💼 Official Mock Test for: TELECOM OPERATOR-II (TELEPHONE / SIGNALS) (SPS-3)
⏱️ Official Timed Examination

TELECOM OPERATOR-II (TELEPHONE / SIGNALS) (SPS-3) - Screening Mock Test 2026

Category: General Competitive

Duration
⏱️ 100 Mins
Question Pool
📝 100 MCQs
Passing Benchmark
🎯 50.0%
Scoring Engine
✓ Instant & Ranked

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Sample Questions & Solved Explanations

Representative sample from official examination pool

Below are representative questions drawn directly from the testing blueprint for this exam. Review these worked examples to understand the question style, difficulty calibration, and grading criteria:

Sample Q1 (Electronics Engineering) Difficulty: Easy
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)
C p >> n
D n = 0 and p = 0
✓ Answer: B - n = p = ni (intrinsic carrier concentration)
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.
Sample Q2 (Electronics Engineering) Difficulty: Easy
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
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
✓ Answer: A - 1.12 eV for Si, and 0.66 eV for Ge
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.
Sample Q3 (Electronics Engineering) Difficulty: Easy
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
C Intrinsic insulator
D Superconductor
✓ Answer: B - N-type semiconductor with electrons as majority carriers and donor ions
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.
Sample Q4 (Electronics Engineering) Difficulty: Easy
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
C Intrinsic semiconductor
D Semimetal
✓ Answer: B - P-type semiconductor with holes as majority carriers and acceptor ions
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.
Sample Q5 (Electronics Engineering) Difficulty: Easy
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)
C n / p = ni
D n * p = constant * T
✓ Answer: B - n * p = ni^2 (independent of doping level)
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.
Sample Q6 (Electronics Engineering) Difficulty: Medium
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
C 0.0 (0%)
D 0.707
✓ Answer: B - 0.5 (50%) at any temperature T > 0 K
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.
Sample Q7 (Electronics Engineering) Difficulty: Medium
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)
C Remains precisely at the center of the bandgap
D Disappears
✓ Answer: B - Shifts upward toward the conduction band (Ec)
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.
Sample Q8 (Electronics Engineering) Difficulty: Medium
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)
B D * mu = k * T
C D / mu = q / (k * T)
D D = mu^2 * V_T
✓ Answer: A - D / mu = V_T = k * T / q (thermal voltage)
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).
Sample Q9 (Electronics Engineering) Difficulty: Easy
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
C Neutral silicon atoms only
D Pure metallic copper
✓ Answer: B - Immobile unneutralized ionized donor and acceptor atoms stripped of their free carriers
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.
Sample Q10 (Electronics Engineering) Difficulty: Easy
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)
C 1.5 V to 2.0 V
D 5.0 V
✓ Answer: B - 0.6 V to 0.8 V (nominal 0.7 V)
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.