MGVCL Exam Paper (30-07-2021 Shift 3) વિષમ' ની સંધી છૂટી પાડો. - કયો વિકલ્પ સાચો છે? વી + સમ વીસ + અમ વિસ્ + અમ વિ + સમ વી + સમ વીસ + અમ વિસ્ + અમ વિ + સમ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) In which of the following situations, there is no need to provide directional overcurrent protection. single end fed, parallel feeder double end fed, single feeder ring main single end fed, single feeder single end fed, parallel feeder double end fed, single feeder ring main single end fed, single feeder ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Directional overcurrent relay protection is used for parallel feeder and ring main system.It is not used for the protection of radial system.
MGVCL Exam Paper (30-07-2021 Shift 3) In public sector, what is the full form of "LIC"? Life Insurers Lively Insurance Corporation Life Insurance Corporation Life Insurance Company Life Insurers Lively Insurance Corporation Life Insurance Corporation Life Insurance Company ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The process of peeling of rocks into layers is called Sublimation Delta Exfoliation Brachans Sublimation Delta Exfoliation Brachans ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Which state is the largest producer of wind energy in India? Maharashtra Andhra Pradesh Madhya Pradesh Tamil Nadu Maharashtra Andhra Pradesh Madhya Pradesh Tamil Nadu ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rank - State - Capaciy1 - Tamilnadu - 7.5 GW2 - Maharastra - 5 GW3 - Karnataka - 4.8 GW4 - Rajsthan - 4.3 GW
MGVCL Exam Paper (30-07-2021 Shift 3) A conductor having surface density is embedded in a dielectric medium of permittivity. The electric field in the medium is E. If it is known that the pressure p on the conductor surface is equal to the electric energy density in the medium, then p (in SI unit)is given by σ²/(2πε) σ²/(2ε) σ/(4πε) σ²/(4π) σ²/(2πε) σ²/(2ε) σ/(4πε) σ²/(4π) ANSWER EXPLANATION DOWNLOAD EXAMIANS APP As the point is near to the conductor will act as infinite sheetElectric field (E) is given by:E = σ²/(2ϵ)