All MCAT Physical Resources
Example Questions
Example Question #1 : Phase Changes
Which of the following will result in the greatest increase in length of a rubber band at its maximum stretch point, as compared to an unaltered control?
Assume the rubber band will experience the same constant force.
Using a rubber band that is twice as long and is twice as elastic
Using a rubber band that is twice as elastic
Using a rubber band that has half the length
Using a rubber band that has twice the cross-sectional area
Using a rubber band that is twice as elastic
To answer this question, we need to know the equation for Young’s modulus, :
Recall the definitions of stress and strain:
Here, is force, is cross-sectional area, is change in length, and is initial length. Plugging these two equations into the definition of Young’s modulus gives us:
Rearranging this equation to solve for change in length gives us:
Since we want to increase we look for answer choices that increase and , or decrease and . Among the answer choices, the greatest increase occurs when you double the elasticity of the rubber band. Recall that Young’s modulus is a measure of the stiffness of a material; therefore, a more elastic (flexible, or less stiff) material will lower Young’s modulus. Having a material twice as elastic will decrease by one half and will give the greatest increase in the change in length of the rubber band.
Using half the length () and using twice the cross-sectional area will decrease the change in length. Using a rubber band twice as long, but twice as elastic, will not alter the change in length.
Example Question #1 : Phase Changes
Which of the following is false regarding Young’s modulus?
Young’s modulus is defined as the slope of the stress vs. strain plot
Young’s modulus is only valid for certain stresses
Young’s modulus has the same units as stress
Young’s modulus increases as stress increases
Young’s modulus increases as stress increases
Young’s modulus is a constant that is calculated by generating a stress vs. strain plot. The slope of the linear region of the stress vs. strain plot is defined as the Young’s modulus. Recall that the stress vs. strain plot is nonlinear past the elastic limit (stress at which the material starts to plastically deform). This means that we cannot calculate the slope past the elastic limit; Young’s modulus is invalid for stresses beyond the elastic limit.
Young’s modulus is defined as follows:
To calculate strain, you divide change in length by initial length. Since both change in length and initial length have the same units, the units cancel and strain becomes a dimensionless quantity. This means that the Young’s modulus will have the same units as stress alone.
Young’s modulus is a material property. This means that the Young’s modulus is constant for a given type of material. It does not depend on the stress experienced by the material. Any changes to stress will be compensated for by a change in strain, which will allow Young’s modulus to be constant for the material. Similarly, Young’s modulus does not depend on strain or on the geometrical properties (length and area) of the material. It will only change if you change the type of material being used.
Example Question #1 : Nuclear Chemistry And Electrons
Which of the following is not accurate regarding spontaneous fission reactions?
The total kinetic energy of the products is less than the kinetic energy of the original atom
The products have smaller binding energies per nucleon than the original atom
The two resulting nuclei are approximately equal to each other in mass and atomic number
The potential energy of the system is greater after the reaction than before
The sum of the atomic weights of the products is less than the atomic weight of the original atom
The products have smaller binding energies per nucleon than the original atom
In fission a large atom, with low binding energy per nucleon, splits into two approximately equal smaller atoms, which are more stable due to greater binding energy per nucleon. All the other properties mentioned are accurate.
Example Question #1 : Nuclear Chemistry And Electrons
In the operation of nuclear reactors, engineers make use of substances known as neutron poisons. These are used to help store nuclear waste and slow nuclear reactions, but are also generated naturally in nuclear chain reactions as a by-product. This natural by-product can stop the desirable chain reaction present in a nuclear reactor used for power generation.
For example, in nuclear power plants, U-235 is used as a fuel. U-235 absorbs a neutron, and subsequently generates neutrons (which power the chain reaction) and Xe-135. Xe-135 is a well-known neutron poison, and thus can impact the continued chain reaction of a nuclear power plant if it becomes over abundant during power generation.
To help account for this, engineers have developed measurements to quantify the impact of Xe-135 on nuclear operations. For instance, the time during which there is an inability to start a reactor due to the buildup of Xe-135 is referred to as the precluded start-up time. Also, the amount of time that the reactor cannot override the effects of built up Xe-135 is called poison outage time. Perhaps the most important measure that engineers have developed is the neutron absorption capacity (σ), which is measured in units of barns and is a function of microscopic cross section. Xe-135 has a neutron absorption capacity of 2.00 * 106 barns, while another common poison, Sm-149, has a neutron absorption capacity of 74,500 barns.
Xe-135 is a fission product of U-235 after it absorbs a neutron. An alternative fission reaction allows U-235 to produce Ba-141 and three neutrons. If all that is produced in this fission reaction is Ba-141, three neutrons, and one other species, which of the following could be that one other species?
Xe-134
Kr-93
Kr-92
Xe-124
Kr-91
Kr-92
This is a tricky question, because it requires two steps. The first step is to recognize that the question specifies that U-235 absorbs a neutron via fission, so it becomes U-236. U-236 then breaks down into three neutrons, Ba-141, and Kr-92, because the total mass must not change.
Uranium has 92 protons, meaning that U-236 has 144 neutrons (236 – 92 = 144). Barium has 56 protons, meaning that Ba-141 must have 85 neutrons.
The identity of the unknown compound is given by the total protons (92) minus the protons in barium (56).
92 – 56 = 36
The unknown compound has 36 protons, meaning it must be Kr. The remaining neutrons must make up the remaining mass of the krypton isotope. There are 144 neutrons total, 85 given to barium, and 3 lost during reaction.
144 – (85 + 3) = 56
Krypton has 56 neutrons and 36 protons, giving it a total mass of 92.
Example Question #2 : Nuclear Chemistry And Electrons
Which of the following is not a characteristic of nuclear fission?
Fission of light elements is exothermic
The combined mass of the resulting nuclei is less than the mass of the original nucleus
Fission is caused when the electromagnetic force dominates the strong nuclear force
Fission can release neutrons as well as new nuclei
Some isotopes of a particular element may perform spontaneous fission, while other isotopes do not
Fission of light elements is exothermic
Fission of heavy elements is exothermic; it releases energy because the products are closer to the peak of the binding energy curve than the original nucleus was. In other words, the products are more tightly bound and have more negative binding energies.
In contrast, fission is endothermic for lighter elements because their fission products are farther from the peak of the binding energy curve (the products are less tightly bound). So, fission of lighter elements would require an energy input in order to occur, making light-element fission endothermic.
All other listed answers are true statements.
Example Question #3 : Nuclear Chemistry And Electrons
Which of the following are correct statements about binding energy?
I. Light nuclei experience an increase in binding energy by fusion, and heavy nuclei by fission
II. Binding energy is the amount of energy required to separate a nucleus into its individual nucleons
III. Radioactive decay processes tend to produce nuclei which have lower binding energies than the original nucleus
II and III only
III only
I and III only
I, II, and III
I and II only
I and II only
The first two statements are true. Statement III is false becuse radioactive decays tend to produce nuclei with higher binding energy per nucleon than the parent nucleus, since nuclei with higher binding energy are held together more tightly and thus are more stable.
Example Question #1 : Atomic Nucleus
In the operation of nuclear reactors, engineers make use of substances known as neutron poisons. These are used to help store nuclear waste and slow nuclear reactions, but are also generated naturally in nuclear chain reactions as a by-product. This natural by-product can stop the desirable chain reaction present in a nuclear reactor used for power generation.
For example, in nuclear power plants, U-235 is used as a fuel. U-235 absorbs a neutron, and subsequently generates neutrons (which power the chain reaction) and Xe-135. Xe-135 is a well-known neutron poison, and thus can impact the continued chain reaction of a nuclear power plant if it becomes over abundant during power generation.
To help account for this, engineers have developed measurements to quantify the impact of Xe-135 on nuclear operations. For instance, the time during which there is an inability to start a reactor due to the buildup of Xe-135 is referred to as the precluded start-up time. Also, the amount of time that the reactor cannot override the effects of built up Xe-135 is called poison outage time. Perhaps the most important measure that engineers have developed is the neutron absorption capacity (σ), which is measured in units of barns and is a function of microscopic cross section. Xe-135 has a neutron absorption capacity of 2.00 * 106 barns, while another common poison, Sm-149, has a neutron absorption capacity of 74,500 barns.
How does a nuclear fusion reaction compare to the nuclear fission reaction described in the passage?
Fusion reactions generate gamma radiation, while fission reactions only generate alpha radiation
Both have unstable intermediates
Fission reactions use less massive species as reactants
Fission reactions generate gamma radiation, while fusion reactions only generate alpha radiation
Fission reactions generate energy, while fusion reactions do not
Both have unstable intermediates
Both reactions have the potential to produce energy, and both reactions have unstable intermediates. Fission reactions have unstable intermediates in the form of species like Xe-135, which the passage describes as being reactive with neutrons to become a different chemical species. Similarly, fusion reactions fuse smaller nuclei into unstable larger nuclei that subsequently give off energy upon their decay.
Example Question #3 : Atomic Nucleus
In the operation of nuclear reactors, engineers make use of substances known as neutron poisons. These are used to help store nuclear waste and slow nuclear reactions, but are also generated naturally in nuclear chain reactions as a by-product. This natural by-product can stop the desirable chain reaction present in a nuclear reactor used for power generation.
For example, in nuclear power plants, U-235 is used as a fuel. U-235 absorbs a neutron, and subsequently generates neutrons (which power the chain reaction) and Xe-135. Xe-135 is a well-known neutron poison, and thus can impact the continued chain reaction of a nuclear power plant if it becomes over abundant during power generation.
To help account for this, engineers have developed measurements to quantify the impact of Xe-135 on nuclear operations. For instance, the time during which there is an inability to start a reactor due to the buildup of Xe-135 is referred to as the precluded start-up time. Also, the amount of time that the reactor cannot override the effects of built up Xe-135 is called poison outage time. Perhaps the most important measure that engineers have developed is the neutron absorption capacity (σ), which is measured in units of barns and is a function of microscopic cross section. Xe-135 has a neutron absorption capacity of 2.00 * 106 barns, while another common poison, Sm-149, has a neutron absorption capacity of 74,500 barns.
In Xe-135, if each neutron and proton were separated and organized into 135 separate particles, then weighed and the masses summed precisely, the total of their masses would be equal to 135 atomic mass units.
Which of the following is true of the nucleus of Xe-135 if it were weighed on its own?
Its total mass would be less than 135amu
The total mass would be less than 135amu, but only if this lost mass could be directly transferred to another chemical species
Its total mass would be exactly equal to 135amu
The total mass would depend on conditions, and could be either more or less than 135amu
Its total mass would be more than 135amu
Its total mass would be less than 135amu
The concept of mass deficit dictates that nuclear subatomic particles typically have lower mass in their bound state than they do individually. If you took the total mass of each unbound constituent, it would be higher than the mass of the bound nucleus. This is the case regardless of whether Xe-135 interacts with another chemical species.
Example Question #116 : Mcat Physical Sciences
The mass of a certain nucleus is . If the nucleus contains 24 protons and 27 neutrons, calculate its mass defect.
When protons and neutrons combine to form the nucleus of an atom, they generally have less total energy than the energies of the individual nucleons combined. This difference is known as the mass defect. In this case, the nucleus has 24 protons, each with a mass of , and 27 neutrons, each with a mass of .
The mass defect will be the difference between this predicted mass and the actual measured mass.
Example Question #4 : Nuclear Chemistry And Electrons
Which of the following cannot be the mass a nucleus containing two protons and two neutrons?
The mass of a proton is and the mass of a neutron is . The mass of the given nucleus would be:
Because of mass deficit, it is not possible to get a mass greater than this value. The mass of a nucleus is always less than the sum of the masses of the parts.
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