• HOME
  • QUIZ
  • CONTACT US
EXAMIANS
  • COMPUTER
  • CURRENT AFFAIRS
  • ENGINEERING
    • Chemical Engineering
    • Civil Engineering
    • Computer Engineering
    • Electrical Engineering
    • Mechanical Engineering
  • ENGLISH GRAMMAR
  • GK
  • GUJARATI MCQ

MGVCL Exam Paper (30-07-2021 Shift 3)

MGVCL Exam Paper (30-07-2021 Shift 3)
The International boundary 'Radcliffe Line’ lies between India and

Afghanistan
China
Japan
Pakistan

ANSWER DOWNLOAD EXAMIANS APP

MGVCL Exam Paper (30-07-2021 Shift 3)
Which is the function of Operating System?

Devices Management
Security
All of these
Process Management

ANSWER DOWNLOAD EXAMIANS APP

MGVCL Exam Paper (30-07-2021 Shift 3)
Which state is the largest producer of wind energy in India?

Madhya Pradesh
Tamil Nadu
Maharashtra
Andhra Pradesh

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Rank - State - Capaciy
1 - Tamilnadu - 7.5 GW
2 - Maharastra - 5 GW
3 - Karnataka - 4.8 GW
4 - Rajsthan - 4.3 GW

MGVCL Exam Paper (30-07-2021 Shift 3)
A 200/400 V, 20 kVA, 2-winding transformer is connected as an auto-transformer to transform 600 V to 200 V. Calculate the kVA rating of the auto-transformer.

50 kVA
30 kVA
60 kVA
25 kVA

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Here auto transformer change voltage from 600 V to 200 V.
Voltage ratio (a) = 600/200 = 3
kVA rating of the auto-transformer = (a/(a - 1))*(kVA rating of auto transformer)
= (3/2)*20
= 30 kVA

MGVCL Exam Paper (30-07-2021 Shift 3)
A DC shunt generator has an induced voltage of 220 V on open circuit. When the machine is on load the terminal voltage is 200 V. Find the load current if the field resistance is 100 Ω and armature resistance is 0.2 Ω.

104 A
96 A
98 A
102 A

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Voltage equation for Dc shunt generator,
Eb = V + Ia*Ra
V = Ish*Rsh
Ish = 200/100
= 2 A
Eb = V + Ia*Ra
20 = Ia*Ra
Ia = 20/0.2
= 100 A
I = 100 - 2
= 98 A

MGVCL Exam Paper (30-07-2021 Shift 3)
A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 2 hours. What is the permissible overload for a duration of two hours. Assume the permissible load kVA as a fraction of rated kVA is 1.43.

715 kVA
1050 kVA
1430 kVA
700 kVA

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Permissible overload for transformer = Full load kVA*factor for permissible overload
= 1000*1.43
= 1430 kVA

MORE MCQ ON MGVCL Exam Paper (30-07-2021 Shift 3)

DOWNLOAD APP

  • APPLE
    from app store
  • ANDROID
    from play store

SEARCH

LOGIN HERE


  • GOOGLE

FIND US

  • 1.70K
    FOLLOW US
  • EXAMIANSSTUDY FOR YOUR DREAMS.
  • SUPPORT :SUPPORT EMAIL ACCOUNT : examians@yahoo.com

OTHER WEBSITES

  • GUJARATI MCQ
  • ACCOUNTIANS

QUICK LINKS

  • HOME
  • QUIZ
  • PRIVACY POLICY
  • DISCLAIMER
  • TERMS & CONDITIONS
  • CONTACT US
↑