MGVCL Exam Paper (30-07-2021 Shift 3) As per Stockholm International Peace Research Institute (SIPRI) report 2019 for the period 2014-2018, which country is the largest arms importer in the world? Germany Saudi Arabia Myanmar India Germany Saudi Arabia Myanmar India ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Data Encryption Standard (DES) was designed by Microsoft None of these Apple IBM Microsoft None of these Apple IBM ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A three phase four pole 50 Hz induction motor has a rotor resistance of 0.02 Ω/phase and stand-still reactance of 0.5 Ω/phase. Calculate the speed at which the maximum torque is developed. 1440 rpm 1500 rpm 1525 rpm 1475 rpm 1440 rpm 1500 rpm 1525 rpm 1475 rpm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Condition for maximum torque,Sm = R2/X2= 0.04Ns = (120*f)/P= (120*50)/4= 1500 rpmS = (Ns - Nr)/NsNr = Ns(1 - Sm)= 1500*(1 - 0.04)= 1440 rpm
MGVCL Exam Paper (30-07-2021 Shift 3) What is the full form of the acronym 'SWIFT', in financial institution? Social World interconnection in Financial Tech process Social Wide in Financial Transaction Society for Worldwide Interbank Financial Telecommunication Society website in Financial Terms Social World interconnection in Financial Tech process Social Wide in Financial Transaction Society for Worldwide Interbank Financial Telecommunication Society website in Financial Terms ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The current flowing in to a balanced delta connected load through line 'a' is 10 A when the conductor of line ’b’ is open. With the current in line 'a' as reference, compute the symmetrical components of the line currents. Assume the phase sequence of ’abc’. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For the given question,Ib = 0Ia = -IcCalculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 1/3*(10 + 10∟-120°)= 5 + j*2.88 AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= (1/3)*(10 + 10∟120°)= 5 - j*2.88 AIao = (1/3)*(Ia + Ib + Ic)= (Ia - Ia)= 0
MGVCL Exam Paper (30-07-2021 Shift 3) A 10-pole induction motor is supplied by a 6-pole alternator, which is driven at 1400 rpm. If the motor runs with a slip of 5%, what is its speed? 882 rpm 1425 rpm 798 rpm 1575 rpm 882 rpm 1425 rpm 798 rpm 1575 rpm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Slip of the induction motor,S = (Ns - Nr)/NsNs = (120*f)/PPoles (P) = 61400 = (120*f)/6f = 70 HzFor P = 10Ns = (120*70)/10= 840 rpmNr = Ns(1 - S)= 840*(1 - 0.05)= 798 rpm
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