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UGC NET Results Formula - What is the procedure

UGC NET Results Formula - What is the procedure

The Central Board of Secondary Education (CBSE) will be conducting the National Eligibility Test (NET) on July 10, 2016(Sunday), on behalf of the University Grants Commission. The UGC NET is conducting for determining the eligibility of Indian nationals for the Eligibility for Assistant Professor only or Junior Research Fellowship and eligibility for Assistant Professor, in Indian universities and colleges.

Procedure and criteria for declaration of result:

The UGC NET result will comprise of the following steps:

Step I: The minimum marks to be obtained in NET for considering a candidate for the award of JRF and eligibility for Assistant Professor are to be acquired separately in Paper-I, Paper-II and Paper-III. The minimum marks for Paper I will be 40 (40 per cent) for General Category and 35 (35 per cent) for SC/ST. Similarly, the minimum marks for Paper II will be 40 (40 per cent) for General Category and 35 (35 per cent) for SC/ST. For Paper III minimum marks to be obtained will be 75 (50 per cent) for General Category and 60 (40 per cent) for SC/ST. 

Step II: Amongst candidates who will clear Step I, a merit list will then be prepared subject-wise and category-wise using theaggregate marks of all the three papers secured by such candidates.

Step III: The top 15 per cent candidates (for each subject and category), which will be taken from the merit list mentioned under step II and then will be declared NET qualified for eligibility for Assistant Professor only. 

Step IV: separate merit list for the award of Junior Research Fellow will be prepared from amongst the NET qualified candidates figuring in the merit list prepared under step III.

UGC NET 2016 Admit Cards available for download at CBSENET.NIC.IN

 

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