Joint Entrance Examination (Main)
With an 80 percentile and a rank around 1,00,000, your chances of getting CSE or ECE in BVRIT (Bharatiya Vidya Bhavan’s Rajiv Gandhi Institute of Technology, Narsapur) depend on the category, gender, and home state quota.
BVRIT is a popular institute in Telangana under the TS EAMCET and JEE Mains admission system (for non-local or management quota). The CSE and ECE branches are in high demand, and closing ranks for these programs under JEE Mains are usually:
CSE (General category): closes around 50,000–80,000
ECE (General category): may go up to 1,00,000+ in some years
With a JEE Main rank around 1 lakh, getting ECE in BVRIT is possible, especially in later rounds.
I think you want to say 86 percentile because JEE Mains releases merit lists based on percentile, not percentages. So, with this percentile, you can get into several government colleges or private colleges in the city of Hyderabad. There many colleges in hyderabad that accept mains score, some popular ones arfe
But there are several colleges in this rank, which you can check with our college predictor in this link from Career360:
https://engineering.careers360.com/jee-main-college-predictor?utm_source=qna&utm_medium=jee_cp
Hello,
There is good news for you. You can get admission to the JECRC Foundation for the B.Tech program with just 12th marks. They can give you admission without a JEE Mains score; they just need your 12th marks, and based on this merit, you can obtain a seat in this institute. While you have 90.20% in your 12th, so it's a strong merit from your end. There is a high possibility that you will get admission to JECRC Foundation.
I hope it's clear your query!!
Hello
NIT Kurukshetra – Civil Engineering (JEE Main Cut-offs)
2024 Round 3 and 4 Final Closing Ranks
All India (AI Other State) Category – Gender Neutral
Opening Rank: Approximately 15,535
Closing Rank: Around 29,292 in Round 1, and up to 33,266 by Round 4
Home State (HS Haryana) Category – Gender Neutral
Opening Rank: Approximately 9,853
Closing Rank: Around 38,638 across various rounds
Expected Cut-Offs for 2025 (Projected)
The estimated closing ranks for Civil Engineering in the early rounds are as follows:
All India (Open): Between 15,520 and 34,638
EWS (All India): Between 4,090 and 46,628
OBC NCL (All India): Between 7,946 and 12,000
SC (All India): Between 1,445 and 5,155
ST (All India): Between 960 and 1,136
Note: These expected cutoffs are only estimates based on previous year trends and are subject to change depending on this year's exam and counselling dynamics.
Hello Srinidhi,
In order to acquire Mechanical Engineering at NIT Kurukshetra, generally the required JEE Main rank will be in the range of:
The cutoffs that I have provided above change from year to year, as well as category and gender. For reserved categories (SC/ST/OBC/EWS), the ranks are much more relaxed.
Hello aspirant,
As per your query, the JEE rank required for NIT Kurukshetra in Computer Science Engineering branch is,
All India Closing Rank (General category) : 6562
Home state Closing Rank (General category) : 8198
For more information, click on the link given below,
https://www.careers360.com/university/national-institute-of-technology-kurukshetra/cut-off
Regards
According to the eligibility criteria for JEE Advanced, the candidates must have appeared in Class 12th and scored 75% aggregate marks. For the first time, you appeared in the CBSE Class 12 board exam in 2024, and in 2025, you appeared in the 12th examination from the NIOS board and scored 75%. Since NIOS is a recognised board, and you fulfil the aggregate marks, you are eligible to appear in JEE Advanced 2026 .
Hii,
If you belong to a PCB background and want to explore fields like Artificial Intelligence and Data Science. But there are some important things to keep in mind. Usually, B.Tech in AI and DS comes under engineering courses, so most colleges prefer students who have studied Mathematics in Class 12 (i.e., PCM or PCMB). That’s because the course involves coding, statistics, algorithms, and problem-solving all of which need a solid foundation in Maths. Some private universities or deemed colleges offer direct admission to PCB students, either by relaxing the Maths requirement or by admitting students through management quota. You may have to take an additional foundation course in Mathematics or coding in the first semester. A few colleges have started B.Sc. in AI & Data Science programs that are more flexible with eligibility. These are not engineering degrees but can still lead to good careers in AI, especially if you're open to further certifications.Since you haven't given the JEE or any entrance exam, you'll have to look for colleges that offer direct admission based on Class 12 marks, have a management quota, or conduct a basic aptitude test or interview, and accept PCB students also.It’s better to call or email the admission desk directly.
To explore more colleges offering AI and DS courses, including eligibility and fee details, you can visit:
https://www.careers360.com/courses/b-tech-in-artificial-intelligence
Thank you and all the best
Yes, the Circle chapter, taught in Class 11 under Coordinate Geometry, is part of the official JEE Main 2026 Mathematics syllabus, so you will definitely encounter questions from it in the exam.
Circles and Conic Sections are explicitly listed in the JEE Main Mathematics syllabus.
You will need to study topics such as the standard and general form of the circle's equation, center and radius, equation when endpoints of a diameter are given, and the intersection of a line with a circle. Tangents and chord-related problems may also be included in the syllabus.
With AIR 23749 in NEET, these are the government medical colleges which you can get in Uttar Pradesh:
1. Autonomous State Medical College - Etah, Fatehpur, Ghazipur, Hardoi, Mirzapur, Pratapgarh, Sidhdharthnagar. 2. Government Medical College - Orai, Banda, Badaun, Bahraich, Basti, Faizabad, Firozabad, Shahjahanpur, Kannauj, Azamgarh, Pilibhit, Sultanpur, Bulandshahr.
Hope this answer helps! Thank You!!!
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