MGVCL Exam Paper (30-07-2021 Shift 2) Vikram Seth is associated with which of the following fields? Music Literature Sports Politician Music Literature Sports Politician ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Replace the underlined phrase grammatically and conceptually with the help of the given options. If the given sentence is correct then select the option 'The given sentence is correct'.The HR will give an topic to test the ability of our employees for the new project will gives a topic to test the ability of will give a topic to test the ability of The given sentence is correct will give a topic to testing the ability of will gives a topic to test the ability of will give a topic to test the ability of The given sentence is correct will give a topic to testing the ability of ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A single phase motor connected to 415 V, 50 Hz supply takes 30 A at a power factor of 0.7 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.9 lagging. 76.32 µF 66.32 µF 96.32 µF 86.32 µF 76.32 µF 66.32 µF 96.32 µF 86.32 µF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = VIcosφ = 8715 wattQ = VAR required to improve PF = P(tanφ₁ - tanφ₂) = 4670 VARCapacitive Reactance, Xc = V²/Q = 36.8790 ohm.C = (2πfXc)⁻¹ = 86.314 μF
MGVCL Exam Paper (30-07-2021 Shift 2) In a short circuit test on circuit breaker, time to reach the peak re-striking voltage is 50 µs with the peak restriking voltage of 100 kV. Determine average RRRV (Rate of rise re-striking voltage)? 3 x 10⁶ kV/sec 1 x 10⁶ kV/sec 4 x 10⁶ kV/sec 2 x 10⁶ kV/sec 3 x 10⁶ kV/sec 1 x 10⁶ kV/sec 4 x 10⁶ kV/sec 2 x 10⁶ kV/sec ANSWER EXPLANATION DOWNLOAD EXAMIANS APP RRRV = Peak value of restriking voltage/time to reach the peak re-striking voltageRRRV = 100*1000/(50 μs)RRRV = 2*10⁶ kV/sec
MGVCL Exam Paper (30-07-2021 Shift 2) નીચે પૈકી કયો 'દાદુર' નો સમાનર્થી નથી મેડક સુમાર દેડકો દર્દુર મેડક સુમાર દેડકો દર્દુર 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. 1.32 Hz 1.42 Hz 1.52 Hz 1.62 Hz 1.32 Hz 1.42 Hz 1.52 Hz 1.62 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