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Test Prep MCAT Test Exam
Medical College Admission Test: Verbal Reasoning, Biological Sciences, Physical Sciences, Writing Sample (Page 9 )

Updated On: 12-Jan-2026
Page 9 of 164
PDF with 811 Questions

A fireman of mass m slides down a vertical pole with an average acceleration A. If the acceleration due to gravity is g, what is the average frictional force exerted by the fireman?

  1. mg
  2. m(g + a)
  3. m(g - a)
  4. ma

Answer(s): C

Explanation:

When the fireman slides down the pole, there are two forces acting on him; the force due to his weight acting vertically downwards, and the force of kinetic friction which acts vertically upwards in the direction opposite to his motion. Since the fireman is sliding downwards, the net force acting on the fireman must act in the downwards direction. This net force Fn is equal to the sum of the forces in the vertical direction. So if we take downwards direction as the positive direction, we get that Fn = the weight of the fireman minus the force due to kinetic friction, and this equals mg ­ fk, where m is the mass of the fireman, g is the acceleration due to gravity, and fk is the force due to kinetic friction. From Newton's second law, the net force Fn = ma, so substituting into our force equation we have that ma - mg - fk. Rearranging this to get an equation for the kinetic friction fk, gives us that fk = mg - ma, or m(g - a). By Newton's third law, this is also the frictional force exerted by the fireman on the pole and the correct answer is choice C.



Which of the following is NOT an intermolecular force?

  1. Dispersion forces
  2. Resonance
  3. Hydrogen bonding
  4. Dipole interactions

Answer(s): B

Explanation:

All intermolecular attractions involve some kind of electrostatic interaction. Hydrogen bonding is a strong dipole interaction between a positive hydrogen dipole in one molecule and another negative dipole. This definitely is an intermolecular attraction, so choice C isn't the answer. Dispersion forces, choice A, are slight momentary dipole
interactions that occur between molecules as the electron clouds around them go through random variations, so this too is incorrect. Dipole interactions characterize the attractions between molecules that have permanent variations in electron density around the molecule, so choice D is also wrong. However, resonance, choice B, is something else entirely; it is the distribution of electrons among atoms within a molecule in more than one possible pattern. Resonance is not a form of intermolecular attraction because it only involves the effects the electron density has on the single molecule. It is an intramolecular phenomenon, so choice B is the correct answer.



A continuous spectrum of light, sometimes called blackbody radiation, is emitted from a region of the Sun called the photosphere. Although the continuous spectrum contains light of all wavelengths, the intensity of the emitted light is much greater at some wavelengths than at others. The relationship between the most intense wavelength of blackbody radiation and the temperature of the emitting body is given by Wien's law, = 2.9 × 106 / T, where is the wavelength in nanometers and T is the temperature in kelvins.
As the blackbody radiation from the Sun passes through the cooler gases in the Sun's atmosphere, some of the photons are absorbed by the atoms in these gases. A photon will be absorbed if it has just enough energy to excite an electron from a lower energy state to a higher one. The absorbed photon will have an energy equal to the energy difference between these two states. The energy of a photon is given by E = hf = hc/ where h = 6.63 × 10-34 J·s, Planck's constant, and c = 3 × 108 m/s, the speed of light in a vacuum.
The Sun is composed primarily of hydrogen. Electron transitions in the hydrogen atom from energy state n = 2 to higher energy states are listed below along with the energy of the absorbed photon:

If the temperature of the Sun's photosphere is 5800 K, what wavelength of radiation does the Sun emit with the greatest intensity?

  1. 2 nm
  2. 50 nm
  3. 500 nm
  4. 4,500 nm

Answer(s): C

Explanation:

To answer this question, you need to use Wien's law, which is given in the passage. It states that = 2.9 × 106 / , where is the most intense wavelength emitted in nm and T is the temperature in kelvins. Since the T temperature of the Sun's photosphere is 5,800 K, the most intense wavelength emitted is given by = 2.9 × 10 / 5,800 = 500 nm, which is choice C.



A continuous spectrum of light, sometimes called blackbody radiation, is emitted from a region of the Sun called the photosphere. Although the continuous spectrum contains light of all wavelengths, the intensity of the emitted light is much greater at some wavelengths than at others. The relationship between the most intense wavelength of blackbody radiation and the temperature of the emitting body is given by Wien's law, = 2.9 × 106 / T, where is the wavelength in nanometers and T is the temperature in kelvins.
As the blackbody radiation from the Sun passes through the cooler gases in the Sun's atmosphere, some of the photons are absorbed by the atoms in these gases. A photon will be absorbed if it has just enough energy to excite an electron from a lower energy state to a higher one. The absorbed photon will have an energy equal to the energy difference between these two states. The energy of a photon is given by E = hf = hc/ where h = 6.63 × 10-34 J·s, Planck's constant, and c = 3 × 108 m/s, the speed of light in a vacuum.
The Sun is composed primarily of hydrogen. Electron transitions in the hydrogen atom from energy state n = 2 to higher energy states are listed below along with the energy of the absorbed photon:

Based on the data in the table, what is the approximate wavelength of a photon emitted in the electron transition from energy state n = 4 to energy state n = 3?

  1. 5 nm
  2. 30 nm
  3. 100 nm
  4. 2,000 nm

Answer(s): D

Explanation:

When an electron makes a transition from a higher energy state to a lower energy state, a photon is emitted with energy equal to the difference between the energies of the two states. The table in the passage provides energies for photon transitions from energy state n = 2 to various higher energy states in hydrogen. The absolute energy difference between states n = 4 and n = 3 is equal to the difference between their energies relative to the n = 2 state. Therefore, the energy of the photon emitted in the transition from n = 4 to n = 3 is 4.08 × 10-19 - 3.02 × 10-19 = 1.06 × 10-19 J.
The energy of a photon E is given by E = hc/, where h is Planck's constant, c is the speed of light in a vacuum, and is the photon's wavelength. Rearranging to solve for gives = hc/E. Using E = 10-19 J from above, we find = (6.6 × 10-34)(3 × 108) / (10-19) m.
Rounding to the nearest integer gives = 20 × 10-34+8+19 m = 2 × 10-6 m. Since 1 nm = 10-9 m, = 2,000 nm, which is choice D.



A continuous spectrum of light, sometimes called blackbody radiation, is emitted from a region of the Sun called the photosphere. Although the continuous spectrum contains light of all wavelengths, the intensity of the emitted light is much greater at some wavelengths than at others. The relationship between the most intense wavelength of blackbody radiation and the temperature of the emitting body is given by Wien's law, = 2.9 × 106 / T, where is the wavelength in nanometers and T is the temperature in kelvins.

As the blackbody radiation from the Sun passes through the cooler gases in the Sun's atmosphere, some of the photons are absorbed by the atoms in these gases. A photon will be absorbed if it has just enough energy to excite an electron from a lower energy state to a higher one. The absorbed photon will have an energy equal to the energy difference between these two states. The energy of a photon is given by E = hf = hc/ where h = 6.63 × 10-34 J·s, Planck's constant, and c = 3 × 108 m/s, the speed of light in a vacuum.
The Sun is composed primarily of hydrogen. Electron transitions in the hydrogen atom from energy state n = 2 to higher energy states are listed below along with the energy of the absorbed photon:

The energy absorbed by a hydrogen atom as its electron undergoes a transition from the n = 1 energy state to the n = state is: (Note: The n = 1 energy state is the ground state of hydrogen.)

  1. infinite.
  2. equal to the binding energy of the electron.
  3. equal to the energy of a zero-frequency photon.
  4. smaller than the energy absorbed in the n = 2 ton = transition.

Answer(s): B

Explanation:

The binding energy of an electron is by definition the energy required to detach an electron from an atom in its ground state. Hydrogen's ground state is the n = 1 state. When the atom is ionized, the electron is separated from the atom and makes a transition to the highest energy state, the n = state. This is not a distinct energy state, but rather it defines the minimum energy that must be transferred to the electron to free it from the atom.
Hence, the energy difference between the n = and the n = 1 state is the binding energy of the electron in the hydrogen atom, and choice B is correct.
Choices A, C, and D can all be ruled out based on the data presented in the passage which indicates that the energy increases as n increases. This implies that the energy of the n = 2 state is greater than that of the n = 1 state by some finite, positive amount E12. Furthermore, according to the table, the energy of the n = state is 5.44 × 10-19 J greater than the energy of the n = 2 state. Since the energy difference between n = and n = 1 is the sum of these two finite, positive quantities (E12 + 5.44 × 10-19 J), it must also be finite and positive.
Therefore, choice A is wrong. Choice C is wrong because a zero frequency photon would correspond to zero energy, as shown by the formula E = hf, but, as discussed above, the energy absorbed must be positive.
Choice D is also wrong because the sum E12+ 5.44 × 10-19 J is clearly greater than 5.44 × 10-19 J, the energy absorbed in the n = 2 to n = transition.



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