JEE Main 2019 Paper 1 Answer Key by Entrance360 (January 9 to 12)

Here are the best engineering colleges for 35,000-50,000 rank in JEE Main 2026 examination -

• NIT Jalandhar
• NIT Jaipur
• NIT Bhopal
• NIT Goa
• NIT Patna
• NIT Surat
• NIT Raipur
• IIIT Trichy
• IIIT Ranchi
• IIIT Nagpur
• BIT Mesra

With 77 marks in JEE Main, you can expect engineering colleges like NIT Mizoram, NIT Meghalaya , and NIT Arunachal Pradesh. Consider private colleges like GITAM, Lovely Professional University, SRM Institute of Science and Technology , Chandigarh University, and Amity University for better branch options.

No, JEE Main Answer Key for 2026 session 1 is not out yet. NTA will release the JEE answer key 2026 on February 4.

With 10 marks in the JEE Main entrance examination, you can get admission in private universities only. 10 marks in JEE Mains equals a 13-19 percentile, which is low to get into any good engineering colleges.

Some of the private universities accepting low JEE Main scores are:

• TIET Patiala
• SASTRA

Students with 68 marks in JEE Main examination can hardly get admission into top IITs and NITs like IIT Gandhinagar , IIT Bombay, NIT Kanpur , and more, although there are many other universities and colleges that accepts 68 marks in JEE Main for admission to Engineering programmes, some of

A score of 32 in JEE Main generally corresponds to a percentile range of about 50 to 60, making it difficult for general category students to secure admission to NITs or IIITs, but you still have good chances in GFTIs and state colleges, especially with the Home State quota, OBC/SC/ST/EWS,

JEE Mains 2026 result for session 1 will be declared by February 12. For more details check

https://engineering.careers360.com/articles/jee-main-result

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

$\begin{array}{rl}& 625=\mathrm{ms}\mathrm{\Delta }\mathrm{T}+\mathrm{mL}\\ & 625=\mathrm{m}[125\times 300+2.5\times {10}^4]\\ & 625=\mathrm{m}[37500+25000]\\ & 625=\mathrm{m}[62500]\\ & \mathrm{m}=\frac{1}{100}\mathrm{kg}\\ & \mathrm{M}=10\text{grams}\end{array}$

Below are the day-wise JEE Main session 1 questions and answers

https://engineering.careers360.com/download/ebooks/jee-main-2026-analysis-january-session