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Turgor pressure is a biology term. It is the pressure that the cell contents exert against the cell wall in plant cells. this is determined by the water content of the vacuole, resulting from the hydrostatic pressure produced by a solution in a space divided by a semipermeable membrane due to a differential in the concentration of solute.
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Dear swetalina,
I guess, the link given by Puneet should answer all your questions regarding the syllabus.
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IITJEE:
Sale at designated branches of Banks and at all IITs : Nov. 20 - Dec. 29, 2006
Postal request for Application Material : Nov. 20 - Dec.15, 2006
Last date for receipt of Completed Application Form at IITs: Dec. 29, 2006
Cost of Application Material : Rs.500/- for SC/ST & women candidates
Rs.1000/- for all other candidates
AIEEE: The All India Engineering Entrance Examination (AIEEE)-2007 will be conducted in the month of April, 2007. The sale of AIEEE application forms is expected to commence from 01 Dec till early Jan. Further details may be obtained from AIEEE or their website http://www.aieee.nic.in/
The last date of receipt of forms at AIEEE is to be ascertained from their advertisements, web site or regional offices located in the states
Kalinga:
Issue of Application Form & Prospectus: 12th Feb, 2007 to 21st March, 2007
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Dear Hari Krishna, According to the officials from IIT, there is a medical board constituted which decides if a person falls under the category of PD (Person with Disability). For any category of disability (viz., locomotor, visual, speech and hearing), benefit would be given to those candidates who have at least 40% permanent physical impairment in relation to a body part / system / extremity / extremities / whole body, etc. The candidates in this category will be required to be certified by a Medical Board. The Medical Board will decide the following: - Whether the candidate qualifies for the benefits under this category, and
- If the disability is likely to interfere in his/her studies.
The decision of the Medical Board shall be final.
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Dear ckmr, Thanks very much for the appreciation. We are currently collecting information about the relative ranking of BITSAT. We will post it as soon as we have it.
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Hi Polymath, Sorry for the delay. The ranking you hv given is approximate. Similarily the ranking provided by the AIEEE will also be approximate as it keeps on changing every year depending upon the marks attained by the students. Moreover the ranking on AIEEE differs for each state. so we would like to know which place you belong to so that u get the closest answer possible.
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but equating change of acceleration rate to t is the worry factor which raises doubts about the question...
since we are trying to prove that in units m/s3 = s
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consider the problem of finding the center of mass for the two particle system shown below. If we try to balance the system on a pivot at a point xp in between x1 and x2 , both particles exert torques that tend to tip the beam, rotating it around the pivot. In order to balance the system on the pivot, we want the torques caused by each of the particles to cancel each other. Since the force exerted by each particle is given by its mass times the acceleration of gravity, xp must be chosen so that m1 g (x1 - xp ) = m2 g (x2 - xp ) Rearranging this equation gives xp (m1 + m2 ) = (x1 m1 + x2 m2 ) This line of reasoning easily generalizes. If a rigid body is composed of n particles connected in a straight line, then the location of the center of mass for the system is xcm where (m1 + m2 + . . . + mn)xcm = m1 x1 + . . . + mn xn cheers
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Dear gayathri , Let me tell you about myself. I took my JEE in 1996 (loooong time back  ). But had not started sooner with friends during 11th, since I was too drawn by all the hard work already put in during 10th boards. i started in full flow after i came in class 12th. But during that time, i had already lost one crucial year or preparation, and my friends had an edge. So i would suggest, that you start early. Dont take it to hard on yourself, but since you have 2 years (as rightly pointed by himani  ), you will have enough time to crack it. JEE rpeparation, once you are in cruise mode, is an amazingly interesting thing.... so dont wait....just go for it...break on through......
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This question makes no sense.  we need more info omcar123. did you type it correctly?
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Point A has to move two dimensionally. One vertically and second horizontally. Consider that A takes a time ' T ' to catch up with B. Let B be moving horizontally . And the vertical distance between A and B is l. Therefore we have two equations in X and Y directions. At first in Y direction we have ( vertically) for A and B 0 T ( v - u cos  ) dt = l -------- ( 1) where '  ' is the angle made by the line joining A and B at any time ' t' with respect to the horizontal. Now horizontally for A and B 0 T v cos dt = uT -------- (2) Solving 1 and 2 we get the required result.
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If it is assumed that the distance between the back wall and the mirror is 2d, then the man is at a distant d from the mirror. If the top of the mirror is placed equal to the height of the man, then he can never see the top of the back wall. So the top of the mirror should be placed at height, which is H above the man?s head, where H is given by H /d = 2h / 3d
 H = 2h / 3; Then if the bottom of the mirror should be at a depth l below the man?s head such that l/d = 2h / 3d
l = 2h/3 ; Therefore the height of the mirror should be (2h + 2h )/ 3 = 4h / 3;similarly horizontally the width of the mirror should be in such a way that w/d = 6h / 3d
w = 2h; There fore the size of the mirror should be A = 4h/3 x 3h
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Q.1. An iron bar of weight W is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle alpha with the horizontal. What is the weight experienced by the man?
Ans. 1. Let the contact force on the shoulder be N. Then balancing the torques about the point on the ground where the rod touches we get N. N x L = mg x 0.5 L x cos  Then the weight experienced by the man will be the vertical force acting on the man, that is N cos  . And N sin will be the horizontal force pushing the man. If there is sufficient frictional force between the shoulder and the rod and between the rod and ground then the rod will not slip. Q.2. If the same iron bar were balanced by two men, what will be the be the weight of the bar as experienced by one of the men at the instant the other man lets go of the bar? Ans.2. Let N be the contact force on the shoulder. Then at that instant we have mg vertically downwards and N vertically upwards. Then mg - N = ma ....... (1). mg L / 2 = I a / ( 0.5 L) .......(2) Solving (1) and (2) we get N.
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dear rohit,
please provide the limits to help us solve the query.
thanks
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Dear nithin,
you will have to be more specific about your queries. optics is too wide a subject to discuss at length in forums. We will have study material on optics put up soon on the site. I'm sure that will help you!
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Dear nithin,
did u ever look at the old man with spectacles and thought how a small lens can change his whole vision..did you ever marvel at the whole mechanism through your own eyes work? did it enchant you to see the objects look closer under water? did a binoculor ever help you see a distant object more clearly? did u look at a rainbow and wondered how the nature could create such a beautiful spectacle?
all of it...ALL of it is optics.. Optics is divided into two fragments Geometric Optics and Wave optics. Geometric deals with reflection, refraction, dispersion, mirrors, lens. Wave optics gets down to particle levels and explains the behaviours of particles through various principles like superposition, huygen's principle and diffraction gratings
its a beautiful subject which can be so easily applied to daily life and understood with ease.
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Considering the figure given in the problem, if mass M moves x distance rightwards then mass m will move 2x downwards. So lets say mass M has acceleration a rightwards and mass m has acceleration 2a downwards. Mass m will have acceleration a rightwards also as all the time it is attached to mass M.
Considering the system the following forces will be acting on mass M :
Mg downwards -- due to gravity
N2 upwards -- normal reaction due to surface
 2N2 left -- due to friction
N1 left -- normal reaciton due to mass m
 1N1 downwards -- due to friction by mass m
T right -- Tension due to string attached with mass M
T right -- Tension due to pully attached with mass M
T downwards -- Tension due to pully attached with mass M
Writing force balance equations for mass M
2 T - N1 - 2N2 = M a
N2 = Mg + 1N1 + T
Now, Forces acting on mass m
T upwards -- Tension due to string attached with mass m
N1 right -- normal reaction due to mass M
1N1 upwards -- friction force
mg downwards -- due to gravity
( 2a acceleration on right )
Writing force balance equations for mass m
N1 = ma
T + 1N1 - mg = m. 2a
By solving these 4 equation you can easily get the answer.
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Velocity along AB is v and  OAB is 30o. Hence velocity along AO is v Cos OAB, i.e. v cos 30.
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just to reexplain the distance AO: suppose the triangle is ABC where A intersects BC at D.
so AB2 = AD2 + BD2
AD2 = AB2 - BD2 =
AD2 = a2 - (a/2)2 = 3a2/4
AD =(a  3)/2
AO = (2/3)AD
= (2/3)*(a 3)/2 =a/ 3
2a / 3v is the correct answer...
good job sonali!
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wow, thats a lot of questions.... let me see how many can i answer in one go..
I need more information about how to get into the iits sir if possible please explain it little more briefly.
you are eligible for JEE if you have done 3 or 4-year Diploma recognized by AICTE or a State Board of Technical Education. Once that is done, you get 2 attempts to clear JEE. If you fail once, you can try the next year but not beyond that.
I want to know how many screening tests are there and how many main exams and also thier dates. Also the total marks of both these exams. Then, how much students do the take through the first screening test.
The JEE pattern changes every year. For this year, there will be two question papers of three hours duration, each consisting of separate sections in Physics, Chemistry and Mathematics. Questions in these papers will be of objective type
Is the pattern of questions in both the screening test and main exam the same?
This year there will be no screening exam. To know the pattern, you can register for the online test goIIT will be conducting on 9th - 10th Dec, on the site, in association with Hindustan Times.
Do we have to write long long answers or just put tick marks. Do they give choices in the answers.
Questions will be objective type this time.
My final exam of diploma will be on april 2008 so if iit-jee comes before that exam what should i do can i attend it?
Yes, ideally you should be able to attend it, but do tell us of your specific case: diploma corse and dates for exact information.
Can i get information regarding online coaching of iit jee. Is correspondence course better than this? Can i get the names of the institutions from which the students got higher ranks in iit jee? I
We cant name specific institutes - which is better and which is not, since its a very subjective issue. You should consult more students over forums to ask about their experiences with the institutes. There are no good online solutions. You can got for a correspondence course, and ping the experts here for any issue you have. However a classroom coaching is highly recommended for JEE if you are starting afresh.
Lateral placements is a separate thread of discussion. That we should cater in a aseparate query.
cheers
hope this helps.
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