Rt Simulation Examination 1
This is a Simulated Examination for Gulf RT Examinations taken from Last Month's HAAD Feedbacks.
This examination contains 100 of the most UPDATED EXAMS from Abu Dhabi, KSA, and UAE.
Take this examination for 120 minutes. You need to get 86% to pass the HAAD. 60% to pass MOH, DHA, or Prometrics.
Please text 0919-286-29-29 in the Philippines or visit our website www. Rtonline. Weebly. Com THIS IS YOUR ASSESSMENT FOR ANY GULF RT EXAMINATIONS INCLUDING HAAD RT, SAUDI PROMETRICS, DUBAI DHA, AND UAE MOH. THE QUESTIONS HERE ARE TAKEN FROM THIS ACTUAL EXAMINATIONS, SO PASSING THIS Read moreASSESSMENT EXAM WILL GIVE YOU A HIGH PROBABILITY OF PASSING THE SAID EXAMINATIONS
- 1.
What are the four types of Radiation?
- A.
Alpha, Beta, Gamma and Radium
- B.
Beta, Gamma, Isotope and Radium
- C.
Beta, Electromagnetic, Ultra violet and Proton
- D.
Alpha, Beta, Gamma and Background
Correct Answer
D. AlpHa, Beta, Gamma and BackgroundExplanation
The four types of radiation are alpha, beta, gamma, and background. Alpha radiation consists of helium nuclei and is the least penetrating form of radiation. Beta radiation consists of high-energy electrons or positrons and is more penetrating than alpha radiation. Gamma radiation is electromagnetic radiation and is the most penetrating form of radiation. Background radiation refers to the low levels of radiation that are present in the environment from natural and man-made sources.Rate this question:
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- 2.
What Radiation is used in smoke detectors?
- A.
Beta
- B.
Alpha
- C.
Gamma
- D.
Background
Correct Answer
B. AlpHaExplanation
This is a hard questionn so dont be worryed if uyou got it wrong.Rate this question:
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- 3.
What piece of equipment is used to measure the radioactive decay count?
- A.
Decay Tube
- B.
Geiger Muller Tube
- C.
Fordy Decay Count
- D.
Muller Rice Tube
- E.
Hodgsonator Decay Count Tube
Correct Answer
B. Geiger Muller TubeExplanation
How can you not get this right.Rate this question:
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- 4.
What radiation is used to kill cancer cells.
- A.
Alpha
- B.
Beta
- C.
Gamma
- D.
Radium
- E.
Hooperium
Correct Answer
C. GammaExplanation
Gamma radiation is used to kill cancer cells because it has high energy and can penetrate deep into tissues. This type of radiation damages the DNA of cancer cells, preventing them from dividing and growing. Gamma radiation is often used in radiation therapy, where targeted doses of radiation are delivered to specific areas of the body to destroy cancer cells. It is a highly effective treatment option for various types of cancer.Rate this question:
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- 5.
Terra / BTX uses X-rays to produce the desired X-ray diffraction pattern or X-ray fluorescence spectrum. These X-rays are produced from:
- A.
A radioactive isotope within the instrument
- B.
A gamma ray neutron source within the instrument
- C.
An X-ray tube operating at 30Kv
- D.
None of the above
Correct Answer
C. An X-ray tube operating at 30KvExplanation
An X-ray tube operating at 30Kv is used to produce the desired X-ray diffraction pattern or X-ray fluorescence spectrum in Terra/BTX. The X-ray tube generates X-rays by accelerating electrons using a high voltage of 30Kv. These accelerated electrons collide with a metal target in the tube, resulting in the emission of X-rays. These X-rays are then used to produce the desired patterns or spectra for analysis.Rate this question:
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- 6.
Radiation which has sufficient energy to remove electrons from atoms in materials through which the radiation passes is called ionizing radiation. Examples of ionizing radiation include all but:
- A.
Alpha particle radiation
- B.
Neutrons
- C.
X-rays
- D.
Radio waves
Correct Answer
D. Radio wavesExplanation
Ionizing radiation refers to radiation that has enough energy to remove electrons from atoms in materials. This process, known as ionization, can have harmful effects on living organisms and materials. Alpha particle radiation, neutrons, and x-rays are all examples of ionizing radiation, as they possess sufficient energy to ionize atoms. On the other hand, radio waves do not have enough energy to remove electrons from atoms and therefore do not fall under the category of ionizing radiation.Rate this question:
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- 7.
The damaging effects of radiation depend not only on its intensity and duration, but also the type of radiation. For instance, 1 millicurie of tritium a centimeter from the body poses a much different hazard than 1 millicurie of phosphorus-32 a centimeter from the body. Consequently, it is often preferable to measure radiation by describing the effect of that radiation on the materials through which it passes. The units for biological effects of radiation are:
- A.
Exposure
- B.
Absorbed Dose
- C.
Dose Equivalent
- D.
REM/Sievert
Correct Answer
D. REM/SievertExplanation
The units for biological effects of radiation are REM/Sievert because they take into account not only the intensity and duration of the radiation, but also the type of radiation and its effect on the materials it passes through. REM (Roentgen Equivalent Man) is a unit that measures the biological effect of radiation on human tissue, while Sievert is the International System of Units (SI) equivalent of REM. These units provide a more comprehensive measurement of the potential harm caused by radiation exposure.Rate this question:
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- 8.
The amount of radiation exposure allowable is defined by the international commission on radiological protection (ICRP).
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
The explanation for the given answer is that the international commission on radiological protection (ICRP) sets the standards for the amount of radiation exposure that is considered safe. These standards are based on extensive research and scientific evidence to ensure that individuals are not exposed to harmful levels of radiation. Therefore, it is true that the amount of radiation exposure allowable is defined by the ICRP.Rate this question:
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- 9.
Radiation control for instruments, such as Terra / BTX is controlled by the US Federal government for instruments operated in the United States.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
The statement is false because radiation control for instruments like Terra/BTX is not controlled by the US Federal government.Rate this question:
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- 10.
It is not necessary to contact your local radiation protection control authority (State, Federal, etc) in order to operate your Terra or BTX.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
The statement is false because it is necessary to contact the local radiation protection control authority in order to operate Terra or BTX. These devices are likely to emit radiation, and therefore, proper authorization and compliance with regulations are required to ensure safety and adherence to legal requirements. Failing to contact the local radiation protection control authority may result in legal consequences and potential harm to individuals and the environment.Rate this question:
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- 11.
Terra / BTX when operated properly does not produce measurable radiation on any external surface of the instrument.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
The statement suggests that when Terra / BTX is operated correctly, it does not emit any detectable radiation on any external surface of the instrument. This implies that the instrument is designed and functioning properly, ensuring that radiation is contained within the instrument and does not pose any risk to the external environment or individuals. Therefore, the correct answer is true.Rate this question:
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- 12.
Terra / BTX feature multiple interlock circuits, which prevent the possible operation of the X-ray generating source when these interlocks are activated.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
The statement is explaining that the Terra/BTX feature multiple interlock circuits that prevent the X-ray generating source from operating when these interlocks are activated. This means that if any of the interlocks are activated, the X-ray generating source will not be able to function. Therefore, the statement is true.Rate this question:
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- 13.
Should an interlock circuit activate, an amber light will display on the main instrument panel.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
If an interlock circuit is activated, it means that there is a safety mechanism in place that prevents certain functions or operations until certain conditions are met. In this case, if the interlock circuit is activated, it will trigger the display of an amber light on the main instrument panel. Therefore, the statement is true.Rate this question:
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- 14.
InXitu expects that the operator will need to remove the acess panel in order to properly use the instrument. That is, the instrument has user-serviced parts inside.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
The explanation for the given answer "False" is that inXitu does not expect the operator to remove the access panel in order to properly use the instrument. The instrument does not have user-serviced parts inside.Rate this question:
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- 15.
Terra / BTX has the ability to perform the following experiments except:
- A.
X-ray Diffraction
- B.
X-ray Fluorescence
- C.
X-ray Time of Flight
Correct Answer
C. X-ray Time of FlightExplanation
Terra / BTX has the ability to perform X-ray Diffraction and X-ray Fluorescence experiments, but it does not have the ability to perform X-ray Time of Flight experiments.Rate this question:
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- 16.
Terra / BTX have been designed such that proper operation will not result in any radiation exposure.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Terra/BTX has been specifically designed to ensure that it operates without causing any radiation exposure. This means that when used correctly, there will be no harmful radiation emitted from Terra/BTX. Therefore, the statement is true.Rate this question:
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- 17.
Which is not an acoustic variable?
- A.
Density
- B.
Pressure
- C.
Distance
- D.
Intensity
Correct Answer
D. IntensityExplanation
Intensity is not an acoustic variable because it refers to the amount of energy carried by sound waves per unit area. Acoustic variables, on the other hand, are physical quantities that describe the characteristics of sound waves, such as density, pressure, and distance.Rate this question:
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- 18.
All of the following are true EXCEPT:
- A.
Two waves with identical frequencies must interfere constructively
- B.
Constructive interference is associated with waves that are in phase
- C.
Out of phase waves interfere destructively
- D.
Waves of different frequencies may exhibit both contructive and destructive interference at different times
Correct Answer
A. Two waves with identical frequencies must interfere constructivelyExplanation
This statement is false because waves with identical frequencies can interfere destructively if they are out of phase. Constructive interference occurs when waves are in phase, meaning their crests and troughs align, resulting in a larger amplitude. However, if the waves are out of phase, their crests and troughs do not align, leading to destructive interference and a smaller overall amplitude.Rate this question:
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- 19.
Put in decreasing order
- A.
Deca
- B.
Deci
- C.
Micro
- D.
Centi
Correct Answer(s)
A. Deca
B. Deci
C. Micro
D. CentiExplanation
The given words are units of measurement prefixes, representing different magnitudes. "Deca" is the largest prefix, representing a factor of 10, followed by "deci" which represents a factor of 0.1. "Micro" represents a factor of 0.000001, making it smaller than both "deca" and "deci". "Centi" represents a factor of 0.01, making it the smallest prefix among the given options. Therefore, the correct order from largest to smallest is "deca, deci, micro, centi".Rate this question:
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- 20.
Which of the following sound waves is ultrasonic and least useful in diagnostic imaging?
- A.
30 KHZ
- B.
8 MHz
- C.
8,000 Hz
- D.
3,000 kHz
- E.
15 Hz
Correct Answer
A. 30 KHZExplanation
The correct answer is 30 KHZ. Ultrasonic sound waves have a frequency above the range of human hearing, typically above 20 kHz. Diagnostic imaging techniques such as ultrasound use high-frequency sound waves to create images of internal body structures. Since 30 KHZ is the lowest frequency among the given options, it is the least useful in diagnostic imaging as it falls below the typical range of ultrasonic frequencies used in this field.Rate this question:
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- 21.
What is the frequency of a wave with 1msec period?
- A.
10,000 Hz
- B.
1,000 kHz
- C.
1kHz
- D.
1MHz
Correct Answer
C. 1kHzExplanation
The frequency of a wave is the number of complete cycles it completes in one second. In this question, it is given that the wave has a period of 1 millisecond. The period is the time taken for one complete cycle. Since there are 1,000 milliseconds in one second, the wave completes 1,000 cycles in one second. Therefore, the frequency of the wave is 1kHz (1,000 Hz).Rate this question:
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- 22.
If the frequency of an US wave is doubled, what happens to the period?
- A.
Doubles
- B.
Halved
- C.
Remains the same
- D.
4 times greater
Correct Answer
B. HalvedExplanation
When the frequency of an ultrasound wave is doubled, the period of the wave is halved. This is because frequency and period are inversely proportional to each other. Frequency is the number of wave cycles per second, while period is the time it takes for one complete wave cycle. When the frequency is doubled, it means that the number of wave cycles per second has increased, so each wave cycle takes less time to complete, resulting in a halved period.Rate this question:
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- 23.
The units of pulse repetition frequency are:
- A.
Per minute
- B.
Msec
- C.
Mm/us
- D.
MHx
Correct Answer
A. Per minuteExplanation
The pulse repetition frequency measures the number of pulses that occur in one minute. It is a unit used to quantify the rate at which pulses are repeated in a given time period. Therefore, the correct answer is "per minute." The other options, msec (milliseconds), mm/us (millimeters per microsecond), and MHx (megahertz), do not accurately represent the units of pulse repetition frequency.Rate this question:
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- 24.
The time from the beginning of a pulse until its end is__________?
- A.
Period
- B.
Pulse duration
- C.
Pulse length
- D.
PRF
Correct Answer
B. Pulse durationExplanation
The time from the beginning of a pulse until its end is referred to as the pulse duration. This term specifically measures the length of time that a pulse of energy or signal is active or present. It is important in various fields such as telecommunications, radar systems, and medical imaging, where the duration of a pulse directly affects the quality and accuracy of the transmitted or received information.Rate this question:
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- 25.
The time that a transducer is pulsing is measured at 18 seconds in one hour of total elapsed time. What is the duty factor?
- A.
0.3
- B.
0.005
- C.
0.5Hz
- D.
0.005msec
Correct Answer
B. 0.005Explanation
60 seconds equal 1 minute. There are 60 minutes in 1 hour so 60 seconds X 60 minutes = 18 divided by 3600 = 0.005Rate this question:
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- 26.
If a wave's amplitude is doubled, what happens to the power?
- A.
Nothing
- B.
Halved
- C.
Dpib;ed
- D.
Quadrupled
Correct Answer
D. QuadrupledExplanation
When a wave's amplitude is doubled, the power of the wave increases by a factor of four, or it becomes quadrupled. This is because power is directly proportional to the square of the amplitude of the wave. Therefore, if the amplitude is doubled, the power will increase by four times.Rate this question:
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- 27.
If the level of an acoustic variable ranges from 55 to 105, what is the amplitude?
- A.
105
- B.
50
- C.
25
- D.
55
Correct Answer
C. 25Explanation
The amplitude of an acoustic variable refers to the maximum displacement or distance from the equilibrium position. In this case, the level of the acoustic variable ranges from 55 to 105. Since the amplitude is the maximum displacement, it would be the difference between the highest and lowest values, which is 105 - 55 = 50. Therefore, the correct answer is 50.Rate this question:
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- 28.
If the intensity of a sound beam remains unchanged while the beam area is reduced in half, what happened to the power?
- A.
Quadrupled
- B.
Doubled
- C.
Halved
- D.
Unchanged
Correct Answer
C. HalvedExplanation
When the beam area is reduced in half, the same amount of power is spread over a smaller area. This means that the power density (power per unit area) increases. Since power is directly proportional to the product of intensity and area, and the intensity remains unchanged, the only way for the power to decrease is if the area is halved. Therefore, the power is halved when the beam area is reduced in half.Rate this question:
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- 29.
Put these intensities in decreasing order
- A.
SATP
- B.
SPTP
- C.
SATA
- D.
SPTP, SATP, SATA
- E.
SPTP SATA, SATP
Correct Answer
D. SPTP, SATP, SATAExplanation
The given answer is in the correct order because it lists the intensities in decreasing order. "SPTP" has the highest intensity, followed by "SATP", and then "SATA" has the lowest intensity.Rate this question:
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- 30.
The duty factor for continuous wave ultrasound is
- A.
10.0
- B.
1%
- C.
1.0
- D.
0.0%
Correct Answer
C. 1.0Explanation
The duty factor for continuous wave ultrasound refers to the percentage of time that the ultrasound machine is emitting ultrasound waves. A duty factor of 1.0 means that the machine is emitting ultrasound waves 100% of the time. This indicates that the ultrasound waves are continuously being transmitted without any interruptions or breaks.Rate this question:
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- 31.
What is the minimum value of the SP/SA factor?
- A.
10.0
- B.
1%
- C.
1.0
- D.
0.0%
Correct Answer
C. 1.0Explanation
The minimum value of the SP/SA factor is 1.0. This means that the selling price (SP) is equal to the selling amount (SA). In other words, there is no markup or discount applied to the selling price.Rate this question:
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- 32.
Which pair of intensities has the same value for continuous wave US?
- A.
Spatial peak & spatial average
- B.
Temporal peak & spatial peak
- C.
Pulse average & temporal average
- D.
Spatial average & temporal average
Correct Answer
C. Pulse average & temporal averageExplanation
The pulse average and temporal average intensities have the same value for continuous wave US. The pulse average intensity is the average intensity over one pulse duration, while the temporal average intensity is the average intensity over a longer time period. Since continuous wave US has a constant intensity, the pulse average and temporal average intensities will be equal.Rate this question:
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- 33.
The fundamental frequency of a transcucer is 2.5 MHz. What is the second harmonic frequency?
- A.
1.25 MHz
- B.
4.5 MHz
- C.
4.8 MHz
- D.
5 cm
Correct Answer
C. 4.8 MHzExplanation
The second harmonic frequency is double the fundamental frequency. Therefore, if the fundamental frequency is 2.5 MHz, the second harmonic frequency would be 5 MHz. However, none of the options provided match this value. Therefore, the correct answer is not available.Rate this question:
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- 34.
A pulse is emitted by a transducer and is traveling in soft tissue. The go-return time, or time of flight, of a sound pulse is 130 microseconds. What is the reflector depth?
- A.
10 cm
- B.
10 mm
- C.
10 m
- D.
130 mm
- E.
13 cm
Correct Answer
A. 10 cmExplanation
PRP(us) = imaging depth x 13 us/cm
130 us = ? x 13 us/cm
130 divide by 13 = 10 cm or
Depth(mm) = 1.54mm/us x go return time(us)/2
= 1.54 x 130/2
= 200.2/2
mm = 100.1 or 10 cmRate this question:
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- 35.
What is the approximate attenuation coefficient of 1MHz US in soft tissue?
- A.
0.5 dBcm
- B.
1cm
- C.
3dBcm
- D.
1dB
Correct Answer
A. 0.5 dBcmExplanation
The approximate attenuation coefficient of 1MHz ultrasound in soft tissue is 0.5 dBcm. Attenuation coefficient refers to the rate at which the ultrasound waves decrease in intensity as they pass through the tissue. In this case, the attenuation coefficient is given in units of decibels per centimeter (dBcm), indicating that the intensity of the ultrasound waves decreases by 0.5 decibels for every centimeter of tissue they pass through.Rate this question:
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- 36.
The relaive output of an US instrument is calibrated in dB and the operator increases the output by 60 dB. The beam intensity is increased by which of the following?
- A.
5%
- B.
Two times
- C.
Twenty times
- D.
One hundred times
- E.
One million times
Correct Answer
E. One million timesExplanation
3dB means 2 x bigger 6dB is 3dB + 3dB, therefore 6dB means 2 x 2 or 4 times bigger. 60dB = 10x10x10x10x10x10 or 1,000,000Rate this question:
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- 37.
The more pixels per inch:
- A.
The better the temporal resolution
- B.
The better is the spatial resolution
- C.
The more shades of gray
- D.
The higher the reliability
Correct Answer
B. The better is the spatial resolutionExplanation
A higher number of pixels per inch indicates a higher spatial resolution. Spatial resolution refers to the level of detail that can be captured and displayed in an image. With more pixels per inch, there are more individual dots of color or shades of gray in the image, allowing for finer details to be represented. Therefore, the better the spatial resolution is.Rate this question:
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- 38.
If we increase the transducer diameter, the beam diameter in the far zone is:
- A.
Increased
- B.
Decreased
- C.
Unchanges
Correct Answer
B. DecreasedExplanation
When the transducer diameter is increased, the beam diameter in the far zone is decreased. This is because the far zone is the region where the beam has already diverged and spread out. Increasing the transducer diameter means that the beam will have a larger initial width, causing it to diverge less as it propagates. As a result, the beam diameter in the far zone will be smaller, leading to a decrease in size.Rate this question:
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- 39.
If the transducer diameter increases, the lateral resolution at its smallest dimension is
- A.
Increased
- B.
Decreased
- C.
Unchanged
Correct Answer
A. IncreasedExplanation
When the transducer diameter increases, the lateral resolution at its smallest dimension is increased. This is because a larger transducer diameter allows for a wider beam width, which results in a larger focal zone. As a result, the ultrasound waves diverge less and can provide better lateral resolution, allowing for more detailed imaging and distinguishing smaller structures.Rate this question:
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- 40.
If we increase the frequency the near zone length is
- A.
Increased
- B.
Decreased
- C.
Unchanged
Correct Answer
A. IncreasedExplanation
When the frequency is increased, the near zone length is increased. This is because the near zone length is inversely proportional to the frequency. As the frequency increases, the wavelength decreases, causing the near zone length to increase.Rate this question:
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- 41.
If the frequency is decreased, the numerical value of the radial resolution is
- A.
Increased
- B.
Decreased
- C.
Unchanged
Correct Answer
A. IncreasedExplanation
When the frequency is decreased, the numerical value of the radial resolution is increased. Radial resolution refers to the ability to distinguish between two closely spaced objects. Decreasing the frequency means that the wavelength becomes longer, resulting in a larger distance between consecutive wave peaks. This larger distance allows for better differentiation between objects, leading to an increased radial resolution.Rate this question:
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- 42.
If the spatial pulse length is 10 mm, what is the axial resolution?
- A.
0.5 cm
- B.
5 cm
- C.
10 mm
- D.
1 cm
Correct Answer
A. 0.5 cmExplanation
The axial resolution is determined by the spatial pulse length, which is the distance between the start and end of a pulse. In this case, the spatial pulse length is given as 10 mm. The axial resolution is typically half of the spatial pulse length, so the axial resolution would be 0.5 cm.Rate this question:
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- 43.
Which of the following is the best lateral resolution?
- A.
15 mm
- B.
6 mm
- C.
0.06 cm
- D.
2 cm
Correct Answer
C. 0.06 cmExplanation
The best lateral resolution is 0.06 cm. Lateral resolution refers to the ability of an imaging system to distinguish between two closely spaced objects. A smaller value indicates a higher resolution, meaning that the system can differentiate between objects that are closer together. Therefore, a lateral resolution of 0.06 cm is the best option among the given choices as it provides the highest level of detail and accuracy in distinguishing objects.Rate this question:
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- 44.
What is the meaning of thermal index = 37
- A.
A tissue temperature may rise 3 degrees Farenheit
- B.
Tissue temperature will rise 3 degrees Celcius
- C.
Tissue temperature may rise 3 degrees Celcius
- D.
Microbubbles 3mm in diameter will burst
Correct Answer
C. Tissue temperature may rise 3 degrees CelciusExplanation
The thermal index is a measure of the potential for heating or mechanical effects in tissue when exposed to ultrasound. A thermal index of 37 indicates that the tissue temperature may rise by 3 degrees Celsius. This means that there is a possibility of heating in the tissue when exposed to ultrasound at this level.Rate this question:
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- 45.
The interaction of microscopic bubbles and ultrasound form the basis for cavitation bioeffects. Which of the following forms of cavitation are most likely to produce microstreaming in the intracellular fluid and shear stresses?
- A.
Stable cavitation
- B.
Normal cavitation
- C.
Transient cavitation
- D.
Inertial cavitation
- E.
Active cavitation
Correct Answer
A. Stable cavitationExplanation
Stable cavitation is the most likely form of cavitation to produce microstreaming in the intracellular fluid and shear stresses. In stable cavitation, the bubbles oscillate in size and shape but do not collapse completely. This oscillation creates microstreaming, which refers to the movement of fluid around the bubbles. The microstreaming effect can cause shear stresses on the surrounding intracellular fluid, leading to various bioeffects.Rate this question:
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- 46.
Which of the following has the greatest output intensity
- A.
B-Mode
- B.
Gray scale imaging
- C.
CW Doppler
- D.
Pulsed Doppler
Correct Answer
D. Pulsed DopplerExplanation
Pulsed Doppler has the greatest output intensity among the given options. Pulsed Doppler is a technique used in ultrasound imaging to measure blood flow velocity. It uses short bursts of ultrasound waves and measures the intensity of the reflected waves to determine the velocity of blood flow. The intensity of the reflected waves is higher in pulsed Doppler compared to other options such as B-Mode or gray scale imaging, which are used for visualizing anatomical structures, or CW Doppler, which uses a continuous wave of ultrasound.Rate this question:
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- 47.
Which of the following best decribes the empirical approach to the study of bioeffects?
- A.
Exposure-response
- B.
Cause-effect
- C.
Risk-benefit
- D.
Causation-reaction
- E.
Effect-microstreaming
Correct Answer
A. Exposure-responseExplanation
The empirical approach to the study of bioeffects involves examining the relationship between exposure to a certain factor and the resulting response or effect. It focuses on gathering and analyzing data to determine the extent and nature of the relationship between exposure and response. This approach aims to provide evidence-based information on the effects of various factors on biological systems.Rate this question:
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- 48.
Which of the following best describes the mechanistic approach to the study of bioeffects?
- A.
Cause-effect
- B.
Exposure-response
- C.
Risk-benefit
- D.
Causation-reaction
- E.
Effect-microstreaming
Correct Answer
A. Cause-effectExplanation
The mechanistic approach to the study of bioeffects focuses on understanding the cause-effect relationship between a specific exposure and its resulting effects. This approach aims to identify and explain the underlying mechanisms by which an exposure leads to a particular biological response or outcome. It seeks to establish a direct link between the cause (exposure) and the effect (biological response), providing a detailed understanding of the processes involved. This approach is important in assessing the potential risks and impacts of various exposures on living organisms.Rate this question:
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- 49.
The propagation speed of US in the AIUM test object is?
- A.
1m/s
- B.
1.54 m/s
- C.
1.54 mm/us
- D.
1 km/s
Correct Answer
C. 1.54 mm/usExplanation
The correct answer is 1.54 mm/us. This is because ultrasound waves travel at different speeds in different materials. In the AIUM test object, the ultrasound waves travel at a speed of 1.54 mm/us.Rate this question:
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- 50.
Enhancement, multipath and side lobes result in:
- A.
Image distortion
- B.
Clearer images
- C.
Artifact
- D.
Resolution problems
Correct Answer
C. ArtifactExplanation
Enhancement, multipath, and side lobes can result in artifacts. These are unwanted distortions or anomalies in an image that can occur due to various factors such as noise, interference, or imperfections in the imaging system. These artifacts can degrade the quality and accuracy of the image, making it less reliable for analysis or interpretation. Therefore, the presence of enhancement, multipath, and side lobes can contribute to the occurrence of artifacts in the image.Rate this question:
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