Messages posted by earpit

it includes btech int.mtech int.msc b.archi ...

FYI, i am

int.M.Tech GPT IIT Roorkee.

This is the diffrence between dual degree and integrated degree course.

Msc at IIT roorkee is probable. You also might get Integrated courses if you get lucky... (talking of roorkee) . Mining at kharagpur is also possible, but with a very thin chance.

depends on your category. If you are GEN, then there is no chance.

B. Tech is 4 year course.

M. Tech is done after btech, so to get btech + mtech it takes 4+2=6 years.

Integrated M.Tech is a 5 year course, which covers the Btech portion and then the specialization, hence you save 1 year.

you do not get a Btech degree, you get just one integrated degree.. value of which is same as btech + mtech.

studying in integrated program is not different, in first year, all the int.Mtech guys study exactly same as what btech guys study..its just that in the later years, integrated Mtech guys go for specialization in their respective branches..

I hope you have knowlege of limits... Ask this to your self: "What is the feild at the exact coordinate of charge ?" HUGE? call it infinite? ...Right, similar is the answer if you look for feild at the surface it self, but as you move infinitsmly away from it,.. shell theorem tells that the feild exists as if the entire chare were present at the center.

who told you this ?? :P ... Later in your course you will learn about binding energy, then you will know what energy is needed to 'pull the proton out'

mv square.

do not opt for bansal at all.... all the head teaches of bansal left this year so i think triVaAg lucknow is the best im frm there only

Well, this is nothing new, it happens every year...Teachers leave and more "expensive" teachers are hired .. :DDont you worry about that,..BTW, Myself from bansal jaipur,and dont drop until you have absolutely no good option left (good~level of NIT's)

one by eight, not six.

exatly what the person above me said. You are right, that Flux is Feild Strength multiplied by Area. Former depends inversly with the square of distance, and the later directly with the square of distance, hence cancelling, and it is a contant. :)

Does not depend on the direction.

See, try to understand like this, as long as you throw with velocity more then V.esc , it will posses enough energy to go out of earth's gravitational feild, to hell with direction, but dont throw it directly at earth , in that case it will surely lose its energy, and wont escape :)

why not? all perpendicular vertors to that line.

@ karna: earth is round, because earth is not square, hence earth is round. lol..

sorry guys, 400K ki image daal dee meine..i was a bit lazy to do the resizing, but anyway, i dont think ive committed any error, if you think, something's fishy, lets discuss.

http://img148.imageshack.us/img148/8613/sol1.jpg

1 sec, i'll post my solution...

look closely, and you find that friction on the bead, and the tension in the string, both have to be equal.

liked it. youve made a valid point, and in a beautiful way.cheers!

an answer.

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