I score 96.03 percentile for jee paper 2.with this score can I get admission into Nit's.i am obc 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 4500. However as the no. of students in April changes, your rank will change according to that.
With this much rank, you will be eligible to get admission in NITs in B.Arch branch. As the cutoffs for B.Arch are not that much high, you will easily get B.Arch branch in one of the top NITs.
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