5Fluids and solids test w solutions

PHYSICS TOPICAL: Fluids and Solids Test 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 H 1.0 He 4.0 Li 6.9 Be 9.0 B 10.8 C 12.0 N 14.0 O 16.0 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 as developed by Fluids and Solids Test 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 + 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.) 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 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% What is the gravitational potential energy of 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 × 10–4 Joules 2.5 × 10–2 Joules 3.1 × 103 Joules × 104 Joules 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 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 8.4 GO ON TO THE NEXT PAGE KAPLAN 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 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 Material bone rectus abdominus muscle resilin rubber common cartoid artery steel tendon Average Young’s Modulus N/m2 × 1010 × 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 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 Elastomers can expand up to three times their original length 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 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 GO ON TO THE NEXT PAGE as developed by Fluids and Solids Test 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 From the information provided in Table 1, which of the following materials are most likely to be elastomers? steel strain 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 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 KAPLAN MCAT Questions 12 through 16 are NOT based on a descriptive passage 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 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 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 A hose of diameter 20 mm is connected to a sprinkler attachment that has 10 holes with a diameter of mm each If the water in the hose travels at m/s, with what velocity does it exit the sprinkler? A B 10 C 20 D 100 m/s m/s m/s m/s 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 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 as developed by Fluids and Solids Test THE ANSWER KEY IS ON THE NEXT PAGE KAPLAN MCAT ANSWER KEY: B D A C B A D B D 10 C 11 12 13 14 15 C C B A A 16 D as developed by Fluids and Solids Test EXPLANATIONS Passage I (Questions 1—5) B When the muscles of the heart relax as they during diastole, the heart is not exerting any force on the blood 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 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, cm3 Converting that into SI units, we note that cm is × 10–2 m, and so cm3 = × 10–6 m 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 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 × m3 × 10 m/s2 × 0.5 m 100 × 100 × 100 cm Note how we have to insert a factor to convert cm3 into m3: m = 100 cm, therefore 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 = J 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 MCAT words, the new blood pressure is the old blood pressure times 1.2 The factor by which the power increases is thus × 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) 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 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: × 1010 N/m2 × 0.01 = × 108 N/m2 A Table 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 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 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 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 × = 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 The velocity of the water coming out of the sprinkler, then, is: v = v1 × A1 100 = 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 The force the nurse must apply, F1, is: A2 F = F2 × A1 0.01 m2 = (100 kg × 10 m/s2) × = 25 kg•m/s2 = 25 N A2 0.4 m 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 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 ... TEST as developed by Fluids and Solids Test THE ANSWER KEY IS ON THE NEXT PAGE KAPLAN MCAT ANSWER KEY: B D A C B A D B D 10 C 11 12 13 14 15 C C B A A 16 D as developed by Fluids and Solids Test. .. 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 GO ON TO THE NEXT PAGE as developed by Fluids and Solids Test Which of the following... (257) 101 Md (258) 102 No (259) 103 Lr (260) GO ON TO THE NEXT PAGE as developed by Fluids and Solids Test Passage I (Questions 1–5) The human circulatory system can be thought of as a closed