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____. Sequential access medium Random access medium expensive storage medium None of these Sequential access medium Random access medium expensive storage medium None of these ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) નીચે માંથી ક્યાં રુઢિપ્રયોગનો અર્થ 'છેતરવું' એવો થાય બે પાંદડે થવું દસ મારવો આડા ઉતરવું હથેળીમાં ચાંદ દેખાડવો બે પાંદડે થવું દસ મારવો આડા ઉતરવું હથેળીમાં ચાંદ દેખાડવો ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Who started the daily newspaper 'National Herald' in 1938? Dadabhai Naoroji Subhas Chandra Bose Jawaharlal Nehru Ram Mohan Roy Dadabhai Naoroji Subhas Chandra Bose Jawaharlal Nehru Ram Mohan Roy ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) The wind speed increases with height because of reduction in lift effect of the air reduction of drag effect of the earth surface enhancement of drag effect on the earth surface reduction in the gravitational force reduction in lift effect of the air reduction of drag effect of the earth surface enhancement of drag effect on the earth surface 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 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 96.32 µF 76.32 µF 86.32 µF 66.32 µF 96.32 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
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 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 kW W1 = -5 kW and W2 = 35 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 kW W1 = -5 kW and W2 = 35 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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