MGVCL Exam Paper (30-07-2021 Shift 1) An electric potential field is produced in air by point 1 μC and 4 μC located at (-2, 1, 5) and (1, 3, -1) respectively. The energy stored in the field is 2.57 mJ 12.50 mJ 10.28 mJ 5.14 mJ 2.57 mJ 12.50 mJ 10.28 mJ 5.14 mJ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Fill in the blanks with suitable Article from the given alternatives.Bread is ___ predominant food prepared from a dough of flour and water an the a No article an the a No article ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) A coil having an inductance of 100 mH is magnetically coupled to another coil having an inductance of 900 mH. The coefficient of coupling between the coils is 0.45. Calculate the equivalent inductance if the two coils are connected in series aiding. 1270 mH 1135 mH 730 mH 932 mH 1270 mH 1135 mH 730 mH 932 mH ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Mutual inductance (M) = k*√(L1*L2)k = 0.45M = 0.45*√(900*100*10⁻⁶) = 135 mHLeq = L1 + L2 + 2MLeq = 900 + 100 + 270Leq = 1270 mH
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 to against since from to against since from ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) The frequency of the emf in the stator of a 4-pole induction motor is 50 Hz, and that in the rotor is 2 Hz. What is the slip and at what speed is the motor running? 3%, 1440 rpm 4%, 1440 rpm 3%, 1455 rpm 4%, 1455 rpm 3%, 1440 rpm 4%, 1440 rpm 3%, 1455 rpm 4%, 1455 rpm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rotor frequency, f_r = s*f_ss = 2/50s = 0.04Ns = 120f/PNs = 1500 rpms = (Ns - Nr)/NsNr = Ns(1 - s)Nr = 1500*(1 - 0.04)Nr = 1440 rpm
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. 150 m 75 m 450 m 225 m 150 m 75 m 450 m 225 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
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