MGVCL Exam Paper (30-07-2021 Shift 2) Magnetic tape is not practical for applications where data must be quickly recalled because tape is a/an____. Random access medium Sequential access medium None of these expensive storage medium Random access medium Sequential access medium None of these expensive storage medium ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) If the velocity of electromagnetic wave in free space is 3 × 10⁸ m/s, then the velocity in a medium with εr of 4.5 and μr of 2 would be 1 × 10⁸ m/s 3 × 10⁸ m/s 27 × 10⁸ m/s 9 × 10⁸ m/s 1 × 10⁸ m/s 3 × 10⁸ m/s 27 × 10⁸ m/s 9 × 10⁸ m/s ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Speed of EM wave in medium = C/√(εr*μr)= (300000000)/√9= 300000000/3= 100000000= 1*10⁸ m/s
MGVCL Exam Paper (30-07-2021 Shift 2) Fill in the blanks with suitable Article from the given alternatives.I heard ____interesting story yesterday an the a No article an the a No article 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. 250 W 300 W 225 W 275 W 250 W 300 W 225 W 275 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) In a short circuit test on circuit breaker, time to reach the peak re-striking voltage is 50 µs with the peak restriking voltage of 100 kV. Determine average RRRV (Rate of rise re-striking voltage)? 3 x 10⁶ kV/sec 2 x 10⁶ kV/sec 4 x 10⁶ kV/sec 1 x 10⁶ kV/sec 3 x 10⁶ kV/sec 2 x 10⁶ kV/sec 4 x 10⁶ kV/sec 1 x 10⁶ kV/sec ANSWER EXPLANATION DOWNLOAD EXAMIANS APP RRRV = Peak value of restriking voltage/time to reach the peak re-striking voltageRRRV = 100*1000/(50 μs)RRRV = 2*10⁶ kV/sec
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. 66.32 µF 96.32 µF 86.32 µF 76.32 µF 66.32 µF 96.32 µF 86.32 µF 76.32 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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