MGVCL Exam Paper (30-07-2021 Shift 2) A THHN conductor will have a 60˚C termination on one end and a 75˚C termination on the other, which ampacity column will be used? 60˚C column None of these 90˚C column 75˚C column 60˚C column None of these 90˚C column 75˚C column ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Conductor with higher temperature ratings, provide the ampacity is determined based on the 60°C ampacity of the conductor.
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 27 × 10⁸ m/s 1 × 10⁸ m/s 9 × 10⁸ m/s 3 × 10⁸ m/s 27 × 10⁸ m/s 1 × 10⁸ m/s 9 × 10⁸ m/s 3 × 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) A short-shunt dc compound generator supplies 50 A at 300 V. If the shunt field resistance, series field resistance and armature resistance are 30 Ω, 0.03Ω, and 0.05 Ω respectively, determine the e.m.f. generated. 304.5 V 308.5 V 306.5 V 302.5 V 304.5 V 308.5 V 306.5 V 302.5 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,Vt = 300 VRse = 0.03 Ω, Rsh = 30 Ω, Ra = 0.05 ΩI_L = 50 AVsh = V + I_L*Rse = 300 + 50*0.03 = 301.5 VIsh = Vsh/Rsh = 301.5/30 = 10.05 AIa = Ish + I_L = 60.05 AEg = Vt + I_L*Rse + IaRaEg = 300 + 50*0.03 + 60.05*0.05 = 304.5025 V
MGVCL Exam Paper (30-07-2021 Shift 2) Which one of the following is the committee on poverty estimation? Ad hoc committe Karve committee Tendulkar committee S.S.Tarapore committee Ad hoc committe Karve committee Tendulkar committee S.S.Tarapore committee ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Fill in the blanks with suitable Preposition from the given alternatives.England skipper Joe Root is excited to bat ____ number three in the Ashes cricket series at in over on at in over on ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A circular loop has its radius increasing at a rate of 2 m/s. The loop is placed perpendicular to a constant magnetic field of 0.1 Wb/m². When the radius of the loop is 2 m, the emf induced in the loop will be zero 0.2π V 0.8π V 0.4π V zero 0.2π V 0.8π V 0.4π V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP emf = dΨ/dt = change of flux linkageemf = dф/dtemf = d(B*Area)/dtemf = B*d(area)/dtemf = B*d(πr²)/dtemf = Bπ*2r*dr/dtemf = Bπ*2r*change of radius per unit timeemf = 0.1*π*2*2emf = 4π
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