What should be my rank in jee advance to get IIT ROORKEE OR IIT KHARAGPUR CS BRANCH being an obc candidate of Jharkhand ?
Answers (2)
Student Expert
13th Jun, 2019
Hello there
To get CSE branch in IIT Kharagpur or IIT Roorkee you need to score a rank below 400.Because both of these IITs are among the top IITs and computer science engineering is most popular B.Tech branch.
Hope I am clear
To get CSE branch in IIT Kharagpur or IIT Roorkee you need to score a rank below 400.Because both of these IITs are among the top IITs and computer science engineering is most popular B.Tech branch.
Hope I am clear
Comments (0)
13th Jun, 2019
Hi aspirant,
CSE is one of the most popular engineering branch in India and both IIT Roorkee and IIT Kharagpur are the best engineering institutes. Hence, they have a high cut-off. As of 2018, the cutoff for CSE for OBC-NCL at IIT Roorkee WAS 225 and at IIT Kharagpur was 137. This year's cut-off are also expected to be similar. So your category rank should be somewhere between 100-150.
Comments (0)
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