Hello Aspirant,

NIT Calicut ranks 23 by NIRF while NIT Surathkal has NIRF rank of 13 which makes it better than NIT Calicut and other NITs.

Talking about faculties, NIT Surathkal has more better faculty where most of them are from thermal backgrounds.

Placements are satisfactory in both the NITs

Hello,

You haven't mentioned whether you belong to OBC-NCL or OBC-CL, if you come under OBC-CL then you are considered as general category. Your overall category rank is expected to be around 6084 and you can calculate it by the formula (100-your total percentile)*112679/100 where 112679 is the total number

Hello student,

Actually you have not clear whether you get passed in 10 + 2 level or not. As you get failed in mathematics, it may be a problem for you in future in this engineering career.

But as per the official notice if you have passing certificate of 10

Hello,

While it is a bit ambiguous with what you are comparing NIT Surathakal Nanotech. But talking in general, these two colleges are more or less in same level. If a company visits either of the two colleges, it generally visits the other too.

Speaking of general placements, past recruiters

Hi Aysha,

You got 98.4 percentile in JEE main exam, your rank would be approximately in between 1500 to 2000. Also, you can check your expected rank using the formula:

Rank = (100-percentile)*(No. of students appeared for the examination)/100

As the no. of students appeared for Paper 2 are 112680,

Hello,

With such score the percentile will be around 99 which is a decent percentile and you will have good chances of getting seat in a top NIT. This year the minimum percentage requirement is scrapped off to get admission in NIT and IIT. You need to pass 10+2 with

Dear aspirant,

Eligibility criteria for NIT Calicut:-

Your 12th PCM marks are above 80% that is enough to make you eligible for admission to NIT Calicut. If you score around 93.88% in JEE Main is also a very good score and you have a higher chance of getting admission to

Hello,

Your overall expected rank will be around 48142. The formula to calculate it is (100-your percentile)*869010/100

where 869010= Total number of candidates appeared in JEE Main January session.You have not mentioned if you belong to OBC-NCL or OBC-CL. If you belong to OBC-CL then you are considered as general