MGVCL Exam Paper (30-07-2021 Shift 2) The Great Barrier Reef is located in which country? Australia India Germany Canada Australia India Germany Canada 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. 932 mH 730 mH 1270 mH 1135 mH 932 mH 730 mH 1270 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
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. 4% 5% 3% 6% 4% 5% 3% 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 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 0.62 0.52 0.72 0.82 0.62 0.52 0.72 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) The wind speed increases with height because of reduction in lift effect of the air reduction in the gravitational force reduction of drag effect of the earth surface enhancement of drag effect on the earth surface reduction in lift effect of the air reduction in the gravitational force reduction of drag effect of the earth surface enhancement of drag effect on the earth surface ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Wind speed increases as the height from the ground increases mainly due to the decrease in the friction produced by land terrain.
MGVCL Exam Paper (30-07-2021 Shift 2) The relation between electric intensity E, voltage applied V and the distance d between the plates of a parallel plate condenser is E = V/(d)² E = V x d E = V/d E = Vx(d)² E = V/(d)² E = V x d E = V/d E = Vx(d)² ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The relation between electric intensity E, voltage applied V and the distance d between the plates of a parallel plate condenser is given by:V = E*LE = V/L
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