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!!!!

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 NIT Kurukshetra.You may have to appear for July session .For checking the probabilities of the colleges,you can use our college predictor.Link is provided below:

https://engineering.careers360.com/jee-main-college-predictor?ici=cp-listing_engineering&icn=JEE-Main-College-Predictor If you are looking for preparing for Jee mains,you can try our knockout package:

https://learn.careers360.com/knockout-jee-main-april/

you can check the previous year cutoff's in the link provided below:

https://engineering.careers360.com/articles/jee-main-cutoff-for-nit-kurukshetra

Hope it helps you!!!

Good luck!!!!

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 electrical it was 5209 and for ECE it was 3984.Considering this you may not get seat, you need to score 92 or above percentile to get rank around 3000 and to get rank around 5000 your percentile must be 86 or above.Cut off is subjected to change it may increase or decrease depending on number of seats available, number of candidates appeared etc.

You may predict the colleges https://engineering.careers360.com/jee-main-college-predictor

• Classical mechanics and theory of relativity
• Electricity, Magnetism and Electromagnetic theory
• Properties of Matter and Kinetic Theory of Gases
• Semiconductor Devices
• Thermodynamics
• Waves and Optics
• Statistical Physics
• Quantum Physics
• Nuclear Physics
• Solid State Physics
• Atomic and Molecular

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 difficult to give the exact percentile.But according to the number of candidates appeared in JEE Main January 2020 session you need to score 90-95 percentile to get such rank as in 2019 round 6 cut off rank.But cut off changes in 2020 depending on number of candidates appeared, seats available in each branch etc.

Please try our JEE main knockout program

https://learn.careers360.com/knockoutfull-jee-main-april/?icn=JEE_Main_knockout&ici=c360_sticky_qna_engineering

Dear Aspirant ,

First of all there was no percentile scheme in 2016 , this is introduced privious year . And privious year closing rank for your category in NIT Kurushetra CSE was 1521 ( SC categoryn rank) in 7th round and 51k (All India Rank) in spot round and 7597 (category rank ) for production engineering in 7th round and 176k in spot round for production engineering .

You can calculate your rank through this link :-

https://engineering.careers360.com/jee-main-rank-predictor

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 need to score 95 or above percentile.Cut off may increase or decrease in 2020.Cut off depends on number of candidates appeared, difficulty level of exam, number of seats available etc.

Hello,

You can calculate your expected rank by the formula (100-your total percentile)*869010/100

where 869010 is the total number of candidates appeared in January session.So by the percentile given by you, your expected rank will be around 330223 and your category rank will be around 28000-30000.With such rank it is difficult to get seat in any NIT.In NIT, Kurukshetra the cut off was as follows in 2019 for SC category

• Civil: 6282
• EE: 5209
• ECE: 3894
• MECH: 5517
• PIE: 7597
• IT: 2004
• CSE: 1521

It is difficult to get seat in NIT, Kurukshetra too.You need to score 85 or above percentile to have decent chances of getting a seat.Cut off may increase or decrease.Our JEE Main Knockout may help you with your preparation

https://learn.careers360.com/knockout-jee-main-april/

Hey Kartik

To get admission into NIT Kurukshetra, considering your category, one needs to score following score:

• Last year, the cutoff rank for Electrical for Home state, SC candidate was AIR 5201
• Accordingly you need to score around 200-220 marks out of 360, and for the new pattern you score should be at least 185-200 marks.

For more details, follow; NIT K