UGC NET Electronic Science December 2024 Question Paper with Answer Key & Detailed Solutions

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UGC NET Electronic Science December 2024 Question Paper with Answer Key and Full Explanations

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Q.1
Answer: Option (3)

Q.2
Answer: Option (3)

Solution

1. JMP SHORT LABEL

Intrasegment (within the same code segment).

Direct jump.

❌ Not intersegment indirect.

2. JMP 5000H:2000H

Intersegment (far jump because both CS and IP are specified).

Direct jump (address is explicitly given).

❌ Not indirect.

3. JMP [2000H]

The destination address is taken from memory location 2000H.

This is indirect.

In 8086, a far indirect jump can use a memory operand containing both IP and CS (depending on operand size), making it an intersegment indirect jump.

✅ Correct answer.

4. JMP [BX]

Indirect jump through memory addressed by BX.

Usually intrasegment (near indirect jump).

❌ Not intersegment.

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:

1. 0.8 pF, 0.2

2. 4 pF, 0.9

3. 1.8 pF, 0.45

4. 2.2 pF, 0.9

Answer: 3

Solution:

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

t_r = 0.35 / Bandwidth
t_r = 0.35 / (10 × 10⁶)
t_r = 35 ns.

Q.24: Effective Mass (m_n*/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 = N_s / P

Computing values for each case:

(A) N_s / P = 2.64 × 10¹⁴ / 1 = 2.64 × 10¹⁴
(B) N_s / P = 5.28 × 10¹⁴ / 10^-1 = 5.28 × 10¹⁵
(C) N_s / P = 2.64 × 10¹⁴ / 10^-8 = 2.64 × 10²²
(D) N_s / 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 = I_enclosed

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 / 2π) ∫ δ(ω - ω₀)e^jωt dω = (1 / 2π) e^jω₀t

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

ω = 2π × 10⁸ ≈ 6.28 × 10⁸
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).

If the individual blocks are non-linear, the overall system is usually non-linear, but it is not a fixed rule. For example, if h1(x)=log⁡(x) and h2(x)=e^x, both are non-linear, but their cascade e^log⁡(x)=x is linear. Therefore, this statement B is not universally true. (FALSE)

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 (K_p) for G(s) = 50 / [(1+0.15s)(s+10)]:

K_p = lim_s→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)

(A) GF = (1/24)(1 + 2*0.02/1.5) = 0.0428
(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: g_m = 2 mA/V, r_d = 10 kΩ, R_D = 50 kΩ.

Total output resistance (R_out):

R_out = (r_d ⋅ R_D) / (r_d + R_D) = (10 × 50) / 60 = 8.33 kΩ

Voltage gain (A_v):

A_v = -g_m ⋅ (r_d || R_D) = -2 × 10⁻³ × 8333.33 = -16.66

Q.45: Digital Logic Fan-out (74HC driving 74LS)

For logic LOW: Max inputs = I_OL(max) / I_IL(max) = 4 mA / 0.4 mA = 10

For logic HIGH: Max inputs = I_OH(max) / I_IH(max) = 4 mA / 20 μA = 200

Since 10 is the limiting case (LOW state), the fan-out is 10.

Q.46: Pinch-off Voltage (V_P) Calculation
Answer: Option (3)

V_P = (q N_D d²) / (2ε)

Given: q = 1.6e-19, N_D = 2e24 m⁻³, d = 40e-9 m, ε = 1.089e-10.

V_P = (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 W_D depends on doping (W_D ∝ √(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, V_BE=0.7V, R_B=100k, R_C=1.2k, V_EE=-9V.

V_B = V_E + V_BE = -9 + 0.7 = -8.3 V

I_B = 8.3 / 100k = 83 μA | I_C = 45 ⋅ 83μA = 3.735 mA

V_C = 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)

v_d = μ_n ⋅ E_x = 0.38 ⋅ 500 = 190 m/s

V_H = v_d ⋅ B_z ⋅ 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 (g_m)
Answer: Option (4)

In saturation:

g_m = dI_D / dV_gs = 2I_D / (V_gs - V_th)

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)

P_rad = 40π² 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 = -∑ p_i log₂(p_i)

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)

A. 0 < K < 1 → Roots are Real & Distinct (IV)
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(Z_diag - Z_series) = 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)

V_oc = (kT/q) ln(I_L/I_s + 1)

V_oc = 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)

MOV A, #0CDH → A = 11001101
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, V_LED = 1.2V, R = 1kΩ.

V_R = 5 - 1.2 = 3.8V

I = V_R / 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: Z_in = Z₀² / Z_L.

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).

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