MGVCL Exam Paper (30-07-2021 Shift 3) Which country leads in the production of biofuel in the world? Brazil Germany United States of America Argentina Brazil Germany United States of America Argentina ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rank - Country - Production (thousands barrel)1 - US - 9402 - Brazil - 4493 - Germany - 684 - China - 68
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 1575 rpm 798 rpm 1425 rpm 882 rpm 1575 rpm 798 rpm 1425 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
MGVCL Exam Paper (30-07-2021 Shift 3) Replace the braketed phrase grammatically and conceptually with the help of the given opation. If the given sentence is correct then select the option ’The given sentence is correct’.The corono virus continued to spread, with the development of infections advancing quickly continued to spreading, with the development of infections The given sentence is correct continued for spread, with the development of infections continue to spread, with the development of infections continued to spreading, with the development of infections The given sentence is correct continued for spread, with the development of infections continue to spread, with the development of infections 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 = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + 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 = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + 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 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 2 hours. What is the permissible overload for a duration of two hours. Assume the permissible load kVA as a fraction of rated kVA is 1.43. 1050 kVA 1430 kVA 715 kVA 700 kVA 1050 kVA 1430 kVA 715 kVA 700 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload for transformer = Full load kVA*factor for permissible overload= 1000*1.43= 1430 kVA
MGVCL Exam Paper (30-07-2021 Shift 3) નીચે પૈકી 'ઓસોડ' નો સમાનર્થી શબ્દ કયો છે? અનરવું ઔષધ વૃંદા અમર અનરવું ઔષધ વૃંદા અમર ANSWER DOWNLOAD EXAMIANS APP
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