MGVCL Exam Paper (30-07-2021 Shift 2) Choose the word which expresses nearly the opposite meaning of the given word "WHIMSICAL" Curious Regular Funny Fantastic Curious Regular Funny Fantastic ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is an application layer service? Mail service Mail service Network virtual terminal File transfer, access and management Mail service Mail service Network virtual terminal File transfer, access and management ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) The wind speed increases with height because of reduction of drag effect of the earth surface reduction in the gravitational force reduction in lift effect of the air enhancement of drag effect on the earth surface reduction of drag effect of the earth surface reduction in the gravitational force reduction in lift effect of the air enhancement of drag effect on the earth surface 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) What is the full form of the acronym 'AIDS' in diseases? Acquired Immune Deficiency Syndrome Amino Intersect Deficiency Syndrome Acquire in Deficiency Syndrome Assert in Dander Syndrome Acquired Immune Deficiency Syndrome Amino Intersect Deficiency Syndrome Acquire in Deficiency Syndrome Assert in Dander Syndrome 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 0.2π V 0.4π V 0.8π V zero 0.2π V 0.4π 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) A 50 Hz synchronous generator is connected to an infinite bus through a line. The p.u. reactances of generator and the line are j0.3 p.u. and j0.2 p.u. respectively. The generator no load voltage is 1.1 p.u. and that of infinite bus is 1.0 p.u. The inertia constant of the generator is 4 MW-sec/MVA. Determine the frequency of natural oscillations if the generator is loaded to 60% of its maximum power transfer capacity and small perturbation in power is given. 1.62 Hz 1.52 Hz 1.32 Hz 1.42 Hz 1.62 Hz 1.52 Hz 1.32 Hz 1.42 Hz ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Frequency of natural oscillation is given by,fn = {((dPe/dδ)at(δo))/M)}dPe/dδ = ((V1*V2)/X*(cosδ))= (11/05)*cosδ= (11/0.5)*0.5M = (H*s)/(πf)= 4/(50π)fn = (1.76*(3/50π))= 9.4 rad/sec = 1.32 Hz
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