MGVCL Exam Paper (30-07-2021 Shift 2) Match the following shown in figure B = (ii) & (1) C = (ii) & (1) A = (i) & (3) A = (iii) & (2) B = (ii) & (1) C = (ii) & (1) A = (i) & (3) A = (iii) & (2) ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In a test by Murray loop for ground fault on 600 m of cable having a resistance of 1.5 Ω/km, the faulty cable is looped with a sound cable of the same length and area of cross section. If the ratio of the other two arms of the testing network at balance is 3 : 1, find the distance of the fault from the testing end of cables. 400 m 200 m 300 m 100 m 400 m 200 m 300 m 100 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,d is distance in mQ is value of variable resistor in ohmP is variable resistance in ohmHere Q/P = 1/3 is given,d = Q/(P + Q)*Loop length= 1/4*(2*600)= 300 m
MGVCL Exam Paper (30-07-2021 Shift 2) In a synchronous motor with field under excited, the power factor will be unity zero leading lagging unity zero leading lagging ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For synchronous motor,Under excitation - Lagging power factorOver excitation - Leading power factorFor synchronous generatorUnder excitation - Leading power factorOver excitation - lagging power factor
MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is a national organisation monitoring the doping control programme in sports in India? Primary Anti-Doping Agency National Anti-Doping Agency Indian Anti-Doping Agency International Anti-Doping Protocol Primary Anti-Doping Agency National Anti-Doping Agency Indian Anti-Doping Agency International Anti-Doping Protocol ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A three-phase, 12 kV, 50 Hz and 300 kW load is operating at 0.8 pf lag. If the pf is to be improved to unity, using star connected capacitor, the per phase value of the reactive VAR required is: 75 kVAR 50 kVAR 100 kVAR 25 kVAR 75 kVAR 50 kVAR 100 kVAR 25 kVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The reactive VAR required = kW*(tanφ1 -tanφ2)Cosφ1 = 0.8φ1 = 36.87 deg.Cosφ2 = 0φ2 = 0 deg.kVAR required = kW*(tan(36.87) - tan(0))= 300*1000(0.75)= 225.00 kVARPer phase value = 225.00/3= 75 kVAR
MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is true regarding access lists applied to an interface? You can apply two access lists to any interface You can place as many access lists as you want on any interface until you run out of memory. You can apply only one access list on any interface One access list may be configured, per direction, for each layer 3 protocol configured on an interface You can apply two access lists to any interface You can place as many access lists as you want on any interface until you run out of memory. You can apply only one access list on any interface One access list may be configured, per direction, for each layer 3 protocol configured on an interface ANSWER DOWNLOAD EXAMIANS APP
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