Kurukshetra University All Questions

Dear Nitish ,

No problem ,but if you have dropped out due to some reason then you must be start from the starting if you wants to take admission again .Fist you have to purse entrance test Jee then at the time of counselling they might be ask whats the

hey,

kurukshetra university , kurukshetra do  not offer any honour course what they do offer is listed below

1. B.E/B.Tech

2. B.S.c

3. M.E/M.Tech

4. B.Com (//B.Com)

5. M.A.

6. B.A.

Hope this helps!

Hello,

Your rank will be around 2,60,703 and the formula is (100-your total percentile)*869010/100 where 869010 is the total number of candidates appeared in January session.And category rank will be in between 15000-20000.In 2019 the round 7 cut off for Home state SC candidate for Mechanical Engineering was 5517, for

Hello!!!!

Hope you are doing great!!!!

Kartik,Your All India Rank would be 104281 approximately,calculated by using the formula 100-88*869010/100 where 88 is your percentile and 869010 is the number of candidates appeared for jee mains in january session.As per the previous year scenario,it would be difficult to get seat in

Hello,

Cut off is given in terms of closing rank and opening rank for each category and branch.In 2019 the round 6 cut off was follows for SC home state category for various branches

• EE: 5209
• Civil: 5581
• Production: 7231
• IT: 2004
• Computer: 1521
• Mechanical: 5432
• ECE: 3894

It is

Hello dear

First of all in 2016 there was no such percentile system. This system is adopted from last year .

And when we calculate your approximately rank it will come out nearly 16000-20000 in school category and it will be more than 2 lakhs in general category and with

Hello,

You have not mentioned your percentile.In 2019 these were the cut off SC home state category branch wise in round 7 JoSAA counselling in NIT, Kurukshetra

• CSE: 1521
• Production and Industry Engineering: 7597
• IT: 2004
• Mechnaical : 5517
• ECE: 3894
• Civil: 6282
• EE: 5209

To get such rank you