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 a
Trang 1PHYSICS 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
Trang 2
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.
Trang 3Passage 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
2 ρv2
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 1 × 10–4 Joules
B 2.5 × 10–2 Joules
C 3.1 × 103 Joules
D 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 5145 J
B 105 J
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 1.2
B 3.5
C 7
D 8.4
GO ON TO THE NEXT PAGE.
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Passage II (Questions 6–11)
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
Table 1
Average Young’s Modulus
rectus abdominus
muscle
3 × 105
common cartoid artery 2 × 106
Rigid solids like bone have a Young’s modulus on
the order of 1010 N/m2 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
cross-links 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
unstretched
stretched
Figure 1
Elastomers have a Young’s modulus given by
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
6 If the stress on an elastomer is doubled, then the strain
will:
A be halved.
B remain the same.
C double.
D change depending on the particular elastomer.
7 What is the approximate maximum stress that bone
can withstand?
A 3 × 106 N/m2
B 6 × 107 N/m2
C 2 × 108 N/m2
D 2 × 109 N/m2
GO ON TO THE NEXT PAGE.
Trang 58 Which of the following graphs of stress versus strain
is consistent with the information presented in Table
1?
A.
B.
C.
D.
strain
strain
strain
strain
steel bone steel
bone
bone
bone steel
steel
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
1 1 From the information provided in Table 1, which of
the following materials are most likely to be elastomers?
I Rectus abdominus muscle
II Tendon III Common cartoid artery
A I only
B II only
C I and III only
D II and III only
GO ON TO THE NEXT PAGE.
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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?
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?
B 10 m/s
C 20 m/s
D 100 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?
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
1 6 An object floats in water with 20% of its volume
above the surface What is the specific gravity of the object?
A 0.16
B 0.2
C 0.64
D 0.8
END OF TEST
Trang 7THE ANSWER KEY IS ON THE NEXT PAGE
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ANSWER KEY:
Trang 9Passage I (Questions 1—5)
When the muscles of the heart relax as they do during diastole, the heart is not exerting any force on the blood
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
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/s2, 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 m3 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
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
100 × 100 × 100 cm3 × 10 m/s2× 0.5 m 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
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
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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)
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:
Young’s modulus = stress
strain stress = Young’s modulus × strain
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
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
Table 1 presents values of Young’s modulus for different materials As indicated in the passage, Young’s modulus is the ratio of the stress to the strain, i.e Y = stress
strain In a graph of stress versus strain, then, it is the y variable over the x 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
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
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