CBSE NET Selection Procedure 2016

CBSE NET Selection Procedure 2016– Central Board of Secondary Education (CBSE) will conduct the National Eligibility Test (NET) on 10 July 2016. The candidates have to appear for three papers as per the mode of examination. The selection of candidates for the award of JRF and Eligibility for Assistant Professor will be done on the basis of minimum marks obtained in NET 2016 Exam. This article on Selection Procedure of CBSE NET has every detail related to the selection of candidates for the award of JRF and Eligibility for Assistant Professor. Scroll down to read more regarding the CBSE NET Selection Procedure 2016.

Before moving to the CBSE NET Selection Procedure 2016, let’s have a look at the CBSE NET Exam Pattern 2016.

Papers	Number of Questions	Compulsory Questions	Time Duration
Paper 1	60	50	1 ¼ hrs.
Paper 2	50	50	1 ¼ hrs.
Paper 3	75	75	2 ½ hrs.

CBSE NET 2016 Minimum Qualifying Marks for Selection

The candidates appearing for CBSE NET 2016 exam need to qualify the exam by obtaining the following marks fixed by the Board. See below.

Category	Paper I	Paper II	Paper III
General	40 (40%)	40(40%)	75 (50%)
OBC (Non-creamy layer) PWD/SC/ST	35 (35%)	35 (35%)	60 (40%)

Procedure & Criteria for Selection in CBSE NET 2016

The procedure & criteria of selection in CBSE NET 2016 is explained in the following steps.

• Step I: The candidates need to obtain minimum marks to be obtained in NET for considering a candidate for the award of JRF and eligibility for Assistant Professor. They are required to obtain minimum marks separately in Paper-I, Paper-II, and Paper –III.
• Step II: Amongst those candidates who have cleared the step I, a merit list will be prepared subject-wise and category-wise using the aggregate marks of all the three papers secured by such candidates.
• Step III: Top 15% candidates (for each subject and category), from the merit list, mentioned under step II, will be declared NET qualified for eligibility for Assistant Professor only. By this method, if the number obtained is less than one, it is taken as at least one, i.e., a minimum of one candidate is qualified wherever a fraction less than one is obtained as per this methodology. However, for fractions greater than one, the figure is rounded off to the nearest whole number.
• Step IV: A separate merit list for the award of JRF will be prepared from amongst the NET qualified candidates figuring in the merit list prepared under step III.

Criteria for Awarding Junior Research Fellowships of CBSE NET 2016

• Out of the candidates who have qualified for Assistant Professor, those candidates who have applied for JRF will be considered for the award of JRF according to the slots available for JRF.
• The methodology for allocation of JRF slots among different subjects is done category-wise.
• The Commission provides a minimum of 3200 fellowships in each UGC-NET.
• As per Government of India reservation policy, a minimum of 27%, 15%, 7.5% and 3% of Fellowships are reserved for the OBC (Non-creamy layer), SC, ST and Persons with disabilities (PWD) respectively.
• The calculation is based on the ratio of the number of candidates who had opted for JRF and were qualified for lectureship eligibility in each subject category-wise to the total number of candidates qualified for lectureship who had opted for JRF in that category overall subjects multiplied by the total number of fellowships to be awarded to that category. The nearest whole number obtained as per this formula provides the number of slots available in any particular subject category-wise.
• If the number of JRF slots obtained for any subject in a particular category is less than one, it is taken as one, i.e., a minimum of one JRF is awarded wherever a fraction less than one is obtained as per this methodology. However, for fractions greater than one, the figure is rounded off to the nearest whole number.