Home / UGC NET PYQ / June 2026 Solved UGC NET Electronic Science June 2026 Question Paper with Answer Key and Full Explanations 📥 Download Question Paper (PDF) 2025 2024 2023 2022 2021 2020 Explanations 1. Answer: Option (3) Total Students = 900 Urban Students = 7/10*400 + 4/5*500 = 680 680/900 * 100 = 75.56% 2. Answer: Option (1) Urban Students = 7/10*400 + 4/5*500 + 4/7*700 + 1/2 * 600 = 1380 3. Answer: Option (3) Male = 700*4/7 = 400 Female = 700*3/7=300 400-300=100 100/400*100 = 25% 4. Answer: Option (2) 400*3/10*40/100 + 500*1/5*60/100+700*3/7*50/100+600*1/3*30/100 = 318 5. Answer: Option (2) Total Female Students in B,C, and D colleges = 600 Female Post-graduates  = 100*60/100 + 300*50/100 + 200*30/100 = 270 270/600 *100 = 45% 📌 Summary Checklist for Success Don't just read the answers. Follow these tips to clear JRF: Solve at least 10 y...

  For a 40-character-wide LCD , the upper address range is: 27 H \boxed{27H} Because the first line has 40 positions: 00 H → 27 H 00H \rightarrow 27H Remember: 27 H = 39 27H = 39 and counting from 0 to 39 = 40 characters . Answer: A (c) only 0100111

What is a BJT? A BJT (Bipolar Junction Transistor) is a three-terminal semiconductor device used for amplification and switching . The three terminals are: Base (B) Collector (C) Emitter (E) The two main types of BJTs are NPN and PNP . NPN BJT Collector (C) Base (B) Emitter (E) NPN In an NPN transistor, a small base current controls a much larger collector current. Basic BJT Current Relationship I E = I C + I B The collector current is approximately related to the base current by: I C = β I B Here, β (beta) is the DC current gain of the transistor. Example: If β = 100 and I B = 20 μA: I C = 100...

  BJT = transistor &  Common-emitter = one way of connecting that transistor BJT as Common Emitter Here's the important diagram: +VCC R C VOUT VIN C 1 B C E GND Input Input voltage is applied between: Base ↔ Emitter Output Output is taken between: Collector ↔ Emitter  why it's called Common Emitter? The emitter is connected to the common reference (ground) . The emitter is common to both the input and output circuits. What actually happens? Suppose we increase the input voltage increases the base current and it increases the Ic Remember: I C = β I B I_C = \beta I_B Now look at the resistor R C R_C . The voltage across R C R_C is: V R C = I C R C V_{RC}=I_C R_C And: V O U T = V C C − I C R C V_{OUT}=V_{CC}-I_C R_C So if I C I_C increases: I C ↑ I_C↑ then: I C R C ↑ I_CR_C↑ therefo...

  Given Circuit The circuit contains: Left resistor: \(2\Omega\) Right resistor: \(2\Omega\) Vertical resistor: \(6\Omega\) Dependent current source: \(2V_2\) A, directed upward \(I_1\) and \(I_2\) enter the two-port network. Step 1: Define the Middle Node Voltage Let the voltage at the middle node be \(V_x\). Apply KCL at the middle node: $$ \frac{V_x-V_1}{2} + \frac{V_x-V_2}{2} = 2V_2 $$ Multiplying by 2: $$ 2V_x-V_1-V_2=4V_2 $$ Therefore: $$ \boxed{ V_x=\frac{V_1}{2}+\frac{5V_2}{2} } $$ Step 2: Find \(I_1\) The current entering port 1 is: $$ I_1=\frac{V_1-V_x}{2} $$ Substitute \(V_x\): $$ I_1= \frac{ V_1- \left( \frac{V_1}{2}+\frac{5V_2}{2} \ri...

Dual-Slope ADC Given C = 0.1 µF R = 100 kΩ V ref = 2 V V o = 10 V Integrator Equation For an integrator: V o = V ref t / RC Therefore: t = V o RC / V ref Calculate RC RC = (100 × 10 3 ) (0.1 × 10 −6 ) = 0.01 s Calculate Conversion Time t = 10(0.01) / 2 t = 0.05 s = 50 ms Therefore, t = 50 ms . Answer: B (50 ms) Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → Further Reading: GATE EC Previous Year Papers with Solutions

Induction Motor Torque Given Frequency, f = 75 Hz Slip, s = 0.04 Rotor resistance, R 2 = 1 Ω Line voltage, V L = 415 V Phase Voltage V ph = 415 / √3 V Since stator voltage drop and rotor reactance are neglected: I 2 = V ph / (R 2 / s) Torque in Synchronous Watts P T = 3V ph 2 / (R 2 / s) P T = 3(415/√3) 2 × 1/0.04 P T = 415 2 / 0.04 ≈ 6889 W Thus, T = 6889 synchronous watts Answer: A (6889 synchronous watts) Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → Furth...

Transistor Power Loss Given β min = 8 R C = 11 Ω V CC = 200 V V CE(sat) = 1 V V BE(sat) = 1.5 V Overdrive factor = 5 Collector Saturation Current I C = (V CC − V CE(sat) ) / R C I C = (200 − 1) / 11 = 18.09 A Forced Beta β f = 8 / 5 = 1.6 Therefore, the base current is: I B = I C / β f I B = 18.09 / 1.6 = 11.31 A Power Loss P = V CE(sat) I C + V BE(sat) I B P = (1)(18.09) + (1.5)(11.31) P ≈ 18.09 + 16.97 = 35.06 W Therefore, P ≈ 35.07 W . Answer: A (35.07 W) Browse All Solved Papers (2012 - 2025) → ...

FDM Bandwidth Given There are 10 channels , each having a bandwidth of 50 kHz . Channel Bandwidth 10 × 50 = 500 kHz There are 10 − 1 = 9 guard bands , with each guard band having a bandwidth of 1 kHz. 9 × 1 = 9 kHz Hence, the total FDM bandwidth is: BW = 500 + 9 = 509 kHz Answer: C (509 kHz) Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → Further Reading: GATE EC Previous Year Papers with Solutions

VSB Transmitter Given VSB transmitted power, P VSB = 0.625 kW Modulation index, m = 0.6 Vestige = 25% = 0.25 VSB Power Equation For a VSB transmitter: P VSB = m 2 / 4 P c (1 + 0.25) Substituting the given values: 0.625 = (0.6) 2 / 4 P c (1.25) Therefore, P c = (0.625 × 4) / (0.36 × 1.25) P c = 5.56 kW Answer: B (5.56 kW) Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → Further Reading: GATE EC Previous Year Papers with Solutions

Charge on the Sphere Given r 1 = 30 cm = 0.3 m Volume charge density: ρ v = ρ 0 r / r 1 ρ 0 = 200 pC/m 3 Total Charge The total charge is obtained by integrating the volume charge density throughout the sphere: Q = ∫ V ρ v dV Using the spherical volume element: dV = r 2 sinθ dr dθ dφ Therefore, Q = ρ 0 r 1 ∫ 0 r 1 r 3 dr ∫ 0 π sinθ dθ ∫ 0 2π dφ Evaluating the integrals: Q = ρ 0 r 1 × r 1 4 / 4 × 2 × 2π Q = πρ 0 r 1 3 Substituting the given values: Q = π(200)(0.3) 3 Q ≈ 16.96 pC Q ≈ 17 pC Answer: C (17 pC) ...

  Given: Z O C = 150 ∠ 25 ∘ Ω Z_{OC}=150\angle25^\circ\ \Omega Z S C = 37.5 ∠ ( − 35 ∘ ) Ω Z_{SC}=37.5\angle(-35^\circ)\ \Omega Use: Z 0 = Z O C Z S C Z_0=\sqrt{Z_{OC}Z_{SC}} Step 1: Multiply Magnitudes: 150 × 37.5 = 5625 150\times37.5=5625 Angles: 25 ∘ + ( − 35 ∘ ) = − 10 ∘ 25^\circ+(-35^\circ)=-10^\circ Therefore, Z O C Z S C = 5625 ∠ ( − 10 ∘ ) Z_{OC}Z_{SC}=5625\angle(-10^\circ) Step 2: Take square root Z 0 = 5625 ∠ − 10 ∘ 2 Z_0=\sqrt{5625}\angle\frac{-10^\circ}{2} Z 0 = 75 ∠ ( − 5 ∘ ) Ω \boxed{Z_0=75\angle(-5^\circ)\ \Omega} Answer: Z 0 = 75 ∠ − 5 ∘ Ω \boxed{Z_0=75\angle-5^\circ\ \Omega} Option B Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → Further Reading: GATE EC Previous Year Papers with Solutions

  If V r e f V_{ref} is the output reference value and V f s V_{fs} is the ideal full-scale output voltage when all digital inputs are 1. Which of the following equations is correct? A. V r e f = V f s ( 1 − 1 2 n ) V_{ref} = V_{fs}\left(1-\frac{1}{2^n}\right) B. V f s = V r e f ( 1 2 n − 1 ) V_{fs} = V_{ref}\left(\frac{1}{2^n}-1\right) C. V f s = V r e f ( 1 − 1 2 n ) V_{fs} = V_{ref}\left(1-\frac{1}{2^n}\right) D. V r e f = V f s ( 1 2 n − 1 ) V_{ref} = V_{fs}\left(\frac{1}{2^n}-1\right) The correct answer is C : V f s = V r e f ( 1 − 1 2 n ) \boxed{V_{fs}=V_{ref}\left(1-\frac{1}{2^n}\right)} Why? For an n-bit DAC , the largest digital input is: 111 … 111 = 2 n − 1 111\ldots111 = 2^n-1 The DAC output is: V o = V r e f Digital input 2 n V_o=V_{ref}\frac{\text{Digital input}}{2^n} At the maximum input: V f s = V r e f 2 n − 1 2 n V_{fs}=V_{ref}\frac{2^n-1}{2^n} Separate the fraction: V f s = V r e f ( 2 n 2 n − 1 2 n ) V_{fs}=V_{ref}\left(\frac{2^n}{2^n}-\frac{1}{2^n}\r...

  Q. The final code after encoding data bits 1101 into 7-bit even parity Hamming Code is: A. 1110101 B. 1011101 C. 1010101 D. 0110101 Hamming (7,4) – Step by Step Data bits: D = [1,1,0,1] Step 1: Parity-Check Matrix H H = [1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 0 1 1 1 1] Parity bits are at positions 1,2,4; data bits at positions 3,5,6,7. Step 2: Extract P from H Using standard form H = [P^T | I], we get: P^T (columns 3,5,6,7) = [1 1 0 1 1 0 1 1 0 1 1 1] Transpose to get P : P = [1 1 0 1 0 1 0 1 1 1 1 1] Step 3: Generator Matrix G = [I | P] G = [1 0 0 0 | 1 1 0 0 1 0 0 | 1 0 1 0 0 1 0 | 0 1 1 0 0 0 1 | 1 1 1] Step 4: Encode Data → Codeword Multiply D × G (mod 2): D = [1 1 0 1] C = D × G = [1 1 0 1 1 0 0] In the above the sequence is D1 D2 D3 D4 P1 P2 P3   (But Transmitted Codeword should be = [P1 P2 D1 P3 D2 D3 D4] or 1010101 ) Answer: C. 1010101 Another Approach Data bits: 1101 1101 For a 7-bit Hamming code , parity bi...

  Q. The A/O gates in which an additional variable or a combination of variables can be included in the logic operation are called: Options: A. AOI Gates B. Expandable Gates C. Variable Gates D. Scalable Gates The correct answer is: B. Expandable Gates \boxed{\text{B. Expandable Gates}} Why? Expandable gates allow additional variables/inputs to be incorporated into the logic operation. Final answer C. 1010101   Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → Further Reading: GATE EC Previous Year Papers with Solutions

Home / UGC NET PYQ / June 2020 Solved UGC NET Electronic Science June 2020 Question Paper with Answer Key and Full Explanations 📥 Download Question Paper (PDF) 2025 2024 2023 2022 2021 2020 Explanations 1.  Answer: Option (2) For Varactor Diode, m = ln(C2/C1) / ln{(1+V2/phi) / (1+V1/phi)} 2.  Answer: Option (4) 3.  Answer: Option (2) 4.  Answer: Option (2) 5.  Answer: Option (4) 6.  Answer: Option (4)  L= (5*10^-3) / P Or, P = (5*10^-3) / (5*10^7) = 10^-10 7.  Answer: Option (4) 8.  Answer: Option (3) 9.  Answer: Option (3) 10.  Answer: Option (1) 11.  Answer: Option (2) 12.  Answer: Option (2) 13.  Answer: Option (1) Slew Rate Slew rate tells us how fast the output voltage of an amplifier/op-amp can change . It is defined as: S R = max ...