Rt Simulation Examination 2
-
The dB is defined as the _______________ of two intensities
-
Sum
-
Difference
-
Product
-
Ratio
-
About This Quiz
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 See morePhilippines 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 ASSESSMENT EXAM WILL GIVE YOU A HIGH PROBABILITY OF PASSING THE SAID EXAMINATIONS
Quiz Preview
- 2.
If the power of a sound wave is increased by a factor of 8 how many decibles is this?
-
3 dB
-
6 dB
-
9 dB
-
8 dB
Correct Answer
A. 9 dBExplanation
3dB means 2 times bigger 6dB is 3dB + 3dB. Therefore 6dB means 2 x 2 or 4 times bigger. Therefore 8 dB 2x2x2=8 or 3dB + 3dB + 3dB = 9 dB biggerRate this question:
-
- 3.
An ultrasound system is set at 0 dB and is transmitting at full intensity. What is the output power when the system is transmitting at 50% of full intensity?
-
-3dB
-
-50 dB
-
10 dB
-
3 dB
-
-10 dB
Correct Answer
A. -3dBExplanation
50% of full intensity is 1/2. 0 dB is full intensity so -3dB would be halfRate this question:
-
- 4.
An ultrasound system is set at 0 dB and is tranmitting at full intensity. What is the output power when the system is transmitting at 10% of full intensity?
-
- 3dB
-
-50 dB
-
40 dB
-
3 dB
-
-10 dB
Correct Answer
A. -10 dBExplanation
0 dB = full intensity 10% of full intensity would be negative or 1/10 or -10dBRate this question:
-
- 5.
What may be a unit of amplitude?
-
Cm
-
Hz
-
Msec
-
Watts
-
None of the above
Correct Answer
A. CmExplanation
The unit of amplitude is typically measured in centimeters (cm). Amplitude refers to the maximum displacement or distance from the equilibrium position in a wave or oscillation. It is a measure of the intensity or strength of the wave and is usually represented by the height or depth of the wave. Therefore, cm is a suitable unit to measure the amplitude as it represents the distance or displacement.Rate this question:
-
- 6.
A Doppler exam is performed with a 5MHz probe and a PRF of 15kHz. Which of these Doppler shifts will create aliasing?
-
2 kHz
-
2.5 kHz
-
7.5 kHz
-
14 kHz
Correct Answer
A. 14 kHz -
- 7.
According to the AIUM statement on bioeffects, there have been no confirmed bioeffects below intensities of __________ watts per square centimeter SPTA.
-
10
-
0.01
-
0.1
-
1
-
100
Correct Answer
A. 0.1Explanation
According to the AIUM statement on bioeffects, there have been no confirmed bioeffects below intensities of 0.1 watts per square centimeter SPTA. This means that any ultrasound exposure below this intensity level has not been found to cause any harmful effects on biological tissues or organisms.Rate this question:
-
- 8.
Which of the following is not a measure of area?
-
Square cm
-
Meters squared
-
Cubic meters
-
Feet x feet
Correct Answer
A. Cubic metersExplanation
Cubic meters is not a measure of area because it is a measure of volume. Area is a two-dimensional measurement that describes the amount of space an object or surface occupies, while volume is a three-dimensional measurement that describes the amount of space an object or substance occupies in all three dimensions. Therefore, cubic meters, which represents the volume of an object or substance, is not a measure of area.Rate this question:
-
- 9.
Which of the following is true of all waves?
-
They travel through a medium
-
All carry energy from one side to another
-
Their amplitudes do not change
-
They travel in a straight line
Correct Answer
A. All carry energy from one side to anotherExplanation
All waves carry energy from one side to another. This is a fundamental characteristic of waves, regardless of the type of wave or the medium through which they travel. As waves propagate, they transfer energy without necessarily transferring matter. This is evident in various types of waves, such as sound waves, light waves, and water waves. The energy carried by waves can be observed through the effects they produce, such as the movement of particles, the sensation of sound or light, or the displacement of water.Rate this question:
-
- 10.
Is Intensity an acoustic variable?
-
True
-
False
Correct Answer
A. FalseExplanation
Intensity is not an acoustic variable. Acoustic variables are properties that describe sound waves, such as frequency, wavelength, and amplitude. Intensity, on the other hand, is a measure of the power or energy carried by a sound wave per unit area. It is related to the amplitude of the wave but is not considered an acoustic variable itself. Therefore, the statement "Intensity is an acoustic variable" is false.Rate this question:
-
- 11.
A force is applied to a surface. If the force is tripled and the surface area over which the force is applied is also tripled, what is the new pressure?
-
Three times larger than the original
-
One third of the original
-
Six times more than the original
-
Unchanged
Correct Answer
A. UnchangedExplanation
When the force applied to a surface is tripled and the surface area over which the force is applied is also tripled, the new pressure remains unchanged. This is because pressure is defined as force divided by surface area. When both the force and surface area are tripled, the ratio between them remains the same, resulting in an unchanged pressure.Rate this question:
-
- 12.
Which of the following units are appropriate to describe the period of an acoustic wave? More than one answer maybe correct.
-
Minutes
-
Microseconds
-
Meters
-
Mm/us
Correct Answer(s)
A. Minutes
A. MicrosecondsExplanation
The period of an acoustic wave refers to the time it takes for one complete cycle of the wave. It is commonly measured in units of time, such as minutes and microseconds. These units are appropriate because they directly relate to the time it takes for the wave to complete a cycle. On the other hand, meters and mm/us are units of distance and velocity respectively, and do not directly describe the period of an acoustic wave.Rate this question:
-
- 13.
Two waves arrive at the same location and interfere. The resultant sound wave is smaller that either of the two original waves. What is this called?
-
Constructive interference
-
Angular interaction
-
Destructive interference
-
In-phase waves
Correct Answer
A. Destructive interferenceExplanation
Destructive interference occurs when two waves combine and their amplitudes cancel each other out, resulting in a smaller amplitude wave. This can happen when the peaks of one wave align with the troughs of the other wave, causing them to subtract from each other. In this case, the resultant sound wave is smaller than either of the two original waves, indicating destructive interference.Rate this question:
-
- 14.
Ultrasound is defined as a sound with a frequency of
-
Greater thatn 20,000 kHz
-
Less thatn 1 kHz
-
Greater than 10 MHz
-
Greater thatn 0.02 MHz
Correct Answer
A. Greater thatn 0.02 MHzExplanation
Ultrasound refers to sound waves with frequencies higher than the upper limit of human hearing, which is typically around 20,000 Hz. The answer "greater than 0.02 MHz" is correct because 0.02 MHz is equivalent to 20,000 Hz. Therefore, any frequency higher than 0.02 MHz would fall within the ultrasound range.Rate this question:
-
- 15.
Compare two sound waves. A and B. The frequency of a wave A is one-third that of wave B. How does the period of a wave A compare the period of wave B?
-
A is one-third as long a B
-
A is the same as wave B
-
A is three times as long as B
-
Cannot be determined
Correct Answer
A. A is three times as long as BExplanation
The period of a wave is the time it takes for one complete cycle of the wave to occur. Since the frequency of wave A is one-third that of wave B, it means that wave A takes three times longer to complete one cycle compared to wave B. Therefore, the period of wave A is three times as long as the period of wave B.Rate this question:
-
- 16.
Which of the following wave is ultrasonic and least useful in diagnotic ultrasound?
-
2 MHz
-
2000 Hz
-
24 kHz
-
7.5 MHz
Correct Answer
A. 24 kHzExplanation
Ultrasonic waves are sound waves with frequencies higher than the upper limit of human hearing, which is typically around 20 kHz. The given options include frequencies of 2 MHz, 2000 Hz, 24 kHz, and 7.5 MHz. Out of these options, 24 kHz is the only frequency that falls within the audible range of human hearing, making it least useful in diagnostic ultrasound. Diagnostic ultrasound typically utilizes frequencies in the range of 2 MHz to 15 MHz, which allows for better resolution and imaging of internal structures.Rate this question:
-
- 17.
True of False: With standard diagnostic imaging instrumentation, the sonographer has the ability to vary the amplitude of a sound wave produced by the transducer.
-
Ture
-
False
Correct Answer
A. TureExplanation
With standard diagnostic imaging instrumentation, the sonographer does have the ability to vary the amplitude of a sound wave produced by the transducer. This is typically done by adjusting the power output of the ultrasound machine, which controls the strength of the sound waves being emitted. By increasing or decreasing the amplitude, the sonographer can adjust the intensity of the sound waves and optimize the image quality for different types of examinations or patient conditions.Rate this question:
-
- 18.
The power of an ultrasound wave can be reported with which of the following units? More than one answer maybe correct.
-
Watts/square centimeter
-
DB/Cm
-
Watts
-
Kg/cm^2
Correct Answer
A. WattsExplanation
The power of an ultrasound wave can be reported in watts because watts is the unit of power. Power is the rate at which energy is transferred or the amount of energy per unit time. Since ultrasound waves carry energy, their power can be measured in watts.Rate this question:
-
- 19.
The maximum value of the density of an acoustic wave is 60 lb/in^2 while the minimum density is 20 lb/in^2. What is the amplitude of the wave?
-
20 lb/in^2
-
40 lb/in^2
-
60 lb/in^2
-
None of the above
Correct Answer
A. 20 lb/in^2Explanation
The amplitude of a wave may be calculated by subracting the minimum value of the acoustic variable from its maximum and then dividing that number in half. In this case, the maximum minus the minimum is 60-20=40. Half of 40 is 20Rate this question:
-
- 20.
The final amplitude of an acoustic wave is reduced to one-half of its original value. The final power is _______ the original power.
-
The same as
-
One-half
-
Double
-
None of the above
Correct Answer
A. None of the aboveExplanation
Changes in the power of a wave are proportional to changes in the waves amplitude squared. 1/2 squared = 1/4 (1/2 + 1/2 = 1/4) When 1/2 of the wave's original amplitude remains, then only 1/4 of the original power remains.Rate this question:
-
- 21.
The amplitude of an acoustic wave decreases from 27 pascals to 9 pascals. If the initial power in the wave was 27 watts, what is the wave final power?
-
3 watts
-
9 watts
-
1 watt
-
None of the above
Correct Answer
A. 3 wattsExplanation
The initial power in the wave is given as 27 watts. The amplitude of the wave decreases from 27 pascals to 9 pascals. The power of a wave is directly proportional to the square of its amplitude. Therefore, if the amplitude decreases by a factor of 3 (from 27 to 9), the power will decrease by a factor of 3 squared, which is 9. Therefore, the final power of the wave is 3 watts.Rate this question:
-
- 22.
What are the units of intensity?
-
Watts
-
Watts/cm
-
Watts/cm^2
-
DB
Correct Answer
A. Watts/cm^2Explanation
The units of intensity are typically measured in watts per square centimeter (watts/cm^2). Intensity refers to the amount of power per unit area, and watts/cm^2 is the appropriate unit to express this measurement. Watts alone do not account for the area over which the power is distributed, and watts/cm is not a valid unit for intensity as it does not include the squared term for area. dB (decibel) is a unit used to measure the relative intensity or power level of a sound or signal, but it is not the standard unit for intensity.Rate this question:
-
- 23.
If the power in an ultrasound beam is unchanged, while at the same time, the beam area doubles, then the beams's intensity
-
Doubles
-
Is halved
-
Is quartered
-
Remains the same
Correct Answer
A. Is halvedExplanation
When the beam area doubles while the power remains the same, the intensity of the ultrasound beam is halved. Intensity is defined as power divided by area, so if the area increases while the power remains constant, the intensity decreases. In this case, since the area doubles, the intensity is halved.Rate this question:
-
- 24.
What happens to an acoustic beams's intensity when the power in the beam increases by 25% while the cross sectional area of the beam remain the same?
-
It increased by 25%
-
Increases by 75%
-
It increases by 50%
-
Decreases by 25%
Correct Answer
A. It increased by 25%Explanation
When the power in the acoustic beam increases by 25% while the cross-sectional area remains the same, the intensity of the beam also increases by 25%. This is because intensity is directly proportional to power, and since the power has increased by 25%, the intensity will also increase by the same percentage.Rate this question:
-
- 25.
When the power in an acoustic beam is doubled and the cross sectional area of the beam is halved, then the intensity of the beam is
-
Doubled
-
Halved
-
Quartered
-
Four time larger
Correct Answer
A. Four time largerExplanation
When the power in an acoustic beam is doubled, it means that the amount of energy transmitted by the beam has increased. However, when the cross-sectional area of the beam is halved, it means that the same amount of energy is now spread over a smaller area. This results in a higher concentration of energy, leading to an increased intensity of the beam. Therefore, the intensity of the beam is four times larger than before.Rate this question:
-
- 26.
What determines the intensity of an ultrasound beam after it has traveled through the body?
-
The sound wave's source
-
The medium throught which the sound travels
-
Both A and B
-
Neither choice A nor B
Correct Answer
A. Both A and BExplanation
The intensity of an ultrasound beam after it has traveled through the body is determined by both the sound wave's source and the medium through which the sound travels. The source of the sound wave determines the initial intensity of the beam, while the medium through which it travels can affect the attenuation or absorption of the sound wave, leading to changes in intensity. Therefore, both factors play a role in determining the intensity of the ultrasound beam after it has passed through the body.Rate this question:
-
- 27.
What happens to the intensity of an ultrasound beam when the beam's cross-sectional area remains unchanged while the amplitude of the wave triples.
-
It triples
-
Increases ninefold
-
Remains the same
-
None of the above
Correct Answer
A. Increases ninefoldExplanation
When the amplitude of an ultrasound wave triples, the intensity of the ultrasound beam increases ninefold. This is because intensity is directly proportional to the square of the amplitude. Therefore, tripling the amplitude will result in the intensity being multiplied by 3^2, which equals 9.Rate this question:
-
- 28.
The wavelength of a cycle in an ultrasound wave can be reported with which units?
-
Units of time (seco, min, etc)
-
Units of distance (feet, cm)
-
Units of area (m^2, etc
-
Mm only
Correct Answer
A. Units of distance (feet, cm)Explanation
Ultrasound waves are a type of mechanical wave that travels through a medium, such as air or water. The wavelength of a wave is the distance between two consecutive points that are in phase, or the distance over which the wave completes one full cycle. In the case of ultrasound waves, the wavelength is measured in units of distance, such as feet or centimeters. This is because the wavelength represents the physical length of the wave, indicating how far the wave travels in one complete cycle. Therefore, the correct answer is units of distance (feet, cm).Rate this question:
-
- 29.
Which of the following terms best describes the relationship between frequency and wavelength for sound traveling in soft tissue?
-
Receiprocal
-
Direct
-
Related
-
Inverse
Correct Answer
A. InverseExplanation
The term "inverse" best describes the relationship between frequency and wavelength for sound traveling in soft tissue. Inverse means that as one variable increases, the other variable decreases, and vice versa. In this case, as the frequency of the sound waves increases, the wavelength decreases, and as the frequency decreases, the wavelength increases. This relationship is consistent with the behavior of sound waves in soft tissue.Rate this question:
-
- 30.
(TRUE OR FALSE) The power in the wave increases as it travels through the mass. (An ultrasound pulse propagates from soft tissue through a mass. Sound's propagation speed in the mass is 1575 m/s.)
-
True
-
False
Correct Answer
A. FalseExplanation
False. The power in the wave does not increase as it travels through the mass. The power of a wave is determined by the amplitude of the wave, which does not change as the wave propagates through a medium. Therefore, the power of the wave remains constant regardless of the medium it travels through.Rate this question:
-
- 31.
The frequency closest to the lower limits of US is:
-
19,000kHz
-
10,000 MHz
-
20,000 msec
-
15,000 Hz
Correct Answer
A. 15,000 HzExplanation
The frequency closest to the lower limits of the US is 15,000 Hz. This is because the given options include frequencies measured in kHz (kilohertz), MHz (megahertz), and msec (milliseconds), while the only option measured in Hz (hertz) is 15,000 Hz. Hertz is the standard unit for measuring frequency, and 15,000 Hz is the closest option to the lower limits of the US frequency range.Rate this question:
-
- 32.
Which of the following sound waves is ultrasonic and least useful in diagnostic imaging?
-
30 KHz
-
8 MHz
-
8,000 Hz
-
3,000 kHz
-
15 Hz
Correct Answer
A. 30 KHz -
- 33.
_____________ is the distance covered by one cycle
-
Pulse length
-
Wavelength
-
Cycle distance
-
Cycle duration
Correct Answer
A. WavelengthExplanation
Wavelength is the distance covered by one cycle. It is a measure of the length between two consecutive points of a wave that are in phase. In other words, it represents the distance between two identical points on a wave, such as two crests or two troughs. Therefore, wavelength is the correct answer as it accurately describes the distance covered by one complete cycle of a wave.Rate this question:
-
- 34.
If the frequency of US is increased from 0.77 MHz to 1.54 MHz, what happens to the wavelength?
-
Doubles
-
Halved
-
Remains the same
-
4 times greater
Correct Answer
A. HalvedExplanation
When the frequency of a wave increases, the wavelength decreases. This is because wavelength and frequency are inversely proportional to each other. Therefore, if the frequency of US is increased from 0.77 MHz to 1.54 MHz, the wavelength will be halved.Rate this question:
-
- 35.
If the frequency is decreased, the numberical value of the radial resolution is
-
Increased
-
Decreased
-
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 two closely spaced objects along the radial axis. A higher numerical value indicates a greater ability to differentiate between these objects, meaning that the resolution is improved. Therefore, when the frequency is decreased, the radial resolution is increased.Rate this question:
-
- 36.
If we increase the frequency, the near zone length is
-
Increased
-
Decreased
-
Unchanged
Correct Answer
A. IncreasedExplanation
When the frequency is increased, the near zone length is increased. The near zone refers to the region close to the source of the waves, where the wavefronts are not yet fully developed. As the frequency increases, the wavelength decreases, causing the wavefronts to become closer together. This results in a longer near zone length, as the waves take more time to fully develop and spread out. Therefore, increasing the frequency leads to an increased near zone length.Rate this question:
-
- 37.
The digital scan converter has a number of pixels assigned to each bit.
-
True
-
False
Correct Answer
A. FalseExplanation
The statement is false because a digital scan converter does not have pixels assigned to each bit. In a digital scan converter, pixels are assigned to each sample or data point, not to each bit. The number of pixels assigned to each sample or data point depends on the resolution of the scan converter.Rate this question:
-
- 38.
The more bits per pixel
-
The better the resolution
-
The higher the pizel density
-
The more shades of gray
-
Higer reliability
Correct Answer
A. The more shades of grayExplanation
The correct answer is "the more shades of gray." This means that as the number of bits per pixel increases, the resolution improves, resulting in a higher pixel density. This allows for a greater range of shades of gray to be displayed, which ultimately leads to higher image quality and more accurate representation of colors and details. Additionally, having more shades of gray enhances the reliability of the image by reducing banding or color artifacts.Rate this question:
-
- 39.
What is the actual time that an US machine is creating a pulse?
-
Duty factor
-
Period
-
Pulse period
-
Pulse duration
Correct Answer
A. Pulse durationExplanation
The pulse duration refers to the actual time that a US (ultrasound) machine is creating a pulse. It measures the length of time that the pulse is being emitted from the machine. The duty factor, period, and pulse period are not directly related to the actual time of pulse creation, making them incorrect options.Rate this question:
-
- 40.
The angle of incidence of an US beam is perpendicular to an interface. The two media have the same propagation speeds. What process cannot occur?
-
Refraction
-
Reflection
-
Transmission
-
Attenuation
Correct Answer
A. RefractionExplanation
When the angle of incidence of an ultrasound (US) beam is perpendicular to an interface and the two media have the same propagation speeds, the beam does not change direction as it passes through the interface. This means that there is no bending of the beam, which is the process of refraction. Therefore, the process that cannot occur in this scenario is refraction.Rate this question:
-
- 41.
What is the most typical Doppler shift measured clinically?
-
3.5 MHz
-
3,500,000 Hz
-
2kHz
-
1,000 kHz
-
20,000 Hz
Correct Answer
A. 2kHzExplanation
The most typical Doppler shift measured clinically is 2kHz. The Doppler effect is used in medical imaging to measure blood flow velocity and direction. In this context, a Doppler shift of 2kHz indicates a moderate flow velocity, which is commonly observed in clinical settings. This frequency shift is within the range that can be easily detected and measured using Doppler ultrasound techniques.Rate this question:
-
- 42.
If the frequency is decreased, the numerical value of the radial resolution is
-
Increased
-
Decreased
-
Unchanges
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 points in an image along the radial direction. A higher numerical value indicates a greater ability to distinguish between these points, which means that the radial resolution is increased. This is because decreasing the frequency results in a longer wavelength, which increases the distance between the points that can be resolved. Therefore, the radial resolution is directly affected by the frequency, and decreasing it leads to an increase in the numerical value of the radial resolution.Rate this question:
-
- 43.
The digital scan converter has a number of pixels assigned to each bit.
-
True
-
False
Correct Answer
A. FalseExplanation
The statement is false because the digital scan converter does not assign a specific number of pixels to each bit. A digital scan converter is a device that converts analog ultrasound signals into digital signals for display on a monitor. It does this by sampling the analog signal at regular intervals and assigning a binary value (bit) to each sample. The number of pixels in the display is determined by the resolution of the monitor, not by the number of bits in the digital scan converter.Rate this question:
-
- 44.
The more bits per pixel
-
The better the resolution
-
The higher the pixel density
-
The more shades of gray
-
Higher reliability
Correct Answer
A. The more shades of grayExplanation
The more bits per pixel refers to the number of bits used to represent each pixel in an image. A higher number of bits allows for a greater range of shades of gray to be displayed. This means that the image will have more levels of gray between black and white, resulting in a more detailed and realistic representation of the image. Therefore, the statement "the more shades of gray" is the correct answer as it accurately reflects the relationship between the number of bits per pixel and the quality of the image.Rate this question:
-
- 45.
Increasing the frequency increases the penetration depth.
-
True
-
False
Correct Answer
A. FalseExplanation
Increasing the frequency does not increase the penetration depth. In fact, higher frequencies result in decreased penetration depth. This is because higher frequencies have shorter wavelengths, which causes the energy to be absorbed and scattered more quickly as it interacts with the material being penetrated. As a result, lower frequencies are generally used when deeper penetration is required, such as in medical imaging or geological surveys.Rate this question:
-
- 46.
All of the following will improve temporal resolution EXCEPT
-
Increased line density
-
Single rather than multi focus
-
Higher frame rate
-
Shallower depth of view
-
Slower speed of sound in a medium
Correct Answer
A. Increased line densityExplanation
Increased line density refers to the number of scan lines used in an ultrasound image. The more scan lines there are, the higher the spatial resolution, which allows for better visualization of small structures. Therefore, increased line density will actually improve temporal resolution, as it allows for more frequent updates of the image over time. Therefore, the correct answer is "increased line density."Rate this question:
-
- 47.
What componet of a transducer changes electrial to mechanical and mechanical back to electrical engery?
-
PZT
-
Scan converter
-
Demodulator
-
Receiver
Correct Answer
A. PZTExplanation
PZT stands for lead zirconate titanate, which is a type of ceramic material commonly used in transducers. PZT is known for its piezoelectric properties, meaning it can convert electrical energy into mechanical energy and vice versa. In the context of a transducer, the PZT component is responsible for converting electrical signals into mechanical vibrations, which are then transmitted into the body or medium being imaged. These mechanical vibrations then bounce back and are converted back into electrical signals by the PZT, allowing the transducer to receive and detect echoes. Therefore, PZT is the component that changes electrical to mechanical and mechanical back to electrical energy in a transducer.Rate this question:
-
- 48.
What determines the frequency of a sound beam from a pulsed transducer?
-
Voltage applied to the PZT
-
PZT thickness
-
Frequency of the system
-
PRF of the pulser
Correct Answer
A. PZT thicknessExplanation
The frequency of a sound beam from a pulsed transducer is determined by the thickness of the PZT material. The PZT thickness affects the resonant frequency of the transducer, which in turn determines the frequency of the sound beam produced.Rate this question:
-
- 49.
What measures the output of the a transducer?
-
Receiver
-
Display
-
Hydrometer
-
Hydrophone
Correct Answer
A. HydrophoneExplanation
A hydrophone is a transducer that measures the output of a transducer. It is specifically designed to detect and measure underwater sound waves, making it the most suitable option for measuring the output of a transducer in an underwater environment. A receiver is a general term and does not specifically measure the output of a transducer. A display is used to show information but does not directly measure the output. A hydrometer is used to measure the specific gravity or density of a liquid, so it is not the correct option for measuring the output of a transducer.Rate this question:
-
Quiz Review Timeline (Updated): Mar 19, 2023 +
Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.
-
Current Version
-
Mar 19, 2023Quiz Edited by
ProProfs Editorial Team -
Jul 21, 2016Quiz Created by
Wantunine
- Accreditation Quizzes
- Advanced Placement Quizzes
- AED Exam Quizzes
- Art Exam Quizzes
- Banking Exam Quizzes
- BISC Exam Quizzes
- CADET Exam Quizzes
- CAE Listening Quizzes
- CHS Exam Quizzes
- Computer Operator Exam Quizzes
- Cosmetology State Board Quizzes
- CSAT Quizzes
- CVS Exam Quizzes
- EHR Exam Quizzes
- Engineering Exam Quizzes
- English Exam Quizzes
- Entrance Exam Quizzes
- FSHN Exam Quizzes
- GATE Quizzes
- IGCSE Quizzes
- IT Certification Quizzes
- JEE Exam Quizzes
- Management Exam Quizzes
- Ministerial Exam Quizzes
- OCA Exam Quizzes
- Online Test Quizzes
- Physical Exam Quizzes
- Police Exam Quizzes
- Practice Exam Quizzes
- Prelim Exam Quizzes
- Professional Certification Quizzes
- SNAP Quizzes
- SSC Quizzes
- SSC CGL Quizzes
- UGC NET Exam Quizzes
- UPSC Quizzes
Back to top