Answers (4)
dhyanabi.ria 11th Feb, 2024

Hey aspirant,

NIT Raipur cutoff 2024 for B.Tech will be available at josaa.nic.in (http://josaa.nic.in/) . With 97 marks, your percentile should be around 93.05600452 -93.89928202  and corresponding rank should be around 76260-66999. NIT Raipur cutoff 2023 for B.Tech for CSE 10067-19729.

Visit : https://engineering.careers360.com/articles/jee-main-cutoff-for-nit-raipur for 2022 cutoff of NIT Raipur.

Hope this helps you. Thank you.

satyamercedes0626 10th Feb, 2024

Hello aspirant,

Hope you are doing great.

So, coming to your question, I regret to inform you that you won't be able to get the CSE Branch with your current marks, as they are very less when it comes to the maximum marks. Try to score more than 200 marks, so that you will be able to get CSE branch in the NIT. Otherwise, you might probably get ECE or EEE branches.

Hope this helps.

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Sajal Trivedi 10th Feb, 2024

Hello aspirant,

IIT is one of the most toughest exams in India and getting into NITs is more tougher than that. So unfortunately but with 97 marks, you can not get into any NITs with any quota. I suggest you to prepare well and try hard next time.

For more information you can visit our website by clicking on the link given below.

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

Thank you

Hope this information helps you.

shivamtrivedi0005 9th Feb, 2024

Hii There,

I hope you are ding well.

JoSAA is set to unveil the JEE Main 2024 cutoff for NIT Raipur, delineating the cutoff for each course individually. Prospective candidates aiming for admission at NIT Raipur will be able to access the JEE Main 2024 cutoff for the institute via the official website. The cutoff criteria for B.Tech 2024 will take into account various factors such as the complexity of the JEE Main exam, seat availability, historical cutoff trends, among others.

You can check the cutoff of NIT Raipur on the link mentioned below:

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


Thanks

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