MGVCL Exam Paper (30-07-2021 Shift 1) Which of the following is the layer of a computer system between the hardware and the user program? Operating System None of these Application Software System Environment Operating System None of these Application Software System Environment ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) At what range of wind speed is the electricity from the wind turbine is generated? 250 – 450 mph 30-35 mph 100 – 125 mph 450 – 650 mph 250 – 450 mph 30-35 mph 100 – 125 mph 450 – 650 mph ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For small turbine starts generating power 12.6 kph (3.5 m/s) is the typical cut-in speed.At 36–54 kph (10–15 m/s) produces maximum generation power.At 90 kph (25 m/s) maximum, the turbine is stopped or braked.
MGVCL Exam Paper (30-07-2021 Shift 1) An RLC series circuit consists of R = 16 Ω, L = 5 mH and C = 2 μF. Calculate the quality factor. 5.125 3.125 4.125 2.125 5.125 3.125 4.125 2.125 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For series RLC circuit,Q = (1/R)*√(L/C)Q = 1/16*√(5 mH/2 μF)Q = 50/16Q = 3.125
MGVCL Exam Paper (30-07-2021 Shift 1) Fill in the blanks with suitable Preposition from the given alternatives.The match was dedicated ___ Indian fast bowler Zaheer, who has announced his retirement from international cricket against from to since against from to since ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) The current flowing to a balanced delta connected load through line 'a' is 15 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 AIa₁ = (7.5 + j4.33) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (-7.5 - j13) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 - j13) AIa₂ = (7.5 + j13) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (7.5 - j 4.33) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (-7.5 - j13) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 - j13) AIa₂ = (7.5 + j13) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (7.5 - j 4.33) A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For the given question,Ib = 0Ia = -IcCalculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 1/3*(15 + 15∟-120°)= 7.5 + j*4.33 AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= (1/3)*(15 + 15∟120°)= 7.5 - j*4.33 AIao = (1/3)*(Ia + Ib + Ic)= (Ia - Ia)= 0
MGVCL Exam Paper (30-07-2021 Shift 1) Murray loop test is performed on a faulty cable 300 m long. At balance, the resistance connected to the faulty core was set at 15 Ω and the resistance of the resistor connected to the sound core was 45 Ω. Calculate the distance of the fault point from the test end. 75 m 225 m 150 m 450 m 75 m 225 m 150 m 450 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,R2 is the resistance connected to the faulty core in ohmR2 is the resistance of the resistor connected to the sound core in ohmR1 Distance of fault location (Lx) = R2/(R1 + R2)*(2*L)= (15/60)*600= 150 m