NTA UGC NET Electronic Science Paper 2 Solved Question Paper 2013 December

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NTA UGC NET Electronic Science Paper 2 Solved Question Paper 2013 December

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1.
NTA UGC NET Electronic Science Paper 2 Solved Question Paper 2013 December qn 1

Answer: (B)

2.

NTA UGC NET Electronic Science Paper 2 Solved Question Paper 2013 December qn 2

Answer: (A)

3. The radiative and non-radiative lifetimes of the minority carriers in the active region of a double heterojunction LED are 60 ns and 100 ns respectively. The internal quantum efficiency is given by:

(A) 60%

(B) 62.5%

(C) 37.5%

(D) 82.5%

Answer: (B)

4. In a p type silicon sample, the whole concentration is 2.25 × 1015/cm3. If the intrinsic carrier concentration is 1.5 × 1010/cm3, the electron concentration is given by

(A) Zero

(B) 1010/cm3

(C) 105/cm3

(D) 1.5 × 1025/cm3

Answer: (C)

5. A signal m(f) = 5 sin (2π 100t) frequency modulates a carrier. The resulting FM signal is

{10 cos (2π 105t) + 15 sin (2π 100 t)}. The approximate bandwidth of FM signal is

(A) 0.1 kHz

(B) 1 kHz

(C) 3.2 kHz

(D) 4 kHz

Answer: (C)

6. What is the octal equivalent of the binary number 11111001?

(A) 761

(B) 762

(C) 371

(D) 372

Answer: (C)

7.
NTA UGC NET Electronic Science Paper 2 Solved Question Paper 2013 December qn 7

Answer: (D)

8. The gauge factor of a resistance wire strain gauge is a measure of

(A) Sensitivity of the gauge

(B) Resolution

(C) Dynamic range of displacement measurement capability

(D) Variable inductance

Answer: (A)

9. In an LVDT, the two secondary wind map are connected in differential to obtain

(A) Higher output voltage

(B) an output voltage which is phase sensitive, i.e., the o/p voltage has a phase which can lead up to a conclusion whether the displacement of the core took place from right to left or from left to right.

(C) In order to establish a null or the reference point for the replacement of the core.

(D) Both (B) and (C)

Answer: (D)

10. Microprocessor is also known as CPU. What is CPU?

(A) Control Processing Unit

(B) Central Processing Unit

(C) Central Progressing Unit

(D) Control Progressing Unit

Answer: (B)

11. If ‘i’ is a variable, i++ denotes

(A) ‘i’ is incremented twice

(B) ‘i’ is incremented once

(C) ‘i’ is replaced with i + 2

(D) ‘i’ is multiplied by 2

Answer: (B)

12. For a SMF, following characteristic are given by

1. 0 ≤ v < 2.405

2. v >> 2.405

3. Core radius is less than 10 μm

4. Attenuation is of the order of 1 dB/km

Which of the following statement is true?

(A) 1 and 4 is correct.

(B) 2 and 3 is correct.

(C) 1 and 3 is correct.

(D) 3 and 4 is correct.

Answer: (C)

13. In a directional coupler, the power is applied at port 1 and output is taken from the port 2. For this following statements are given:

1. Coupling coefficient is given by C = 10 log10  P1/ P4.

2. Coupling coefficient is C = 10 log10  P4 /P1.

3. Directivity is given by D = 10 log10  P4/ P3.

4. Directivity is given by D = 10 log  P2 /P1.

Which of the following statement is true?

(A) 1 and 3 is correct.

(B) 2 and 3 is correct.

(C) 2 and 4 is correct.

(D) 1 and 4 is correct.

Answer: (A)

14. An AM demodulator can be implemented with

1. A linear multiplier followed by low pass filter.

2. A linear multiplier followed by high-pass filter.

3. A diode followed by low pass filter.

4. A linear multiplier followed by band-stop filter.

Which one of the following is correct?

(A) 1 only

(B) 3 only

(C) 1 and 3

(D) 4 only

Answer: (C)

15. NAND Gate can also be represented by

1. AND – invert Gate

2. OR – Invert Gate

3. Bubbled OR

4. Bubbled AND

Which of the following is correct?

(A) 1, 2 and 3

(B) 1 & 2

(C) 2 & 4

(D) 1 & 3

Answer: (D)

16. In 8085 processor,

1. AD0 – AD7 are treated as input as well as output signals.

2. HLDA is an output signal.

3. HOLD is an output signal.

4. SOD is an output signal.

(A) 1, 2 and 3 are output signals.

(B) 2, 3 and 4 are output signals.

(C) 1, 2 and 4 are output signals.

(D) 1, 3 and 4 are output signals

Answer: (C)

17. A variable type ‘char’ indicates the range and the no. of bytes.

1. 1 byte

2. 2 bytes

3. –127 to + 128

4. –128 to +127

(A) 1 & 3

(B) 2 & 3

(C) 1 & 4

(D) 2 & 4

Answer: (C)

18. The hygrometer is an instrument used to measure

1. Rainfall

2. Altitude

3. Relative humidity

4. Temperature

Which of the following is true?

(A) 1 and 2 is true.

(B) 1 and 4 is true.

(C) 3 is true.

(D) 3 and 4 is true

Answer: (C)

19. Hall effect can be used to measure

1. Conductivity of the charge carrier.

2. Mobility of the charge carrier.

3. The number of valence electrons per atom.

4. Band gap of the material.

Which of the following is correct?

(A) 1 and 2 are true

(B) 2 and 3 are true

(C) 3 and 4 are true

(D) 1 and 4 are true

Answer: (A)

20. In a JFET, the amplification factor μ, transconductance gm., and the dynamic drain resistance (rd) are related as

(A) μ = gm/ rd

(B) μ = rd/ gm

(C) μ = rd × gm

(D) μ = gm ⋅ rd2

Answer: (C)

21. Consider the following materials used for fabrication of Photonic Ics.

1. Ga P

2. In As

3. Ga As

4. Si

The correct sequence of the ascending order in terms of band gap energies is

(A) 1, 4, 2 and 3

(B) 4, 1, 2 and 3

(C) 2, 3, 1 and 4

(D) 2, 4, 3 and 1

Answer: (D)

22. Consider the following:

1. Quantization

2. Sampling

3. Encoding

4. Low-pass filter

The correct sequence for converting a low pass analog signal to Digital signal is

(A) 4, 3, 1, 2

(B) 4, 1, 2, 3

(C) 4, 2, 1, 3

(D) 4, 3, 2, 1

Answer: (C)

23. Consider the following circuit configurations:

1. Common emitter

2. Common base

3. Emitter follower

4. Emitter follower using Darlington pair

The correct sequence in increasing order of the input resistances of these configurations is

(A) 2, 1, 4, 3

(B) 1, 2, 4, 3

(C) 2, 1, 3, 4

(D) 1, 2, 3, 4

Answer: (C)

24. 8085 microprocessor has a flag register. The following flags are there, keep them in the ascending order starting from LSB to MSB.

1. Parity flag

2. Zero flag

3. Carry flag

4. Sign flag

(A) 1, 2, 3 and 4

(B) 2, 4, 1 and 3

(C) 3, 2, 1 and 4

(D) 3, 1, 2 and 4

Answer: (D)

25. The following logic families have their propagation delay. Arrange them from lowest propagation delay to highest propagation delay.

1. TTL (Standard)

2. ECL

3. Low power CMOS

4. DTL

(A) 2, 1, 4 and 3

(B) 2, 4, 1 and 3

(C) 4, 2, 3 and 1

(D) 1, 2, 3 and 4

Answer: (A)

26. Match List-I with List-II and select the correct answer using codes given below:

List – I                                    List – II

(a) Indirect band gap              (i) Ga As

(b) Direct band gap                 (ii) Cavity

(c) Laser diode                        (iii) Spontaneous emission

(d) LED                                  (iv) Si

Codes:

(a)        (b)        (c)        (d)

(A)       (ii)        (i)         (iii)       (iv)

(B)       (iii)       (iv)       (i)         (ii)

(C)       (i)         (iv)       (ii)        (iii)

(D)       (iv)       (i)         (ii)        (iii)

Answer: (D)

27. List – I                              List – II

(a) Poisson’s equation             (i) ∇2V = O

(b) Laplace equation               (ii) φH.dl = I

(c) Ampere’s Law                   (iii) ∇⋅E = ρυ/∈o

(d) Divergence of the field     (iv) ∇2V = – ρυ/∈

Codes:

(a)        (b)        (c)        (d)

(A)       (iv)       (ii)        (i)         (iii)

(B)       (ii)        (iii)       (i)         (iv)

(C)       (iv)       (i)         (ii)        (iii)

(D)       (i)         (iv)       (ii)        (iii)

Answer: (C)

28. List – I                                                      List – II

(a) Frequency Modulation                              (i) Balanced Modulator

(b) Double sideband suppressed carrier          (ii) Companding

(c) PCM                                                          (iii) Pre-emphasis and Deemphasis

(d) AM                                                            (iv) Envelope Detection

Codes:

(a)        (b)        (c)        (d)

(A)       (i)         (ii)        (iii)       (iv)

(B)       (iv)       (iii)       (ii)        (i)

(C)       (iii)       (i)         (ii)        (iv)

(D)       (ii)        (iv)       (i)         (iii)

Answer: (C)

29. List – I                              List – II

(a) Non-Weighted Code         (i) 8421

(b) Error-Detecting Code        (ii) Gray Code

(c) Weighted Code                 (iii) Hamming Code

(d) Error-Correcting                (iv) Parity Check

Codes:

(a)        (b)        (c)        (d)

(A)       (ii)        (i)         (iii)       (iv)

(B)       (ii)        (iv)       (i)         (iii)

(C)       (i)         (ii)        (iii)       (iv)

(D)       (i)         (iii)       (iv)       (ii)

Answer: (B)

30. List – I                              List – II

(Transducers)                          (Measured Quantity)

(a) Capacitive transducer        (i) Temperature

(b) Thermocouple                    (ii) Power

(c) Bourdon gauge                  (iii) Displacement

(d) Solar cell                            (iv) Pressure

Codes:

(a)        (b)        (c)        (d)

(A)       (i)         (iii)       (ii)        (iv)

(B)       (iii)       (i)         (iv)       (ii)

(C)       (ii)        (iv)       (i)         (iii)

(D)       (iv)       (i)         (iii)       (ii)

Answer: (B)

31. List – I                              List – II

(a) Voltage regulated IC         (i) Figure of merit

(b) Bistable multivibrator        (ii) CMRR

(c) OPAMP                             (iii) Capture range and locking range

(d) Phase Locked Loop          (iv) flip-flop

Codes:

(a)        (b)        (c)        (d)

(A)       (i)         (iii)       (iv)       (ii)

(B)       (i)         (iv)       (ii)        (iii)

(C)       (iv)       (iii)       (i)         (ii)

(D)       (iii)       (ii)        (iv)       (i)

Answer: (B)

32. List – I      List – II

(a) ASK           (i) Uses two frequencies, mark and space.

(b) FSK           (ii) Baud rate is same as bit rate.

(c) BPSK         (iii) On-off keying

(d) QPSK        (iv) Baud rate is half of bit rate

Codes:

(a)        (b)        (c)        (d)

(A)       (iii)       (i)         (ii)        (iv)

(B)       (ii)        (iii)       (i)         (iv)

(C)       (iii)       (ii)        (i)         (iv)

(D)       (i)         (iv)       (ii)        (iii)

Answer: (A)

33. List – I                  List – II

(a) if statement            (i) case

(b) do statement          (ii) return

(c) switch statement    (iii) while

(d) main statement      (iv) else

Codes:

(a)        (b)        (c)        (d)

(A)       (iii)       (iv)       (ii)        (i)

(B)       (iv)       (iii)       (ii)        (i)

(C)       (iii)       (i)         (iv)       (ii)

(D)       (iv)       (iii)       (i)         (ii)

Answer: (D)

34. List – I                              List – II

(Amplifier type)                      (Circuit/property)

(a) RC coupling                       (i) Higher voltage gain and impedance matching.

(b) Inductive coupling            (ii) Ability to amplify dc and low frequency signal.

(c) Transformer coupling         (iii) Minimum possible non-linear distortion

(d) Direct coupling                  (iv) Low collector supply voltages can be used.

Codes:

(a)        (b)        (c)        (d)

(A)       (iv)       (i)         (iii)       (ii)

(B)       (iii)       (iv)       (i)         (ii)

(C)       (i)         (ii)        (iii)       (iv)

(D)       (iv)       (iii)       (ii)        (i)

Answer: (D)

35. List – I                  List – II

(a) assembler               (i) executes program line by line.

(b) compiler                 (ii) creates machine code.

(c) interpreter              (iii) finds faults during execution.

(d) debugger               (iv) creates object file.

Codes:

(a)        (b)        (c)        (d)

(A)       (iv)       (iii)       (i)         (ii)

(B)       (ii)        (i)         (iv)       (iii)

(C)       (ii)        (iv)       (i)         (iii)

(D)       (iii)       (iv)       (i)         (ii)

Answer: (B)

Directions: Q. Nos. 36 to 45: The following items consist of two statements, one labelled the

“Assertion” (A) and other labelled the “Reason” (R). You are to examine these two statements carefully and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the Reason is a correct explanation of the Assertion. Select your answers to these items using the codes given below and mark your answer accordingly.

Codes:

(A) Both (A) and (R) are true and (R) is the correct explanation of (A).

(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

(C) (A) is true, but (R) is false.

(D) (A) is false, but (R) is true.

36. Assertion (A): A reflex klystron makes use of a single cavity.

Reason (R): The single cavity performs the functions of buncher as well as catcher cavity.

Answer: (A)

37. Assertion (A): LP01 is basically HE11 mode.

Reason (R): Linearly polarised modes are strongly guided modes, in which TE and TM modes do not have identical propagation constants.

Answer: (C)

38. Assertion (A): Combinational circuits have no memory, timing or feedback loop.

Reason (R): The output of combinational circuit depends on their input and previous output state.

Answer: (C)

39. Assertion (A) Digital base band modulation is also referred to as TDM.

Reason (R): Pulse analog modulation is first step in converting analog signal to Digital Signal.

Answer: (D)

40. Assertion (A): Under certain conditions, one gets peak response at high frequency in transformer coupled amplifiers.

Reason (R): High frequency model of transformer coupled amplifiers is a resonant circuit.

Answer: (C)

41. Assertion (A): At high temperatures, the avalanche breakdown voltage of zener diode remains constant.

Reason (R): At high temperatures, mean free paths of electrons and holes are shorter, therefore a larger field is required to cause ionization.

Answer: (C)

42. Assertion (A): Piezoelectric transducers are not suitable for ultra-low frequency vibration measurement.

Reason (R): Piezoelectric materials have a high dielectric constant.

Answer: (A)

43. Assertion (A): The contents of Random Access memory is a volatile in nature. As soon as the power is removed to the IC, the contents are erased from the memory.

Reason (R): The contents of Read only Memory are a non-volatile in nature.

Answer: (B)

44. Assertion (A): Logarithms are used to change a multiplication or division problem into a simpler addition and subtraction problem.

Reason (R): The Laplace transform enables the circuit analyst to convert the set of integro differential equations describing a circuit to the complex frequency domain.

Answer: (B)

45. Assertion (A): char str[] = “paper” ‘str’ acts as a constant pointer to a string.

Reason (R): *P = “Paper” “P” acts as a pointer to a constant string.

Answer: (B)

Read the passage and answer the Question Nos. 46 to 50 that follows based on your understanding of passage.

Power devices with pnpn layers such as SCR, SCS and GTO belong to thyrister series. However UJT is also a member of this group. The silicon controlled rectifier (SCR) is three terminal pnpn semiconductor switching device, which is probably most important circuit element after the diode and the transistor. An SCR can be used as a controlled switch to perform various functions such as rectification inversion and regulation of power flow. It is unidirectional power switch and being extensively used in switching ac and dc. It has capability to control power as high as 10 MW. The frequency range of its application has also been extended up to 50 kHz. SCR is a rectifier with control action. The third element called gate enables the controlled rectification gate is used to turn SCR ‘ON’ but not turn it ‘OFF’. SCR is made of Si because it has to handle high power and high temperature.

There are many advantages of SCR as a switch i.e. the switching speed is very high up to 109 operation per second and it permits control over a large current (30- 100 A) in the load by means of a small gate current of few mA. SCR has two states turn ON and turn OFF state. There are two ways to turn ON (i) keep the gate open and made the supply voltage equal to the break over voltage (ii) operate SCR with supply voltage less than break over voltage. The general methods used for turning the SCR OFF are (i) anode current interruption and (ii) force commutation. It is not advisable to turn the SCR ‘ON’ with zero gate signals.

Some of the applications of SCR are motor controls, relay controls, inverters, preventive circuits, regulated power supply and phase control. Another pnpn device have characteristic similar to that of SCR is SCS (Silicon Controlled Switch).

A device that can control ac power to the load during the positive and negative cycle of input is called ‘triac’. It is basically a diac with a gate terminal for bilateral turn ON. Gate Turn OFF (GTO) switch is one more pnpn device which can be turned ON or OFF with cathode gate. Some application of GTO is counters, pulse generators and multivibrators.

Light activated SCR (SASCR) is turned ON by the light falling on the gate. It is used as a relay, optical light controller, phase controller and motor control device.

46. If the applied voltage is increased further when an SCR is forward biased, then

(A) It becomes more forward biased.

(B) It becomes more reverse biased.

(C) It will come to ‘ON’ state.

(D) It will come to ‘OFF’ state.

Answer: (C)

47. The di/dt if an SCR in an ac voltage controller is high if the load is

(A) A pure inductance

(B) An induction motor

(C) A pure resistance

(D) A combination of resistance and inductance.

Answer: (D)

48. Triac can be considered as

(A) Two SCRs connected in antiparallel with common gate.

(B) Two SCRs connected in parallel with common gate.

(C) Two Resisters connected in anti-parallel.

(D) Two GTOs connected in parallel with common cathode gate.

Answer: (A)

49. Forced commutation is employed when anode voltage is

(A) AC

(B) DC

(C) Pulsating

(D) Triangular

Answer: (B)

50. Diac is

(A) Two terminal bidirectional switch.

(B) Two terminal Unilateral switch.

(C) Three terminal bidirectional switch.

(D) Three terminal unilateral switch.

Answer: (A)


Year Wise Solved UGC NET Electronic Science Paper 2 Previous Year Question Papers

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