Question 24

What is the difference between the gauge pressures at a depth of 20 cm and 50 cm in a large, round-bottomed flask filled with water?

Answer 24

The pressure at a depth of 50 cm is 2.5 times the pressure at a depth of 20 cm.

The equation for gauge, or hydrostatic, pressure is P_H = (ρ_fluid)(g)(z), which states that pressure is proportional to depth. Note that pressure is independent of vessel shape; the only variables involved are fluid density and depth beneath the surface.

Question 25

Vessel 1 is filled with water, while Vessel 2 is filled with an unknown fluid. The pressure 10 cm beneath the surface of the liquid in Vessel 2 is 3 times the pressure at a depth of 10 cm in Vessel 1. What is the density of the unknown fluid?

Answer 25

3 g/cm³

Since hydrostatic pressure is measured as P_H = (ρ_fluid)(g)(z), pressure is proportional to fluid density. If the pressure in one fluid is higher than that of a second at equal depths, the density of the first fluid must also be higher, by the same proportion.

Question 26

What property distinguishes an ideal fluid as it flows through a pipe?

Answer 26

Ideal fluids undergo laminar flow, a form of motion in which the fluid flows in parallel layers, with no disruption between them, and the speed varies across the cross-section of the pipe.

This is opposed to turbulent flow, where the fluid can have eddies, vortices, and local disruption to the fluid’s speed and direction of flow.

Question 27

How does the flow of a viscous fluid differ from that of an ideal fluid?

Answer 27

A viscous fluid undergoes Poiseuille flow, in which the fluid near the walls of the pipe flows much more slowly due to viscous interactions with the walls.

Above a certain fluid speed and pipe diameter, viscous fluids begin to flow turbulently as well.

Question 28

Under what conditions does fluid flow become turbulent?

Answer 28

It occurs in non-ideal fluids when the cross-sectional area of the flow becomes large or when the fluid velocity increases significantly.

For any given system, threshold values of cross-sectional area and velocity exist, but the specifics won’t be tested on the MCAT.

Question 29

How does the speed of a fluid flowing through a pipe change as the diameter of the pipe decreases?

Answer 29

It increases in speed.

This is akin to placing your thumb partially over the end of a water hose, causing the water to flow through faster.

Question 30

Water flowing through a pipe reaches a region where the pipe’s cross-sectional area is halved. How does the fluid velocity change in this region of the system?

Answer 30

As the area of the pipe is reduced by half, the fluid velocity doubles.

The continuity equation describes the relationship between cross-sectional area and velocity:

A₁v₁ = A₂v₂

Where:

A = cross-sectional area of the pipe (m²)
v = fluid velocity (m/s)

Question 31

According to the continuity equation, what value remains constant through all points of the same pipe system?

Answer 31

The volume flow rate, sometimes called simply flow rate.

The continuity equation states that A₁v₁ = A₂v₂. Since area is measured in m² and velocity is measured in m/s, A*v has units of m³/s. This equation, then, is simply a mathematical way to say that flow rate is constant.

Question 32

Water flowing through a pipe reaches a region where the radius of the pipe increases by a factor of 3. How does the fluid velocity change in this region of the system?

Answer 32

It decreases by a factor of 9.

According to the continuity equation (A₁v₁ = A₂v₂), cross-sectional area and velocity are inversely proportional.

However, the cross-sectional area of a round pipe is proportional to the square of the radius, so if the radius triples, the area increases by a factor of 9. Therefore, the velocity decreases by the same amount.

Question 33

Define:

surface tension

Answer 33

This is the ability of a fluid’s surface to resist an external force. It can also be thought of as a fluid’s tendency to interact with its own particles.

Surface tension is caused by attractive forces between the molecules of the fluid. It is responsible for the spherical shape of soap bubbles and the ability to skip a stone off the surface of a lake.

Question 34

What properties cause fluids to have high surface tension?

Answer 34

Fluids with high intermolecular attractive forces have large surface tension values.

For instance, fluids in which the molecules are bound by hydrogen bonding will have more surface tension than nonpolar fluids, where the molecules are only attracted by dispersion forces.

Question 35

Which fluid has a higher surface tension, water (H₂O) or carbon tetrachloride (CCl₄)?

Answer 35

water (H₂O)

Water molecules are attracted by hydrogen bonding, while carbon tetrachloride molecules are nonpolar and only attracted by dispersion forces. Hydrogen bonding is a much stronger interaction, and leads to surface tension that is nearly four times as strong.

Question 36

Describe the purpose of Bernoulli’s equation.

Answer 36

This relates the pressure exerted by a fluid to its speed and depth. It can be thought of as conservation of energy for fluid systems.

Question 37

What is Bernoulli’s equation?

In other words, please write out (or mentally think of) Bernoulli’s equation and the variables involved.

Answer 37

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

Where:

P = pressure exerted by the fluid (in Pa)
ρ = density of the fluid (in kg/m³)
v = speed of the fluid (in m/s)
h = depth of the fluid (in m)

Question 38

According to Bernoulli’s equation, how does the pressure exerted by a fluid change as its velocity increases?

Answer 38

The pressure it exerts decreases.

Bernoulli’s equation reads:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

Assuming that the depth does not change, the third term on each side is constant and cancels out. If the velocity increases, v₂ > v₁, meaning that P₂ < P₁.

Question 39

According to Bernoulli’s equation, how does the pressure exerted by a fluid change as height increases?

Answer 39

The pressure it exerts decreases.

Bernoulli’s equation reads

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

Assuming a static fluid, or constant velocity, the second term on each side is constant and cancels out. Here, h₂ > h₁, so P₂ must be less than P₁.

Question 40

Why do some people put boards on the outside of their windows during a hurricane?

Answer 40

During a hurricane, the wind speed increases significantly. According to Bernoulli’s equation, this results in a severe drop in pressure outside the house.

The pressure difference between the inside and outside leads to the danger of the windows being blown out; boards on the outside can prevent this damage.

Question 41

How does the air pressure on top of a mountain compare to that at the base of the mountain?

Answer 41

The air pressure on top of the mountain is lower.

Bernoulli’s equation predicts that pressure decreases as height, or altitude, increases.

Question 42

Define:

solid

Answer 42

It is an object of a particular shape and volume, neither of which change in the absence of external forces.

Question 43

What are the properties of an ideal solid?

Answer 43

• perfectly elastic
• smooth
• homogenous

Question 44

An ideal bar of iron is exposed to an external force, which causes it to flex. What happens to its shape when the force is released?

Answer 44

The bar assumes the same shape it held before it was exposed to the force.

Since ideal solid objects are perfectly elastic, they can be deformed in any way and will return to their original shape.

Question 45

Define:

elastic limit

Answer 45

A solid’s elastic limit is the maximum force to which it can be exposed and still return to its original shape.

An object which is exposed to a force greater than its elastic limit will be permanently deformed. Different substances have different elastic limits.

Question 46

Define:

coefficient of thermal expansion

Answer 46

An object’s coefficient of thermal expansion is a measure of how much the object’s size changes due to temperature change.

Most objects expand when they are heated, and shrink when they are cooled.

Question 47

What formula describes how the length of an object changes when its temperature changes?

Answer 47

An object’s length change is given by the following equation:

ΔL/L = α_L ΔT

Where:

L = object’s original length (m)
ΔL = change in the object’s length (m)
α_L = object’s coefficient of thermal expansion (1/K)
ΔT = change in temperature (K)

Question 48

What is stress, as it refers to a solid?

Answer 48

The force applied to a solid divided by the area over which the force is exerted.

Stress has SI units of N/m², identical to the units of pressure, and can be thought of in a similar fashion.

Question 49

What is strain, as it refers to a solid?

Answer 49

• It is the change in length of an object that is placed under stress.
• It is also defined as the change in length divided by the original length, or ΔL/L.

Since strain is given in units of length/length, it is a unitless quantity.

Question 50

What information can be discerned from the modulus of a solid?

Answer 50

• The modulus of a solid is a measure of how resistant the solid is to stress.
• The higher the modulus, the less the solid changes (with respect to volume, for example) when stressed.

Of the several kinds of modulus for a given solid, all are defined as stress divided by strain.

Question 51

Define:

Young’s modulus

Answer 51

The modulus that measures a solid’s response to a compressive force (one that makes the object shorter).

As with all moduli, Young’s modulus is defined as stress divided by strain. In this case, the stress is a compressive force, and the strain is the proportional length change under compression.

Y = (F/A)/(ΔL/L)

Question 52

By what proportion is a bronze bar compressed when it is exposed to a stress of 10 MPa?

The Young’s modulus of bronze is 100 GPa.

Answer 52

0.01% of its original length

Young’s modulus is defined as:

Y = stress/strain
100 x 10⁹ Pa = 10 x 10⁶ Pa / strain
strain = ΔL/L = 10^-4

Question 53

Define:

shear forces

Answer 53

• It is a force exerted parallel to the surface of an object.
• It causes the object to temporarily deflect sideways in response to the stress.

Question 54

Define:

shear modulus

Answer 54

It is the modulus that measures a solid’s response to a shear force.

As with all moduli, shear modulus is defined as stress divided by strain. In this case, the stress is a shear force, and the strain is the proportional deflection under shear.

S = (F/A)/(Δx/h)

Question 55

By how much is a 1 m steel cube deflected when it is exposed to a stress of 80 MPa?

The Young’s modulus of bronze is 80 GPa.

Answer 55

It is deflected sideways by 1 mm.

Shear modulus is defined as:

S = stress/strain
80 x 10⁹ Pa = 80 x 10⁶ Pa / strain
strain = Δx/h = 10^-3
Δx / (1 m) = 10^-3 m

Question 56

A cube of styrofoam with a density of 0.5 g/cm³ is dropped into water, quickly popping back up to the surface. What fraction of the cube will be submerged in the water?

Answer 56

One-half of the cube will be submerged.

The proportion of an object submerged is simply equal to ρ_obj/ρ_fluid, where ρ_obj is the object’s density and ρ_fluid is the fluid’s density. In this case, the object has one-half of the density of the liquid, so one-half of it will be submerged when it floats. Note that if ρ_obj > ρ_liq, this fraction will be greater than 1 and the object will sink to the bottom.