62 Views

What is minimum marks or rank is required to get IIT DELHI in CSE in OBC category .


Ravinder Kashyap 26th May, 2019
Answer (1)
S. Saha Student Expert 5th Jul, 2019


The official authority institute will publish the JEE Advanced 2019 cutoff for IIT Delhi after the announcement of the result.

As of previous year's cutoff of Indian Institute of Technology, Delhi, for OBC-NCL candidate, the opening rank for computer science and engineering was 38 and closing rank was 65.

Know all about JEE Advanced

Candidates can get access to all the details about JEE Advanced including eligibility, syllabus, exam pattern, sample papers, cutoff, counselling, seat allotment etc.

Download Now

Know More About

Related Questions

Amity University | B.Sc Admis...
Apply
Ranked amongst top 3% universities globally (QS Rankings)
JSS University Mysore 2025
Apply
NAAC A+ Accredited| Ranked #24 in University Category by NIRF | Applications open for multiple UG & PG Programs
TMV, Pune | M.Sc Computer App...
Apply
2-year postgraduate program focusing on advanced computing and IT applications
Sanskaram University M.design...
Apply
100+ Industry collaborations | 10+ Years of legacy
TMV, Pune | B.Sc Computer App...
Apply
3-year undergraduate program focusing on practical computing skills.
Graphic Era (Deemed to be Uni...
Apply
NAAC A+ Grade | Among top 100 universities of India (NIRF 2024) | 40 crore+ scholarships distributed
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books