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15 B, Hill Crest Residency

Ramakrishna Mutt Road

Ooty – 643001

January 15, 2022

The Principal

SNS College of Arts and Science

R K Puram

Ooty – 643006

Subject: Request Letter for Transfer Certificate

Dear Sir,

I am Adithya Srinivasan, a second-year student of B Sc Computer Science. I am writing this letter to request for a TC as my family is relocating to Coimbatore next month. I have already applied to colleges in Coimbatore, and I have been shortlisted as well. I would be required to submit the Transfer Certificate as soon as I receive a confirmation of my admission.

I request you to kindly consider my request and issue me a Transfer Certificate by the end of this month.

Thank you

Yours sincerely,

Signature of the sender

ADITHYA SRINIVASAN C

B Sc Computer Science II Year

Roll No. xxxx

1. The van der Waals’ equation. 

2. \(P+\Big(\frac{an^2}{V^2}\Big)\)(V — nb) = nRT 

where p – pressure, V – volume, n – no. of moles of the gas, a & b – van der Waals’ constants, R – universal gas constant and T – absolute temperature.

Z = R + jX = 6 + j8 = 10 ∠ 53.1°

It shows that current lags behind the applied voltage by 53.1°. Let V be taken as the reference quantity. Then v = (141.4 x √2) × sinωt = 200 sin ω t; i = (V_m/Zsinωt) − 30° = 20sin(ωt − 53.1°). 

(i) When the voltage is + 100 V and increasing; 100 = 200 sin ωt; sin ωt = 0.5 ; ω t = 30° 

At this instant, the current is given by i = 20 sin (30° − 53.1°) = − 20 sin 23.1° = − 7.847A.

(ii) drop across resistor = iR = − 7.847 × 6 = − 47 V.

(iii) Let us first find the equation of the voltage drop V_L across the inductive reactance. Maximum value of the voltage drop = I_mX_L = 20 × 8 = 160 V. It leads the current by 90°. Since current itself lags the applied voltage by 53.1°, the reactive voltage drop across the applied voltage by (90° − 53.1°) = 36.9°.

Hence, the equation of this voltage drop at the instant when ωt = 30° is

V_L = 160 sin(30° + 36.9°) = 160sin66.9° = 147.2V.

a. False. This results in more photoelectric absorption of the low-energy photons and therefore a reduction in the total energy per unit area passing through per unit time (beam intensity). 

b. True. Increasing the tube current results in increased numbers of electrons colliding with the target and therefore increasing the number of photons produced by both characteristic and Bremsstrahlung radiation. 

c. False. According to the inverse square law, beam intensity is inversely proportional to distance, as long as the radiation is from a point source and there is no absorption or scatter. 

d. True. Increasing the tube voltage results in an increase in the kinetic energy of electrons colliding with the target, resulting in an increase in photon beam intensity. 

e. False.

a. False. The number of electrons (tube current, mA) is controlled by the filament voltage. 

b. True. Although the number of filament electrons produced are unaffected by kV, the number of photons created in the target per electron decreases with a reduction in kV, e.g. number of photos at 10 mA, 90 kV > number of photons at 10 mA, 60 kV.

c. True. Beam intensity is the total amount of energy per unit area passing through per unit time. 

d. False. HVL reduces as beam penetration reduces. 

e. True. Due to a reduction in photon energy, the rate of photoelectric attenuation increases more than the rate of Compton attenuation.

Here, N = 10/I = I/0.5 = 2; Q = ω₀L/R = 2π × 75 × 9/50 = 84.8 

Let f₁ and f₂ be the frequencies at which current falls to half its maximum value at resonance frequency. Then, as seen from above

f₀/f₁ - f₁/f₀ = √3/Q or 75/f1 - f1/75 = √3/84.4

or (752 - f²₁)/75f₁  = 0.02 or f²₁ + 1.5f₁ - 5625 = 0 or f₁ = 74.25 Hz

Also f₂/75 - 75/f₂ = √3/84.4 or f²₂ - 1.5f₂ - 5625 = 0 or f₂ = 75.75Hz.

The applied voltage lags behind the current which, in other words, means that current leads the voltage. 

(i) ∴ p.f. is leading 

(ii) p.f. = cos φ = cos 30° = 0.866 (lead) 

(iii) Circuit is capacitive 

(iv) Active power = VI cos φ = 100 × 10 × 0.866 = 866 W 

Reactive power = VI sin φ = 100 × 10 × 0.5 = 500 VAR (lead)

or VAR = √((VA)² - W²) = √((100 x 10)² - 866²) = 500(lead)

a. True. IRR99 Regulation 13: every employer must consult a Radiation Protection Adviser on compliance with the regulations. 

b. True. IRR99 Regulation 17: the employer should provide the local rules describing the controlled area and the working practice for these areas. 

c. False. Quality assurance of radiation equipment is a requirement of the IRR99, not procedures and protocols. The quality assurance requirements of the Ionising Radiation (Medical Exposure) Regulations 2000 (IRMER) are concerned with procedures and protocols and periodic audit of compliance. 

d. False. To have expert advice and ensure that an MPE is involved with medical exposure is a part of the IRMER. 

e. True. Incidents involving equipment faults are covered under the IRR99.

a. False. The dose to the patient should be as low as reasonably achievable (ALARA principle) in order to make sure that the patient receives maximum benefit with minimum risk. 

b. True. Dose limits apply to workers, such that a dose in excess of these values is deemed not to be justified, no matter how great the benefit. 

c. False. The IRR99 is not concerned with radiation protection of patients. Patient protection was introduced into UK law in the Ionising Radiation (Medical Exposure) Regulations 2000 (IRMER) 

d. True. 

e. True.

a. False. Approximately 80% of X-rays emitted by a tube are Bremsstrahlung radiation.

b. True. 

c. True. The kinetic energy of filament electrons needs to exceed the binding energy of K-shell electrons (70 keV for tungsten). 

d. False. For tungsten: K_α radiation = K-shell binding energy (70 keV) – L-shell binding energy (12 keV) = 58 keV. K_β radiation = K-shell binding energy (70 keV) – M-shell binding energy (2 keV) = 68 keV. 

e. False. As the binding energy of L-shell electrons is equal to 12 keV, the photon energy produced when an electron from an outer shell occupies the gap is too small to leave the tube (i.e. less than 10 keV). 

a. False. X-ray absorption and scatter are stochastic processes. 

b. True. Attenuation is the total number of photons that have been removed, as a result of scatter or absorption, from the primary beam after passing through the attenuated material. 

c. False. The X-rays transmitted through the patient form the primary image, while the scattered X-rays obscure it. 

d. False. The HVL is the thickness of a material that will reduce the intensity of a mono-energetic beam to half its value, not the number of photons. 

e. True. The LAC is the probability that a photon interacts (absorbed or scattered) per unit length it travels in a specific material. Hence, the greater the LAC, the lower the HVL of the material.

a. True. Characteristic radiation results in the end production of photons that have the same energy, constituting a line spectrum, i.e. the difference in binding energies between the two shells, which is constant for a given material. 

b. False. It mainly involves filament electrons dislodging K-shell electrons.

c. False. The tube voltage and filament voltage increase the rate of production of photons, but do not influence the photon energy, which is dependent on the atomic number of the target material. 

d. True. 

e. True. An increase in atomic number results in an increase in the binding energy of electrons and hence K-radiation. 

Correct option: d) In-charge Sanitation Inspector

Correct option: b) A- IV, B-III, C- II, D- I

a. True. The dose to an organ increases in proportion to the following: 

i. The effective half-life 

ii. The fraction taken up by the organ 

iii. The activity administered to the patient 

iv. The energy of alpha and beta radiation emitted. 

b. False. 

c. False. Breast-feeding should be avoided for 24 h. 

d. True. 

e. False. The number of images taken is independent of the dose to the patient in nuclear medicine examination.

a. False. Leakage radiation from the X-ray tube is than 2% of the dose received by staff in the room. Scatter radiation from Compton interaction within the patient is the main radiation dose to staff. 

b. False. The principles of radiation protection are: 

i. Time: the shorter the exposure time, the lower the dose received 

ii. Distance: the inverse square law states that the intensity of the beam reduces from a source as distance increases 

iii. The thicker/denser the material, the better the shielding it provides. 

c. False. 

d. False. The Ionising Radiation Regulations 1999 (IRR99) are concerned with setting dose limits for worker, not the IRMER. The equivalent dose limit for the lens of a classified worker should not exceed 150 mSv per annum. 

e. True. Lead goggles are often used by interventionists.

a. True. It is proportional to the thickness of the imaged object. 

b. True. It is proportional to the difference between linear attenuation coefficients of the tissues involved. 

c. False. With increasing peak kV (kVp), the relative probability of the Compton effect increases, which compromises contrast. 

d. False. In the range of photon energies used in general radiography, the contrast between low-atomic-number tissues is low and only minimally dependent on kVp. 

e. False. It is due to a large difference in density. The effective atomic numbers of air and soft tissue are very similar.

a. False. Attenuation is due to Compton scattering and photoelectric absorption. Bremsstrahlung relates to the production of X-rays. 

b. True. This is due to the different attenuation properties of air, fat, soft tissue and bone (the main four ‘radio graphic densities’).

c. False. As they improve contrast, their effect is analogous to decreasing the peak kV (kVp). 

d. False. Negative-contrast media (for example, air or carbon dioxide) are radio lucent. 

e. True.

a. True. 

b. False. This is the value for rhodium. Barium has an absorption edge of 37 keV. 

c. True. The atomic numbers of iodine and barium are 53 and 56, respectively. 

d. False. As the K-absorption edge of iodine is 33 keV, it is most effective in attenuating photons of this energy. The answer to this question would be true for barium. 

e. True. This happens when an electron within the K-shell is ejected and replaced by an electron from another valence band. 

a. True. 

b. True. This orientates the internal dipoles in one direction, giving the crystal its piezoelectric properties. Once the crystal is cooled below the Curie temperature, this orientation is preserved, even in the absence of an external voltage. Heating above the Curie point in the absence of an external voltage would result in destruction of this polarization and elimination of the piezoelectric properties. 

c. False. The wavelength produced at the resonant frequency is double the crystal thickness. 

d. False. The thickness of the matching plate equals a quarter of the wavelength. 

e. False. In this case, all the sound energy would be completely absorbed in the backing layer. Therefore, a backing layer of impedance slightly lower than that of the PZT is used. 

a. False. As the Q factor is the ratio of the resonant frequency to the transducer bandwidth, a low Q value indicates a wide bandwidth. 

b. True. A heavily damped transducer produces a pulse of short duration. The shorter the pulse, the wider the spectrum of frequencies produced (bandwidth). Therefore, a heavily damped transducer has a low Q factor. 

c. True. A low Q transducer is highly damped and produces a pulse of short duration and therefore of low spatial pulse length (SPL). The shorter the SPL, the better the axial resolution. 

d. True. Continuous-wave imaging requires a transducer with high persistence of sound (‘ringing’). Such a transducer needs to be lightly damped, and therefore of a high Q factor. 

e. True. A high Q transducer has a narrow bandwidth, which is optimal for detection of the relatively small frequency changes caused by the blood flow.

a. True. This is because every point on the transducer surface produces a spherical wavelet (as described by Huygens’ principle); these wavelets undergo constructive and destructive interference producing a planar wave front. This part of the beam is called a ‘near field’. 

b. False. It is proportional to the squared radius of the transducer and inversely proportional to wavelength; the near field extends further in the case of big transducers emitting high-frequency ultrasound. 

c. False. The beam starts to diverge in the far field (Fraunhofer region). The Fresnel region is another name for the near field. 

d. True. The sine of this angle is proportional to the ratio of wavelength and transducer diameter. Therefore, a large transducer of high frequency has a less divergent beam in the far field. 

e. False. It cannot be focused electronically; however, it can be focused to one set depth using a curved piezoelectric crystal or a plastic acoustic lens at the transducer face.

The Correct option is D. 1.8 Wm^-2

a. True. Note that the thermal index for soft tissue (TIS) should be monitored. 

b. True. 

c. True. Note that these values are the same as for neonatal ultrasound. The thermal index for cranial imaging (TIC) and thermal index for bone (TIB) should be used for central nervous system and general scanning, respectively. 

d. False. Consent is required of all patients, with verbal consent being generally acceptable. The British Medical Ultrasound Society recommends that, in the case of healthy volunteers, consent should ideally be expressed in a written form. 

e. False. The time should not exceed double that needed for a diagnostic examination.

a. True. 

b. False. The Royal College of Radiologists recommends in its ‘Standards for Ultrasound Equipment’ that testing should be performed annually or biannually. 

c. False. Resolution is usually tested using a test object containing a set of filament targets in a tissue-mimicking material phantom. A string phantom is used for testing the velocity in a Doppler mode. 

d. False. They need to be filled with a material in which the speed of sound (1540 m s^-1 ), attenuation (0.5 dB cm^-1 MHz^-1 ) and density (1050 kg m^-3 ) are close to that in a soft tissue.

e. True. The ultrasound emitted from a transducer exerts a force proportional to the output power, which can be ‘weighed’ on a special balance. A power of 1 W corresponds to approximately 68 mg.

a. True. 

b. True. This relates to differences in the DQE. 

c. False. When using high-resolution image plates and a narrow laser beam for stimulated emission, a higher resolution is achievable with computed radiography. 

d. False. They are greater with digital radiography, as each radio graphic machine needs its own detector, whereas with computed radiography the image plates may be used between existing machines. Costs may be saved later, however, as the patient throughput may be greater with digital radiography. 

e. True.

a. False. The thermal index for soft tissue (TIS) should be monitored. The TIS assumes that only soft tissue is insonated. At 10 weeks after the last menstrual period, ossification of the fetal spine starts and the TIB should be monitored. 

b. False. Scanning with a TI between 0.7 and 1.0 is restricted to 60 min. There are no time restrictions for a TI <0.7. 

c. True. 

d. False. Scanning of the central nervous system (transcranial or spinal ultrasound) is not recommended with a TI >3.0. In this case, the thermal index for cranial imaging (TIC) should be monitored. General neonatal scanning is not recommended with a TI >6.0. Note that, although allowed, scanning with a TI just below these values is unlikely to be of clinical use due to the time restrictions (e.g. general scanning with a TI between 4 and 5 is allowed for 15 s, and with a TI between 5 and 6 for 5 s only). 

e. False. This possibility exists with an MI >0.3.

a. False. The Compton effect is independent of the atomic number of the material as it only involves free electrons whose binding energy is negligible. 

b. False. As photon energy increases, the rate of photoelectric absorption decreases more than the Compton effect. Hence, at high photon energies, the Compton effect still predominates. 

c. True. The Compton effect predominates in low-atomic-number materials e.g. air, water and tissue. 

d. False. Photoelectric absorption predominates in high-atomic-number materials, e.g. lead and contrast medium. 

e. False.

a. False. K-edge binding energy is higher than L-edge binding energy. 

b. False. As photon energy increases, photoelectric attenuation decreases between the L-shell and the K-shell. 

c. False. The atomic number of tungsten is 74. Its K-shell binding energy is 70 keV. 

d. True. When photon energy reaches EL, interaction with L-shell electrons becomes possible and photoelectric absorption increases. 

e. True.

a. False. The DQE of indirect conversion is greater. 

b. True. The K-edge of selenium is only 13 keV, which is less than the mean photon energy range used in general radiography. Much less of the incident X-ray beam will therefore be absorbed than with caesium iodide, which has K-edges of 36 keV and 33 keV. 

c. False. For the reasons mentioned in (b), the opposite is true. To achieve the same image quality, a higher patient dose is required with direct detectors. 

d. True. Again, this is related to differences in the DQE. 

e. True. Another measure of this is the modulation transfer factor, which is greater with direct than indirect conversion detectors. The differences relate to the direct transfer of electrical charge within the selenium, driven by the potential difference applied. With indirect detectors there is the possibility of light scatter within the scintillator, which increases unsharpness.