MGVCL Exam Paper (30-07-2021 Shift 2) Encrypted security payload extension header is new in IPv5 IPv4 IPv6 IP IPv5 IPv4 IPv6 IP ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In the circuit shown in figure, the voltage across 8 Ω resistor is 20 V. Calculate the supply voltage. 47 V 38 V 27 V 18 V 47 V 38 V 27 V 18 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equivalent resistance of the circuit,Req = (12||18) + 8Req = 7.2 + 8Req = 15.2 ohmSource current = current through 8 ohm resistor, I = Vr/R8I = 20/8I = 2.5 ASupply votage V = I*ReqV = 2.5*15.2V = 38 V
MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is not a Font style in Microsoft Word? Regular Italic Bold Superscript Regular Italic Bold Superscript ANSWER DOWNLOAD EXAMIANS APP
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 0.2π V 0.8π V zero 0.4π V 0.2π V 0.8π 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) Operating system is a program that runs during log off a program that runs during log off a layer between the user and the hardware a program that runs for short duration a program that runs during log off a program that runs during log off a layer between the user and the hardware a program that runs for short duration ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 3-phase motor load has a p.f. of 0.397 lagging. The two wattmeter method used to measure power show the input as 30 kW. Find the reading on each wattmeter. W1 = -5 kW and W2 = 35 kW W1 = -10 kW and W2 = 40 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -5 kW and W2 = 35 kW W1 = -10 kW and W2 = 40 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,Cosφ = 0.397P = 30 kW = √3*V_L*I_L*cosφV_L*I_L = 30kW/(√3*0.397) = 43628.48 VAReadings of wattmeter W1 = V_L*I_L*cos(30+φ)Readings of wattmeter W2 = V_L*I_L*cos(30-φ)
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