MGVCL Exam Paper (30-07-2021 Shift 3) Which state is the largest producer of wind energy in India? Andhra Pradesh Maharashtra Madhya Pradesh Tamil Nadu Andhra Pradesh Maharashtra Madhya Pradesh Tamil Nadu ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rank - State - Capaciy1 - Tamilnadu - 7.5 GW2 - Maharastra - 5 GW3 - Karnataka - 4.8 GW4 - Rajsthan - 4.3 GW
MGVCL Exam Paper (30-07-2021 Shift 3) 1. મોટેભાગે 'જાણે' શબ્દ હોય ત્યારે,ઉત્પ્રેક્ષા અલંકાર બને છે.૨. જયારે ઉપમેય અને ઉપમાન એક જ હોય ત્યારે ઉપમા અલંકાર બને છે.સરખાવવામાં આવેલ બે શબ્દોની વચ્ચે 'જયારે', 'જેવો', 'જેવી' જેવા શબ્દો આવે ત્યારે રૂપક અલંકાર બને.જયારે ટીકા કે નિંદા કે વ્યંગના રૂપે પ્રશંસા કરાય ત્યારે અતિશયોક્તિ અલંકાર બને.ઉપરોક્ત વિધાનોમાંથી કયું/ક્યાં વિધાનો સાચા છે. માત્ર 1 માત્ર 4 2, 4 1, 3 માત્ર 1 માત્ર 4 2, 4 1, 3 ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The elements are specialized computers used to connect two or more transmission lines. Transfering Switching Networking Broadcasting Transfering Switching Networking Broadcasting ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The International boundary 'Radcliffe Line’ lies between India and Afghanistan Japan Pakistan China Afghanistan Japan Pakistan China 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 = 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 = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 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 = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) 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? 1425 rpm 882 rpm 1575 rpm 798 rpm 1425 rpm 882 rpm 1575 rpm 798 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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