Class 11
Hello,
PEstored = Work done on the wire during stretching= integral of (Fdx) [F=force, dx = small displacement]
But by hooke's lawY = FL/Ax [F=force, L=length of wire, A= area, x = extension]So, F= (AY/L)x
Thus PEstored = Integral of (AY/L)*(xdx)= (AYl2)/2L [l = maximum
Hello Hitesh,
To get admission in B.J Medical College , Ahmedabad through NEET exam the candidates with All India General Quota need to score 560+ this would be a decent score to get in this college.
So, score above 560 would safe marks for getting admission in this college.
Hope
Hi nasreen,
Physics 60 chemistry 60 mathematics 60
Physics syllabus
*General dimension , modern physics , thermal physics , mechanics , optics , oscillation , hooked law and force , centrifugal and centripetal forces ,electricity and magnetism , thermal and chemical effect of current .
Chemistry syllabus
* Alcohol ,
Hi Shaheen,
Apart from NCERT for Aiims
We suggest you that
GRB publication , modern ABC of chemistry for class 11 ( vol 1 and 2).
Objective chemistry by Dinesh.
Apart from this I have some lecture note .
https://drive.google.com/drive/folders/0B6tUgUuYohuYTG9wY2k0ck5yM00
This contains the lecture videos ..
If you need lecture
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