69. What will be contents of AL and status of CF in 8086 after the execution of the following instruction? AL : 73 CL : 29 ADD AL, CL DAA Step 1: Addition Operation Perform hexadecimal addition: 73H + 29H = 9CH . After ADD AL, CL , AL = 9CH , CF = 0, and AF = 0. Step 2: Decimal Adjust (DAA) Logic Lower Nibble Check: The lower nibble (CH) is > 9. Therefore, add 06H to AL. 9CH + 06H = A2H Upper Nibble Check: The new upper nibble (AH) is > 9. Therefore, add 60H to AL. A2H + 60H = 102H Step 3: Final Register State The 8-bit register AL takes the lower two digits, and the overflow sets the carry flag. AL = 02 , CF = 1 Correct Option: 1. AL = 02, CF = 1 Browse All Solved Papers (2012 - 2025) → ...

59. Arrange the following in the order of decreasing gain: A. Monopole Antenna B. Dipole Antenna C. Yagi-Uda Antenna D. Horn Antenna Choose the correct answer from the options given below: 1. C, B, D, A 2. D, B, C, A 3. D, C, B, A 4. A, B, C, D Solution: 1. Horn Antenna (D): Highest gain (~10-25 dBi). It is an aperture antenna with high directivity. 2. Yagi-Uda Antenna (C): Medium-High gain (~7-15 dBi). Uses parasitic elements to focus the signal. 3. Dipole Antenna (B): Low gain (2.15 dBi). A basic half-wave resonant antenna. 4. Monopole Antenna (A): Lowest gain. Simplest structure, essentially half of a dipole. Decreasing Order: Horn (D) > Yagi-Uda (C) > Dipole (B) > Monopole (A) Correct Op...

Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → Further Reading: GATE EC Previous Year Papers with Solutions

Power Electronics MCQ Solution Match List-I with List-II. List-I (Power Device) List-II (Operation Type) A. DIAC I. Bipolar mode MOSFET B. TRIAC II. Bidirectional Diode Thyristor C. IGBT III. Unilateral Thyristor D. SCR IV. Bidirectional Triode Thyristor Options: A-II, B-I, C-IV, D-III A-I, B-III, C-IV, D-II A-I, B-II, C-III, D-IV A-II, B-IV, C-I, D-III Solution Step 1: Identify DIAC DIAC stands for Diode for Alternating Current . It conducts in both directions after breakover voltage is reached. Hence it is a Bidirectional Diode Thyristor . A → II Step 2: Identify TRIAC TRIAC stands for Triode for A...

Optical Fiber Communication MCQ Solution The three windows in case of optical fibres to be used for communication are: A. 850 nm B. 1310 nm C. 1410 nm D. 1110 nm E. 1550 nm Choose the correct answer from the options given below: A, B and C only A, B and D only B, C and D only A, B and E only Solution Step 1: Recall the standard optical communication windows Optical fiber communication systems commonly operate in three low-loss wavelength windows: First Window = 850 nm Second Window = 1310 nm Third Window = 1550 nm Step 2: Compare with the given options Option Wavelength Communication Window? A 850 nm ✔ Yes B 1310 nm ...

Communication Systems MCQ Solution Arrange the following in the order of decreasing transmission power requirement: A. FM B. AM C. DSB-SC D. VSB-SC E. SSB-SC Choose the correct answer: B, C, D, E, A B, C, E, D, A A, B, C, D, E E, D, C, B, A Solution Step 1: Compare AM and DSB-SC AM transmits a carrier along with two sidebands. A large portion of power is wasted in the carrier. Therefore, AM requires more transmission power than DSB-SC. Step 2: Compare DSB-SC, VSB-SC, and SSB-SC DSB-SC transmits both sidebands. VSB-SC transmits one complete sideband and a vestige of the other. SSB-SC transmits only one sideband. Hence: DSB-SC > VSB-SC > SSB-SC Step 3: Compare FM with Amplitude Modulation Schemes FM has constan...

50. Determine V C and V B for the network given in the figure with β = 45, V BE = 0.7V. 10 μF V B R B 100 kΩ I B V EE = − 9V V C I C R C = 1.2 K C 2 = 10 μF V out 1. V C = -4.48 V, V B = -8.3 V 2. V C = -8.3 V, V B = -4.48 V 3. V C = -3.52 V, V B = -8.3 V 4. V C = -4.84 V, V B = -4.77 V Answer: 1 Solution VE = VB - VBE or, VB = VE + VBE = -9 + 0.7 = 8.3 V Ic = 45*Ib Ib = {0 - (-8.3)} / {100 *10^(-3)} = 83 micro-ohm Ic = 45* 83 micro-ohm = 3.74 milli-ampere Vc = 3.7 * 10^(-3) * 1.2 * 10^(3) = 4.48 V Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → ...

49. For a p-n junction diode. A. The condition of zero net electron and hole currents requires that the Fermi level must be constant throughout the sample. B. Depletion layer width at thermal equilibrium for a one-sided abrupt junction (W D ) is independent of doping concentration. C. 1 ⁄ C 2 versus V plot can yield doping density (N) and extrapolation to 1 ⁄ C 2 = 0 gives (ψ bi - 2KT/q) for one-sided abrupt junction. D. For Zener diode, if Breakdown voltage (V BD ) is larger than 6 E g /q then breakdown mechanism is band-to-band tunnelling and temperature coefficient of V BD is negative. Choose the correct answer from the options given below: 1. A and B only 2. A and C only 3. B and C only 4. B and D only 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 dens...