NTA UGC NET 2019: Previous Year's Question Papers (2012 to 2018)

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Thank you for your question. As of now, the National Testing Agency (NTA) has not yet officially announced the exact application dates for the UGC NET December 2024 cycle. Typically, the application process begins a few months before the exam date, so it is advisable to regularly check the NTA's official website or stay updated through trusted sources. This will help you avoid missing important deadlines. Should you have any other queries, feel free to ask. We are here to assist you!

Hello there,

It is important to carry all the documents that you have been told to bring on your exam day. Every piece of document in the exam hall matters. NTA guidelines on national level exams states that a candidate must have 2 passport size photographs that is there on their online application. This is solely because the purpose of identification of a candidate. One photograph will be pasted on attendance sheet. You need to carry what else is the guidelines to avoid any issue in exam hall.

I hope this information will be helpful to you.

Hello aspirant,

Solving previous year question papers immensely helps you in enhancing your preparation for a competitive exam like NET UGC. It gives you the idea of exam pattern, level of questions asked in exam etc.

To get the UGC NET last year question papers, you can visit our website by clicking on the link given below.

https://competition.careers360.com/download/ugc-net-sample-papers-previous-years

Thank you

Hope this information helps you.

Hello there,

If you have mistakenly clicked on the wrong gender, you need not to be worried. The NTA (National Testing Agency) conducts UGC-NET Exam twice a year- June and December cycle and always open a correction window for those candidates who have filled any of their details incorrectly. This correction window will give you a chance to correct your mistakes in the application form. But, it is very important to know that this correction window may be or may not be opened for a longer period of time. like for a week or so.

To keep update, carefully watch NTA UGC-NET website.

https://ugcnet.nta.ac.in/

GOOD LUCK!

Hello,

Yes, candidate who has completed their Master's from IGNOU can apply for UGC-NET. The eligibility criteria of appearing in UGC-NET Exam is to score an equivalent or more than 55% in their Master's CGPA.

UGC-NET Exam happens twice a year. The June cycle has been re-scheduled. The second cycle will take place in the month of December.

Hello Aspirant,

As per the guidelines issued by UGC NET, a candidate enrolled in a Master's program, whether in the 1st semester or the last, is eligible to sit for the UGC NET exam of the subject in which they are doing masters in. Please don't let people misguide you, you may even check the eligibility criteria mentioned on the UGC NET website.

Hope this helps!

UGC considers degrees from universities recognized by the Association of Indian Universities (AIU).  For confirmation, it is best to  directly contact the UGC or check their website for a list of recognized universities.

The standard UGC NET eligibility criteria require a Master's degree completed within two years of the exam date.

https://competition.careers360.com/exams/ugc-net

I hope it helps!

You can Download UGC NET Question Paper from Eduncle UGC NET Question Paper Blog

Hello aspirant,

Before applying, candidates must review the qualifying requirements for UGC NET 2023. Twice a year, the UGC NET is held to ascertain the eligibility of applicants seeking to become Assistant Professors and JRFs.

The age limit for JRF is 30 years but there is no agr limit for assistant professor. So you are eligible for assistant professor job.

Thank you

Hope this information helps you.

For calculating the percentile from the raw score,we need to know to few things which are.

1. First we should know the number of candidates appeared for exam in the shift. Let us denote this but by N.

2. Then all the candidates of the shift are sorted in decreasing order of their marks.

3. Then raw score of each Candidate is noted. Let denoted by T. Then number of candidates in that shift whose raw score are equal or less than T are counted . Suppose this number is m .

4. Then percentile score of calculated as

P =[ m÷N]×100

Hope it would help you