MGVCL Exam Paper (30-07-2021 Shift 2) જે ક્રિયા દર્શાવે પણ વાક્યનો અર્થ પૂર્ણ કરે એ એને ____કહેવાય. - આપેલ વિકલ્પોમાંથી યોગ્યનું ચયન કરી વાક્ય પૂર્ણ કરો નીપાત ક્રિયા-વિભક્તિ વિભક્તિ કૃદંત નીપાત ક્રિયા-વિભક્તિ વિભક્તિ કૃદંત 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 0.8π V 0.4π V 0.2π V zero 0.8π V 0.4π V 0.2π V zero 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) If all the devices are connected to a central hub, then the topology is called Tree Topology Bus Topology Tree Topology Star Topology Tree Topology Bus Topology Tree Topology Star Topology ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A three-phase, 12 kV, 50 Hz and 300 kW load is operating at 0.8 pf lag. If the pf is to be improved to unity, using star connected capacitor, the per phase value of the reactive VAR required is: 50 kVAR 75 kVAR 100 kVAR 25 kVAR 50 kVAR 75 kVAR 100 kVAR 25 kVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The reactive VAR required = kW*(tanφ1 -tanφ2)Cosφ1 = 0.8φ1 = 36.87 deg.Cosφ2 = 0φ2 = 0 deg.kVAR required = kW*(tan(36.87) - tan(0))= 300*1000(0.75)= 225.00 kVARPer phase value = 225.00/3= 75 kVAR
MGVCL Exam Paper (30-07-2021 Shift 2) સ્વાધીન' નો વિરોધીભાસી શબ્દ કયો છે? અસ્વાધીન નીસ્વધીન કુવાધીન પરાધીન અસ્વાધીન નીસ્વધીન કુવાધીન પરાધીન ANSWER DOWNLOAD EXAMIANS APP
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. 730 mH 932 mH 1135 mH 1270 mH 730 mH 932 mH 1135 mH 1270 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
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