I85PREP
⌘ K
Maharashtra HSC Radiology Technician (Vocational) study materials
Maharashtra HSC Radiology Technician (Vocational) study resources

Maharashtra HSC Radiology Technician (Vocational) practice questions

One question at a time. Build your Maharashtra HSC Radiology Technician (Vocational) understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Radiation Physics and X-ray Equipment

In a diagnostic X-ray tube, what is the primary function of the rotating anode?

  1. To absorb scattered radiation leaving the collimator
  2. To rectify the high-voltage supply before it reaches the tube
  3. To focus the electron beam onto the cathode cup
  4. To spread the heat produced at the focal spot over a large area Correct answer.

Explanation

The rotating anode presents a moving track to the electron beam, so the heat generated by electron bombardment is distributed over a large circular area while the actual focal spot stays small. This allows higher tube currents and longer exposures without melting the target. Rectification and beam focusing are handled by other components.

Question 2Radiation Physics and X-ray Equipment

Which property of the X-ray beam increases when the tube potential (kVp) is raised while mA and exposure time are held constant?

  1. Penetrating power of the beam Correct answer.
  2. Thermal rating of the anode disc
  3. Size of the effective focal spot
  4. Number of electrons boiled off the filament

Explanation

kVp determines the kinetic energy gained by electrons before they strike the target, so a higher kVp produces photons of higher maximum and average energy, giving greater penetrating power. Filament emission depends on mA and filament heating, while focal spot size and anode heat capacity are fixed by tube design.

Question 3Radiation Physics and X-ray Equipment

A radiographer selects 100 mA and 0.2 s on the control panel. What is the resulting mAs?

  1. 0.5 mAs
  2. 20 mAs Correct answer.
  3. 200 mAs
  4. 500 mAs

Explanation

mAs is the product of tube current in milliamperes and exposure time in seconds, so 100 mA multiplied by 0.2 s equals 20 mAs. This value governs the total number of X-ray photons produced and hence the receptor dose for a given kVp.

Question 4Radiation Physics and X-ray Equipment

What is the normal direction of travel of X-ray photons in the useful beam after they leave the tube window?

  1. Back toward the cathode filament
  2. Straight ahead in a diverging cone from the focal spot Correct answer.
  3. Radially in all directions from the focal spot
  4. Parallel to the anode disc surface

Explanation

Useful X-ray photons emerge through the tube window and form a diverging cone whose apex is at the focal spot, which is why the beam covers a larger field at greater distances. Photons emitted in other directions are absorbed by the tube housing and its lead lining.

Question 5Radiation Physics and X-ray Equipment

Which material is most commonly used for the anode target in a general-purpose diagnostic X-ray tube?

  1. Tungsten Correct answer.
  2. Aluminium
  3. Copper
  4. Graphite

Explanation

Tungsten is chosen because of its very high melting point and high atomic number, which give good heat tolerance and efficient X-ray production. Copper is used for the anode stem to conduct heat away, and graphite may be used for the disc body, but the target track itself is tungsten.

Question 6Radiation Physics and X-ray Equipment

Which interaction in the patient produces the characteristic X-rays seen in a diagnostic spectrum?

  1. Photoelectric absorption in tissue Correct answer.
  2. Pair production in the collimator
  3. Compton scattering in the patient
  4. Bremsstrahlung in the anode

Explanation

Characteristic radiation arises when an incoming photon ejects an inner-shell electron and an outer-shell electron drops down to fill the vacancy, releasing a photon of energy equal to the shell difference. In the patient this follows photoelectric absorption, and the emitted characteristic photons are of low energy and mostly absorbed locally.

Question 7Radiation Physics and X-ray Equipment

An X-ray tube has a target angle of 12 degrees and a filament length that projects a 3 mm apparent focal spot. What is the actual focal spot length on the anode track?

  1. About 36 mm
  2. About 14 mm Correct answer.
  3. About 0.6 mm
  4. About 3 mm

Explanation

The apparent focal spot is the actual focal spot projected onto a plane perpendicular to the beam, so apparent size equals actual size multiplied by the sine of the target angle. Dividing 3 mm by sin 12 degrees, about 0.208, gives roughly 14 mm on the track, which is why a small target angle allows a small effective focal spot with a large heat-loading area.

Question 8Radiation Physics and X-ray Equipment

What is the main purpose of adding aluminium filtration to the X-ray beam?

  1. To increase the number of photons reaching the receptor
  2. To reduce the size of the focal spot
  3. To increase the maximum photon energy
  4. To remove low-energy photons that would only add patient dose Correct answer.

Explanation

Filtration preferentially absorbs low-energy photons, which are too weak to penetrate to the receptor but are absorbed in superficial tissue and add to skin dose. Removing them hardens the beam, raising the average energy without changing the peak kVp.

Question 9Radiation Physics and X-ray Equipment

Which device is placed between the patient and the image receptor to reduce the amount of scattered radiation reaching the receptor?

  1. Antiscatter grid Correct answer.
  2. Ionisation chamber
  3. Compensating filter
  4. Light-beam diaphragm

Explanation

An antiscatter grid consists of alternating lead strips and radiolucent spacers that absorb photons travelling at an angle while transmitting those aligned with the primary beam. This improves contrast by preventing scattered photons from fogging the image.

Question 10Radiation Physics and X-ray Equipment

In a single-phase, full-wave rectified generator, how many pulses of X-ray production occur per second at a 50 Hz supply?

  1. 50
  2. 100 Correct answer.
  3. 25
  4. 150

Explanation

Full-wave rectification converts both halves of the alternating supply cycle into usable tube current, so the number of pulses equals twice the supply frequency. At 50 Hz this gives 100 pulses per second, which is why such generators are described as 100-pulse units.

Question 11Radiation Physics and X-ray Equipment

What is the principal advantage of a three-phase, twelve-pulse generator over a single-phase generator?

  1. Reduction of the focal spot size
  2. Removal of the need for collimation
  3. Higher and more constant tube voltage throughout the exposure Correct answer.
  4. Elimination of the need for beam filtration

Explanation

Three-phase generators produce a nearly constant potential across the tube, so photons have a higher average energy for the same kVp setting and more photons are produced per mAs. This shortens exposure times and reduces patient dose compared with the voltage ripple of single-phase units.

Question 12Radiation Physics and X-ray Equipment

Which statement correctly describes the inverse square law as applied to X-ray beam intensity?

  1. Intensity is inversely proportional to distance
  2. Intensity is inversely proportional to the square of the distance Correct answer.
  3. Intensity is directly proportional to the square of the distance
  4. Intensity is independent of distance

Explanation

Because the beam diverges from a point source, the same number of photons spreads over an area that grows with the square of the distance, so intensity falls as one over distance squared. Doubling the distance therefore reduces intensity to one quarter, which is why source-to-receptor distance changes must be compensated by mAs.

Question 13Radiation Physics and X-ray Equipment

An exposure of 40 mAs at 100 cm source-to-receptor distance produces acceptable density. What mAs is needed at 200 cm to maintain the same receptor exposure?

  1. 10 mAs
  2. 20 mAs
  3. 80 mAs
  4. 160 mAs Correct answer.

Explanation

Doubling the distance reduces intensity to one quarter, so the mAs must be increased fourfold to compensate. Multiplying 40 mAs by four gives 160 mAs, following the inverse square law relationship between distance and required exposure.

Question 14Radiation Physics and X-ray Equipment

What is the purpose of the cathode focusing cup in an X-ray tube?

  1. To confine the electron beam to a small area on the anode Correct answer.
  2. To cool the anode disc during exposure
  3. To rectify the high voltage applied to the tube
  4. To filter low-energy photons from the beam

Explanation

The focusing cup surrounds the filament and is maintained at a negative potential relative to it, so electrostatic repulsion shapes the electron cloud into a narrow beam directed at the focal spot. This constrains the area of the anode that is bombarded and hence the size of the focal spot.

Question 15Radiation Physics and X-ray Equipment

A radiographer must select the small focal spot for a routine extremity examination. Which benefit is expected?

  1. Shorter exposure time for the same mAs
  2. Higher contrast from reduced Compton scatter
  3. Sharper geometric detail in the image Correct answer.
  4. Greater anode heat capacity

Explanation

A smaller focal spot reduces geometric unsharpness because the penumbra around each structure is narrower, improving spatial resolution. The trade-off is a lower maximum mA, since the heat load is concentrated on a smaller area of the track.

Question 16Radiation Physics and X-ray Equipment

What does the anode heel effect describe?

  1. Variation in beam intensity across the field because photons emerge through different thicknesses of target Correct answer.
  2. Reduction in focal spot size at high mA
  3. Loss of tube vacuum at high kVp
  4. Increased scatter production at the anode side of the field

Explanation

Photons emitted toward the anode side must travel through more target material and are therefore more attenuated, so intensity is lower on that side of the beam. The effect is greater for small target angles and short source-to-receptor distances, and it is why thicker body parts are placed under the cathode side.

Question 17Radiation Physics and X-ray Equipment

Which of the following best explains why the X-ray spectrum has a maximum photon energy equal to the kVp?

  1. Some electrons transfer their full kinetic energy to a single photon in one interaction Correct answer.
  2. Characteristic photons always exceed the kVp energy
  3. All electrons lose their entire energy in a single bremsstrahlung event
  4. The filter removes all photons above the kVp

Explanation

In bremsstrahlung the electron can lose any fraction of its kinetic energy, and in the limiting case it gives up all of it to one photon. That photon therefore has energy equal to the electron's kinetic energy, which is set by the kVp, giving the spectrum its sharp high-energy cutoff.

Question 18Radiation Physics and X-ray Equipment

What is the function of the automatic exposure control (AEC) system in a diagnostic X-ray unit?

  1. To terminate the exposure when a preset receptor dose is reached Correct answer.
  2. To rotate the anode at the correct speed during exposure
  3. To adjust the collimator field size to the receptor
  4. To select the kVp automatically from the patient's thickness

Explanation

The AEC uses ionisation chambers or solid-state detectors in front of the receptor to monitor radiation reaching it and cuts off the exposure once a predetermined dose is reached. This gives consistent receptor exposure despite variations in patient thickness, provided the correct chambers are selected.

Question 19Radiation Physics and X-ray Equipment

A lead diaphragm collimator is used on a mobile X-ray unit. What is its main radiological purpose?

  1. To measure the exposure reaching the image receptor
  2. To increase the beam's average photon energy
  3. To increase the anode rotation speed
  4. To restrict the irradiated field and reduce scatter and patient dose Correct answer.

Explanation

Collimation limits the primary beam to the area of clinical interest, which reduces the volume of tissue irradiated and the amount of scattered radiation reaching the receptor. It also improves image contrast and reduces unnecessary patient dose.

Question 20Radiation Physics and X-ray Equipment

Which factor directly determines the maximum energy of photons produced in an X-ray tube?

  1. Tube potential difference in kVp Correct answer.
  2. Exposure time in seconds
  3. Focal spot size
  4. Tube current in mA

Explanation

The kinetic energy gained by an electron is the product of its charge and the accelerating potential difference, so the kVp sets the upper limit of photon energy. mA and time affect the number of photons produced, while focal spot size affects image sharpness.

Question 21Radiation Physics and X-ray Equipment

Why is the X-ray tube envelope evacuated to a high vacuum?

  1. To increase the atomic number of the target material
  2. To prevent electrons from colliding with gas molecules before reaching the anode Correct answer.
  3. To allow visible light to pass through the tube window
  4. To reduce the heat capacity of the anode

Explanation

A high vacuum removes gas molecules that would otherwise scatter or absorb electrons travelling from filament to target, and it prevents ionisation and arcing at high voltage. This allows a controlled electron beam and stable operation at diagnostic potentials.

Question 22Radiation Physics and X-ray Equipment

Which statement about the line-focus principle is correct?

  1. It increases the apparent focal spot relative to the actual focal spot
  2. It applies only to stationary anode tubes
  3. It requires the target angle to be 90 degrees
  4. It allows a large actual focal spot with a small apparent focal spot Correct answer.

Explanation

By angling the target, the actual area bombarded by electrons is projected onto a smaller apparent area in the direction of the beam. This spreads heat over a larger track while preserving fine detail, and it operates in rotating and stationary anode tubes alike.

Question 23Radiation Physics and X-ray Equipment

A radiographer needs to maintain receptor exposure after increasing the source-to-receptor distance from 90 cm to 180 cm. Which adjustment is correct?

  1. Halve the mAs
  2. Double the kVp
  3. Reduce the field size by half
  4. Quadruple the mAs Correct answer.

Explanation

Doubling the distance reduces beam intensity to one quarter under the inverse square law, so the mAs must be increased fourfold to restore receptor exposure. kVp is kept constant because changing it would also alter contrast and penetrating power.

Question 24Radiation Physics and X-ray Equipment

What is the primary reason for using a rotating anode rather than a stationary anode in a high-output diagnostic tube?

  1. The rotating anode distributes heat over a larger track, permitting higher mA Correct answer.
  2. The rotating anode allows the tube to operate without filtration
  3. The rotating anode removes the need for a focusing cup
  4. The rotating anode produces higher-energy photons

Explanation

Rotation spreads the electron bombardment over a circular track, greatly increasing the area available for heat dissipation while keeping the focal spot small. This permits higher tube currents and longer exposures without damaging the target, which a stationary anode cannot sustain.

Question 25Radiation Physics and X-ray Equipment

A single-phase generator with a 10-degree ripple produces a kVp of 80. Which statement about the resulting spectrum is most accurate?

  1. The spectrum contains no bremsstrahlung component
  2. The spectrum has a higher maximum energy than 80 keV
  3. The spectrum contains more low-energy photons than a constant-potential unit at the same kVp Correct answer.
  4. The spectrum is identical to that of a constant-potential generator at 80 kVp

Explanation

Voltage ripple means the tube potential falls below the peak for part of each cycle, so many electrons are accelerated by less than the kVp and produce lower-energy photons. Compared with constant potential, the spectrum is shifted toward lower energies, giving a lower average energy and more low-energy photons for the same kVp setting.

Question 26Radiation Physics and X-ray Equipment

During a radiograph exposure, the radiographer hears a loud continuous hum from the single-phase generator and notices the mAs reading is far higher than set. Which waveform description fits a single-phase, full-wave rectified generator operating correctly?

  1. A bidirectional waveform with 50 pulses per second at 50 Hz supply
  2. A constant potential waveform with negligible ripple
  3. A unidirectional waveform with 25 pulses per second at 50 Hz supply
  4. A unidirectional waveform with 100 pulses per second at 50 Hz supply Correct answer.

Explanation

Full-wave rectification converts both halves of the AC cycle into useful unidirectional pulses, so a 50 Hz supply yields 2 pulses per cycle and therefore 100 pulses per second. The ripple remains 100 percent, and the waveform never reverses polarity.

Question 27Radiation Physics and X-ray Equipment

A radiographer selects a higher mA station while keeping exposure time and kVp unchanged. What is the direct effect on the quantity and quality of the emitted X-ray beam?

  1. Photon quantity increases and the maximum photon energy also increases
  2. Photon quantity increases while the maximum photon energy stays the same Correct answer.
  3. Photon quantity stays the same while the maximum photon energy rises
  4. Photon quantity decreases and the maximum photon energy falls

Explanation

mA controls the tube current, so a higher mA increases the number of electrons hitting the target and therefore the number of X-ray photons produced. The kVp, not the mA, determines the kinetic energy of the electrons and hence the maximum photon energy, so the spectrum's high-energy limit is unchanged.

Question 28Radiation Physics and X-ray Equipment

A fixed X-ray installation uses a tube with total filtration equivalent to 2.5 mm aluminium. The radiographer adds a 0.5 mm copper filter for a high-kVp chest technique. What is the expected effect on the beam?

  1. All photons are removed equally, leaving the mean energy unchanged
  2. Low-energy photons are preferentially removed, lowering the mean energy of the remaining beam
  3. Low-energy photons are preferentially removed, raising the mean energy of the remaining beam Correct answer.
  4. High-energy photons are preferentially removed, lowering the beam's penetrating power

Explanation

Added filtration absorbs low-energy photons more strongly than high-energy ones, so the beam becomes harder and its mean energy rises while overall intensity falls. Copper is effective at high kVp because it removes the low-energy tail without excessive reduction of useful photons.

Question 29Radiation Physics and X-ray Equipment

An X-ray tube's anode disc has a focal track radius of 75 mm and rotates at 3000 revolutions per minute. Compared with a stationary anode under identical exposure factors, what is the principal operational consequence?

  1. The emitted beam becomes more penetrating because of the higher anode speed
  2. Heat is concentrated in one spot, increasing heat capacity per unit area
  3. The effective focal spot size becomes smaller, improving spatial resolution
  4. Heat is spread over a longer focal track, allowing higher mA or longer exposures Correct answer.

Explanation

A rotating anode moves the electron beam along a circular focal track, so heat is distributed over a much larger area than a stationary focal spot. This raises the permissible tube loading and allows higher mA and longer exposures without melting the target.

Question 30Radiation Physics and X-ray Equipment

A radiographer must reduce the radiation dose to a patient's gonads during an AP pelvis while keeping sufficient diagnostic information. Which combination best achieves this?

  1. Open collimation to include the whole abdomen with a gonad shield
  2. Increase filtration to 5 mm aluminium and remove all collimation
  3. Reduce mAs to one quarter and open collimation to compensate
  4. Close collimation to the region of interest with a gonad shield and appropriate kVp Correct answer.

Explanation

Tight collimation reduces the volume of tissue irradiated and the amount of scattered radiation produced, while a gonad shield attenuates the primary beam over the radiosensitive organs. Selecting an appropriate kVp maintains adequate penetration so that repeat exposures are avoided.

Question 31Radiographic Techniques and Positioning

In a PA projection of the chest, which anatomical structure is positioned closest to the image receptor?

  1. The anterior ribs
  2. The heart
  3. The sternum
  4. The thoracic spine Correct answer.

Explanation

In a PA chest projection the patient faces the upright detector, so the posterior structures lie nearest the receptor. The thoracic spine therefore sits closest to the image receptor, while the sternum and heart are anterior. This arrangement minimises cardiac magnification and is the standard reason PA is preferred over AP for chest radiography.

Question 32Radiographic Techniques and Positioning

A radiographer is asked to demonstrate the acromioclavicular joint with minimal overlap of the joint space. Which projection is best suited to this request?

  1. AP shoulder with the arm internally rotated
  2. Lateral scapula projection
  3. Axial (superoinferior) shoulder projection
  4. AP projection with a 10-15 degree cephalic tube angulation Correct answer.

Explanation

A cephalic angulation of the tube compensates for the sloping orientation of the AC joint and projects the joint space clear of the acromion and clavicle. The internally rotated AP shoulder mainly demonstrates the humeral head in profile. A lateral scapula or axial shoulder does not isolate the AC joint for this purpose.

Question 33Radiographic Techniques and Positioning

For a standard lateral projection of the lumbar spine, the patient is positioned so that the vertebral column is parallel to the image receptor. What is the main purpose of this positioning?

  1. To show the spinous processes in profile Correct answer.
  2. To demonstrate the intervertebral foramina open
  3. To visualise the pedicles as circles
  4. To reduce the radiation dose to the gonads

Explanation

In a true lateral lumbar spine the vertebral bodies are seen in profile and the spinous processes project posteriorly clear of the vertebral bodies. The intervertebral foramina are demonstrated open on oblique projections, not on a true lateral. Pedicles appear as circles on an AP view where they are seen end-on.

Question 34Radiographic Techniques and Positioning

Which projection of the hand best demonstrates the first carpometacarpal joint space free of overlap?

  1. Ball-catcher's (Norgaard) projection
  2. PA hand
  3. Lateral hand
  4. Robert's (AP oblique) projection Correct answer.

Explanation

Robert's projection places the hand in a fully pronated AP oblique position, which opens the first carpometacarpal joint and demonstrates it free of superimposition. The standard PA hand shows the joint with overlap from the trapezium and first metacarpal base. Lateral and ball-catcher's views do not isolate this joint.

Question 35Radiographic Techniques and Positioning

A patient is unable to stand for an erect abdominal radiograph. Which change in positioning best preserves diagnostic value while accommodating the patient?

  1. Perform the examination supine with a vertical beam and the cassette under the patient
  2. Perform the examination supine with a horizontal beam directed at the anterior abdomen Correct answer.
  3. Perform the examination prone with the cassette under the abdomen
  4. Defer the examination until the patient can stand

Explanation

When the patient cannot stand, a supine or decubitus view with a horizontal beam is used so that free air and fluid levels can still be detected. A vertical-beam supine film is a routine abdominal view but will not demonstrate air-fluid levels. Deferring the examination may delay diagnosis of a serious condition.

Question 36Radiographic Techniques and Positioning

In an AP projection of the knee, the patient is positioned with the patella directed straight forward. Which structure is demonstrated in profile at the medial and lateral margins of the joint?

  1. The patella
  2. The tibial plateau
  3. The fibular head
  4. The femoral condyles Correct answer.

Explanation

With the knee in a true AP position, the femoral condyles are seen in profile along the medial and lateral margins of the joint. The patella is superimposed over the distal femur and is not seen in profile. The tibial plateau is demonstrated as a horizontal line, and the fibular head lies laterally.

Question 37Radiographic Techniques and Positioning

A radiographer is performing a lateral projection of the cervical spine and the patient's shoulders obscure the lower cervical vertebrae. Which action is most appropriate?

  1. Ask the patient to depress the shoulders and use a caudal tube angulation if required Correct answer.
  2. Increase the exposure factors to penetrate the shoulders
  3. Use a grid with a higher grid ratio
  4. Repeat the projection with the patient supine

Explanation

Depressing the shoulders pulls the soft tissue clear of the lower cervical spine, and a caudal tube angulation may help project the lower vertebrae clear of the shoulders. Increasing exposure alone does not resolve the anatomical superimposition. Changing the patient's position to supine or altering the grid does not address shoulder overlap.

Question 38Radiographic Techniques and Positioning

Which projection of the pelvis is best for demonstrating the sacroiliac joints?

  1. Lateral pelvis
  2. Inlet view of the pelvis
  3. AP oblique (frog-leg) pelvis Correct answer.
  4. AP pelvis

Explanation

An AP oblique projection of the pelvis rotates the sacroiliac joints clear of the superimposed iliac bones, demonstrating them more clearly. The standard AP pelvis shows the joints with significant overlap. Lateral and inlet views are used for other purposes such as assessing the pelvic ring and sacrum.

Question 39Radiographic Techniques and Positioning

During a PA projection of the paranasal sinuses, the patient's head is positioned with the orbitomeatal line perpendicular to the image receptor. Which sinuses are best demonstrated by this positioning?

  1. Frontal and ethmoidal sinuses Correct answer.
  2. Mastoid air cells
  3. Sphenoidal sinuses
  4. Maxillary sinuses

Explanation

With the orbitomeatal line perpendicular to the receptor, the frontal and ethmoidal sinuses are projected clear of the petrous ridges. The maxillary sinuses are better shown with the head tilted back so the petrous ridges fall below the sinus floor. The sphenoidal sinuses and mastoid air cells require different projections.

Question 40Radiographic Techniques and Positioning

Which of the following is the standard method for demonstrating the temporomandibular joint in an open-mouth position?

  1. Submentovertex projection
  2. Panoramic radiograph
  3. Transcranial projection Correct answer.
  4. Townes projection

Explanation

The transcranial projection is a conventional method for demonstrating the temporomandibular joint, and it can be performed with the mouth open or closed. A panoramic radiograph shows the joint but is not primarily designed to demonstrate the open-mouth position. The submentovertex and Townes projections are used for other skull and facial structures.

Question 41Radiographic Techniques and Positioning

A radiographer needs to demonstrate the patellofemoral joint space. Which projection is most appropriate?

  1. Skyline (tangential) projection of the patella Correct answer.
  2. Medial oblique knee
  3. Lateral knee with the knee flexed at 30 degrees
  4. AP knee with the patient weight-bearing

Explanation

The skyline or tangential projection directs the beam along the patellofemoral joint space, demonstrating the patella and femoral trochlea free of overlap. A lateral knee shows the patella in profile but not the joint space. AP and oblique knee projections superimpose the patella on the femur.

Question 42Radiographic Techniques and Positioning

In a lateral projection of the thoracic spine, which breathing technique is commonly used to improve visualisation?

  1. Suspended full expiration
  2. Suspended full inspiration
  3. Rapid breathing during exposure
  4. Slow shallow breathing during exposure Correct answer.

Explanation

Slow shallow breathing during the exposure blurs the ribs and lung markings across the thoracic vertebrae, improving visualisation of the spine. Suspended inspiration or expiration does not blur the ribs and leaves them superimposed. Rapid breathing may cause excessive motion blur.

Question 43Radiographic Techniques and Positioning

Which projection of the foot best demonstrates the metatarsophalangeal joints free of overlap?

  1. Medial oblique projection Correct answer.
  2. Weight-bearing dorsoplantar projection
  3. Dorsoplantar projection
  4. Lateral projection

Explanation

A medial oblique projection of the foot separates the metatarsals and opens the metatarsophalangeal joints, demonstrating them free of overlap. The dorsoplantar and weight-bearing dorsoplantar projections superimpose the metatarsals. The lateral foot superimposes all metatarsals on each other.

Question 44Radiographic Techniques and Positioning

A radiographer is performing an AP projection of the shoulder. To demonstrate the greater tuberosity in profile, how should the humerus be positioned?

  1. Neutral position
  2. Externally rotated Correct answer.
  3. Abducted to 90 degrees
  4. Internally rotated

Explanation

External rotation of the humerus brings the greater tuberosity into profile on the lateral aspect of the humeral head. Internal rotation profiles the lesser tuberosity instead. Neutral and abducted positions do not reliably profile the greater tuberosity.

Question 45Radiographic Techniques and Positioning

Which projection is used to demonstrate the optic foramina?

  1. Rhese (optic foramen) projection Correct answer.
  2. Water's projection
  3. Towne's projection
  4. Caldwell projection

Explanation

The Rhese projection positions the optic foramen in the centre of the orbit and demonstrates it within the lower outer quadrant of the orbit. Water's and Caldwell projections demonstrate the sinuses and facial bones, while Towne's demonstrates the occipital region and foramen magnum.

Question 46Radiographic Techniques and Positioning

In a lateral projection of the sacrum, which positioning adjustment helps separate the sacrum from the superimposed iliac bones?

  1. Flexing the patient's hips
  2. Angulating the tube 15 degrees caudally
  3. Rotating the patient slightly Correct answer.
  4. Using a larger field of view

Explanation

A slight rotation of the patient separates the sacrum from the iliac bones so the sacral profile is seen more clearly. Tube angulation and hip flexion do not address the superimposition of the iliac bones. A larger field of view does not change the anatomical overlap.

Question 47Radiographic Techniques and Positioning

Which of the following best describes the purpose of a decubitus abdominal projection?

  1. To demonstrate the bony pelvis
  2. To reduce the radiation dose to the patient
  3. To demonstrate air-fluid levels in a patient who cannot stand Correct answer.
  4. To assess the position of the diaphragm in expiration

Explanation

A decubitus projection uses a horizontal beam with the patient lying on one side, allowing free air and air-fluid levels to be demonstrated in a patient who cannot stand. It is not designed to assess the bony pelvis or diaphragm position, and it does not reduce dose.

Question 48Radiographic Techniques and Positioning

A patient with a suspected hip fracture is positioned for an AP hip projection. Which positioning aid best ensures the femoral neck is demonstrated without foreshortening?

  1. Internal rotation of the foot by about 15 degrees Correct answer.
  2. External rotation of the foot by about 30 degrees
  3. Flexion of the hip to 90 degrees
  4. Abduction of the limb to 45 degrees

Explanation

Internal rotation of the foot brings the femoral neck parallel to the image receptor, demonstrating it without foreshortening. External rotation foreshortens the neck and profiles the greater trochanter. Abduction and flexion change the projection and do not correct foreshortening.

Question 49Radiographic Techniques and Positioning

Which projection of the wrist is used to demonstrate the scaphoid with the wrist in ulnar deviation?

  1. Lateral wrist
  2. AP oblique wrist
  3. PA wrist with ulnar deviation Correct answer.
  4. Carpal tunnel projection

Explanation

Ulnar deviation extends the scaphoid and brings it into a more elongated profile, demonstrating it free of overlap on a PA wrist projection. The lateral and oblique wrist projections do not provide this elongation. The carpal tunnel projection is designed to demonstrate the carpal tunnel rather than the scaphoid.

Question 50Radiographic Techniques and Positioning

In an AP projection of the ankle, the foot is positioned in slight internal rotation. What is the purpose of this rotation?

  1. To demonstrate the calcaneus in profile
  2. To demonstrate the lateral malleolus free of the tibia
  3. To open the tibiotalar joint space Correct answer.
  4. To demonstrate the base of the fifth metatarsal

Explanation

Slight internal rotation of the foot in an AP ankle projection opens the tibiotalar joint space by aligning the joint surfaces parallel to the beam. It does not demonstrate the lateral malleolus free of the tibia or the calcaneus in profile. The base of the fifth metatarsal is demonstrated on foot projections.