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The JAC Chandigarh College Predictor 2025 designed by Careers360 enables students to know their likely engineering colleges based on their JEE Main score. JAC Chandigarh organises counselling sessions every year to distribute seats in different B.Tech courses of engineering colleges in Chandigarh. This tool eases this by providing students with the probable colleges that they will get into as determined by previous years data. It also includes college wise placement statistics, cutoffs and rankings so that the candidates can choose college wisely.
Counselling type
JEE Main Paper 1 percentile
Class 12th passing state
Seat Type
Here are the steps you’ll need to follow in order to find out which possible institutes you can apply for through the predictor:
Visit the Careers360 JAC Chandigarh College Predictor 2025 webpage.
Select the type of counselling you wish to participate in.
Enter your JEE Percentile in “Overall JEE Main Paper 1 Percentile.” section.
Did you complete your class 12th in Chandigarh?Yes or No.
Select the seat type.
Click on the “Predict my colleges” button.
Even though the JAC Chandigarh 2025 College Predictor will be most helpful for those students who already have a dream about the engineering college where they want to get their education. Taking into account the marks that students secure in JEE Main, they do get an idea regarding college in which they can take an admission.
The JAC Chandigarh 2025 College Predictor is useful for every engineering student.
Based on JEE Main scores, it also offers individual college suggestions to students so they can find colleges that are best suited to them.
Average user can obtain more college placements, cutoffs, and rankings information, and other related decision-making information.
It is also very easy to navigate around the tool so none of the students, no matter their aptitude in handling technology, will find it hard to use the tool.
Employing past records and counting patterns, the predictor raises the estimation of accuracy of colleges predicting giving likely students a better understanding of counselling of their chances of being admitted to college.
Candidates can know the JAC Chandigarh 2025 participating institutes list here to get an idea of their expected colleges:
JAC Chandigarh B.Tech 2025 Participating Institutes
S.No
Institutes Names
1
University Institute of Engineering and Technology, Panjab University, Chandigarh
2
Dr. S.S. Bhatnagar University Institute of Chemical Engineering and Technology, Panjab University, Chandigarh
3
University Institute of Engineering and Technology, Panjab University SSG Regional Centre, Hoshiarpur
4
Chandigarh College of Engineering and Technology
5
Chandigarh College of Architecture
Institute Name
Round 1 Closing Rank
Round 2 Closing Rank
Round 3 Closing Rank
Round 4 Closing Rank
Round 5 Closing Rank
University Institute of Engineering and Technology, PUSSG Regional Centre, Hoshiarpur
59266
76907
85640
91894.0
174183.0
Chandigarh College of Engineering & Technology, Chandigarh
70983
82042
86758
89506.0
127278.0
31588
38149
39898
42817.0
51049.0
Below are the day-wise JEE Main session 1 questions and answers
https://engineering.careers360.com/download/ebooks/jee-main-2026-analysis-january-session
The complete analysis for JEE Main Jan 29 shift 1 exam 2026 will be updated soon at https://engineering.careers360.com/articles/jee-main-2026-january-29-question-paper-with-solutions-pdf
As of now, January 23 shift 2 is the toughest shift of JEE Mains 2026. Get the difficulty level of all shifts and days here
https://engineering.careers360.com/articles/toughest-shift-of-jee-main-2026-session-1-exam
JEE Main 28 Jan shift 2 exam will end at 6 PM. The complete analysis and memory-based questions will solution will be updated in the below article. Keep checking the page-
https://engineering.careers360.com/articles/jee-main-2026-january-28-shift-2-question-paper-with-solutions-pdf
The JEE Main Jan 28 paper for shift 1 was moderate. Although Chemistry was a bit tricky. As expected, Maths was time consuming. But Physics was on the easier side.
You can find the answers for Jan 28 shift 1 online. They are memory based. I found this article helpful
Hi Lucky, Please refer to this link and you can download the free pdf.
https://engineering.careers360.com/download/ebooks/jee-main-2026-memory-based-questions-and-analysis-of-21st-january-shift-1
625 = ms Δ T + mL 625 = m [ 125 × 300 + 2.5 × 10 4 ] 625 = m [ 37500 + 25000 ] 625 = m [ 62500 ] m = 1 100 kg M = 10 grams
HI Manisha Maharana
You can download the JEE Mains 10 Free Mock Test with Detailed solutions. Its a feely downloadable pdf. https://engineering.careers360.com/download/sample-papers/jee-main-10-full-mock-test-and-explanations-pdf Also, you can check ad attemp the online mock test on our platform. https://learn.careers360.com/test-series-jee-main-free-mock-test/
A general equation of a circle is
$ x^2+y^2+2 g x+2 f y+c=0 $
Since it passes through $(0,0)$,
$ c=0 $
So the equation becomes
$ x^2+y^2+2 g x+2 f y=0 $
It cuts the x -axis at ( $a, 0$ ).
Substituting:
$ a^2+2 g a=0 $
g=-a/2
Hi Ritesh Here is the article for your reference https://hospitality.careers360.com/articles/best-books-for-nchmct-jee-preparation
Jan. 29, 2026, 7:21 p.m. IST
Jan. 29, 2026, 7:19 p.m. IST
Jan. 29, 2026, 3:12 p.m. IST
Jan. 29, 2026, 9:17 a.m. IST
Jan. 28, 2026, 9:18 p.m. IST
Jan. 28, 2026, 7:16 p.m. IST
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Jan. 26, 2026, 10:44 p.m. IST
Jan. 25, 2026, 8:16 a.m. IST
Jan. 24, 2026, 7:35 p.m. IST
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