Medical Tests Medical Tests Certifications MCAT-TEST Questions & Answers

• Question 231:

  There are two opposing theories of light: the particle theory and the wave theory. According to the particle theory, light is composed of a stream of tiny particles that are subject to the same physical laws as other types of elementary particles.

  One consequence of this is that light particles should travel in a straight line unless an external force acts on them. According to the wave theory, light is a wave that shares the characteristics of other waves. Among other things, this means

  that light waves should interfere with each other under certain conditions.

  In support of the wave theory of light, Thomas Young's double slit experiment proves that light does indeed exhibit interference. Figure 1 shows the essential features of the experiment. Parallel rays of monochromatic light pass through two

  narrow slits and are projected onto a screen. Constructive interference occurs at certain points on the screen, producing bright areas of maximum light intensity. Between these maxima, destructive interference produces light intensity minima.

  The positions of the maxima are given by the equation dsin = n, where d is the distance between the slits, is the angle shown in Figure 1, the integer n specifies the particular maxima, and is the wavelength of the incident light. (Note: sin tan

  for small angles.)

  

  Figure 1

  A beam of electrons can also produce an interference pattern. Which one of the following expressions gives a consistent definition of an electron's "wavelength" if it has a total energy given by E? (Note: h = 6.6 ?10-34 J?s is Planck's constant and v is the speed of the electrons.)

  A. hvE

  B. hE/v

  C. hv/E

  D. E/hv

  Correct Answer: C

  The easiest way to solve the problem is to check the units of each of the answers, a process called dimensional analysis. Since wavelength is measured in meters, the correct answer will be in meters as well. First, note that h has the units of

  J?s, and v, a speed, has the units of m/s. Then, choice C has the units of hv/E = (J?s)(m/s)/J = m, so it is correct. The other choices are incorrect.

  Choice A has the units of hvE = (J?s)(m/s)(J) = J2m. Choice B has =

  

  Choice D has the units of E/hv = J/[(J?s)(m/s)] = m-1

  Another way to solve the problem is to recall that the photon energy E is given by E = hc/, where c is the speed of light, and is the photon's wavelength. Solving for , we obtain = hc/ E. To find the "wavelength" of an electron, replace the speed of light c with the speed of the electron v, and obtain = hv/E, where E is now the energy of the electron. Again, this is choice C.

• Question 232:

  There are two opposing theories of light: the particle theory and the wave theory. According to the particle theory, light is composed of a stream of tiny particles that are subject to the same physical laws as other types of elementary particles.

  One consequence of this is that light particles should travel in a straight line unless an external force acts on them. According to the wave theory, light is a wave that shares the characteristics of other waves. Among other things, this means

  that light waves should interfere with each other under certain conditions.

  In support of the wave theory of light, Thomas Young's double slit experiment proves that light does indeed exhibit interference. Figure 1 shows the essential features of the experiment. Parallel rays of monochromatic light pass through two

  narrow slits and are projected onto a screen. Constructive interference occurs at certain points on the screen, producing bright areas of maximum light intensity. Between these maxima, destructive interference produces light intensity minima.

  The positions of the maxima are given by the equation dsin = n, where d is the distance between the slits, is the angle shown in Figure 1, the integer n specifies the particular maxima, and is the wavelength of the incident light. (Note:

  sin tan for small angles.)

  

  Figure 1

  

  What is the angle for the third maximum (n = 3)?

  

  A. Option A

  B. Option B

  C. Option C

  D. Option D

  Correct Answer: B

  To solve this problem, apply the formula given in the passage which quantifies the positions of the intensity maxima. The formula is dsin = n, where d is the distance between the slits, is the angle, and is the wavelength. The note in the passage says that sin when is small. You have to know that this approximation is only valid when is measured in radians. Making this approximation, we obtain d = n, and solving for we obtain = n /d. Note that the distance units of and d can be anything as long as they

  

  are the same. n is given in the question stem, and and d are given in Figure 1. Substituting, we obtain

  which is choice B.

• Question 233:

  There are two opposing theories of light: the particle theory and the wave theory. According to the particle theory, light is composed of a stream of tiny particles that are subject to the same physical laws as other types of elementary particles. One consequence of this is that light particles should travel in a straight line unless an external force acts on them. According to the wave theory, light is a wave that shares the characteristics of other waves. Among other things, this means that light waves should interfere with each other under certain conditions.

  In support of the wave theory of light, Thomas Young's double slit experiment proves that light does indeed exhibit interference. Figure 1 shows the essential features of the experiment. Parallel rays of monochromatic light pass through two narrow slits and are projected onto a screen. Constructive interference occurs at certain points on the screen, producing bright areas of maximum light intensity. Between these maxima, destructive interference produces light intensity minima. The positions of the maxima are given by the equation dsin = n, where d is the distance between the slits, is the angle shown in Figure 1, the integer n specifies the particular maxima, and is the wavelength of the incident light. (Note: sin tan

  for small angles.)

  

  Figure 1

  Which of the following supports the particle theory of light?

  A. The energy of light is quantitized.

  B. Light exhibits interference.

  C. Light is subject to the Doppler effect.

  D. No particle can have a speed greater than the speed of light.

  Correct Answer: A

  If a quantity is quantized, it means that there is a fundamental unit of that quantity which cannot be further divided. For example, charge is quantized, so charges only appear in nature as multiples of the fundamental charge e. There is not a continuous range of values for the possible charge on an object. Similarly, light energy is quantized. The fundamental unit of light energy is called a photon. The photon, because it cannot be subdivided, is similar to other elementary particles. Thus, quantization implies particle-like characteristics, and choice A is correct.

  The passage states that the wave theory, not the particle theory, predicts that light will exhibit interference, so choice B is incorrect. The Doppler effect refers to the change in frequency observed when a wave source or detector is in motion. That the Doppler effect occurs with light as well as sound waves indicates that light has wave, not particle, characteristics and choice C is wrong. Neither particles nor waves can travel faster than the speed of light, so the fact that particles cannot travel faster than light does not support or weaken the particle theory and choice D is wrong.

• Question 234:

  When softball players take batting practice, they often use a machine called an "automatic pitcher," which is essentially a cannon that uses air pressure to launch a projectile. In a prototype automatic pitcher, a softball is loaded into the barrel of the cannon and rests against a flat disk. That disk is locked into place, and a high air pressure is built up behind it. When the disk is released, the softball is pushed along the barrel of the cannon and ejected at a speed of V0. Figure 1 shows the batter and automatic pitcher. The angle of the barrel to the horizontal is . The unit vectors I and j point in the horizontal and vertical directions respectively.

  

  Figure 1

  The height above the ground y of the softball as a function of time t is shown in Figure 2, where t = 0 at Point A, t = tB at Point B, and t = tC at Point C. The softball is ejected from the barrel of the cannon at Point A; it reaches its maximum height at Point B; and the batter hits the softball at Point C. (Note: Assume that the effects of air resistance are negligible unless otherwise stated.)

  

  Figure 2

  What is the ratio of the horizontal distance traveled by the softball at Point B to the horizontal distance traveled at Point C?

  A. 5:1

  B. 4:1

  C. 3:3

  D. 1:2

  Correct Answer: D

  Figure 2 shows that the curve is symmetric around t = tB. Since the axes of graphs are always linear unless otherwise stated, the time elapsed at Point C is twice the time elapsed at Point B. There are no horizontal forces because air resistance is negligible. Therefore, the horizontal speed is the constant V0 throughout the softball's flight. The horizontal distance traveled is then given by x = V0 t, where t is the time elapsed. Since the time elapsed at Point C is twice the time elapsed at Point B, the horizontal distance traveled at Point C is twice the horizontal distance traveled at Point B, and choice D is correct.

• Question 235:

  When softball players take batting practice, they often use a machine called an "automatic pitcher," which is essentially a cannon that uses air pressure to launch a projectile. In a prototype automatic pitcher, a softball is loaded into the barrel of the cannon and rests against a flat disk. That disk is locked into place, and a high air pressure is built up behind it. When the disk is released, the softball is pushed along the barrel of the cannon and ejected at a speed of V0. Figure 1 shows the batter and automatic pitcher. The angle of the barrel to the horizontal is . The unit vectors I and j point in the horizontal and vertical directions respectively.

  

  Figure 1

  The height above the ground y of the softball as a function of time t is shown in Figure 2, where t = 0 at Point A, t = tB at Point B, and t = tC at Point C. The softball is ejected from the barrel of the cannon at Point A; it reaches its maximum height at Point B; and the batter hits the softball at Point C. (Note: Assume that the effects of air resistance are negligible unless otherwise stated.)

  

  Figure 2

  How does the work done by the automatic pitcher change as the angle of the barrel to the horizontal increases?

  A. The work done increases, because the softball's maximum height increases.

  B. The work done decreases, because the softball lands closer to the cannon.

  C. The work done does not change, because the air pressure behind the disk is unchanged.

  D. The work done does not change, because gravity is a conservative force.

  Correct Answer: C

  The softball starts off at rest and acquires a speed v0 as it is launched from the cannon. The work-energy theorem states that the work done equals the change in the kinetic energy. Since the softball acquires a kinetic energy equal to (1/2) mv0 , the automatic pitcher must have done work on it. The pitcher uses air pressure, which builds up behind a disk, to do the work when the disk is released. The angle of the barrel to the horizontal will not affect this mechanism, and the softball will still be ejected with the same kinetic energy. Hence, the work done by the pitcher does not change and choice C is correct.

  Although it is true that the softball's maximum height increases and that the distance it lands from the cannon decreases, the work done by the pitcher does not change, so choices A and B are wrong. Although it is also true that gravity is a conservative force, it is irrelevant because the question asks about the work done by the pitcher, not the work done by gravity. Hence, choice D is incorrect as well.

• Question 236:

  When softball players take batting practice, they often use a machine called an "automatic pitcher," which is essentially a cannon that uses air pressure to launch a projectile. In a prototype automatic pitcher, a softball is loaded into the barrel of the cannon and rests against a flat disk. That disk is locked into place, and a high air pressure is built up behind it. When the disk is released, the softball is pushed along the barrel of the cannon and ejected at a speed of V0. Figure 1 shows the batter and automatic pitcher. The angle of the barrel to the horizontal is V0. The unit vectors I and j point in the horizontal and vertical directions respectively.

  

  Figure 1

  The height above the ground y of the softball as a function of time t is shown in Figure 2, where t = 0 at Point A, t = tB at Point B, and t = tC at Point C. The softball is ejected from the barrel of the cannon at Point A; it reaches its maximum height at Point B; and the batter hits the softball at Point C. (Note: Assume that the effects of air resistance are negligible unless otherwise stated.)

  

  Figure 2

  What is the acceleration of the softball t seconds after it exits the barrel?

  

  A. Option A

  B. Option B

  C. Option C

  D. Option D

  Correct Answer: A

  Newton's second law F = ma states that a force produces an acceleration in the same direction as the force. In the absence of a force, there is no acceleration. In this situation, the only force present is the force of gravity, which is the same at

  all times during the softball's flight. Near the Earth's surface, the gravitational force is directed downward and its magnitude is given by F = mg, where m is the softball's mass and g is the acceleration due to gravity. Putting the gravitational

  force into Newton's second law, we obtain ma = mg. Canceling the masses, we obtain a = g. In other words, the acceleration of the softball is the same as the acceleration due to gravity.

  Figure 1 shows that the unit vector j points upwards, which means that 璲 points downwards.

  Because j is a unit vector, its magnitude is 1. Since the softball accelerates downward with a magnitude of g, the acceleration is symbolically given by g(璲) = 璯j. Thus, choice A is correct. Choice C is incorrect because it has the wrong

  magnitude. Choices B and D are wrong because they both include an acceleration in the horizontal direction 璱.

• Question 237:

  When softball players take batting practice, they often use a machine called an "automatic pitcher," which is essentially a cannon that uses air pressure to launch a projectile. In a prototype automatic pitcher, a softball is loaded into the barrel of the cannon and rests against a flat disk. That disk is locked into place, and a high air pressure is built up behind it. When the disk is released, the softball is pushed along the barrel of the cannon and ejected at a speed of V0. Figure 1 shows the batter and automatic pitcher. The angle of the barrel to the horizontal is . The unit vectors I and j point in the horizontal and vertical directions respectively.

  

  Figure 1

  The height above the ground y of the softball as a function of time t is shown in Figure 2, where t = 0 at Point A, t = tB at Point B, and t = tC at Point C. The softball is ejected from the barrel of the cannon at Point A; it reaches its maximum height at Point B; and the batter hits the softball at Point C. (Note: Assume that the effects of air resistance are negligible unless otherwise stated.)

  

  Figure 2

  How will V0 change if the impulse on the softball remains the same but its mass is doubled?

  A. It will decrease by a factor of 4.

  B. It will decrease by a factor of 2.

  C. It will not change.

  D. It will increase by a factor of 2.

  Correct Answer: B

  You should know that the change in momentum of a body is equal to the impulse applied to that body. The magnitude of the momentum of a body is given by mv, where m is its mass and v is its speed. In this situation an impulse is applied to the softball to launch it from the cannon. Before the impulse the speed of the softball is zero, so we can write J = (mv) = mv0 ?m(0) = mv0, where J is the impulse. Solving for V0 , we obtain V0 = J/m. Since m is inversely proportional to V0 , doubling the mass will halve V0 , so choice B is correct.

• Question 238:

  Which of the following is true when ice melts?

  A. The changes in both enthalpy and entropy are positive.

  B. The changes in both enthalpy and entropy are negative.

  C. The charge in enthalpy is positive; the change in entropy is negative.

  D. The change in enthalpy is negative; the change in entropy is positive.

  Correct Answer: A

  When heat is absorbed during a reaction, H is positive; when heat is released, H is negative. The melting of ice requires absorption of heat to disrupt the attractions between molecules, so the change in enthalpy is positive, and you can eliminate choice B and choice D. Now, entropy is the measurement of the disorder in a system; the change in entropy is expressed as S. When the disorder of a system increases, S is positive; when disorder decreases, S is negative. When ice melts, the individual molecules of water become freer to move around, so there is an increase in disorder and S is positive. Therefore, choice C is wrong and choice A is the correct answer.

• Question 239:

  When softball players take batting practice, they often use a machine called an "automatic pitcher," which is essentially a cannon that uses air pressure to launch a projectile. In a prototype automatic pitcher, a softball is loaded into the barrel of the cannon and rests against a flat disk. That disk is locked into place, and a high air pressure is built up behind it. When the disk is released, the softball is pushed along the barrel of the cannon and ejected at a speed of V0. Figure 1 shows the batter and automatic pitcher. The angle of the barrel to the horizontal is . The unit vectors I and j point in the horizontal and vertical directions respectively.

  

  Figure 1

  The height above the ground y of the softball as a function of time t is shown in Figure 2, where t = 0 at Point A, t = tB at Point B, and t = tC at Point C. The softball is ejected from the barrel of the cannon at Point A; it reaches its maximum height at Point B; and the batter hits the softball at Point C. (Note: Assume that the effects of air resistance are negligible unless otherwise stated.)

  

  Figure 2

  What physical quantity is NOT the same at Point C as at Point A?

  A. The velocity of the softball

  B. The speed of the softball

  C. The gravitational potential energy of the softball

  D. The horizontal component of the velocity of the softball

  Correct Answer: A

  The velocity of the softball is a vector, which means that it has both a magnitude and direction. At Point A, the velocity vector points up and to the right, because the softball is traveling in his direction. At Point C, however, the velocity vector points down and to the right. Since the velocity's direction has changed, it is not the same at Points A and C, and choice A is correct.

  You should know that the gravitational potential energy of the softball is given by U = mgy, where m is the mass, g is the acceleration due to gravity, and y is the height above the Earth. Since Points A and C are at the same height y, the gravitational potential is the same and choice C is incorrect.

  The passage says to ignore air resistance, so the total mechanical energy, which is the kinetic plus potential energy, is conserved in this system. Since the gravitational potential is the same at Points A and C, the kinetic energy (1/2)mv2 must be the same as well. Because the mass m does not change, the speed v must be the same at both points also, and choice B is wrong.

  Since air resistance is negligible, the only force present is that of gravity, which acts in the vertical direction. Consequently, there is no horizontal force on the softball, which means there can be no horizontal acceleration. Therefore, the horizontal component of the velocity must remain the same throughout the softball's flight, and choice D is wrong.

• Question 240:

  A deep-sea research module has a volume of 150 m3. If ocean water has an average density of 1,025 kg/m3, what will be the buoyant force on the module when it is completely submerged in the water? (Note: The acceleration due to gravity is 9.8 m/s2.)

  

  A. Option A

  B. Option B

  C. Option C

  D. Option D

  Correct Answer: D

  Archimedes' principle states that the buoyant force exerted by a fluid on a submerged object is equal to the weight of the fluid displaced. In this case, the submerged object is the research module and the fluid is ocean water. Mathematically, the buoyant force is given by F = mwg, where mw is the mass of the water and g is the acceleration due to gravity. Since we are not given the mass of the water displaced, we express it as mw = pwVw, where pw is the density of the water and Vw is the volume of the water displaced. But the volume of the water displaced is identical to the volume of the submerged module, V0. Hence, mw = pwVw, = pwV0, and F = mwg = pwV0g. Since the answer choices are far apart, we round off the values given in the question and calculate F = pwV0g (1,000 kg/m3 )(150 m3)(10 m/S2 ) = 1.5 X 106 N.