MGVCL Exam Paper (30-07-2021 Shift 2) If the velocity of electromagnetic wave in free space is 3 × 10⁸ m/s, then the velocity in a medium with εr of 4.5 and μr of 2 would be 9 × 10⁸ m/s 1 × 10⁸ m/s 3 × 10⁸ m/s 27 × 10⁸ m/s 9 × 10⁸ m/s 1 × 10⁸ m/s 3 × 10⁸ m/s 27 × 10⁸ m/s ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Speed of EM wave in medium = C/√(εr*μr)= (300000000)/√9= 300000000/3= 100000000= 1*10⁸ m/s
MGVCL Exam Paper (30-07-2021 Shift 2) Nangal Wildlife Sanctuary is located in which state? Tamil Nadu Sikkim Punjab Mizoram Tamil Nadu Sikkim Punjab Mizoram ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Calculate the inductance per meter of a 50 Ω load cable that has an capacitance of 20 pF/m. 25 nH 50 nH 50 mH 25 mH 25 nH 50 nH 50 mH 25 mH ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Characteristic impedance, Z = √(L/C)50 = √(L/20*10⁻¹²)L = (50*50)*(20*10⁻¹²)L = 50000*10⁻¹²L = 50 nH
MGVCL Exam Paper (30-07-2021 Shift 2) The operating time of an inverse definite minimum time (IDMT) overcurrent relay is ____ proportional to the plug-setting multiplier (PSM) and____proportional to the time-multiplier setting (TMS). inversely, directly inversely, inversely directly, inversely directly, directly inversely, directly inversely, inversely directly, inversely directly, directly ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In a synchronous motor with field under excited, the power factor will be unity leading lagging zero unity leading lagging zero ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For synchronous motor,Under excitation - Lagging power factorOver excitation - Leading power factorFor synchronous generatorUnder excitation - Leading power factorOver excitation - lagging power factor
MGVCL Exam Paper (30-07-2021 Shift 2) A single phase motor connected to 415 V, 50 Hz supply takes 30 A at a power factor of 0.7 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.9 lagging. 96.32 µF 66.32 µF 76.32 µF 86.32 µF 96.32 µF 66.32 µF 76.32 µF 86.32 µF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = VIcosφ = 8715 wattQ = VAR required to improve PF = P(tanφ₁ - tanφ₂) = 4670 VARCapacitive Reactance, Xc = V²/Q = 36.8790 ohm.C = (2πfXc)⁻¹ = 86.314 μF
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