MGVCL Exam Paper (30-07-2021 Shift 2) Which one of the following is the largest windfarm located in India? Vindhayachal Talcher Muppandal Mundra Vindhayachal Talcher Muppandal Mundra 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) If all the devices are connected to a central hub, then the topology is called Tree Topology Star Topology Bus Topology Tree Topology Tree Topology Star Topology Bus Topology Tree Topology ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A THHN conductor will have a 60˚C termination on one end and a 75˚C termination on the other, which ampacity column will be used? 75˚C column 60˚C column None of these 90˚C column 75˚C column 60˚C column None of these 90˚C column ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Conductor with higher temperature ratings, provide the ampacity is determined based on the 60°C ampacity of the conductor.
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 3 × 10⁸ m/s 9 × 10⁸ m/s 1 × 10⁸ m/s 27 × 10⁸ m/s 3 × 10⁸ m/s 9 × 10⁸ m/s 1 × 10⁸ m/s 27 × 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) સ્વાધીન' નો વિરોધીભાસી શબ્દ કયો છે? પરાધીન અસ્વાધીન નીસ્વધીન કુવાધીન પરાધીન અસ્વાધીન નીસ્વધીન કુવાધીન ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 4 hours. What is the permissible overload for a duration of four hours. Assume the permissible load kVA as a fraction of rated kVA is 1.24. 930 kVA 2480 kVA 1240 kVA 620 kVA 930 kVA 2480 kVA 1240 kVA 620 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload = Full load kVA*factor for permissible load= 1000*1.24= 1240 kVA
EXAMIANS