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Yeah, Biotechnology is a very popular program.
Right now, there is not even a single NRI doing full 4 year program at IITM.
Foreign students have a special facility: Direct Admission of Students from Abroad.
It is mandatory for students to stay in hostel. Any way there will be no one to check if u really stay in your room.
The job prospects are also good. Its very easy to get a decent job in India.

For more details, visit:
http://himanshu.wordpress.com/2006/07/03/dqindia-indias-top-t-schools-complete-survey/

Cheers

Use formula for "refraction at spherical surfaces".

First consider refraction at the outer spherical surface. If the image forms before it reaches the cavity, then it is the answer. If the image crosses teh cavity, then consider this image as object and find the location of object. If the position of image is within the cavity, then this will be the answer. If the image is beyond the other extreme of the cavity, then consider another refraction at the other spherical surface of the cavity. If this image is formed before it crosses the outer surface of sphere, then that will be the answer. If not, then consider the reflection also at the other outer surface of the sphere. If u find an image out side the sphere, then that is the answer. If not, then no real image is formed.

I hope I've confused u well. :-)

I strongly believe that there is a mistake in the problem. It should be "2 angles of transmission", instead of incidence.

Now, going ahead with this point, given a ray going towards on a surface interface, the ray gets partially reflected and partially refracted (2 angles of transmission). But, if thh angle of incidence is greater than a critical value, then only reflection takes place (total internal reflection, hence, 1 angle of transmission).

The solution problem turn extremley tedious had the length of the string been constant... In this case, as the disc rotates, the point to which teh string is tied comes down, so the string is no more horizontal... then the tension is not horizontal.... going on and on... the problem turns in to a huge demon... i tried to solve this way, when i was preparing for JEE, but never solved it successfully.

In any problem like this, unless other wise specifically stated, u can assume that the sring is wound around the disk.

"Strings tied to fixed point" problems can be solved only in linear motion...

Yeah..
Here the string is not tied to the top most point of the disc... it is wound around the disc. So, as the disc rotates, the string gets unwound and the length of the string increases... here length of the string is not constant..

As I stated an earlier reply, friction acts even during rolling, as no body is perfectly rigid.

The disk has two horizontal forces acting on it.
1) Tension towards right
2) Friction towards right.

In this case, both turn too balance. (try checking ur self.) So, net force acting on the disc is zero... which means, the disc velocity remains constant...
Here initial velocity is zero... hence the disc's centre of mass continues to stay at rest.

Weight of a body in free fall is "mg" itself.

The apparent weight will be zero.

Ideally speaking, rolling friction will be zero on any perfectly rigid body.
But no body is perfectly rigid. There will be deformation at the point of contact with ground. So, the lines of action of normal force and weight are different. this results in rolling friction.

For detailed discussion, refer to:
http://webphysics.davidson.edu/faculty/dmb/PY430/Friction/rolling.html

The answer is

=v/r.
But the argument is as follows:
Let linear velocity of the centre of the disc be 'v1' towards left. Let the angular velocity of the disc be

.
The disc is moving to right with speed v. so the string moves with speed v to left.So, the velocity of the top most point of the disc should be v --------------1
velocity of the top most point wrt centre is

r towards left. the centre is moving to left with v1.
So the speed of top most point is v1+

r ---------------------2
by 1 and 2, we get,
v1 +

r = v -------------------3
Similarly we can say that the speed of bottom most point of the disc is

r-v1 to right ---------------------4
For the disc to roll on the block, the bottom most point should have the same speed as th eblock ---------------5
so, by 4 and 5, we haev

r-v1 = v ---------------------6

By 3 and 6, we get,

=v/r.

Method 1:

For every differential mass "dm" there is another identical "dm" exactly radially opposite. The centre of mass of these two exist at the centre of sphere. Arguing this way, the center of mass of sphere is at its centre.

method 2:

We can use Spherical polar coordinates to sole this.
Refer to
http://surendranath.tripod.com/Applets/Math/Coordinates/SphCoordApplet.html
for a visual description of spherical polar coordinates. Its just a matter of a triple integral to end up with the centre of mass to be at centre of sphere. Buzz if u find it difficult to integrate.

As stated by kmmankad, the man has a tendency to topple.
So, lets neglect the height of the man (a lilliput). So, no toppling!!

Now the question asks if the man can pull the block. The max possible friction is
f= (mu)*(M+m)*g, where mu is coeff of friction, m and m are masses of block and man respectively. So, if the man is sufficiently strong and can apply this much force on the srting (the same force will be transmitted to block in the form of tension), then he can definitely pull the block!!

kmmankad is wrong. The expression he stated is correct only when the block is constrained to move vertically. but here that is not the case in this problem.

waterdemon is perfectly right but the value of 'a' we got finally is not the absolute acceleration of block. It is the acceleration of block with reference to wedge. So, all this rel acceleration vector to the wedge's acceleration vector to get block's acceleration vector. It is essential to add the accelerations vectorially, not algebraically.