Brother either you have not completed the question ans you have not mentioned

y(5x^4y+1)dx+(x+2y+2x^5y)dy =?        what is it equal to is it equal to zero or anything else kindly reply

hey

Since you will fall into All India Quota for NIT surthakal, For ECE the closing and opening ranks approximately based on previous years  were 2000-4000.

For your rank to fall in that range, you have to score around 220-240marks in the exam.

Hope this helped

all the best

hey viseswara rao.

since you did not mention if you belong to All India quota or home state or what branch you want to get in.

here are the general opening and closing ranks for All Inida and Home State 920-18678 and 250-12236 for GENERAL category based on previous years.

hope this helped

thank you

Hello Aspirant,

None of the NIT offer B.Tech in Aerospace so you will have to clear JEE Advanced and only 4 IITs and IIEST Shibpur provide this programme.

IIT Bombay

IIT Madras

IIT Kharagpur

IIT Kanpur

Good Luck!

hi,

see artificial intelligence  and internet of things are new cutting edge technologies. any stream of btech students can pursue this. as it is new in industry thats why it may not included in any colleges cse syllabus. now come to  computer networking, it is included in cse, ece, electrical, IT, applied electronics branches syllabus of btech. there is no computer science courses in india with the specialization in the mentioned subjects of yours. computer science engineering is a 4 year course and you will get opportunity to learn many things including AI, IOT, computer networking etc.

hope you may understand.

Hello dear aspirant .

As this year JEE mains pattern completely changed hence it is a little difficult to predict exact stats but yes approximately 88-90 percentile will be ok in your case.

As per new trend we can simply say that in order to get this much percentile you really have to study hard and score at least 150 for it.

Ok good luck