5Fluids and solids test w solutions

PHYSICS TOPICAL:
Fluids and Solids
Test 1
Time: 21 Minutes*
Number of Questions: 16

* The timing restrictions for the science topical tests are optional. If you
are using this test for the sole purpose of content reinforcement, you
may want to disregard the time limit.


MCAT

DIRECTIONS: Most of the questions in the following test
are organized into groups, with a descriptive passage
preceding each group of questions. Study the passage,
then select the single best answer to each question in the
group. Some of the questions are not based on a
descriptive passage; you must also select the best answer
to these questions. If you are unsure of the best answer,
eliminate the choices that you know are incorrect, then
select an answer from the choices that remain. Indicate
your selection by blackening the corresponding circle on
your answer sheet. A periodic table is provided below for
your use with the questions.

PERIODIC TABLE OF THE ELEMENTS
1
H
1.0

2
He
4.0

3
Li
6.9

4
Be
9.0

5
B
10.8

6
C
12.0

7
N
14.0

8
O
16.0

9
F

19.0

10
Ne
20.2

11
Na
23.0

12
Mg
24.3

13
Al
27.0

14
Si
28.1

15
P
31.0

16
S
32.1

17
Cl
35.5

18
Ar
39.9

19
K
39.1

20
Ca
40.1

21
Sc
45.0

22
Ti
47.9

23
V
50.9

24
Cr

52.0

25
Mn
54.9

26
Fe
55.8

27
Co
58.9

28
Ni
58.7

29
Cu
63.5

30
Zn
65.4

31
Ga
69.7

32
Ge
72.6

33
As
74.9

34
Se
79.0

35
Br
79.9

36
Kr
83.8

37
Rb
85.5

38
Sr
87.6

39
Y

88.9

40
Zr
91.2

41
Nb
92.9

42
Mo
95.9

43
Tc
(98)

44
Ru
101.1

45
Rh
102.9

46
Pd
106.4

47
Ag
107.9

48
Cd
112.4

49
In
114.8

50
Sn
118.7

51
Sb
121.8

52
Te
127.6

53
I
126.9

54
Xe

131.3

55
Cs
132.9

56
Ba
137.3

57
La *
138.9

72
Hf
178.5

73
Ta
180.9

74
W
183.9

75
Re
186.2

76
Os
190.2

77
Ir
192.2

78
Pt
195.1

79
Au
197.0

80
Hg
200.6

81
Tl
204.4

82
Pb
207.2

83
Bi

209.0

84
Po
(209)

85
At
(210)

86
Rn
(222)

87
Fr
(223)

88
Ra
226.0

89
Ac †
227.0

104
Rf
(261)

105
Ha
(262)

106
Unh
(263)

107
Uns
(262)

108
Uno
(265)

109
Une
(267)

*

58
Ce
140.1

59
Pr
140.9

60
Nd
144.2

61
Pm
(145)

62
Sm
150.4

63
Eu
152.0

64
Gd
157.3

65
Tb
158.9

66
Dy
162.5

67
Ho

164.9

68
Er
167.3

69
Tm
168.9

70
Yb
173.0

71
Lu
175.0

†

90
Th
232.0

91
Pa
(231)

92
U

238.0

93
Np
(237)

94
Pu
(244)

95
Am
(243)

96
Cm
(247)

97
Bk
(247)

98
Cf
(251)

99
Es
(252)

100
Fm
(257)

101
Md
(258)

102
No
(259)

103
Lr
(260)

GO ON TO THE NEXT PAGE.

2

as developed by


Fluids and Solids Test 1
Passage I (Questions 1–5)
The human circulatory system can be thought of as a
closed system of interconnecting pipes through which fluid
is continuously circulated by two pumps. The two
synchronous pumps, the right and left ventricles of the
heart, work as simple two-stroke force pumps. The

muscles of the heart regulate the force by contracting and
relaxing. The contraction (systole) lasts about 0.2 seconds,
and a complete systole/diastole (contraction/relaxation)
cycle lasts about 0.8 seconds.
For blood pressures and speeds in the normal range,
the volume flow rate of blood through a blood vessel is
directly proportional to the pressure difference over a length
of the vessel and to the fourth power of the radius of the
vessel.
The total mechanical energy per unit volume of blood
just as it leaves the heart is:
E/V = ρgh + P +

1
ρv2
2

The first term on the right side of the equation is the
gravitational potential energy per unit volume of blood,
where ρ is the density, g is the acceleration due to gravity,
and h is the height of the blood with respect to the height
of the heart. The second term is the blood pressure just as
the blood leaves the heart. The work done by the heart
creates the blood pressure. The third term is the kinetic
energy per unit volume of blood, where v is the velocity of
the blood as it leaves the heart. (Note: The density of blood
equals 1050 kg/m3 and the acceleration due to gravity is
9.8 m/s2.)

1 . Why is diastolic blood pressure much lower than

systolic blood pressure? (Note: A typical
systole/diastole reading in mmHg is 120/80.)
A . Because the heart exerts more force on the blood
during diastole
B . Because the heart exerts no force on the blood
during diastole
C . Because the radii of the blood vessels increase
during diastole, while the force exerted by the
heart on the blood remains the same
D . Because the radii of the blood vessels decrease
during diastole, while the force exerted by the
heart on the blood remains the same

2 . Which of the following is a way to achieve
approximately a 50% increase in the volume flow rate
of blood through a blood vessel?
A . Increase the radius by 10%
B . Increase the cross-sectional area of the vessel by
10%
C . Decrease the change in pressure by 50%
D . Decrease the speed of flow by 25%

3 . What is the gravitational potential energy of 8 cm3 of
blood in a 1.8-meter tall man, in a blood vessel 0.3 m
above his heart? (Note: The man’s blood pressure is
1.3 × 104 N/m2.)
A.
B.
C.
D.

1 × 10–4 Joules
2.5 × 10–2 Joules
3.1 × 103 Joules
4 × 104 Joules

4 . The blood pressure in a capillary bed is essentially
zero, allowing blood to flow extremely slowly through
the tissues in order to maximize exchange of gases,
nutrients, and waste products. What is the work done
on 200 cm3 of blood against gravity to bring it to the
capillaries of the brain, 50 cm above the heart?
A.
B.
C.
D.

5145 J
105 J
10 J
1J

5 . During intense exercise, the volume of blood pumped
per second by an athlete’s heart increases by a factor of
7, and his blood pressure increases by 20%. By what
factor does the power output of the heart increase
during exercise?
A.
B.
C.

D.

1.2
3.5
7
8.4

GO ON TO THE NEXT PAGE.

KAPLAN

3


MCAT
Passage II (Questions 6–11)

Elastomers have a Young’s modulus given by

When a force F is applied along the axis of a rod
having length l and cross-sectional area A , it experiences a
stress equal to F / A . The rod’s response to stress is called
strain, which is the fractional change in length of the rod
(∆l/l). A particular material’s response to stress can be
characterized by Young’s modulus, Y, which is the ratio of
the stress to the strain. Table 1 lists Young’s moduli for
various materials.

Y = 3ρRT/M,
where ρ is the density of the material, R is the universal

gas constant (8.3 J/mole • K), T is the temperature, and M
is the mean molecular weight of the molecule between
cross-links. Unlike that of rigid solids, the Young’s
modulus of an elastomer varies with the applied stress
between 105 – 107 N/m2.

Table 1

Material
bone
rectus abdominus
muscle
resilin
rubber
common cartoid artery
steel
tendon

Average
Young’s Modulus
N/m2
2 × 1010
3 × 105
2×
6×
2×
2×
7×

106

106
106
1011
108

Rigid solids like bone have a Young’s modulus on
the order of 10 10 N/m 2 and can lengthen by no more than
1% without breaking. When stretched, the distance between
the atoms in these solids changes, creating a restoring
force. In this way, changes in internal energy are
responsible for the restoring force of rigid solids.
The elastic properties of soft biological materials
such as tissue differ from those of rigid solids. These soft
biological materials are called elastomers. Elastomers
consist of long molecular chains connected by sulfur crosslinks. Stretching the material elongates the molecular
chains until they are almost parallel as shown in Figure 1.
Elastomers can expand up to three times their original
length.

6 . If the stress on an elastomer is doubled, then the strain
will:
A.
B.
C.
D.

be halved.
remain the same.
double.
change depending on the particular elastomer.

7 . What is the approximate maximum stress that bone
can withstand?
A.
B.
C.
D.

3×
6×
2×
2×

106
107
108
109

N/m2
N/m2
N/m2
N/m2

stretched
unstretched
Figure 1

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4

as developed by


Fluids and Solids Test 1
8 . Which of the following graphs of stress versus strain
is consistent with the information presented in Table
1?
C.

stress

steel

bone

steel

A.
B.
C.
D.

strain

strain
B.

D.

strain

I only
II only
I and III only
II and III only

bone
stress

stress

steel
bone

I. Rectus abdominus muscle
II. Tendon
III. Common cartoid artery

bone
stress

A.

1 1 . From the information provided in Table 1, which of
the following materials are most likely to be
elastomers?

steel

strain

9 . Which of the following would lower the Young’s
modulus of an elastomer?
A . Increasing the temperature of the elastomer
B . Increasing the mean molecular weight of the
molecules between cross-links
C . Increasing the density of the elastomer
D . Replacing the sulfur cross-links with oxygen
cross-links

1 0 . A rock climber is suspended by a harness of negligible
mass that is connected to a rope by a metal clamp. The
clamp experiences a stress producing a strain that is
below the threshold of breaking. The clamp exerts a
restoring force equal in magnitude to:
A . the product of the strain and Young’s modulus for
the metal.
B . the difference between the stress and the weight of
the individual.
C . the weight of the individual.
D . the product of Young’s modulus for the metal and
the change in length of the metal.

GO ON TO THE NEXT PAGE.

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5

MCAT
Questions 12 through 16 are NOT
based on a descriptive passage.
1 2 . Liquids A and B are standing in separate columns from
which the air has been removed. Liquid A has a gauge
pressure (pressure above atmospheric pressure) of X Pa
5 m below the surface. Liquid B is three times as dense
as liquid A. What is the gauge pressure, in Pa, 15 m
below the surface of liquid B?
A.
B.
C.
D.

1 6 . An object floats in water with 20% of its volume
above the surface. What is the specific gravity of the
object?
A.
B.
C.
D.

0.16
0.2
0.64
0.8

X
3X

9X
12X

1 3 . A hose of diameter 20 mm is connected to a sprinkler
attachment that has 10 holes with a diameter of 2 mm
each. If the water in the hose travels at 1 m/s, with
what velocity does it exit the sprinkler?
A.
1
B.
10
C . 20
D . 100

m/s
m/s
m/s
m/s

1 4 . A nurse uses a hydraulic lift to help a heavy patient in
and out of bed. What force must the nurse apply to a
100 cm2 piston to lift a 100 kg patient lying on a 0.4
m2 platform?
A.
25 N
B.
1,000 N
C . 40,000 N
D . 250,000 N
1 5 . An object of unknown density is thrown into a deep

lake and sinks to the bottom. How does the buoyant
force change as the object is sinking?
A . It remains the same because the object displaces
the same amount of water.
B . It increases because the pressure increases with
depth.
C . It decreases to sustain net acceleration downward.
D . The change in buoyant force cannot be determined
without knowing the density of the object.

END OF TEST

6

as developed by


Fluids and Solids Test 1

THE ANSWER KEY IS ON THE NEXT PAGE

KAPLAN

7


MCAT
ANSWER KEY:
1. B
6. D

2. A
7. C
3. B
8. A
4. D
9. B
5. D
10. C

8

11.
12.
13.
14.
15.

C
C
B
A
A

16. D

as developed by


Fluids and Solids Test 1
EXPLANATIONS

Passage I (Questions 1—5)
1.

B
When the muscles of the heart relax as they do during diastole, the heart is not exerting any force on the blood.

2.

A
In the passage we are told that the volume flow rate of blood in a blood vessel is directly proportional to the pressure
difference over a length of the vessel and to the fourth power of the radius of the vessel. Before performing any actual
calculations it is worthwhile to see if we can eliminate some answer choices just by looking at the direction of change of the
parameter. Choice C can be eliminated because decreasing the pressure difference would decrease the volume flow rate. Choice D
is likewise incorrect: The volume flow rate is simply the speed at which a unit volume of blood moves in a blood vessel. It is
thus reasonable to infer that decreasing the speed of flow would also decrease the volume flow rate.
To choose between choices A and B we will have to perform a calculation. If, as is indicated in choice A, we increase
the radius of the vessel by 10%, the new radius would be 1.1 r, where r is the old radius. Since the volume flow rate is
proportional to the fourth power of the radius, the new volume flow rate would be increased by a factor of (1.1)4 = (1.21)2 ≅ 1.2
× 1.2 = 1.44. In order to see whether this is close enough to be the correct answer, let us also consider for a moment choice B.
The cross-sectional area of a blood vessel is proportional to the radius squared (A = r2), and since the volume flow rate is
proportional to the radius to the 4th power, it must be proportional to the area squared. If the cross-sectional area is increased by
10%, the volume flow rate would thus increase by a factor of 1.12 = 1.21, which falls far short of the 50% increase we are after.
Choice A is therefore the correct answer.
3.

B
A glance at the answer choices reveals that they differ by orders of magnitude. We therefore have quite a bit of leeway
in making approximations when performing calculations. In the passage we are given a formula for the total mechanical energy
per unit volume of blood. It is also stated in the passage that the first term on the left hand side, ρgh, is the gravitational
potential energy term, where ρ, the density of the blood, is 1050 kg/m3, g is 9.8 m/s 2, and h is the height with respect to the

heart, in the case of this question 0.3 m. Substituting in these numbers, and keeping in mind what we said about
approximations, the gravitational potential energy per unit volume is 1000 kg/m3 × 10 m/s2 × 0.3 m = 3000 kg/ms2. before
leaping to answer choice C, however, we have to keep in mind that this value is only the energy per unit volume, and that
to obtain the gravitational potential energy itself we need to multiply that by the volume under consideration, 8 cm3.
Converting that into SI units, we note that 1 cm is 1 × 10–2 m, and so 1 cm3 = 1 × 10–6 m 3. The gravitational potential
energy is (3 × 103 kg/ms2) × (8 × 10–6 m3) = 24 × 10–3 kg•m2/s2 = 2.4 × 10–2 kg•m2/s2 = 2.4 × 10–2 J, which is closest to
choice B.
4.

D
The work done against gravity is simply the change in the gravitational potential energy of the blood: mgh. The
mass of the blood is its density times its volume, and the change in height is 50 cm, or 0.5 m to keep the units consistent.
Substituting in numbers from the passage and the question stem and approximating:
W = 1000 kg/m3 × 200 cm3 ×

1 m3
× 10 m/s2 × 0.5 m
100 × 100 × 100 cm 3

Note how we have to insert a factor to convert cm3 into m3: 1 m = 100 cm, therefore 1 m3, which is 1m × 1m × 1m,
is equal to 100 cm × 100 cm × 100 cm = (100 × 100 × 100) cm3. Carrying out the arithmetic, we arrive at W = 1 J.
5.

D
Power is change in energy per unit time. In this case, the change in energy is the work that the heart does in pumping
the blood. This work creates the pressure in the circulatory system. The expression for this work is equal to the pressure times
the volume of blood that the heart pumps. The power associated with this work, i.e. the power generated by the heart, is thus:
power =

blood pressure × the volume of the blood pumped

time over which blood is pumped

We are told in the question stem that the volume of blood increases by a factor of seven. If this were the only change,
then choice C would be the correct answer. But this is not the full story. The blood pressure also increases by 20%. In other

KAPLAN

9


MCAT
words, the new blood pressure is the old blood pressure times 1.2. The factor by which the power increases is thus 7 × 1.2 =
8.4.
Notice also that only choice D is larger than 7, and so if we know that the blood pressure would also increase power,
choice D has to be the correct answer without doing the last multiplication.
Passage II (Questions 6—11)
6.

D
We need first to relate the strain to the stress. In the first paragraph of the passage, we are told that Young’s modulus is
the ratio of the stress to the strain. Rearranging, we find that the stress equals the strain times Young’s modulus:
stress
strain
stress = Young’s modulus × strain
Young’s modulus =

If Young’s modulus were a constant, the strain would double as you double the stress, since the two would be directly
proportional. Remember, however, that we are dealing with elastomers. The passage states that the Young’s modulus for an
elastomer varies with the applied stress, i.e. even for a particular elastomer, the Young’s modulus is not a constant. Therefore,
we really can’t know what effect doubling the stress would have on the strain.

7.

C
In the second paragraph we are told that bone can lengthen by no more than 1% of its original length. The question
then is: what is the stress that will cause the bone to lengthen by this maximum amount? Since stress divided by strain is
Young’s modulus, the maximum stress that bone can withstand is the product of its Young modulus and the maximum strain
of 1% or 0.01:
2 × 1010 N/m2 × 0.01 = 2 × 108 N/m2.
8.

A
Table 1 presents values of Young’s modulus for different materials. As indicated in the passage, Young’s modulus is
stress
the ratio of the stress to the strain, i.e. Y =
. In a graph of stress versus strain, then, it is the y variable over the x
strain
variable. In other words, the slope of a stress vs. strain curve is Young’s modulus. According to the table, bone has a lower
Young’s modulus than steel. Its slope would therefore be smaller. Choices B and D show both to have a Young’s modulus of
zero: regardless of the pressure one exerts on the material, they undergo the same deformation. That is clearly unreasonable.
9.

B
Towards the end of the passage, we are given a formula for Young’s modulus of elastomers:
Y=

3ρRT
M

Young’s modulus is therefore directly proportional to temperature (T) and the density (ρ): i.e., if either of these
quantities increases, the elastomer’s Young’s modulus increases by the same proportion. Choices A and C are therefore

incorrect. Y, however, is inversely proportional to M, the mean molecular weight. Increasing M then would decrease Y, thus
making B the correct choice. Choice D, replacing the sulfur cross-links with oxygen cross-links, is incorrect because the
passage mentions nothing about the dependence of Young’s modulus on the crosslinks.
10.

C
This question is an example of the importance of keeping basic concepts in mind and being able to apply them without
being distracted by the seeming complexity of the topic or question. If the individual is being suspended, her weight (the force
of gravity acting on her) is pointing downward, and since the individual is not falling, this force is balanced by the tension of
the rope, which is transmitted through the metal clamp. The other choices can be eliminated by dimensional analysis alone. We
are looking for something that would give us force. Choice A, strain × Y, would give us stress which is defined in the passage
as force/area. Instead of units of force, it will have units of force per unit area. Choice B is not meaningful as it is a difference
between two things that do not even have the same units: one cannot subtract a force (in Newtons) from stress (in N/m2).
Choice D, Y × l, would give us something in N/m rather than N.

10

as developed by


Fluids and Solids Test 1
11.

C
First we need to determine the defining characteristics of elastomers. The second paragraph of the passage tells us that
Young’s modulus of rigid solid materials is on the order of 1010 N/m2. For elastomers, however, Young’s modulus is much
lower: on the order of 105 to 107 N/m2 according to the last paragraph. So our strategy should be to assume that materials with
a Young’s modulus on the order of 105 to 107 N/m2 will be called elastomers. Looking at the answer choices and consulting
Table 1, we see that I, the rectus abdominus muscle, and III, the common cartoid artery have Young’s moduli that fall within
the range we want, but tendon has a Young’s modulus that is too high.

Independent Questions (Questions 12–16)
12.

C
The pressure exerted by a fluid is dependent on its density and on the depth below the surface. The precise formula is P
= gρz, where g is the acceleration due to gravity, ρ is the density of the fluid, and z the distance from the surface. This pressure,
however, is only the gauge pressure, or the pressure above the atmospheric pressure felt at the surface. In this question, liquid B
is three times as dense as liquid A, and we are asked to find the gauge pressure 15 m below the surface, which is three times
deeper than the depth at which the pressure for liquid A is reported. The expected gauge pressure is thus 3 × 3 = 9 times higher
than that of liquid A, or 9X.
13.

B
This question calls for an understanding of the continuity equation: v1A1 = v2A2. This implies that when water flows
from a pipe with a larger cross-sectional area to a pipe with smaller cross-sectional area the flow velocity increases, and vice
versa. The hose has a diameter of 20mm; its cross-sectional area is thus (20/2)2 = 100 mm2. The water flows into a sprinkler
that has 10 holes of diameter 2mm each, thus giving us an effective area A2 of 10 × (2/2)2 = 10 mm 2. The velocity of the
water coming out of the sprinkler, then, is:
v 2 = v1 ×

A1
100
= 1 m/s ×
A2
10

= 10 m/s

Note that we did not need to bother converting the mm2 into SI units: since we are only after the ratio of the two areas,
it is sufficient to have the two consistent with each other without worrying about the units for v1.

14.

A
The hydraulic lift is an application of Pascal’s principle, which states that the pressure applied to any part of a liquid
F1
will be transmitted undiminished to every other part of the liquid. Quantitatively, the relationship describes this principle is
A1
F2
=
, since pressure is force divided by area. Let the nurse’s side be labeled 1. The force the nurse must apply, F1, is:
A2
F 1 = F2 ×

A1
0.01 m2
= (100 kg × 10 m/s2) ×
= 25 kg•m/s2 = 25 N.
A2
0.4 m 2

If you chose choice D, you probably neglected to make sure that the two area quantities are in the same units.
15.

A
The buoyant force that a fluid exerts on an object is equal to the weight of the fluid that has been displaced by the
object. Once the object is completely submerged, the amount of water it displaces is unchanged. The buoyant force is therefore
constant as it sinks. Choice B is incorrect because the buoyant force exerted by water, presumed to be incompressible, is not
dependent on pressure. Choice C is incorrect because the object will sink to the bottom as long as the buoyant force is less than
the object’s weight: the buoyant force does not have to decrease as it sinks. Choice D is incorrect because the buoyant force will
remain constant as long as the volume of the object does not change: the precise value of the object’s density is not necessary to

conclude this fact.
16.

D

KAPLAN

11


MCAT
First we must remember the definition of the term specific gravity: the specific gravity of an object is the ratio of the
density of the object to the density of water. As a ratio, it is hence dimensionless.
specific gravity =

ρ0
ρw

where ρ0 is the density of the object and ρw is the density of water. These two need to be expressed in the same units.
When an object is placed in a liquid, two forces act on it. There is of course the weight of the object, which is just the
gravitational force pulling it down. There is also the buoyant force pointing up, with a magnitude given by Archimedes’
principle:
Fb = weight of fluid displaced
= mass of fluid displaced × g
= density of fluid × volume displaced × g
= density of fluid × volume of object submerged × g
In water, then, the buoyant force on an object would be Fb = ρwVsg, where Vs = volume of the part of the object that
is submerged. In equilibrium, the two forces are equal, and we can write:
ρwVsg = m0g

, where m0 = mass of object

= ρ0V0g , where V0 = total volume of object
Note that V0 is not necessarily equal to Vs: they both have to do with the volume of the object, but one is the total
volume, and the other is just the volume that is submerged. In this particular case, 80% of the object is submerged, or Vs = 0.8
Vs
V0, or equivalently,
= 0.8. From this we can rearrange the equation and obtain the specific gravity:
V0
ρ0 Vs
=
= 0.8
ρw V0
Incidentally, we can generalize the result we just obtained to obtain the following rule of thumb: If an object floats on
water so that x% of its volume is submerged, then its density is x% of that of the density of water. This is valid up until the
two densities are equal, beyond which the relation breaks down.

12

as developed by