i have 88.82 percentile in jee paper2 exam in januvary 2020.can i get admission in nit calicut? my hometown in kerala. or can i get admission in any other best engineering college in kerala.
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You can calculate your rank in JEE Main Paper 2 using the formula:Rank = (100-percentile)*(No. of students appeared for the examination)/100.
As the no. of students appeared for Paper 2 (Jan) are around 112679, your rank will be around 12600. However as the no. of students in April changes, your rank will change according to that.
With this rank, there are very less chances of getting Architecture branch at NIT Calicut. As the closing cutoff rank for this branch is around 1200-1300 which is far away from your expected rank.
To check the list of all colleges which you may get with this percentile, kindly use the link provided below:
For more information regarding the same, kindly use the B.Tech companion from the link given:
Hi dear,
You have scored 88.82 percentile in JEE main paper 2 that means your rank should be around 12597. This is not your final rank.Your final rank will depend on number of student appear in April month exam. You can also calculate your rank as
Rank=(100-your percentile)×112679÷100
With this rank you have low chance to get NIT CALICUT. You can check cutoff of all college here
Hope it helps
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