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 zero 0.4π V 0.8π V 0.2π V zero 0.4π V 0.8π V 0.2π 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) In transformer, bushing failure occurs due to(i) transformer vibrations(ii) dielectric losses(ii) partial discharge (ii) and (iii) only (i) and (iii) only (i), (ii) and (iii) (i) and (ii) only (ii) and (iii) only (i) and (iii) only (i), (ii) and (iii) (i) and (ii) only ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Choose the word which expresses nearly the opposite meaning of the given word "WHIMSICAL" Fantastic Funny Curious Regular Fantastic Funny Curious Regular 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 250 W 275 W 225 W 300 W 250 W 275 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) Which of the following is true regarding access lists applied to an interface? You can apply two access lists to any interface You can place as many access lists as you want on any interface until you run out of memory. You can apply only one access list on any interface One access list may be configured, per direction, for each layer 3 protocol configured on an interface You can apply two access lists to any interface You can place as many access lists as you want on any interface until you run out of memory. You can apply only one access list on any interface One access list may be configured, per direction, for each layer 3 protocol configured on an interface ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A single phase motor connected to 415 V, 50 Hz supply takes 30 A at a power factor of 0.7 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.9 lagging. 86.32 µF 66.32 µF 76.32 µF 96.32 µF 86.32 µF 66.32 µF 76.32 µF 96.32 µF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = VIcosφ = 8715 wattQ = VAR required to improve PF = P(tanφ₁ - tanφ₂) = 4670 VARCapacitive Reactance, Xc = V²/Q = 36.8790 ohm.C = (2πfXc)⁻¹ = 86.314 μF
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