in how much marks NIT Delhi got for computer science branch
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The minimum marks required to get admission to NIT Delhi in computer science engineering is the cutoff, which the institute releases for all category candidates. The cutoff is released with opening and closing ranks for candidates belonging to home state that means Delhi, in this case and other state, candidates from outside Delhi. The opening rank means that admission starts with it and closing rank means admission ends with it. You will not get admission in NIT Delhi if you have marks less than its closing rank for your particular category and state.
The cutoff for NIT Delhi 2017 is mentioned below :
Home State
General Category
Opening Rank - 3287
Closing Rank - 11564
OBC-NCL Category
Opening Rank - 2076
Closing Rank - 3822
SC Category
Opening Rank - 491
Closing Rank - 3403
ST Category
Opening Rank - 818
Closing Rank - 4014
Other State
General Category
Opening Rank - 3421
Closing Rank - 7816
OBC-NCL Category
Opening Rank - 1764
Closing Rank - 6742
SC Category
Opening Rank - 785
Closing Rank - 1530
ST Category
Opening Rank - 342
Closing Rank - 716
You can use this cutoff and estimate it for this year.
All the best!
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