MGVCL Exam Paper (30-07-2021 Shift 2) Who started the daily newspaper 'National Herald' in 1938? Subhas Chandra Bose Dadabhai Naoroji Jawaharlal Nehru Ram Mohan Roy Subhas Chandra Bose Dadabhai Naoroji Jawaharlal Nehru Ram Mohan Roy ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) What is the full form of the acronym 'AIDS' in diseases? Acquired Immune Deficiency Syndrome Amino Intersect Deficiency Syndrome Acquire in Deficiency Syndrome Assert in Dander Syndrome Acquired Immune Deficiency Syndrome Amino Intersect Deficiency Syndrome Acquire in Deficiency Syndrome Assert in Dander Syndrome ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In transformer, bushing failure occurs due to(i) transformer vibrations(ii) dielectric losses(ii) partial discharge (i), (ii) and (iii) (i) and (iii) only (i) and (ii) only (ii) and (iii) only (i), (ii) and (iii) (i) and (iii) only (i) and (ii) only (ii) and (iii) only ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A wire pilot is generally economical for distances up to 75 km to 100 km 10 km to 15 km 20 to 25 km 30 km to 60 km 75 km to 100 km 10 km to 15 km 20 to 25 km 30 km to 60 km ANSWER EXPLANATION DOWNLOAD EXAMIANS APP A wire pilot is generally economical for distances up to 5 or 10 miles, beyond which a carrier-current pilot usually becomes more economical.
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. 3% 4% 5% 6% 3% 4% 5% 6% 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) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 4 hours. What is the permissible overload for a duration of four hours. Assume the permissible load kVA as a fraction of rated kVA is 1.24. 1240 kVA 2480 kVA 930 kVA 620 kVA 1240 kVA 2480 kVA 930 kVA 620 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload = Full load kVA*factor for permissible load= 1000*1.24= 1240 kVA
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