Hello sir,

Thanks for reaching out

With a JEE main rank of 91.5 percentile, you have good chances in branches like ECE, Mechanical, or Civil in lower NITs such as NIT, Agartala.NIT, Sikkim;NIT, Mizoram.

Likely government options include

DCRUST,Murthal

UIET Kurukshetra

HBTU Kanpur

(Through state counselling)

Private colleges:

Thapar University Patiala (Non-CSE)

Chandigarh university

LPU, Jalandar

UPES, Dehradun

These are some great options offering good infrastructure and placements at your score.

Apply through multiple state councelling,fill max choices to improve your chances.

Explore more:

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

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

1. IIIT Hyderabad
2. University of Hyderabad
3. Woxsen University

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 aspirant,

The syllabus of JEE MAINS PAPER 2A for degree B.Arch inculdes three subjects: maths, aptitude test, and drawing.

The exam is of 400 marks with 77 questions and of 3 hours. -1 negative marking will be given for every wrong answer of maths or numerical value question. For drawing test, there will be no negative marking.

There are many platforms online providing sample papers, one of which is careers360. The link of which, I am attaching here,

https://engineering.careers360.com/articles/jee-main-syllabus-2026

Best of luck for your preparation.

That's great! MA in English Literature is a wonderful choice if you love reading and writing. It opens up many career options like teaching, writing, editing, and more. Best wishes for your journey!

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:

• Home State Quota (Haryana): Up to 40,000 (general category)
• Other State Quota: Up to 20,000-25,000 (general category)

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 .