MGVCL Exam Paper (30-07-2021 Shift 1) The Prime Minister Rozgar Yojana (PMRY) was launched in which year? 2003 1993 2005 1995 2003 1993 2005 1995 ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) An alternator is supplying load of 300 kW at a power factor of 0.5 lagging. If the power factor is raised to unity, how many more kilowatts can alternator supply for the same kVA loading? 300 kW 450 kW 600 kW 150 kW 300 kW 450 kW 600 kW 150 kW ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = S*cosφS = P/cosφS = 300/0.5S = 600 kVAFor same kVA rating and unity power factor,P = 600*1P = 600 kWPextra = 600 - 300Pextra = 300 kW
MGVCL Exam Paper (30-07-2021 Shift 1) Choose the word which best expresses the similar meaning of the given word " CALLOW" Calm Immature Sophisticated Mature Calm Immature Sophisticated Mature ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) જ્વાળા: ધુમાડો:: બરફ : ___ ગરમી તાપમાન બાષ્પ શીતળતા ગરમી તાપમાન બાષ્પ શીતળતા ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Fill in the blanks with suitable Preposition from the given alternatives.The match was dedicated ___ Indian fast bowler Zaheer, who has announced his retirement from international cricket against from to since against from to since ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Determine maximum permissible load which a 15 MVA, ONAN cooled transformer to IS : 2026 can take for 6 hr, if the initial load on it was 12 MVA. The weighted ambient temperature is 20˚C. Assume the permissible load kVA as a fraction of rated kVA is 1.25. 15 MVA 12 MVA 18.75 MVA 9.6 MVA 15 MVA 12 MVA 18.75 MVA 9.6 MVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Maximum permissible load = Rated MVA rating*fraction of rated kVAMaximum permissible load = 15*1.25Maximum permissible load = 18.75 MVA
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