JEE Normalization Process

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!!

With 88 percentile in JEE Mains, chances for CS/ECE/EEE in top NITs or IIITs are low. However, you may get these branches in private universities like Amity, Sharda, Galgotias, or lower-tier state government colleges through state counselling.

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.

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.

Dear candidate ,
Cutoffs for JEE Main 2024 for Tatyasaheb Kore Engineering college ( TKIET ) . Warananagar :
- CSE : 65,000 ( All India ) , 36,000 ( Home state ) .
- Electronics and Telecom : 40,000
- Chemical Engineering : 47,000 .
- Cyber security : up to - 85,000 .

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.

For JEE Main 2024, the closing ranks at Rajarambapu Institute of Technology, Islampur (General category, All India) were approximately:

• Computer Engineering: 2,10,000 – 2,40,000

• CSE (AI/ML or IT): 2,30,000 – 2,80,000

• Electronics & Telecommunication: 3,60,000 – 3,85,000

• Electrical Engineering: 4,70,000 – 4,80,000

• Civil Engineering: 4,15,000 – 4,85,000

• Mechanical Engineering: 6,05,000 – 6,15,000

• Robotics & Automation: 5,50,000 – 6,00,000

These are approximate ranges based on Round 3 or final round cutoffs. Some variation can occur based on seat availability and category.

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

Yes, you are eligible for JEE Advanced 2026.

You passed Class 12 in 2024 (CBSE) but didn’t meet the 75% requirement, so you reappeared through NIOS in 2025 and scored 75%. Since NIOS is a recognized board and you now meet the 75% eligibility, you’re allowed to appear for JEE Advanced in 2026. Also, JEE Advanced allows attempts in the year of passing 12th and the following year. So as long as you qualify JEE Main 2026 and are in the top 2.5 lakh candidates, you’re eligible.