UGC NET Electronic Science December 2024 Question Paper with Answer Key and Full Explanations
📥 Download Question Paper (PDF)Q.1
Answer: Option (3)
Q.2
Answer: Option (3)
Q.3
Answer: Option (3)
Q.4
Answer: Option (4)
For inverting OP-AMP
V0 = -Rf/Rin * Vin
or, V0 = -10/(16/8) * 2 = -10 V
Q.5
Answer: Option (3)
6. The feedback network used in Colpitts oscillator is given below:
The value of \(C_2\) and feedback factor '\(\beta\)' for oscillating frequency of 1 MHz is:
The oscillating frequency \(f_0\) of a Colpitts oscillator is given by:
\(f_0 = \frac{1}{2\pi\sqrt{L_{eq}C_{eq}}}\)
where \(L_{eq}\) is the total inductance and \(C_{eq}\) is the equivalent capacitance.
In this case, \(L = 20 \text{ mH}\) and \(C_1 = 4 \text{ pF}\).
The equivalent capacitance for two capacitors in series is \(C_{eq} = \frac{C_1 C_2}{C_1 + C_2}\).
The given frequency is \(f_0 = 1 \text{ MHz}\).
Substituting the values:
\(1 \times 10^6 = \frac{1}{2\pi\sqrt{20 \times 10^{-3} \times \frac{4 \times 10^{-12} \times C_2}{4 \times 10^{-12} + C_2}}}\)
Solving for \(C_2\):
Squaring both sides and rearranging, we get:
\((2\pi f_0)^2 L = \frac{C_1 C_2}{C_1 + C_2}\)
\(\frac{C_1 C_2}{C_1 + C_2} = \frac{1}{(2\pi \times 10^6)^2 \times 20 \times 10^{-3}}\)
Let's calculate the right-hand side value first:
\(\frac{1}{(2\pi \times 10^6)^2 \times 20 \times 10^{-3}} \approx \frac{1}{(39.48 \times 10^{12}) \times 20 \times 10^{-3}} \approx \frac{1}{789.6 \times 10^9} \approx 1.266 \times 10^{-12} \text{ F} = 1.266 \text{ pF}\)
So, \(\frac{4 \times C_2}{4 + C_2} = 1.266\) (where \(C_2\) is in pF)
\(4 C_2 = 1.266 (4 + C_2)\)
\(4 C_2 = 5.064 + 1.266 C_2\)
\(2.734 C_2 = 5.064\)
\(C_2 = \frac{5.064}{2.734} \approx 1.85 \text{ pF}\)
This is closest to 1.8 pF.
The feedback factor \(\beta\) is given by \(\beta = \frac{C_2}{C_1}\).
\(\beta = \frac{1.8 \text{ pF}}{4 \text{ pF}} = 0.45\)
Therefore, the correct option is 1.8 pF and 0.45.
Q.7
Answer: Option (4)
Q.8
Answer: Option (3)
Q.9
Answer: Option (3)
Q.10
Answer: Option (4)
Q.11
Answer: Option (3)
Solution Hint: 20 * 0.8 = 16KW ; 16 / 0.85 = 18.82
Q.12
Answer: Option (2)
Solution Hint: E = V/d; j = epsilon * dE/dt
Q.13
Answer: Option (3)
Q.14
Answer: Option (3)
Q.15
Answer: Option (2)
Solution Hint: 1 / (2 * pi * r)
Q.16
Answer: Option (3)
Solution Hint: 1 / RC
Q.17
Answer: Option (3)
Q.18
Answer: Option (2)
Solution Hint: Ftotal = (F1 + (F2 - 1)/G)
Q.19
Answer: Option (3)
Q.20
Answer: Option (3)
Solution Hint: V(t) = V0(1 - e^(-t/RC))
Q.21: Oscillator Matching
Answer: Option (4)
C. LC Oscillator → I: Uses inductors and capacitors for feedback and resonance. It is mainly used for radio frequency (RF) applications like Hartley and Colpitts oscillators.
D. Crystal Oscillator → II: Uses a quartz crystal for high stability. Crystals have sharp resonance and thus are used where high frequency stability is required.
Matching is: A-III, B-IV, C-I, D-II.
Q.22
Answer: Option (1)
To find the open-loop transfer function, we trace the forward and feedback paths step-by-step.
Forward Path: G1G2
Feedback Loops: G1H1, G2H2, and the nested loop G1G2H3 + G1G2H1H2.
Full transfer function becomes:
G1G2 / (G1H1 + G2H2 + G1G2H3 + G1G2H1H2)
Q.23: Rise Time Calculation
Bandwidth = 10 MHz
tr = 0.35 / Bandwidth
tr = 0.35 / (10 × 106)
tr = 35 ns.
Q.24: Effective Mass (mn*/m0)
Answer: Option (4)
- InSb: 0.014 | GaAs: 0.067 | AlAs: 0.15 | BN: 0.20 | GaP: 0.34
Ascending Order: InSb < GaAs < AlAs < BN < GaP (C, A, D, B, E).
Q.25: RTD Temperature Calculation
Answer: Option (1)
Given: R0 = 100 Ω, R = 150 Ω, α = 0.004/°C, T0 = 25°C
150 = 100(1 + 0.004(T - 25))
1.5 = 1 + 0.004(T - 25)
0.5 / 0.004 = T - 25
125 = T - 25 → T = 150°C
Q.26: SCR Conduction State
Answer: Option (2)
In an SCR (P-N-P-N) during conduction:
- J1 & J3: Forward biased.
- J2: Initially reverse biased, switches to forward bias once triggered.
Correct Condition: J1 and J3 forward biased, J2 reverse biased (blocking mode boundary).
Q.27: Priority of interrupts in 8086
Answer: Option (B)
Order from highest to lowest:
- Reset (B) - absolute highest, restarts processor
- Internal Interrupts (A) - traps and exceptions
- Software Interrupts (D) - user-defined INT instructions
- Non-maskable Interrupt (E) - cannot be disabled
- Hardware Interrupt (C) - INTR, triggered externally
Correct descending order: B > A > D > E > C.
Q.28: LVDT Application
Answer: Option (2)
An LVDT (Linear Variable Differential Transformer) measures displacement by providing a variable output voltage in proportion to the position of a movable core.
Q.29: Monolayer Formation Time
Answer: Option (2)
Formula: t = Ns / P
Computing values for each case:
(B) Ns / P = 5.28 × 1014 / 10-1 = 5.28 × 1015
(C) Ns / P = 2.64 × 1014 / 10-8 = 2.64 × 1022
(D) Ns / P = 7.29 × 1014 / 1 = 7.29 × 1014
Ascending order of time: A, D, B, C.
Q.30: Ampere's Circuital Law
Answer: Option (3)
∮ H • dL = Ienclosed
Relates the line integral of magnetic field intensity H to the total current enclosed, based on Maxwell's equations.
Q.31: Inverse Fourier Transform
Answer: Option (4)
Substitute X(ω) = δ(ω - ω0) into the inversion integral:
x(t) = (1 / 2Ï€) ∫ δ(ω - ω0)ejωt dω = (1 / 2Ï€) ejω0t
Q.32: Laplace & System Response
Answer: Option (2)
System Equation: s2Y(s) + 3sY(s) + 2Y(s) = X(s)
H(s) = 1 / (s2 + 3s + 2) = 1 / ((s + 1)(s + 2))
For unit step input X(s) = 1/s, applying partial fractions:
Y(s) = 1/2s - 1/(s+1) + 1/(2(s+2))
Inverse Laplace: y(t) = (1/2 - e-t + (1/2)e-2t) u(t)
Q.33: Conductivity Computation
Answer: Option (3)
Formula: σ = ωε tan δ
ε = 18 × 8.85 × 10-12 = 1.593 × 10-10
σ = (6.28e8) × (1.593e-10) × (9.85e-3) = 9.86 × 10-4 S/m
Q.34: System Cascade Properties
Answer: Option (3)
- Statement A: Cascade of linear systems is linear (True).
- Statement C: Cascade of causal systems is causal (True).
- Statement D: Cascade of time-invariant systems is time-invariant (True).
Q.35: Barkhausen Criterion
Answer: Option (2)
- Statement A: Loop gain magnitude must be 1 (True).
- Statement D: Total phase shift must be 0° or 360° (True).
Q.36: Error Constants
Answer: Option (2)
Evaluating Position Error Constant (Kp) for G(s) = 50 / [(1+0.15s)(s+10)]:
Kp = lims→0 G(s) = 50 / 10 = 5
Q.37: 8086 Control Flags
Answer: Option (1)
Machine control flags include: Direction Flag (DF), Trap Flag (TF), and Interrupt Flag (IF).
Q.38: Communication Systems
Answer: Option (2)
- Shannon-Hartley: Capacity formula.
- Huffman Coding: Source coding/Lossless compression.
- OFDM: 4G/5G Physical Layer.
Q.39. Option (4)
Q.40: 8051 Stack Pointer (SP)
Answer: Option (4)
SP is an 8-bit register. It is initialized to 07H on reset. Data is stored at 08H on the first PUSH.
Q.41: Z-Transform ROC
Answer: Option (3)
- Causal finite → All Z except 0.
- Anticausal finite → All Z except infinity.
- Causal infinite → |z| > r.
Order: A, D, B, C.
Q.42: Inverse Transducers
Answer: Option (4)
Piezoelectric crystals act as inverse transducers by converting electrical voltage into mechanical motion (vibration).
Q.43: Gauge Factor (GF) Analysis
Answer: Option (2)
Formula: GF = (Δl / l) ⋅ (1 + 2 ⋅ ΔD / D)
(B) GF = (1/15)(1 + 2*0.02/1.5) = 0.0685
(D) GF = (2/20)(1 + 2*0.02/0.5) = 0.1080
(C) GF = (2/15)(1 + 2*0.01/0.5) = 0.1387
Ascending Order: A, B, D, C.
Q.44: Common-Source MOSFET Amplifier
Parameters: gm = 2 mA/V, rd = 10 kΩ, RD = 50 kΩ.
Total output resistance (Rout):
Rout = (rd ⋅ RD) / (rd + RD) = (10 × 50) / 60 = 8.33 kΩ
Voltage gain (Av):
Av = -gm ⋅ (rd || RD) = -2 × 10⁻³ × 8333.33 = -16.66
Q.45: Digital Logic Fan-out (74HC driving 74LS)
For logic LOW: Max inputs = IOL(max) / IIL(max) = 4 mA / 0.4 mA = 10
For logic HIGH: Max inputs = IOH(max) / IIH(max) = 4 mA / 20 μA = 200
Since 10 is the limiting case (LOW state), the fan-out is 10.
Q.46: Pinch-off Voltage (VP) Calculation
Answer: Option (3)
VP = (q ND d2) / (2ε)
Given: q = 1.6e-19, ND = 2e24 m⁻³, d = 40e-9 m, ε = 1.089e-10.
VP = (1.6e-19 ⋅ 2e24 ⋅ (40e-9)²) / (2 ⋅ 1.089e-10) ≈ 2.35 V.
Q.47: 3-bit Asynchronous Binary Counter
Answer: Option (3)
- A. Correct: Pulse 5 high phase indicates FF2 and FF3 in hold mode.
- B. Incorrect: FF3 toggles every 4 pulses, not on the 6th pulse edge.
- C. Correct: During pulse 5 high, FF1 is ready to toggle while FF3 holds.
- D. Incorrect: FF2 does not toggle on pulse 4 edge.
Correct statements: A and C only.
Q.48: Multivibrator & Circuit Matching
Answer: Option (2)
- A. Monostable → III (Delay circuit)
- B. Bistable → IV (Flip-flop/Counter)
- C. Clamping → II (DC Level shifter)
- D. Schmitt trigger → I (Noise removal/Hysteresis)
Q.49: P-N Junction Diode Statements
Answer: (2)
A. True: Equilibrium requires a flat Fermi level.
B. False: Depletion width WD depends on doping (WD ∝ √(1/N)).
C. True: 1/C² vs V plot yields doping density and built-in potential.
D. False: Avalanche breakdown (> 6V) has a positive temperature coefficient.
Correct statements: A and C only.
Q.50: BJT Circuit Voltages
Answer: Option (1)
Given: β=45, VBE=0.7V, RB=100k, RC=1.2k, VEE=-9V.
VB = VE + VBE = -9 + 0.7 = -8.3 V
IB = 8.3 / 100k = 83 μA | IC = 45 ⋅ 83μA = 3.735 mA
VC = 0 - (3.735mA ⋅ 1.2k) = -4.48 V
Q.51: Scaling Principles
Answer: Option (3)
Statements A, C, and D are correct. Scaling is systematic (A), miniaturization alters characteristics (C), and supply voltages are often kept constant for I/O compatibility (D).
Q.52: Modulation Power Requirements
Answer: 3 (Correct order: A, B, C, D, E)
Order of decreasing power: FM > AM > DSB-SC >VSB-SC> SSB-SC. SSB-SC is the most efficient.
Q.53: Optical Fiber Windows
Answer: 4 (A, B, E)
Standard windows: 850 nm (First), 1310 nm (Second), and 1550 nm (Third).
Q.54: Microscope & Spectroscopy Matching
Answer: Option (2)
Matchings: A → II (SEM-Surface), B → I (XRD-Crystal), C → III (TEM-Internal), D → IV (EDS-Composition).
Q.55: Power Device Matching
Answer: Option (4)
A → II (DIAC-Bidirectional Diode), B → IV (TRIAC-AC Control), C → I (IGBT-Bipolar Mode), D → III (SCR-Unidirectional).
Q.56: Superheterodyne Receiver
Answer: Option (4)
A. True: Mixer stage generates sum and difference frequencies.
C. True: IF amplifier operates at a fixed frequency for better gain/selectivity.
Q.57: 8086 Physical Address Calculation
Answer: Option (2)
Physical Address = (Segment × 10H) + Offset
(1005H × 10H) + 5555H = 10050H + 5555H = 155A5H
Q.58: D-to-A Converter Weights
Answer: Option (3)
Weight is inversely proportional to resistance: P (1k) > R (2k) > Q (4k) > S (8k).
Order: C, D, A, B.
Q.59: Antenna Gain Ranking
Answer: Option (3)
Decreasing order: Horn (D) >Yagi-Uda (C) > Dipole (B) > Monopole (A)
Typical gains: Yagi (~10dBi), Horn (~15dBi), Monopole (lowest gain), Dipole (~2dBi).
Q.60: MOS Fabrication Steps
Answer: Option (1)
Correct order: B, A, C, D. (Thick oxide growth → Gate oxide/PSG deposition → n-type doping → Planarization/Metallization).
Q.61: Silicon Growth (FZ vs CZ)
Answer: Option (3)
FZ process offers lower contamination compared to CZ because it does not use a crucible.
Q.62: Cellular Channel Splitting
Answer: Option (3)
Logic: 12 macrocells × 3 microcells per macrocell = 36 total cells.
36 cells × 10 channels = 360 channels.
Q.63: Hall Voltage Calculation (n-type Ge)
Answer: Option (3)
vd = μn ⋅ Ex = 0.38 ⋅ 500 = 190 m/s
VH = vd ⋅ Bz ⋅ w = 190 ⋅ 0.1 ⋅ 4e-3 = 76 mV
Direction: -Y direction (electrons accumulate on -y side).
Q.64: Transducer Definitions
Answer: Option (4)
Statements A, B, and C are true. Current-carrying coils in magnetic fields (motors) are inverse transducers, not primary sensors.
Q.65: Thin Film Vacuum Environment
Answer: Option (1)
Vacuum increases the mean free path of atoms, allowing them to reach the substrate without collisions.
Q.66: MOSFET Transconductance (gm)
Answer: Option (4)
In saturation:
gm = dID / dVgs = 2ID / (Vgs - Vth)
Q.67: 8051 Port 3 Pins
Answer: Option (2)
Port 3 matching: P3.0 (RxD), P3.1 (TxD), P3.2 (INT0), P3.3 (INT1), P3.4 (T0), P3.5 (T1).
8051 Port 3 Alternate Functions
| Pin | Function | Description |
|---|---|---|
| P3.0 | RXD | Serial Data Input |
| P3.1 | TXD | Serial Data Output |
| P3.2 | INT0 | External Interrupt 0 |
| P3.3 | INT1 | External Interrupt 1 |
| P3.4 | T0 | Timer 0 Input |
| P3.5 | T1 | Timer 1 Input |
| P3.6 | WR | Data Memory Write |
| P3.7 | RD | Data Memory Read |
Matching: A-III (P3.0-RXD), B-I (P3.3-INT1), C-IV (P3.6-WR), D-II (P3.7-RD).
Q.68: Hertzian Dipole Radiated Power
Answer: Option (1)
Prad = 40Ï€2 I02 • (dl / λ)2
Where I0 is peak current, dl is length, and λ is wavelength.
Q.69: 8086 ADD and DAA Instructions
Answer: Option (1)
Initial: AL = 73H, CL = 29H.
ADD AL, CL: 73H + 29H = 9CH (CF=0)
DAA Adjustment: Lower nibble (C > 9) → add 6: 9C + 6 = A2. Upper nibble (A > 9) → add 60: A2 + 60 = 102H.
Final result: AL = 02H, CF = 1.
Q.70: Entropy Calculation
Answer: Option (1)
H = -∑ pi log2(pi)
If p=1, H = -1 • log2(1) = 0. (No uncertainty).
Q.71: Root Type Analysis for Parameter K
Answer: Option (3)
Given S = -1 ± √(1 - K)
B. K = 0 → Roots are S1=0, S2=-2 (III)
C. K = 1 → Roots are Real & Equal (I)
D. K > 1 → Roots are Complex Conjugates (II)
Matching: A-IV, B-III, C-I, D-II.
Q.72: Gunn Diode Characteristics
Answer: Option (1)
B. True: Operates on hot electron transfer to higher energy valleys.
C. True: Relies on bulk negative resistance, not junctions.
Q.73: Bandgap Descending Order (0 K)
Answer: Option (2)
Values: Diamond (5.5 eV) > GaN (3.4 eV) > AlP (2.45 eV) > CdSe (1.84 eV) > PbTe (0.32 eV).
Order: B, D, A, C, E.
Q.74: Lattice Network Z-Parameters
Answer: Option (3)
Lattice with series arms 4 Ω and diagonal arms 8 Ω.
Z11 = Z22 = (4+8) || (8+4) = 12 || 12 = 6 Ω
Z12 = Z21 = ½(Zdiag - Zseries) = ½(8 - 4) = 2 Ω
Q.75: Switch and Logic Gate Circuit
Answer: Option (3)
AND gate output = A. OR gate output = Y.
If S1 is Open (1) and S2 is Closed (0):
AND(1,0) → A = 0 (Low). OR(1,0,0) → Y = 1 (High).
Q.76: TTL Speed-Power Product (Descending)
Answer: Option (2)
Order: 74 Standard (100pJ) > 74S (60pJ) > 74LS (18pJ) > 74ALS (4pJ).
Order: B, C, A, D.
Q.77: Amplifier Conduction Angles
Answer: Option (3)
Order: Class A (360°) > Class AB (180-360°) > Class B (180°) > Class C (<180 br=""> Order: B, A, D, C.
Q.78: Nyquist Stability Criterion
Answer: Option (1)
A. True: Critical point is -1 + j0.
C. True: Stable if no zeros (closed-loop poles) in the right-half S plane.
Q.79: Solar Cell Open-Circuit Voltage
Answer: Option (4)
Voc = (kT/q) ln(IL/Is + 1)
Voc = 0.025875 ⋅ ln(108) ≈ 0.48 V
Q.80: Antenna Property Matching
Answer: Option (3)
A-IV (λ/2 Dipole: 73 Ω), B-I (Wire: l > λ/2), C-II (Reflector: Parabolic), D-III (Arrays: High Directivity).
Q.81: K-map Function Minimization
Answer: Option (3)
Y = ∑m(7, 9, 10, 11, 12, 13, 14, 15).
Filling the K-map for variables AB and CD allows for grouping and term reduction.
K-map Grouping
- Group 1 (m12-15): AB = 11 → term = AB
- Group 2 (m10, 11, 14, 15): A=1, D=1 → term = AD
- Group 3 (m9, 11): A=1, B=0, D=1 → term = A·B'·D
- Group 4 (m7, 15): B=1, C=1, D=1 → term = BCD
- Group 5 (m10, 14): A=1, C=1, D=0 → term = AC
Simplified Boolean expression: Y = AB + AC + AD + BCD
Correct Answer: Option 3.
Q.82: Metal-Semiconductor Systems
Answer: Option (3)
Statement A: (Correct) Covalent surfaces have high surface states near the neutral level (Eg/3).
Statement C: (Correct) Optical dielectric constant dependence on wavelength is similar for Si, Ge, and GaAs.
Note: Statement B is an oversimplification for semiconductors, and Statement D is false because higher barrier height increases resistance.
Q.83: Brillouin Zone Symmetry (FCC/Diamond)
Answer: Option (3)
- A. Γ (Center): Degeneracy = 1 (II)
- B. X (100): Degeneracy = 6 (I)
- C. K (110): Degeneracy = 12 (IV)
- D. L (111): Degeneracy = 8 (III)
Q.84: 8051 Instruction Trace
Answer: Option (4)
RR A (Rotate Right) → A = 11100110
CPL A (Complement) → A = 00011001
SWAP A (Nibble Swap) → A = 10010001 = 91H
Q.85: 7445 Decoder LED Current
Answer: Option (2)
Supply = 5V, VLED = 1.2V, R = 1kΩ.
VR = 5 - 1.2 = 3.8V
I = VR / R = 3.8V / 1kΩ = 3.8 mA
Q.86: Reduction-Projection Lithography
Answer: Option (1)
Resolution (R) = K1(λ / NA)
Depth of Focus (DOF) = K2(λ / NA2)
Q.87: Reflex Klystron Characteristics
Answer: Option (1)
- A. True: It is a single cavity klystron.
- C. True: Bunching parameter has a negative sign.
- E. True: Space charge effects are neglected in basic analysis.
Q.88: Peripheral IC Matching
Answer: Option (4)
- A. 8259 → IV (Interrupt Controller)
- B. 8251 → III (USART)
- C. 8279 → II (Keyboard/Display)
- D. 8254 → I (Timer)
Q.89: N-channel FET Statements
Answer: Option (3)
B. True: Larger transconductance than P-channel due to electron mobility.
D. True: Electrons provide larger mobility than holes used in P-channels.
Q.90: Communication System Matching
Answer: Option (1)
- A. FM → III (Carson's Rule)
- B. DSB-SC → I (Costas Receiver)
- C. PIN → II (Photodetector)
- D. Optical → IV (WDM)
Q.91: 4x1 Multiplexer Output
Answer: Option (4)
Select Lines: S1=A=0, S0=B=1 (Binary 01).
The MUX selects input I1. Given I1 = C̅.
Result: F = C̅.
Q.92: 4-bit Even Parity Checker
Answer: Option (3)
Output C = x ⊕ y ⊕ z ⊕ P
For Option 3: x=1, y=0, z=1, P=1.
C = 1 ⊕ 0 ⊕ 1 ⊕ 1 = 1. (Correct).
Q.93: Mealy State Diagram Analysis
Answer: Option (4)
Analyzing transitions (Input/Output):
From state d, if input = 0, the transition goes to b with output 0. (Matches Option 4).
Q.94: 74ALS164 Shift Register (6th Pulse)
Answer: Option (3)
Initial: 00000000. Input A·B = 1.
After 6 clocks (shifting in 1s): Q7 to Q0 = 00111111.
Represented as 11111100 (Option 3).
Q.95: PLD Logic Output
Answer: Option (1)
Connections for O1 OR gate:
AND1 = A̅B | AND2 = AB̅.
Result: O1 = A̅B + AB̅ (XOR Logic).
Q.96: Lossless Line Phase Constant
Answer: Option (2)
β = ω√(LC)
Q.97: Quarter-Wavelength (λ/4) Line
Answer: Option (3)
The normalized impedance inverts: Zin = Z02 / ZL.
Q.98: VSWR Definition
Answer: Option (1)
VSWR = |V|max / |V|min
Q.99: Characteristic Impedance (Z0)
Answer: Option (3)
Z0 = √((R + jωL) / (G + jωC))
Q.100: Transmission Line Modeling
Answer: Option (1)
Transmission lines use the Distributed Method (R, L, G, C parameters over infinitesimally small lengths).