Is 92.3 in jee mains paper 2 a good percentile to get into any nit for general category?
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Rank in JEE Main Paper 2 can be calculated by the formula:
Rank = (100-percentile)*(No. of students appeared for the examination)/100
As the no. of students appeared for Paper 2 are 112679, your rank will be around 8700. However as the no. of students in April changes, your rank will change according to that.
With this much rank, there are very less chances of getting a NIT through open category. However, you may get into NIT Hamirpur or NIT Raipur if home state quota for any of the two applies to you.
You can check the last years' cutoffs using the link provided below:
https://engineering.careers360.com/articles/jee-main-cutoff-for-b-arch-b-planning
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