36) If the in-phase and quadrature components in an M-ary PSK system are permitted to be independent, then this scheme becomes a: 1. QAM [Option ID = 2973] 2. DPSK [Option ID = 2974] 3. M-ary QAM [Option ID = 2975] 4. FSK [Option ID = 2976] Answer: 1 Solution  QAM = Amplitude + Phase Modulation Previous yr Question papers with Full Explanations → Electronics and Communiaction Study Materials →

  35) If an amplitude modulated wave 10[1+0.6 cos 2π10 3 t] cos 2π.10 6 .t is to be detected by a linear diode detector, then the time constant will be: 1. 1.7 msec [Option ID = 2969] 2. 0.17 msec [Option ID = 2970] 3. 17 msec [Option ID = 2971] 4. 0.17 sec [Option ID = 2972] Answer: 2 Solution  to detect envelope the time constant, t will be 1/carrier frequency << t << 1/message frequency Here, 1/10^6 << t << 1/10^3 It should be greater than 1 microsec but less than 1 millisecond   so, correct answer is 0.17 msec Previous yr Question papers with Full Explanations → Electronics and Communiaction Study Materials →

  34) A 10kW carrier is simultaneously modulated by two modulating signals corresponding to a modulation index of 40% and 30%, respectively. The total radiated power will be: 1. 10 kW [Option ID = 2965] 2. 12.5 kW [Option ID = 2966] 3. 11.25 kW [Option ID = 2967] 4. 10.25 kW [Option ID = 2968] Answer: 3 Soution Effective modulation index = sqrt(0.4^2 + 0.3^2) = 0.5 now , Pt = Pc(1+(Effective modulation index)^2/2) or, Pt = 10kW * 1.125 = 11.25 KW Previous yr Question papers with Full Explanations → Electronics and Communiaction Study Materials →

33) If f m is modulating frequency and m f is modulation index, then by the Carson's rule, the bandwidth of an FM signal at the input of a conventional discriminator will be: 1. 2 f m (m f +1) Hz [Option ID = 2961] 2. 2 f m (m f -1) Hz [Option ID = 2962] 3. f m (2m f +1) Hz [Option ID = 2963] 4. f m (2m f -1) Hz [Option ID = 2964] Answer: A Solution  As per Carson's law the bandwidth of FM is 2*fm(mf+1) where, fm is message frequency; mf = modulation index Previous yr Question papers with Full Explanations → Electronics and Communiaction Study Materials →

  32.  A single phase 220V, 1kW electric room radiator is connected across 220V supply through a triac. For a delay angle of 90°, the value of power dissipated by the radiator is: 1. 49.985 W 2. 2500 W 3. 398.9 W 4. 500 W Answer: 4  Power Dissipated by a Radiator Controlled by a TRIAC Given Supply Voltage, V = 220 V Rated Power of Radiator, P full = 1000 W Delay (Firing) Angle, α = 90° Load Type = Purely Resistive Formula For a single-phase AC voltage controller using a TRIAC feeding a resistive load: P = P full × [1 − (α/π) + sin(2α)/(2π)] Calculation Convert the firing angle into radians: α = 90° = π/2 Substitute into the power equation: P = 1000 × [1 − (π/2)/π + sin(π)/(2π)] Since: (π/2)/π = 1/2 sin(π) = 0 Therefore: P = 1000 × (1 − 1/2) P = 1000 ×...

  31.  The absolute maximum operating frequency of a converter grade SCR whose turn on time and turn off time are 3 µs and 200 µs respectively is: 1. 9.4 Khz 2. 5.2 khz 3. 4.9 Khz 4. 300 KHz Answer: C Solution 1/{(3+200)*10^(-6)} = 4.9 khz Previous yr Question papers with Full Explanations → Electronics and Communiaction Study Materials →

30. In optical fibers, the Rayleigh scattering is proportional to: 1.  λ 2. 1/ λ^2 3.1/ λ^3 4.1/ λ^4 Answer: D Solution. Why Rayleigh Scattering is Proportional to 1/λ⁴ An accelerating charge emits electromagnetic radiation. The radiated power from an oscillating dipole is proportional to the square of its acceleration: P ∝ a² If the oscillation frequency is f , then the acceleration scales as: a ∝ f² Therefore: P ∝ f⁴ Convert Frequency to Wavelength Since: f = c / λ substituting into the power relation gives: P ∝ (1/λ)⁴ or P ∝ 1/λ⁴ This is known as the Rayleigh scattering law . Intuitive Way to Remember Suppose the wavelength doubles: λ → 2λ Then the scattering becomes: 1/(2λ)⁴ = 1/(16λ⁴) Therefore, doubling the wavelength reduces Rayleigh scattering by 16 t...

  29) In a multimode fiber (step index), number of modes passing at an operating wavelength of 1300 nm are 1000, the refractive index of the core is 1.50 and that of the cladding is 1.48. The value of core diameter is: 1. 600 nm [Option ID = 2945] 2. 1.4 μm [Option ID = 2946] 3. 125 μm [Option ID = 2947] 4. 4.25 μm [Option ID = 2948] Answer: C Solution number of multimode, M = V^2/2 and multimode density V = pi*d/ λ * sqrt(n1^2 - n2^2) given, M = 1000, n1 = 1.50, n2 =1.48 M = V^2/2 or, V^2 = 2000 V = pi*d/1300*10^(-9) *  sqrt(1.50^2 - 1.48^2) or, 2000 = { pi*d/1300*10^(-9) * sqrt(1.50^2 - 1.48^2)}^2 or, d = 75.6 micrometer Previous yr Question papers with Full Explanations → Electronics and Communiaction Study Materials →