MGVCL Exam Paper (30-07-2021 Shift 2) Which one of the following is the largest windfarm located in India? Talcher Mundra Muppandal Vindhayachal Talcher Mundra Muppandal Vindhayachal ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rank - Windfarm name - Capacity1 - Muppandal windfarm - 1500 MW2 - Jaisalmer Wind Park - 1064 MW3 - Brahmanvel Wind Farm Dhule - 964 MW4 - Damanjodi Wind Farm - 99 MW5 - Tuppadahalli Wind Farm - 56.1 MW
MGVCL Exam Paper (30-07-2021 Shift 2) In the context of smart meters, the term 'SMETS' stand for Smart Metering Equipment Technical Specifications Standard Monitoring Equipment Technical Specifications Smart Monitoring Equipment Technical Specifications Standard Metering Equipment Technical Specifications Smart Metering Equipment Technical Specifications Standard Monitoring Equipment Technical Specifications Smart Monitoring Equipment Technical Specifications Standard Metering Equipment Technical Specifications ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) If the velocity of electromagnetic wave in free space is 3 × 10⁸ m/s, then the velocity in a medium with εr of 4.5 and μr of 2 would be 9 × 10⁸ m/s 27 × 10⁸ m/s 3 × 10⁸ m/s 1 × 10⁸ m/s 9 × 10⁸ m/s 27 × 10⁸ m/s 3 × 10⁸ m/s 1 × 10⁸ m/s ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Speed of EM wave in medium = C/√(εr*μr)= (300000000)/√9= 300000000/3= 100000000= 1*10⁸ m/s
MGVCL Exam Paper (30-07-2021 Shift 2) A wire pilot is generally economical for distances up to 20 to 25 km 75 km to 100 km 30 km to 60 km 10 km to 15 km 20 to 25 km 75 km to 100 km 30 km to 60 km 10 km to 15 km ANSWER EXPLANATION DOWNLOAD EXAMIANS APP A wire pilot is generally economical for distances up to 5 or 10 miles, beyond which a carrier-current pilot usually becomes more economical.
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 0.4π V zero 0.2π V 0.8π V 0.4π 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) 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 B A C D B A ANSWER DOWNLOAD EXAMIANS APP
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