MGVCL Exam Paper (30-07-2021 Shift 2) What is the full form of the acronym 'AIDS' in diseases? Amino Intersect Deficiency Syndrome Acquired Immune Deficiency Syndrome Acquire in Deficiency Syndrome Assert in Dander Syndrome Amino Intersect Deficiency Syndrome Acquired Immune Deficiency Syndrome Acquire in Deficiency Syndrome Assert in Dander Syndrome 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. 730 mH 1135 mH 1270 mH 932 mH 730 mH 1135 mH 1270 mH 932 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, 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: 25 kVAR 50 kVAR 100 kVAR 75 kVAR 25 kVAR 50 kVAR 100 kVAR 75 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) Vikram Seth is associated with which of the following fields? Politician Music Sports Literature Politician Music Sports Literature ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Choose the word which expresses nearly the opposite meaning of the given word "WHIMSICAL" Regular Fantastic Funny Curious Regular Fantastic Funny Curious ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 3-phase motor load has a p.f. of 0.397 lagging. The two wattmeter method used to measure power show the input as 30 kW. Find the reading on each wattmeter. W1 = -5 kW and W2 = 35 kW W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 kW W1 = 5 kW and W2 = 25 kW W1 = -5 kW and W2 = 35 kW W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 kW W1 = 5 kW and W2 = 25 kW ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,Cosφ = 0.397P = 30 kW = √3*V_L*I_L*cosφV_L*I_L = 30kW/(√3*0.397) = 43628.48 VAReadings of wattmeter W1 = V_L*I_L*cos(30+φ)Readings of wattmeter W2 = V_L*I_L*cos(30-φ)
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