i got 88.82 percentile in jee paper2 in januvary 2020.can i get admission in nit calicut? i'm obc category.my home state kerala. or can i get admission in best 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:
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