MGVCL Exam Paper (30-07-2021 Shift 2) Who started the daily newspaper 'National Herald' in 1938? Dadabhai Naoroji Ram Mohan Roy Subhas Chandra Bose Jawaharlal Nehru Dadabhai Naoroji Ram Mohan Roy Subhas Chandra Bose Jawaharlal Nehru ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 50 Hz synchronous generator is connected to an infinite bus through a line. The p.u. reactances of generator and the line are j0.3 p.u. and j0.2 p.u. respectively. The generator no load voltage is 1.1 p.u. and that of infinite bus is 1.0 p.u. The inertia constant of the generator is 4 MW-sec/MVA. Determine the frequency of natural oscillations if the generator is loaded to 60% of its maximum power transfer capacity and small perturbation in power is given. 1.42 Hz 1.52 Hz 1.62 Hz 1.32 Hz 1.42 Hz 1.52 Hz 1.62 Hz 1.32 Hz ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Frequency of natural oscillation is given by,fn = {((dPe/dδ)at(δo))/M)}dPe/dδ = ((V1*V2)/X*(cosδ))= (11/05)*cosδ= (11/0.5)*0.5M = (H*s)/(πf)= 4/(50π)fn = (1.76*(3/50π))= 9.4 rad/sec = 1.32 Hz
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 = 5 kW and W2 = 25 kW W1 = -10 kW and W2 = 40 kW W1 = 10 kW and W2 = 20 kW W1 = -5 kW and W2 = 35 kW W1 = 5 kW and W2 = 25 kW W1 = -10 kW and W2 = 40 kW W1 = 10 kW and W2 = 20 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)
MGVCL Exam Paper (30-07-2021 Shift 2) The Great Barrier Reef is located in which country? Germany Australia India Canada Germany Australia India Canada ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) If the velocity of electromagnetic wave in free space is 3 × 10⁸ m/s, then the velocity in a medium with εr of 4.5 and μr of 2 would be 27 × 10⁸ m/s 3 × 10⁸ m/s 9 × 10⁸ m/s 1 × 10⁸ m/s 27 × 10⁸ m/s 3 × 10⁸ m/s 9 × 10⁸ m/s 1 × 10⁸ m/s ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Speed of EM wave in medium = C/√(εr*μr)= (300000000)/√9= 300000000/3= 100000000= 1*10⁸ m/s
MGVCL Exam Paper (30-07-2021 Shift 2) Fill in the blanks with suitable Preposition from the given alternatives.England skipper Joe Root is excited to bat ____ number three in the Ashes cricket series in on over at in on over at ANSWER DOWNLOAD EXAMIANS APP
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