Dear student,
Greetings
The CBSE UGC NET 2018 Result are declared, there are some steps which should be followed to get detailed information:
Minimum marks to be obtained in NET for considering a candidate for the award of JRF and eligibility for Assistant Professor:

General- 
Paper 1 and 2 - 40%
ST/SC/PWD/OBC-
Paper 1 and 2 - 35%

Then, Amongst those candidates who have cleared step I, a merit list will be prepared subject-wise and category-wiseusing the aggregate marks of all the three papers secured by such candidates.

Lastly, Top 6% 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.

I Hope this helps, Thank you!

UGC NET Cut off for July Session has been declared on July 31. You can check Computer Science Cut Off from the link below:

https://competition.careers360.com/articles/ugc-net-cutoff

Hello Prakash, The last year cutoff of Computer Science's for JRF was - Unreserved (65.71), OBC (60.57), SC (56.57), ST (54.87)

The examination authority CBSE decides UGC NET cut off - candidates are shortlisted on the basis of qualifying marks (category wise) followed by preparation of merit list as per the scored marks. Based on UGC NET cut off, only top 15% candidates selected for Assistant/Associate Professorship.

This year expecting cutoff for unreserved is between 63-65.5

All the best

I would suggest you to go through the below provided link to know about the cut off in detail -

UGC -NET Cut off

Hope this helps!!!