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The college infrastructure is above average. the college space is 100 acres with one active canteen and an inhouse bank overall no major setback can be seen in terms of the infrastructure of the college.hostel allocation facilities are also present.
the college infrastructure is above average. the college space is 100 acres with one active canteen and an inhouse bank overall no major setback can be seen in terms of the infrastructure of the college.hostel allocation facilities are also present
Hello,
Admission in INSTITUTE OF ENGINEERING & TECHNOLOGY,LUCKNOW is based on merit in JEE Main. The Common Rank List (CRL) will be considered for admission under the category of the applicant. It means that if you belong to BC category then you will be allotted a seat under BC category based on your CRL. The counselling will be  You can check the previous year cut off for reference by clicking on the link given below
Hello Santosh,
IET Lucknow started taking admission through JEE Mains last year. So, not much data is available about the cutoffs of this college. However, IET is one of the top colleges in UP. The cutoff will be on the higher side. So, you can expect a percentile around 94 to be the cutoff. I am sorry to say you will not get a seat in this college at your rank. Although you can go for other options.
You can use the college predictor tool to predict your college:
https://engineering.careers360.com/jee-main-college-predictor?utm_source=qna&utm_medium=jee_cp
Not much data about IET's cutoff is available so it is advisable to sit for the counselling. Also, you can try for other state engineering colleges.
Thankyou!
Hello,
Yes, there are fair chances for you to get computer science branch at IET Lucknow considering this rank to be your all india rank. In the other two colleges there are chances for you to get electronics and communication branch.
If you want to check out the colleges you can get ,you can check out the link given below -
https://www.careers360.com/ts-eamcet-college-predictor?utm_source=qna&utm_medium=ini-cet_cp