UGC NET Electronic Science June 2020 Question Paper with Answer Key & Detailed 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:

$$\boxed{SR=\max\left|\frac{dV_o}{dt}\right|}$$

Unit:

$$\boxed{\text{V/}\mu s}$$

For example, if an op-amp has

$$SR=1\ V/\mu s$$

it can change its output by at most 1 volt every microsecond.

For given, SR >=2*π*f*Signal Amplitude

For, 10V peak, SR>=6.28*f*10/(10^-6) V/microsecond

Or, 0.5>= 6.28*f*10/(10^-6) V/microsecond

Or, f = 8000 or 8 KHz

Similarly, for 1 V peak, answer is 80 KHz

14. Answer: Option (3)

15. Answer: Option (2)

16. Answer: Option (3)

For an ideal ADC

V0 = Vref * (Digital Input/2^n)

where, 2^n = number of levels

and Digial Input is (2^n - 1) as all digital inputs are 1

So, Vfs = Vref*{(2^n - 1)/2^n)

Or, Vfs = Vref*(1 - 1/2^n)

17. Answer: Option (2)

18. Answer: Option (1)

Data bits:

$$1101$$

For a 7-bit Hamming code, parity bits are at positions:

$$1,\;2,\;4$$

Data bits go to positions:

$$3,\;5,\;6,\;7$$

Place $1101$:

Position	1	2	3	4	5	6	7
Bit	$P_1$	$P_2$	1	$P_4$	1	0	1

Using even parity:

$P_1$: positions 1,3,5,7

$$P_1+1+1+1=\text{even}$$

There are three 1s, so:

$$P_1=1$$

$P_2$: positions 2,3,6,7

$$P_2+1+0+1=\text{even}$$

Already two 1s:

$$P_2=0$$

$P_4$: positions 4,5,6,7

$$P_4+1+0+1=\text{even}$$

So:

$$P_4=0$$

Therefore:

$$\boxed{1010101}$$

Answer: C. 1010101

19. Answer: Option (2)

20. Answer: Option (4)

21. Answer: Option (1)

A <== 85H

CPL A ==> 01111010

ADD A, # ==>

01111010

+

10000101

= FFH

22. Answer: Option (3)

23. Answer: Option (1)

24. Answer: Option (2)

25. Answer: Option (2)

26. Answer: Option (4)

For TE10 mode:

Cut-off frequency, fc = c/2a = 3 * 10^8 / 2 * 7 * 10^(-2)

Phase velocity = c / √{1 - (fc/f)^2}

= 4*78 * 10^8 m/s (approx.)

27. Answer: Option (4)

28. Answer: Option (3)

Q = ∫_V ρ_v dV

Using the spherical volume element:

dV = r² sinθ dr dθ dφ

Therefore,

Q = ρ₀ r₁ ∫₀^r₁ r³ dr ∫₀^π sinθ dθ ∫₀^2π dφ

Evaluating the integrals:

Q = ρ₀ r₁ × r₁⁴ / 4 × 2 × 2π

Q = πρ₀r₁³

Substituting the given values:

Q = π(200)(0.3)³

Q ≈ 16.96 pC

28. Answer: Option (2)

For, Single Sideband (SSB)

Pt = Pc(1+m^2/4)

where, m = modulation index

Vestigial Sideband (VSB) keeps:

• one complete sideband

• a small portion of the other sideband

Given, Pc{(m^2)/4 + (m^2)/4*(0.25)} = 0.625

Or, Pc{5*(m^2)/16} = 0.625

Or, Pc = 5.56 kW

30. Answer: Option (4)

Dispersion is the phenomenon in which light is separated into its component frequencies or wavelengths.

31. Answer: Option (3)

Given, a FDM system, 10 channels are multiplexed having bandwidth of 50KHz each.

Guard band between the channels are 1KHz.

Minimum bandwidth required for transmission is:

(50*10) + 1*(10-9) = 509KHz

32. Answer: Option (3)

 The fundamental purpose of dividing a geographical area into cells in cellular communication is to enable frequency reuse.

33. Answer: Option (1)

 Ic = Vcc - Vce / Rc 

= (200-1)/11

=18.09A

forced β = 8/5 = 1.6

β.Ib = Ic

Ib = Ic / β = 18.09/1.6 = 11.31A

Power loss in the transistor

Vce*Ic + Vbe*Ib

= 1 * 18.09 + 1.5*11.31

= 18.09 + 16.97

= 35.06W

34. Answer: Option (1)

Phase Voltage = 415/√3

Torque developed by motor

= 3 * (Phase Voltage)^2 / (R / s)

= 3* (415/√3)^2 / (1/0.04)

= 6889

35. Answer: Option (2)

36. Answer: Option (1)

37. Answer: Option (2)

RC = 100 kilo-ohm *0.1 micro-farad = 0.01

V0 = Vref * t/RC

Or, t = 50 m Second

38. Answer: Option (4)

39. Answer: Option (4)

40. Answer: Option (4)

41. Answer: Option (2)

42. Answer: Option (3)

43. Answer: Option (2)

44. Answer: Option (4)

45. Answer: Option (4)

46. Answer: Option (3)

Step 1: Determine the Middle Node Voltage

Let the voltage at the middle node be \(V_x\).

Applying KCL at the middle node:

$$ \frac{V_x-V_1}{2} + \frac{V_x-V_2}{2} = 2V_2 $$

Multiplying both sides by 2:

$$ 2V_x-V_1-V_2=4V_2 $$

Thus:

$$ \boxed{ V_x=\frac{V_1}{2}+\frac{5V_2}{2} } $$

Step 2: Determine \(I_1\)

The current entering port 1 is given by:

$$ I_1=\frac{V_1-V_x}{2} $$

Substituting the value of \(V_x\):

$$ I_1= \frac{ V_1- \left( \frac{V_1}{2}+\frac{5V_2}{2} \right) }{2} $$

Therefore:

$$ I_1= \frac{V_1}{4} - \frac{5V_2}{4} $$

Comparing with the Y-parameter equation:

$$ I_1=Y_{11}V_1+Y_{12}V_2 $$

Hence:

$$ \boxed{Y_{11}=\frac14} \qquad \boxed{Y_{12}=-\frac54} $$

Step 3: Determine \(I_2\)

The current entering port 2 is given by:

$$ I_2=\frac{V_2-V_x}{2} $$

Substituting the value of \(V_x\):

$$ I_2= \frac{ V_2- \left( \frac{V_1}{2}+\frac{5V_2}{2} \right) }{2} $$

Therefore:

$$ I_2= -\frac{V_1}{4} - \frac{3V_2}{4} $$

Comparing with:

$$ I_2=Y_{21}V_1+Y_{22}V_2 $$

Hence:

$$ \boxed{Y_{21}=-\frac14} \qquad \boxed{Y_{22}=-\frac34} $$

Final Y-Matrix

$$ \boxed{ [Y]= \begin{bmatrix} \frac14 & -\frac54 \\[2mm] -\frac14 & -\frac34 \end{bmatrix} \text{ S} } $$

Correct option: (C) (a) & (e) only

\(Y_{21}=-\frac14\) and \(Y_{22}=-\frac34\)

47. Answer: Option (4)

48. Answer: Option (1)

49. Answer: Option (4)

50. Answer: Option (4)

📌 Summary Checklist for Success

Don't just read the answers. Follow these tips to clear JRF:

• Solve at least 10 years of PYQs (2012-2025).
• Focus on numericals from Op-Amps, Control Systems, and EMFT.
• Master the 8086 Instruction set and 8051 SFRs.

Browse All Solved Papers (2012 - 2025) →

UGC-NET Electronics Study Material (Subject 88) →

Further Reading: GATE EC Previous Year Papers with Solutions