MGVCL Exam Paper (30-07-2021 Shift 2) A circular loop has its radius increasing at a rate of 2 m/s. The loop is placed perpendicular to a constant magnetic field of 0.1 Wb/m². When the radius of the loop is 2 m, the emf induced in the loop will be 0.8π V 0.2π V zero 0.4π V 0.8π V 0.2π V zero 0.4π V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP emf = dΨ/dt = change of flux linkageemf = dф/dtemf = d(B*Area)/dtemf = B*d(area)/dtemf = B*d(πr²)/dtemf = Bπ*2r*dr/dtemf = Bπ*2r*change of radius per unit timeemf = 0.1*π*2*2emf = 4π
MGVCL Exam Paper (30-07-2021 Shift 2) Who started the daily newspaper 'National Herald' in 1938? Jawaharlal Nehru Dadabhai Naoroji Subhas Chandra Bose Ram Mohan Roy Jawaharlal Nehru Dadabhai Naoroji Subhas Chandra Bose Ram Mohan Roy ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A single phase 240 V, 1 kW heater is connected across single phase 240 V, 50 Hz supply through an SCR. For firing angle delay of 90˚, calculate the power absorbed in the heater element. 300 W 275 W 250 W 225 W 300 W 275 W 250 W 225 W ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Resistance of heater, R = V²/PR = 57.6 ohmRMS value of voltage through controlled SCR = (Vm/2π)*(1 + cosα)= 120 VPower absorb by heater = Vrms²/R= 120²/57.6= 250 W
MGVCL Exam Paper (30-07-2021 Shift 2) Transistors were used in ____ of Computers. First Generation Second Generation Fourth Generation Third Generation First Generation Second Generation Fourth Generation Third Generation ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In transformer, bushing failure occurs due to(i) transformer vibrations(ii) dielectric losses(ii) partial discharge (i) and (iii) only (i) and (ii) only (ii) and (iii) only (i), (ii) and (iii) (i) and (iii) only (i) and (ii) only (ii) and (iii) only (i), (ii) and (iii) ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) 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 opposing. 730 mH 1135 mH 932 mH 1270 mH 730 mH 1135 mH 932 mH 1270 mH ANSWER EXPLANATION DOWNLOAD EXAMIANS APP M = k√(L₁L₂)M= 0.45√(100*900)M = 135 mHL_eq for opposite connection of coils = L₁ + L₂ - 2ML_eq = 900 + 100 - 2*135L_eq = 730 mH
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