MGVCL Exam Paper (30-07-2021 Shift 2) A three-phase, 4-pole, 50 Hz induction motor runs at 1425 rpm. Find out the percentage slip of the induction motor. 6% 4% 3% 5% 6% 4% 3% 5% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Slip, s = (Ns - Nr)/NsNs = 120f/PNs = 1500 rpms = (1500 - 1425)/1500s = 0.05%s = 5%
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 C A B D C A B D ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Which one of the following is the largest windfarm located in India? Vindhayachal Mundra Muppandal Talcher Vindhayachal Mundra Muppandal Talcher ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rank - Windfarm name - Capacity1 - Muppandal windfarm - 1500 MW2 - Jaisalmer Wind Park - 1064 MW3 - Brahmanvel Wind Farm Dhule - 964 MW4 - Damanjodi Wind Farm - 99 MW5 - Tuppadahalli Wind Farm - 56.1 MW
MGVCL Exam Paper (30-07-2021 Shift 2) હુ ગીત ગાઈશ' ને કર્મણીમાં ફેરવો. હુ ગીતને ગાઈશ ગીતને હુ ગાઈશ ગીતને હુ જ ગાઇશ મારાથી ગીત ગવાશે હુ ગીતને ગાઈશ ગીતને હુ ગાઈશ ગીતને હુ જ ગાઇશ મારાથી ગીત ગવાશે ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A single phase 240 V, 1 kW heater is connected across single phase 240 V, 50 Hz supply through an SCR. For firing angle delay of 90˚, calculate the power absorbed in the heater element. 275 W 300 W 225 W 250 W 275 W 300 W 225 W 250 W ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Resistance of heater, R = V²/PR = 57.6 ohmRMS value of voltage through controlled SCR = (Vm/2π)*(1 + cosα)= 120 VPower absorb by heater = Vrms²/R= 120²/57.6= 250 W
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. 730 mH 1270 mH 932 mH 1135 mH 730 mH 1270 mH 932 mH 1135 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
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