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...

  Given, 𝑧^2= 2𝑥^2− 𝑦^2 F(x, y, z) = 2𝑥^2− 𝑦^2 - 𝑧^2 normal vector, ∇F = (4x, -2y, -2z) Evaluating at P = (1,0,√2) ∇F(1,0,√2) = (4, 0, 2√2) Normalizing the vector we get √2 i / √3 - 1 k /√3 Answer D

  Given,  𝑤⃗⃗ ̇ =𝐴𝑤⃗⃗ ,  𝑤⃗⃗ = e^t *  ux + e^-2t *  uy  𝑤⃗⃗ ̇ = e^t * ux - 2* e^-2t * uy with 𝑤⃗⃗ (𝑡=0)= ux + uy [ 1 ]   1 A.[e^t     ]     =   [  e^t     ]      e^-2t               -2e^-2t Or, A = [1   0  ]               0   -2 Answer A

  GATE Electronics and Communication (EC) Questions Paper With Answer Key Download Pdf [2026] Download Question Paper                See Answers   2026 | 2025 | 2024 | 2023 | 2022 | 2021 | 2020 GATE - EC 2026 Answers with Explanations  Q.1 Answer B Q.2 1st = 100 4th = 76 So the four top students are somewhere between 100 and 76. That's only a 24-mark range: 100 - 76 = 24 < 25 Hence those four students automatically satisfy S3. Answer A Q.3 Answer D Q.4 (log 𝑝^1/𝑛 𝑦)(log 𝑦^1/𝑛 𝑝)=16 Or, n.log p y * n.log y P = 16 Or, n^2 = 16 Or, n = 4 Answer B Q.5 Answer B Q.6 Answer A Q.7 Answer C Q.8 P k = 4*k P 1 +P 2 +P 3 +...+P 10 = 4 + 8 + 12 + ... +40 =4(1+2+3+...+10) =4*10*(10+1)/2 =220 Answer C Q.9 Answer C Q.10 Answer C Q.11 Answer A  See solution Q.12 Answer D  See solution Q.13 x(t) = u(t-2) * tu(t) L{x(t)} = (e^-2s) / s  . 1/s^2 = (e^-2s)/s^3 Answer D Q.14 Answer C Q.15 x1[n] = cos(2* π*10*n/...

  For T= 0,  Delay = (2+1) = 3ns For T= 1 Delay = (2+1) + (2+1) = 6ns For worst case it is 6 ns Answer (c) 6 ns

Flat vs Frequency-Selective Fading Simulator Flat vs Frequency-Selective Fading Interactive Multipath Channel Simulator Change the channel mode, delay spread, signal bandwidth, Doppler frequency, and SNR. Observe how the channel response changes. Parameters Channel Mode Flat Fading Frequency-Selective Fading Signal Type Pure Sine Wave Wideband Multitone Signal Center Frequency 1.00 MHz Signal Bandwidth 1.00 MHz Maximum Doppler 30 Hz SNR 25 dB Reflected Path Power -3 dB Maximum Path Delay 2.0 μs Number of Paths 4 Run Simulation Channel -- Signal BW -- Coherence BW -- RMS Delay Spread -- 1. Transmitted Signal 2. Channel Impulse Response 3. Channel Frequency Response 4. Received Signal 5. Signal Spectrum 6. Received Envelope 7. Doppler ...

  1. At node X The source current i s ( t ) i_s(t) splits into capacitor current and resistor current: i s ( t ) = i C ( t ) + i R ( t ) i_s(t)=i_C(t)+i_R(t) For the capacitor, i C = C d v C d t = d v C d t i_C=C\frac{dv_C}{dt} =\frac{dv_C}{dt} because C = 1 F C=1F . For the 1 Ω 1\Omega resistor, i R = v C 1 = v C i_R=\frac{v_C}{1}=v_C Therefore, i s = d v C d t + v C i_s=\frac{dv_C}{dt}+v_C so v ˙ C = i s − v C \boxed{\dot v_C=i_s-v_C} This gives the second state equation. 2. For the inductor branch The inductor 1 H 1H and resistor 1 Ω 1\Omega are parallel . Therefore, they have the same voltage . For the inductor, v L = L d i L d t v_L=L\frac{di_L}{dt} Since L = 1 H L=1H , v L = d i L d t v_L=\frac{di_L}{dt} For the parallel 1 Ω 1\Omega resistor, v R = i R v_R=i_R Since their voltages are equal, d i L d t = i R \frac{di_L}{dt}=i_R Now the source current splits between the inductor and resistor: i s = i L + i R i_s=i_L+i_R Therefore, i R = i s − i L i_R=i_s-i_L Hence, i ˙ L =...

Start with a simple baseband tone Take m ( t ) = cos ⁡ ( ω m t ) m(t)=\cos(\omega_m t) and carrier cos ⁡ ( ω c t ) . \cos(\omega_c t). If we simply multiply them: s ( t ) = m ( t ) cos ⁡ ( ω c t ) s(t)=m(t)\cos(\omega_ct) then s ( t ) = cos ⁡ ( ω m t ) cos ⁡ ( ω c t ) . s(t)=\cos(\omega_mt)\cos(\omega_ct). Using cos ⁡ A cos ⁡ B = 1 2 [ cos ⁡ ( A + B ) + cos ⁡ ( A − B ) ] , \cos A\cos B =\frac12[\cos(A+B)+\cos(A-B)], we get s ( t ) = 1 2 cos ⁡ ( ( ω c + ω m ) t ) + 1 2 cos ⁡ ( ( ω c − ω m ) t ) \boxed{s(t)=\frac12\cos((\omega_c+\omega_m)t)+\frac12\cos((\omega_c-\omega_m)t)} So we get two frequencies : ω c + ω m \boxed{\omega_c+\omega_m} and ω c − ω m . \boxed{\omega_c-\omega_m}. These are the upper sideband (USB) and lower sideband (LSB) . We want only one Suppose we want the upper side...