SSC Junior Engineer 2014 recruitment exam was conducted on Sunday of 25th May 2014. The SSC JE Electrical Answer key for Evening/ afternoon shift test paper can be downloaded from this webpage. We also includes full set of question paper along with their four objective choices and correct answer represented in “green tick” mark.
There are four sections in this years SSC JE 2014 exam paper. First part consists questions of general ability and reasoning question while other three parts contains core engineering branch questions. SSC JE Jobs 2014 aspirant students may have to answer one core paper in which they have finished their engineering degree in academics.
SSC JE Electrical answer key 2014 are available to download from this page. There are 4 different sets of SSC Paper with question booklet number: DB-2014. These sets have different key as 321GK 4, 414 LI 5, 542 PK 6 and 612 OL 7.
Here we choose SSC Junior Engineer paper booklet no. 321GK 4 as a reference to these questions. Please cross check questions in your booklet and this answer key with solution paper.
All SSC Junior Engineer 25th May 2014 Exam Answer Keys:
- SSC JE 25th May 2014 – Morning Paper – Answer Keys
- SSC JE 25th May 2014 – Afternoon/Evening Paper – Answers Keys
SSC JE Electrical Engg 2014 Evening Paper Answer Keys:
The SSC JE 2014 notification was issues by Staff Service Commission of India in the month of February and March. Commission issued admit card for this prestigious exam in April and May 2014.
Now as the exam had conducted and we have given you the official answer keys. Now its your turn to share your marks obtained in electrical paper as well as other paper. The SSC Result for Junior Engineer will be uploaded on official website of SSC after few weeks.
If you finds answer of any question as wrong then you can file a appeal before 19th June 2014 in front of commission. appeals after this due date shall not be entertained.
qno.121 since the north pole is moving away from metal ring the induced current in the ring will such that it will oppose the motion according to LENZ’s law and region of the ring nearer to magnet’s north pole will act as south pole.now according to right hand thumb rule the direction of induced current will be clockwise.(if seen from magnet side and anticlockwise if seen from opposite side)but it is not clear whether we are standing on magnet side or opposite please reply sir..
q no 137 the answer will be 8000w since impedance (z)=sqrt r square+(XL-XC) square =5ohms now power = Vsquare/Z
200*200/5=8000w
please chech q no 137 the answer will be 8000w
answer of q factor of parallel resonance cirucit is wronge.
there is no correct option.
Answer you tick as correct is Q factor of series resonant network.
what can we do about this?
Nice…super
The answer key is given as a photo ..super