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Instructions / Reading Passage
11 Questions in this set
COMPUTER APTITUDE TEST (Basic Computer Knowledge): Write the correct answer for the provided multiple-choice questions involving expert systems, industrial computer applications (POS, CIM, Actuarial accounting), and management information system (MIS) modeling.
Q. 1
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
Expert systems are part of the general area of research known as
A
AI
✓ Correct
B
ES
C
STUDENT
D
RAND
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Artificial Intelligence (AI) is the broad field that encompasses the development of expert systems.
Q. 2
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
In the future, users of a computer system may identify themselves by entering a:
A
Hard print
B
Soft print
C
Voice print
✓ Correct
D
Digital
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Biometric identification methods include voice recognition (voice prints) for user authentication.
Q. 3
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
The most prominent system in the retail sales industry is the ________ system:
A
POS (point-of-scale)
✓ Correct
B
COBOL
C
ACM
D
PC/XT
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Point of Sale systems are standard for managing transactions and inventory in retail.
Q. 4
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
The integration of the computer with manufacturing is called CIM or
A
CAM
B
CAD
C
Computer manufacturing
✓ Correct
D
All the above
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Computer Integrated Manufacturing (CIM) refers to the full automation of a manufacturing facility.
Q. 5
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
A ________ purpose computer is designed for a specific application:
A
Special
✓ Correct
B
Analog
C
Hybrid
D
Digital
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Special-purpose computers are built to perform one specific set of tasks or applications.
Q. 6
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
Actuarial accounting system is associated with the ________ industry:
A
Insurance
✓ Correct
B
Reliable
C
Consistent
D
Exhaustive
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Actuarial science is primarily used by the insurance industry to assess risk and uncertainty.
Q. 7
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
A variable that has no physical meaning and is used to obtain an initial basic feasible solution to a linear programming problem is known as:
A
Basis
B
Algorithm
C
Artificial variable
✓ Correct
D
Basic variable
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
In linear programming, artificial variables are used to facilitate finding an initial feasible solution.
Q. 8
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
QUB is an example of ________ systems:
A
Problem-solving
B
Inventory
C
Electronic shopping
✓ Correct
D
Algebra
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
QUB (historically standing for Queen's University Belfast Electronic Shopping System experiment or similar early regional academic network paradigms referenced in localized commerce and MIS curriculums) is categorized as an early or specialized example of an Electronic shopping system or transactional network.
Q. 9
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
Slack is the calculated time span within which the event must occur:
A
True
✓ Correct
B
False
C
Cannot be said
D
All the above
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Slack in project management represents the amount of time a task can be delayed without affecting the deadline.
Q. 10
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
The latest time that the event can be delayed without delaying the completion of the entire project in PERT chart is:
A
Earliest allowable time
B
Latest allowable time
✓ Correct
C
Earliest pass
D
Latest pass
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
This term defines the absolute deadline for a specific node in a PERT network.
Q. 11
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
Anthony suggested that the area of management planning and control be segmented into:
A
Operational control
B
Strategic planning
C
Management control
D
All of the above
✓ Correct
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Anthony's framework includes Strategic Planning, Management Control, and Operational Control.
📜
Instructions / Reading Passage
14 Questions in this set
SYSTEMS AND DECISION-MAKING: Solve questions regarding transaction processing, data collection terminals, operations research history, and the files required for receiving and engineering control systems.
Q. 12
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
Which of the following does not fall in the personal Decision-Making Area?
A
Pay-rolls
B
Cash flow analysis
✓ Correct
C
Income-tax assessment and recovery
D
Person's experience
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Cash flow analysis is typically an organizational or corporate financial function rather than personal.
Q. 13
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
How much should an organisation spend for information can be determined by some type of a:
A
Cost analysis
B
Benefit analysis
C
Cost-benefit analysis
✓ Correct
D
Any of the above
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Organisations weigh the cost of data acquisition against the value/benefit it provides.
Q. 14
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
A transportation problem in which the total supply available at the origins exactly satisfies the total demand required at the destinations is known as:
A
Degenerate solution
B
Balanced transportation problem
✓ Correct
C
Unbalanced transportation problem
D
All the above
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
A "Balanced" problem occurs when total supply equals total demand.
Q. 15
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
Operations Research came into being due to:
A
Medical reasons
B
Military reasons
✓ Correct
C
Educational reasons
D
To improve transpiration
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Operations Research originated for tactical and logistical planning during World War II.
Q. 16
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
The funds management subsystems attempts to:
A
Increase cash input
B
Decrease cash output
C
Balance cash inflow with outflow
✓ Correct
D
All the above
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Effective funds management ensures liquidity by matching timing of inflows and outflows.
Q. 17
Computer Aptitude Test
Difficulty: Medium
(1 Mark)
A turnaround document can be a(n):
A
Punched card
B
OCR document
C
Neither nor (b)
✓ Correct
D
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Based on the specific classification of turnaround documents in the source answer key.
Q. 18
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
In the financial Decision-Making system which of the following is not a transaction processing system input:
A
Cash receipts
B
Cash returns
C
Cash issues
D
Warehouse requisition ship
✓ Correct
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Requisition slips are inventory/logistics inputs rather than direct financial transaction inputs.
Q. 19
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
Header labels contain:
A
Creation data
✓ Correct
B
Control totals
C
Both and (b)
D
Either nor (b)
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
File headers typically store metadata such as the creation date and file ID.
Q. 20
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
Detected money errors are corrected with:
A
Debit entries
B
Credit entries
C
Both and (b)
✓ Correct
D
Neither nor (b)
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Adjustments for errors require offsetting debit or credit entries depending on the nature of the error.
Q. 21
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
The model base contains programs:
A
Written by the firm's programmers
B
Provided by the computer vendor
C
Purchased from outside firms, such as software houses
D
All of the above
✓ Correct
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
A Decision Support System's model base can incorporate software from various origins.
Q. 22
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
Management has become more complex because:
A
There is less time to react to competitive actions
B
The equipment used in many of the firm's activities has become more complex
C
Firms have become larger
D
All of the above
✓ Correct
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Global competition, technological complexity, and organizational scale contribute to management difficulty.
Q. 23
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
What data most likely would not be keyed into the order entry subsystem?
A
Item quantity
B
Customer number
C
Item number
D
Item cost
✓ Correct
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Costs are usually master data retrieved from a file rather than manually keyed per order.
Q. 24
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
The engineering change control system makes changes in the:
A
Operations file
✓ Correct
B
Production schedule
C
Work force data
D
All the above
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Design changes directly impact the steps and specifications in the operations file.
Q. 25
Computer Aptitude Test
Difficulty: Hard
(1 Mark)
Corporate modeling software can be used for:
A
Cash planning
B
Operational budgeting
C
Capital budgeting
D
All the above
✓ Correct
E
None of the above
💡
Step-by-Step Explanation & Concept Rationale
Modeling software assists in various strategic and operational planning tasks.
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Q. 1
Electronics Engineering
Difficulty: Easy
(1 Mark)
In an intrinsic (pure) semiconductor at room temperature (300 K), the relationship between electron concentration (n) and hole concentration (p) is:
A
n >> p
B
n = p = ni (intrinsic carrier concentration)
✓ Correct
C
p >> n
D
n = 0 and p = 0
💡
Step-by-Step Explanation & Concept Rationale
In an intrinsic semiconductor, thermal energy excites valence electrons across the bandgap, creating equal numbers of free conduction electrons and holes. Thus, n = p = ni. For Silicon at 300 K, ni ≈ 1.5 * 10^10 cm^-3.
Q. 2
Electronics Engineering
Difficulty: Easy
(1 Mark)
What is the forbidden energy bandgap (Eg) at 300 K for Silicon (Si) and Germanium (Ge), respectively?
A
1.12 eV for Si, and 0.66 eV for Ge
✓ Correct
B
0.7 eV for Si, and 0.3 eV for Ge
C
1.42 eV for Si, and 1.12 eV for Ge
D
5.0 eV for Si, and 1.12 eV for Ge
💡
Step-by-Step Explanation & Concept Rationale
At room temperature (300 K), the forbidden energy bandgap of Silicon is approximately 1.12 eV, while for Germanium it is approximately 0.66 eV (0.67 eV). Note that 0.7 V and 0.3 V represent the forward threshold/barrier voltages of their respective PN junctions.
Q. 3
Electronics Engineering
Difficulty: Easy
(1 Mark)
Doping intrinsic Silicon with a pentavalent impurity (such as Phosphorus, Arsenic, or Antimony) creates a(n):
A
P-type semiconductor with holes as majority carriers
B
N-type semiconductor with electrons as majority carriers and donor ions
✓ Correct
C
Intrinsic insulator
D
Superconductor
💡
Step-by-Step Explanation & Concept Rationale
Pentavalent atoms have 5 valence electrons. Four form covalent bonds with adjacent Si atoms, while the fifth electron is loosely bound (donor level ~0.05 eV below conduction band) and easily liberated at room temperature, creating an N-type semiconductor where electrons are majority carriers.
Q. 4
Electronics Engineering
Difficulty: Easy
(1 Mark)
Doping Silicon with a trivalent impurity (such as Boron, Gallium, or Indium) creates a(n):
A
N-type semiconductor with donor impurities
B
P-type semiconductor with holes as majority carriers and acceptor ions
✓ Correct
C
Intrinsic semiconductor
D
Semimetal
💡
Step-by-Step Explanation & Concept Rationale
Trivalent atoms have 3 valence electrons, leaving one covalent bond vacancy (a hole). They readily accept electrons from nearby bonds, creating mobile holes as majority carriers and fixed negative acceptor ions, forming a P-type semiconductor.
Q. 5
Electronics Engineering
Difficulty: Easy
(1 Mark)
The Mass Action Law for a semiconductor under thermal equilibrium states that:
A
n + p = ni
B
n * p = ni^2 (independent of doping level)
✓ Correct
C
n / p = ni
D
n * p = constant * T
💡
Step-by-Step Explanation & Concept Rationale
Under thermal equilibrium, the product of electron concentration n and hole concentration p is constant and equals the square of the intrinsic carrier concentration: n * p = ni^2. If donor doping increases n, hole concentration p decreases proportionately through recombination.
Q. 6
Electronics Engineering
Difficulty: Medium
(1 Mark)
The Fermi-Dirac distribution function f(E) gives the probability that an energy state E is occupied by an electron. At energy E = Ef (Fermi energy level), the probability f(Ef) is:
A
1.0 (100%)
B
0.5 (50%) at any temperature T > 0 K
✓ Correct
C
0.0 (0%)
D
0.707
💡
Step-by-Step Explanation & Concept Rationale
f(E) = 1 / [1 + exp((E - Ef) / (k * T))]. When E = Ef, exp(0) = 1, so f(Ef) = 1 / (1 + 1) = 1/2 = 0.5 (or 50%) at all temperatures above absolute zero.
Q. 7
Electronics Engineering
Difficulty: Medium
(1 Mark)
In an N-type semiconductor as donor doping concentration Nd increases, the Fermi level (Ef):
A
Shifts downward toward the valence band
B
Shifts upward toward the conduction band (Ec)
✓ Correct
C
Remains precisely at the center of the bandgap
D
Disappears
💡
Step-by-Step Explanation & Concept Rationale
Ef - Ei = k * T * ln(Nd / ni). As donor concentration Nd increases, the density of electrons in the conduction band rises, shifting the Fermi level closer to the conduction band edge Ec. In degenerate semiconductors, Ef enters the conduction band.
Q. 8
Electronics Engineering
Difficulty: Medium
(1 Mark)
Einstein's relation relating carrier diffusion coefficient (D) to mobility (mu) in a semiconductor is:
A
D / mu = V_T = k * T / q (thermal voltage)
✓ Correct
B
D * mu = k * T
C
D / mu = q / (k * T)
D
D = mu^2 * V_T
💡
Step-by-Step Explanation & Concept Rationale
Einstein's relation states that Dn / mun = Dp / mup = V_T = k * T / q, where thermal voltage V_T ≈ 25.86 mV (approx. 26 mV) at room temperature (300 K).
Q. 9
Electronics Engineering
Difficulty: Easy
(1 Mark)
The depletion region (space-charge layer) formed at an unbiased PN junction consists of:
A
Free mobile electrons and holes
B
Immobile unneutralized ionized donor and acceptor atoms stripped of their free carriers
✓ Correct
C
Neutral silicon atoms only
D
Pure metallic copper
💡
Step-by-Step Explanation & Concept Rationale
When P and N regions meet, electrons diffuse from N to P and holes from P to N. This leaves behind uncovered, fixed positive donor ions on the N-side and fixed negative acceptor ions on the P-side, creating a built-in electric field that opposes further carrier diffusion.
Q. 10
Electronics Engineering
Difficulty: Easy
(1 Mark)
The built-in contact potential barrier (V0) of a Silicon PN junction at 300 K is typically around:
A
0.1 V to 0.2 V
B
0.6 V to 0.8 V (nominal 0.7 V)
✓ Correct
C
1.5 V to 2.0 V
D
5.0 V
💡
Step-by-Step Explanation & Concept Rationale
V0 = V_T * ln[(Na * Nd) / ni^2]. For typical doping levels in Silicon at 300 K, the contact barrier potential is approximately 0.7 V. For Germanium, it is approximately 0.3 V.
Q. 11
Electronics Engineering
Difficulty: Easy
(1 Mark)
When a PN junction is forward-biased (P-side positive, N-side negative):
A
The depletion layer width widens, and barrier height increases
B
The depletion layer width narrows, the potential barrier decreases (V0 - Vf), and majority carriers cross the junction in large numbers
✓ Correct
C
Current flow drops to zero
D
Breakdown immediately occurs
💡
Step-by-Step Explanation & Concept Rationale
Forward bias opposes the built-in electric field, reducing barrier height to (V0 - Vf) and narrowing the space-charge width. This allows majority carriers to overcome the barrier, resulting in exponential diffusion current: I = Is * [exp(V / (eta*V_T)) - 1].
Q. 12
Electronics Engineering
Difficulty: Easy
(1 Mark)
When a PN junction is reverse-biased (P-side negative, N-side positive):
A
Depletion layer narrows
B
Depletion layer widens, barrier height increases (V0 + Vr), and only a tiny reverse saturation current (Is) due to minority carriers flows
✓ Correct
C
Current increases exponentially
D
Resistance drops to zero
💡
Step-by-Step Explanation & Concept Rationale
Reverse bias aids the built-in field, pulling majority carriers away from the junction and widening the depletion layer. Only thermally generated minority carriers swept across by the field constitute the reverse saturation current Is.
Q. 13
Electronics Engineering
Difficulty: Easy
(1 Mark)
The reverse saturation current (Is) of a Silicon PN diode approximately doubles for every:
A
1 degree Celsius rise in temperature
B
10 degrees Celsius rise in temperature
✓ Correct
C
50 degrees Celsius rise in temperature
D
100 degrees Celsius rise in temperature
💡
Step-by-Step Explanation & Concept Rationale
Because thermal generation of electron-hole pairs increases exponentially with temperature, the reverse saturation current Is of a diode approximately doubles for every 10 °C rise in operating junction temperature: Is(T2) = Is(T1) * 2^((T2 - T1) / 10).
Q. 14
Electronics Engineering
Difficulty: Medium
(1 Mark)
The forward voltage drop (Vf) of a Silicon PN diode at constant current decreases with temperature at a rate of approximately:
A
-2.0 to -2.5 mV per degree Celsius (-2 mV/°C)
✓ Correct
B
+10 mV/°C
C
-1.0 V/°C
D
Zero (temperature independent)
💡
Step-by-Step Explanation & Concept Rationale
Due to the rapid increase in intrinsic carrier concentration ni with temperature, the forward diode voltage at fixed current decreases by roughly -2 mV/°C to -2.5 mV/°C. This negative temperature coefficient is frequently exploited in semiconductor temperature sensors.
Q. 15
Electronics Engineering
Difficulty: Medium
(1 Mark)
Transition (or depletion layer) capacitance (CT) in a reverse-biased PN junction dominates over diffusion capacitance because:
A
Charge is stored as immobile ions across the expanding depletion dielectric like a parallel-plate capacitor (CT ∝ 1 / sqrt(V0 + Vr))
✓ Correct
B
Minority carrier diffusion is very high
C
Current is leading voltage by 180 degrees
D
Barrier potential drops to zero
💡
Step-by-Step Explanation & Concept Rationale
Under reverse bias, diffusion current is negligible. The immobile ions on either side of the depletion layer act as charged plates separated by a dielectric of width W, giving transition capacitance CT = epsilon * A / W. Because W ∝ sqrt(V0 + Vr), CT varies inversely with reverse voltage.
Q. 16
Electronics Engineering
Difficulty: Easy
(1 Mark)
A Varactor diode (varicap) operates in which region and is primarily used for:
A
Forward bias as a light source
B
Reverse bias, utilizing its voltage-variable transition capacitance for electronic tuning in RF local oscillators and phase-locked loops (PLLs)
✓ Correct
C
Breakdown region as a voltage regulator
D
High-current power rectification
💡
Step-by-Step Explanation & Concept Rationale
A Varactor diode is specifically designed to maximize the voltage-dependent transition capacitance under reverse bias. Varying reverse voltage controls CT smoothly, providing electronic tuning for LC resonant tanks without mechanical variable capacitors.
Q. 17
Electronics Engineering
Difficulty: Medium
(1 Mark)
Zener breakdown occurs in PN junctions that are:
A
Lightly doped, with wide depletion layers, breaking down above 6 V via impact ionization
B
Heavily doped, with extremely narrow depletion layers (W < 10 nm), where high electric field (> 10^6 V/cm) pulls valence electrons directly into the conduction band by quantum mechanical tunneling
✓ Correct
C
Forward biased with high current
D
Fabricated exclusively from Germanium
💡
Step-by-Step Explanation & Concept Rationale
Zener breakdown occurs in heavily doped junctions at low reverse voltages (< 5-6 V). The narrow depletion width creates an intense electric field that directly ruptures covalent bonds via quantum tunneling. It exhibits a negative temperature coefficient of breakdown voltage.
Q. 18
Electronics Engineering
Difficulty: Medium
(1 Mark)
Avalanche breakdown in a reverse-biased PN junction occurs in:
A
Lightly doped junctions with breakdown voltages greater than 6 V, where thermally generated carriers gain high kinetic energy to ionize lattice atoms by impact ionization (carrier multiplication)
✓ Correct
B
Heavily doped junctions at 2 V
C
Forward-biased diodes
D
Photodiodes in complete darkness only
💡
Step-by-Step Explanation & Concept Rationale
In lightly doped diodes, the depletion layer is relatively wide. High reverse voltage accelerates minority carriers to high velocities, dislodging additional valence electrons upon impact. These secondary carriers cause cumulative carrier multiplication (avalanche), having a positive temperature coefficient of breakdown voltage.
Q. 19
Electronics Engineering
Difficulty: Medium
(1 Mark)
A Zener diode with a nominal breakdown voltage of 5.6 V has a temperature coefficient that is:
A
Strongly positive
B
Strongly negative
C
Nearly zero, because the negative temperature coefficient of Zener tunneling cancels the positive temperature coefficient of Avalanche multiplication
✓ Correct
D
Infinite
💡
Step-by-Step Explanation & Concept Rationale
Below 5 V, Zener breakdown (negative temp coeff) dominates. Above 6 V, Avalanche breakdown (positive temp coeff) dominates. Around 5.6 V, both mechanisms contribute equally, cancelling each other's temperature drift to yield superior voltage reference stability.
Q. 20
Electronics Engineering
Difficulty: Medium
(1 Mark)
A Tunnel diode (Esaki diode) exhibits negative differential resistance (dV/dI < 0) over a portion of its forward characteristic due to:
A
Extreme degenerate doping (1 in 10^3 atoms) that causes quantum mechanical electron tunneling across an ultra-thin depletion barrier
✓ Correct
B
Thermal heating of the junction
C
Avalanche carrier multiplication
D
External magnetic field
💡
Step-by-Step Explanation & Concept Rationale
Degenerate doping causes the conduction band on the N-side to overlap the valence band on the P-side. Under small forward bias, electrons tunnel directly through the thin barrier (~10 nm). As forward bias increases, band overlap diminishes, causing forward current to drop from peak current (Ip) to valley current (Iv), creating negative resistance used in high-frequency microwave oscillators.
Q. 21
Electronics Engineering
Difficulty: Medium
(1 Mark)
A Schottky barrier diode is formed by a metal-semiconductor junction (e.g., gold or platinum on N-type silicon). Its key operational advantages are:
A
Majority-carrier device with virtually zero reverse recovery time (ultra-fast switching) and lower forward voltage drop (approx. 0.2 to 0.3 V)
✓ Correct
B
Extremely high reverse breakdown voltage exceeding 20 kV
C
High diffusion capacitance under reverse bias
D
Negative dynamic resistance
💡
Step-by-Step Explanation & Concept Rationale
Because conduction is by majority carriers (electrons in N-type silicon), there is no minority carrier charge storage in the depletion region. Reverse recovery time is virtually zero (picoseconds), and forward drop is only 0.2-0.3 V, making Schottky diodes ideal for high-efficiency switch-mode power supplies (SMPS) and RF mixers.
Q. 22
Electronics Engineering
Difficulty: Hard
(1 Mark)
A PIN diode contains an undoped intrinsic (I) semiconductor layer sandwiched between heavily doped P and N layers. At microwave radio frequencies, it behaves as a:
A
Linear variable RF resistor whose resistance is controlled by DC bias current, useful as an RF switch and attenuator
✓ Correct
B
Zener regulator
C
Digital flip-flop
D
Constant voltage source
💡
Step-by-Step Explanation & Concept Rationale
At low frequencies, a PIN diode acts as a standard rectifier. At RF/microwave frequencies, carriers cannot follow fast AC cycles; the intrinsic layer behaves as an almost pure resistance whose RF value varies continuously from thousands of ohms to under 1 ohm depending on applied DC forward bias current.
Q. 23
Electronics Engineering
Difficulty: Easy
(1 Mark)
A Photodiode is always operated in which mode to detect incident optical radiation with linear response and high bandwidth?
A
Forward bias mode
B
Reverse bias mode (photoconductive mode)
✓ Correct
C
Zero bias (photovoltaic mode) only
D
Thermal breakdown mode
💡
Step-by-Step Explanation & Concept Rationale
In reverse bias, dark current is minimal. Incident photons of energy h*nu >= Eg generate electron-hole pairs in the wide depletion layer, which are swept by the strong built-in electric field. Reverse photocurrent is strictly linear with optical power and switching speed is maximal.
Q. 24
Electronics Engineering
Difficulty: Medium
(1 Mark)
Light Emitting Diodes (LEDs) are fabricated from direct bandgap compound semiconductors (such as GaAs, GaN, InGaN) rather than indirect bandgap semiconductors like Silicon because:
A
In direct bandgap materials, electrons recombine directly across the bandgap releasing energy predominantly as photons (light) without requiring lattice phonon momentum assistance
✓ Correct
B
Silicon is opaque to electricity
C
Direct bandgap materials have zero thermal resistance
D
Silicon cannot be doped P-type
💡
Step-by-Step Explanation & Concept Rationale
In direct bandgap semiconductors, the conduction band minimum and valence band maximum align at the same crystal momentum (k = 0). Radiative electron-hole recombination is direct and highly efficient, producing photons: lambda = h * c / Eg. In Silicon (indirect bandgap), recombination requires a phonon (lattice vibration), dissipating energy as heat.
Q. 25
Electronics Engineering
Difficulty: Easy
(1 Mark)
What is the peak inverse voltage (PIV) rating across each diode in a single-phase center-tapped full-wave transformer rectifier supplying peak secondary voltage Vm across each half?
A
PIV = Vm
B
PIV = 2 * Vm
✓ Correct
C
PIV = Vm / 2
D
PIV = sqrt(2) * Vm
💡
Step-by-Step Explanation & Concept Rationale
In a center-tapped full-wave rectifier, during the non-conducting half cycle, the reverse-biased diode must withstand its own half-winding peak voltage Vm plus the peak voltage delivered by the other conducting half to the load, resulting in PIV = 2 * Vm.
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Instructions / Reading Passage
25 Questions in this set
Select the best option/antonym for the following English Vocabulary questions:
Q. 1
English Language Comprehension
Difficulty: Medium
(1 Mark)
Autonomy
A
submissiveness
B
dependence
✓ Correct
C
subordination
D
slavery
💡
Step-by-Step Explanation & Concept Rationale
Autonomy refers to self-governance; its opposite is relying on others.
Q. 2
English Language Comprehension
Difficulty: Medium
(1 Mark)
Assert
A
agree
B
acquiesce
C
abjure
✓ Correct
D
abdicate
💡
Step-by-Step Explanation & Concept Rationale
To assert is to state firmly; abjure means to solemnly renounce or deny.
Q. 3
English Language Comprehension
Difficulty: Medium
(1 Mark)
Accord
A
solution
B
act
C
dissent
✓ Correct
D
concord
💡
Step-by-Step Explanation & Concept Rationale
Accord signifies agreement or harmony; dissent represents disagreement.
Q. 4
English Language Comprehension
Difficulty: Medium
(1 Mark)
Alive
A
passive
B
dead
✓ Correct
C
asleep
D
drowsy
💡
Step-by-Step Explanation & Concept Rationale
The direct biological opposite of being alive is being dead.
Q. 5
English Language Comprehension
Difficulty: Medium
(1 Mark)
Ancestors
A
supporters
B
disciples
C
followers
D
descendants
✓ Correct
💡
Step-by-Step Explanation & Concept Rationale
Ancestors are those from whom one is descended; descendants are one's progeny.
Q. 6
English Language Comprehension
Difficulty: Medium
(1 Mark)
Abdicate
A
claim
✓ Correct
B
snatch
C
plunder
D
seize
💡
Step-by-Step Explanation & Concept Rationale
To abdicate is to renounce a throne or power; to claim is to assert a right to it.
Q. 7
English Language Comprehension
Difficulty: Medium
(1 Mark)
Abhorrence
A
aversion
B
liking
✓ Correct
C
appreciation
D
fear
💡
Step-by-Step Explanation & Concept Rationale
Abhorrence is extreme hatred; liking indicates a feeling of fondness or pleasure.
Q. 8
English Language Comprehension
Difficulty: Medium
(1 Mark)
Acquisitive
A
miserly
B
frugal
✓ Correct
C
simple
D
austere
💡
Step-by-Step Explanation & Concept Rationale
Acquisitive means eager to get wealth; frugal implies being economical or sparing.
Q. 9
English Language Comprehension
Difficulty: Medium
(1 Mark)
Audacious
A
timid
✓ Correct
B
vulgar
C
low
D
unpractised
💡
Step-by-Step Explanation & Concept Rationale
Audacious refers to being bold and daring; timid means showing a lack of courage.
Q. 10
English Language Comprehension
Difficulty: Medium
(1 Mark)
Acquitted
A
entrusted
B
convicted
✓ Correct
C
freed
D
burdened
💡
Step-by-Step Explanation & Concept Rationale
To be acquitted is to be cleared of a charge; to be convicted is to be found guilty.
Q. 11
English Language Comprehension
Difficulty: Medium
(1 Mark)
Absolute
A
scarce
B
limited
✓ Correct
C
faulty
D
efficient
💡
Step-by-Step Explanation & Concept Rationale
Absolute means total or unconditional; limited implies having fixed bounds.
Q. 12
English Language Comprehension
Difficulty: Medium
(1 Mark)
Acute
A
sharp
B
critical
C
dull
✓ Correct
D
sensitive
💡
Step-by-Step Explanation & Concept Rationale
Acute describes something sharp or intense; dull describes something blunt or weak.
Q. 13
English Language Comprehension
Difficulty: Medium
(1 Mark)
Anathematise
A
radiate
B
regulate
C
deceive
D
bless
✓ Correct
💡
Step-by-Step Explanation & Concept Rationale
To anathematise is to curse or denounce; its spiritual opposite is to bless.
Q. 14
English Language Comprehension
Difficulty: Medium
(1 Mark)
Addition
A
multiplication
B
subtraction
✓ Correct
C
enumeration
D
division
💡
Step-by-Step Explanation & Concept Rationale
Subtraction is the mathematical operation that removes, while addition adds.
Q. 15
English Language Comprehension
Difficulty: Medium
(1 Mark)
Attract
A
repulse
B
reject
C
repel
✓ Correct
D
distract
💡
Step-by-Step Explanation & Concept Rationale
To attract is to pull toward; to repel is to push away or drive back.
Q. 16
English Language Comprehension
Difficulty: Medium
(1 Mark)
Avoidance
A
possession
B
pursuit
✓ Correct
C
passion
D
power
💡
Step-by-Step Explanation & Concept Rationale
Avoidance is the act of staying away; pursuit is the act of following or chasing.
Q. 17
English Language Comprehension
Difficulty: Medium
(1 Mark)
Alleviation
A
exaggeration
B
exasperation
C
magnification
D
intensification
✓ Correct
💡
Step-by-Step Explanation & Concept Rationale
While "alleviation" means to make suffering, deficiency, or a problem less severe (to ease or mitigate), "intensification" means to make something grow more acute, strong, or severe.
Q. 18
English Language Comprehension
Difficulty: Medium
(1 Mark)
Alienate
A
gather
B
identify
C
assemble
D
unite
✓ Correct
💡
Step-by-Step Explanation & Concept Rationale
To alienate is to isolate or cause to become unfriendly; to unite is to join together.
Q. 19
English Language Comprehension
Difficulty: Medium
(1 Mark)
Adequate
A
profuse
B
abounding
C
scanty
✓ Correct
D
abundant
💡
Step-by-Step Explanation & Concept Rationale
Adequate means sufficient; scanty means small or insufficient in quantity.
Q. 20
English Language Comprehension
Difficulty: Medium
(1 Mark)
Appointment
A
disappointment
B
suspension
C
dismissal
✓ Correct
D
discharge
💡
Step-by-Step Explanation & Concept Rationale
"Appointment" refers to the act of officially assigning a job, position, or office to someone. Therefore, dismissal (the act of officially removing someone from a job or position) is its direct structural and professional antonym.
Q. 21
English Language Comprehension
Difficulty: Medium
(1 Mark)
Amalgamate
A
generate
B
repair
C
materialize
D
separate
✓ Correct
💡
Step-by-Step Explanation & Concept Rationale
Amalgamate means to combine into one; separate means to part or divide.
Q. 22
English Language Comprehension
Difficulty: Medium
(1 Mark)
Acclamation
A
denunciation
✓ Correct
B
suppression
C
termination
D
applause
💡
Step-by-Step Explanation & Concept Rationale
Acclamation is loud approval; denunciation is public condemnation.
Q. 23
English Language Comprehension
Difficulty: Medium
(1 Mark)
Ambiguous
A
obscure
B
secular
C
explicit
✓ Correct
D
equivocate
💡
Step-by-Step Explanation & Concept Rationale
Ambiguous means vague or unclear; explicit means stated clearly and in detail.
Q. 24
English Language Comprehension
Difficulty: Medium
(1 Mark)
Antipathy
A
obedience
B
admiration
C
agreement
D
fondness
✓ Correct
💡
Step-by-Step Explanation & Concept Rationale
Antipathy is a deep-seated feeling of dislike; fondness is affection or liking.
Q. 25
English Language Comprehension
Difficulty: Medium
(1 Mark)
Amicable
A
cunning
B
shy
C
hostile
✓ Correct
D
crazy
💡
Step-by-Step Explanation & Concept Rationale
Amicable refers to friendliness and goodwill; hostile refers to unfriendliness/opposition.
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