MGVCL Exam Paper (30-07-2021 Shift 2) The wind speed increases with height because of reduction of drag effect of the earth surface enhancement of drag effect on the earth surface reduction in lift effect of the air reduction in the gravitational force reduction of drag effect of the earth surface enhancement of drag effect on the earth surface reduction in lift effect of the air reduction in the gravitational force ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Wind speed increases as the height from the ground increases mainly due to the decrease in the friction produced by land terrain.
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. 932 mH 1135 mH 1270 mH 730 mH 932 mH 1135 mH 1270 mH 730 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
MGVCL Exam Paper (30-07-2021 Shift 2) Recently, India and Benin signed MoUs on education, health and ____ Against Corruption e-visa facilities Trading Defence Against Corruption e-visa facilities Trading Defence 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. 66.32 µF 96.32 µF 76.32 µF 86.32 µF 66.32 µF 96.32 µF 76.32 µF 86.32 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
MGVCL Exam Paper (30-07-2021 Shift 2) પાણી : પાણીગ્રહણ :: હસ્ત : ??? પ્રેમાલાપ હસ્તમેળાપ વિવાહ છુટાછેડા પ્રેમાલાપ હસ્તમેળાપ વિવાહ છુટાછેડા 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 = -10 kW and W2 = 40 kW W1 = 10 kW and W2 = 20 kW W1 = -5 kW and W2 = 35 kW W1 = 5 kW and W2 = 25 kW W1 = -10 kW and W2 = 40 kW W1 = 10 kW and W2 = 20 kW W1 = -5 kW and W2 = 35 kW W1 = 5 kW and W2 = 25 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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