Hello,

IIT Kanpur offer Bachelor of Science programme (B.S). It is a 4 year Undergraduate programme in Chemistry, Earth Sciences, Economics, Mathematics and Scientific Computing, Physics. 
Admission to the programme is done through JEE Advanced entrance exam. JEE Advanced is a national level entrance exam. You will first have to

In Institutes like IITs, NITs, and event in many colleges they have, Departments of Biological Science and Bioengineering(BSBE), Which offers BTech in BSBE and Biotechnology, and the admission criterion remains same. If you want to do Engineering in BSBE from any IIT then you need to qualify JEE Advance nothing

Its recommended to solve from best suggested books for each chapger shared below. 

https://university.careers360.com/articles/how-prepare-jam-physics

You should also go for writing down important concepts and revise regularly and solve maximum mock exams. 

Hello,
Eligibility criteria vary for every course. Kindly mention you are seeking admission in which course so I can help you.
Good luck!

Dear student.

Greetings.

Yes, a student with MSc in physics can pursue for M.tech in environmental engineering. 

Thank you. All the best.

Eligibility to appear in NEET (UG) - 2019 

 Eligibility for appearing in NEET (UG) is as stipulated in Indian Medical Council Act-1956 and the Dentists Act-1948 as amended in 2018.   Eligibility as per related Regulations is as follows:- 

i. He/she has completed age of 17 years at the time of

Eligibility to appear in NEET (UG) - 2019 

 Eligibility for appearing in NEET (UG) is as stipulated in Indian Medical Council Act-1956 and the Dentists Act-1948 as amended in 2018.   Eligibility as per related Regulations is as follows:- 

i. He/she has completed age of 17 years at the time of

 Magnitude of Sum of Two vectors Vector P and Vector q is given by

R^2 =p^2+q^2+2pqcos($)   __________________________(i)

Here $ indicates angle between them.

We are given 

R=p+q

So Squaring  both the sides we  get

R^2=p^2+q^2+2pq    ______________________(ii)

From (i) and (ii) we can write

P^2+q^2+2pqcos($)=p^2+q^2+2pq

Therefore 2pq=2pqcos($)

Cos($)=1

Therefore $=0

So