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Test Prep MCAT Test Exam
Medical College Admission Test: Verbal Reasoning, Biological Sciences, Physical Sciences, Writing Sample (Page 26 )

Updated On: 30-Jan-2026
Page 26 of 164
PDF with 811 Questions

The periodic beating of the heart is controlled by electrical impulses that originate within the cardiac muscle itself. These pulses travel to the sinoatrial node and from there to the atria and the ventricles, causing the cardiac muscles to contract. If a current of a few hundred milliamperes passes through the heart, it will interfere with this natural system, and may cause the heart to beat erratically. This condition is known as ventricular fibrillation, and is life-threatening. If, however, a larger current of about 5 to 6 amps is passed through the heart, a sustained ventricular contraction will occur. The cardiac muscle cannot relax, and the heart stops beating. If at this point the muscle is allowed to relax, a regular heartbeat will usually resume.
The large current required to stop the heart is supplied by a device known as a defibrillator. A schematic diagram of a defibrillator is shown below. This device is essentially a "heavy-duty" capacitor capable of storing large amounts of energy. To charge the capacitor quickly (in 1 to 3 seconds), a large DC voltage must be applied to the plates of the capacitor. This is achieved using a step-up transformer, which creates an output voltage that is much larger than the input voltage. The transformer used in this defibrillator has a step-up ratio of 1:50.


The AC voltage that is obtained from the transformer must then be converted to DC voltage in order to charge the capacitor. This is accomplished using a diode, which allows current flow in one direction only. Once the capacitor is fully charged, the charge remains stored until the switch is moved to position B and the plates are placed on the patient's chest. To cut down the resistance between the patient's body and the defibrillator, the electrodes are covered with a wetting gel before use. Care must be taken to insure that the patient is not in electrical contact with the ground while the defibrillator is in use.
The plates of the capacitor are originally separated by a vacuum. If a dielectric K > 1 is introduced between the plates of the capacitor, and the capacitor is allowed to charge up, which of the following statements is/are true?
I) The capacitance of the capacitor will increase.
II) The voltage across the capacitor plates will increase.
III) The charge stored on the capacitor will increase.

  1. I only
  2. I and II only
  3. II and III only
  4. I and III only

Answer(s): D

Explanation:

The capacitance of a parallel-plate capacitor is given by

where K is the dielectric constant, 0 is the permittivity of free space, A is the area of overlap of the two the plates, and d is the separation of the two plates. This implies that when a material with a dielectric constant K is introduced between the plates of a capacitor, the capacitance increases by a numerical factor K. In the question we are told that a dielectric with a value of K > 1 is introduced between the plates which means that the capacitance will increase. So statement I is correct and choice C can be ruled out. It is important to note at this point that the voltage across the plates does not increase with the introduction of a dielectric. This is because the voltage across the plates of a fully charged capacitor is equal to the voltage applied across the plates when it was initially charged. Since this is held constant at 8,000 volts, the voltage across the plates will remain at 8,000 volts. So statement II is false and choice B can be eliminated.
To choose between choices A and D, look at how the charge on the plates of a capacitor is affected by the introduction of a dielectric. The capacitance of a capacitor is related to the voltage across the plates and the charge stored by the equation C = Q/V, where C is the capacitance, Q is the charge stored, and V is the voltage. As mentioned previously, the voltage across the capacitor remains constant when the dielectric is
introduced. This implies that the capacitance of the capacitor is directly proportional to the charge stored. Since this is the case, as the capacitance increases, the charge stored on the plates must also increase. So statement III is true and choice D is the correct answer.



The periodic beating of the heart is controlled by electrical impulses that originate within the cardiac muscle itself. These pulses travel to the sinoatrial node and from there to the atria and the ventricles, causing the cardiac muscles to contract. If a current of a few hundred milliamperes passes through the heart, it will interfere with this natural system, and may cause the heart to beat erratically. This condition is known as ventricular fibrillation, and is life-threatening. If, however, a larger current of about 5 to 6 amps is passed through the heart, a sustained ventricular contraction will occur. The cardiac muscle cannot relax, and the heart stops beating. If at this point the muscle is allowed to relax, a regular heartbeat will usually resume.
The large current required to stop the heart is supplied by a device known as a defibrillator. A schematic diagram of a defibrillator is shown below. This device is essentially a "heavy-duty" capacitor capable of storing large amounts of energy. To charge the capacitor quickly (in 1 to 3 seconds), a large DC voltage must be applied to the plates of the capacitor. This is achieved using a step-up transformer, which creates an output voltage that is much larger than the input voltage. The transformer used in this defibrillator has a step-up ratio of 1:50.

The AC voltage that is obtained from the transformer must then be converted to DC voltage in order to charge the capacitor. This is accomplished using a diode, which allows current flow in one direction only. Once the capacitor is fully charged, the charge remains stored until the switch is moved to position B and the plates are placed on the patient's chest. To cut down the resistance between the patient's body and the defibrillator, the electrodes are covered with a wetting gel before use. Care must be taken to insure that the patient is not in electrical contact with the ground while the defibrillator is in use.
Why is it important to insure that the patient is not in electrical contact with the ground while the defibrillator is in use?

  1. Contact with the ground will decrease the resistance across the patient's body.
  2. The doctor administering the treatment will be in greater danger of receiving an electric shock if the patient is in electrical contact with the ground.
  3. Contact with the ground will cause a smaller current to pass through the patient's heart.
  4. The patient receiving the treatment will be in greater danger of receiving burns due to the high current density if he is in electrical contact with the ground.

Answer(s): C

Explanation:

When there is no ground connection, all the current is forced through the patient's heart, despite the relatively large resistance of the patient's body. However, if the patient is in contact with the ground, this provides a path of far less resistance than the patient's body, resulting in most of the current flowing through the ground, rather than through the patient. This will render the defibrillator ineffective. Choice A is incorrect because the
resistance across the patient's body cannot change. Choice B is incorrect. If the defibrillator is not in contact with the patient when the doctor receives a shock, then the shock hazard has nothing to do with the patient being grounded or not. If, however, the electrodes are in contact with the patient when the doctor receives a shock, the ground will take most of the current, and thereby reduce the shock that the doctor experiences. So the doctor administering the treatment will be in less danger of receiving a shock when the patient is in contact with the ground.
Choice D is incorrect since the burns received by the patient are due to the application of large currents over a small area of skin, and have nothing to do with the presence of a ground. Therefore, in order to reduce the burns that the patient receives, the area of the electrodes is increased to distribute the current and reduce the current density. The correct answer is choice C, contact with the ground will cause a smaller current to pass through the patient's heart.



The periodic beating of the heart is controlled by electrical impulses that originate within the cardiac muscle itself. These pulses travel to the sinoatrial node and from there to the atria and the ventricles, causing the cardiac muscles to contract. If a current of a few hundred milliamperes passes through the heart, it will interfere with this natural system, and may cause the heart to beat erratically. This condition is known as ventricular fibrillation, and is life-threatening. If, however, a larger current of about 5 to 6 amps is passed through the heart, a sustained ventricular contraction will occur. The cardiac muscle cannot relax, and the heart stops beating. If at this point the muscle is allowed to relax, a regular heartbeat will usually resume.
The large current required to stop the heart is supplied by a device known as a defibrillator. A schematic diagram of a defibrillator is shown below. This device is essentially a "heavy-duty" capacitor capable of storing large amounts of energy. To charge the capacitor quickly (in 1 to 3 seconds), a large DC voltage must be applied to the plates of the capacitor. This is achieved using a step-up transformer, which creates an output voltage that is much larger than the input voltage. The transformer used in this defibrillator has a step-up ratio of 1:50.

The AC voltage that is obtained from the transformer must then be converted to DC voltage in order to charge the capacitor. This is accomplished using a diode, which allows current flow in one direction only. Once the capacitor is fully charged, the charge remains stored until the switch is moved to position B and the plates are placed on the patient's chest. To cut down the resistance between the patient's body and the defibrillator, the electrodes are covered with a wetting gel before use. Care must be taken to insure that the patient is not in electrical contact with the ground while the defibrillator is in use.
If a dielectric was inserted between the plates of the capacitor in the defibrillator when the switch is in position
A:

  1. the energy stored in the capacitor would increase.
  2. the energy stored in the capacitor would decrease.
  3. the electric field between the plates would increase.
  4. the electric field between the plates would decrease.

Answer(s): A

Explanation:

Since the capacitor is, in effect, connected to a constant voltage source, the potential difference doesn't change when a dielectric is inserted between the plates. Therefore, the charge on the plates must increase to compensate for the effect of the polarization of charge in the dielectric. The polarization of charge in the dielectric sets up an electric field opposite in direction and weaker in magnitude than the electric field due to the charge on the capacitor plates. This would decrease the net electric field between the plates if the capacitor were not connected to a constant voltage source. However, since it is connected to a constant voltage source when the dielectric is inserted, additional charge builds up on the plates to keep the potential difference constant. Therefore, the net electric field between the plates with the dielectric in place is equal to the net electric field between the plates with no dielectric and choices C and D are wrong. Since the charge on the plates increases, the energy stored will increase because the energy stored is given by (1/2)QVk, where Q is the charge on the plates and V is the potential difference between the plates. Therefore, choice A is correct.



In response to period of extreme psychological trauma, a patient begins experiencing a feeling of detachment. He says, "I felt like it wasn't real while it was happening. I was just watching myself do it without any control. I mean, you know, I knew it was happening but I didn't feel like it was." The patient is describing:

  1. dissociative identity disorder.
  2. an anxiety disorder.
  3. depersonalization disorder.
  4. a schizophrenic episode.

Answer(s): C

Explanation:

A: Dissociative identity disorder involves 2 or more personalities with imperfect recall between them.
B: Anxiety disorders are mood disorders characterized by excessive levels of arousal.
D: Schizophrenia is characterized by hallucinations, delusions, and disordered thinking.



As a result of substance abuse throughout adolescence, a young adult suffers from a number of psychological symptoms reflecting diminished executive functioning. Which of the following are likely true of this patient?
I). Pathological changes to the prefrontal cortex.
II). Increased susceptibility to auditory hallucinations.
III). Reduced behavioral impulse control.

  1. I only
  2. III only
  3. I and III only
  4. II and III only

Answer(s): C

Explanation:

Executive functioning, or the ability to formulate and stick to long term plans, exercise judgment and impulse control with respect to those plans, is controlled primarily by the prefrontal cortex. In a patient with diminished executive functioning, both (I) and (III) would likely be seen. Thus, (C) is the right answer.
II: The prefrontal cortex is not part of the auditory processing pathway. Thus, there is no reason why a patient with diminished executive functioning would likely have auditory hallucinations.



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