MGVCL Exam Paper (30-07-2021 Shift 2) When different types of consumers are charged at different uniform per unit rates, it is called as ____ two-part tariff block rate tariff maximum demand tariff flat rate tariff two-part tariff block rate tariff maximum demand tariff flat rate tariff ANSWER EXPLANATION DOWNLOAD EXAMIANS APP When different types of consumers are charged at different uniform per unit rates, it is said to be Flat Rate Tariff.In this type, the consumers are grouped into different classes.Each class is charged at different uniform rate.
MGVCL Exam Paper (30-07-2021 Shift 2) A delta-connected balanced three-phase load is supplied from a three-phase, 400 V supply. The line current is 20 A and the power taken by the load is 10 kW. Find the power factor. 0.82 0.72 0.52 0.62 0.82 0.72 0.52 0.62 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = 10 kWVL = 400 VIL = 20 A3- phse power, P = √3*VL*IL*cosφcosφ = 10*1000/(√3*400*20)= 0.7216
MGVCL Exam Paper (30-07-2021 Shift 2) In the following question, one part of the sentence may have an error. Find out which part of the sentence has an error and select the option corresponding to it. If the sentence contains no error, Select "No error'' option. (Avoid punctuation errors).(A) The Manager, with / (B) all his employee, was / (C) fired from the company /(D) NO ERROR B A C D B A C D ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Which one of the following is the committee on poverty estimation? S.S.Tarapore committee Karve committee Ad hoc committe Tendulkar committee S.S.Tarapore committee Karve committee Ad hoc committe Tendulkar committee ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A coil having an inductance of 100 mH is magnetically coupled to another coil having an inductance of 900 mH. The coefficient of coupling between the coils is 0.45. Calculate the equivalent inductance if the two coils are connected in series opposing. 1135 mH 1270 mH 932 mH 730 mH 1135 mH 1270 mH 932 mH 730 mH ANSWER EXPLANATION DOWNLOAD EXAMIANS APP M = k√(L₁L₂)M= 0.45√(100*900)M = 135 mHL_eq for opposite connection of coils = L₁ + L₂ - 2ML_eq = 900 + 100 - 2*135L_eq = 730 mH
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 25 mH 50 mH 25 nH 50 nH 25 mH 50 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
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