MGVCL Exam Paper (30-07-2021 Shift 2) સોહેલ ખાન : મલાયકા અરોર :: સૈફ અલી ખાન : ?? પરવીન બાબી કિરણ ખેર માધુરી દિક્ષિત અમૃતા સિંહ પરવીન બાબી કિરણ ખેર માધુરી દિક્ષિત અમૃતા સિંહ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is not a Font style in Microsoft Word? Bold Superscript Italic Regular Bold Superscript Italic Regular 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. 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) The relation between electric intensity E, voltage applied V and the distance d between the plates of a parallel plate condenser is E = Vx(d)² E = V/(d)² E = V/d E = V x d E = Vx(d)² E = V/(d)² E = V/d E = V x d ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The relation between electric intensity E, voltage applied V and the distance d between the plates of a parallel plate condenser is given by:V = E*LE = V/L
MGVCL Exam Paper (30-07-2021 Shift 2) In the following question, one part of the sentence may have an error. Find out which part of the sentence has an error and select the option corresponding to it. If the sentence contains no error, Select "No error'' option. (Avoid punctuation errors).(A) The Manager, with / (B) all his employee, was / (C) fired from the company /(D) NO ERROR C D A B C D A B 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 0.2π V 0.8π V zero 0.4π V 0.2π V 0.8π V zero 0.4π 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
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