Firstly, I am assuming that the two bodies are in vaccum. In that case, the other body doesn't remain stationary. It will also move. The calculation based on a circular motion of the sphere to which velocity is given is incorrect. Let us solve the problem in the reference frame of the sphere which isn't given a velocity. It also accelerates. So its an non inertial reference frame. Then we must apply the modified form of Newtons second law

where is the reduced mass of the system, and as the masses are equal, its equal to m/2

Rearrange and integrate this equation and integrate for v from V to 0, and for x from 6R to 4R

The third round of counselling is the only hope, but top branches are almost impossible

The direction that you show for angular velocity is it correct?

For that direction (clockwise), the right sphere will never collide with the left one as the friction at its point of contact is rightwards.

However if the friction is leftwards, then friction will be rightwards and the right sphere can approach the left one. In that case we will need the distance between the spheres to find with what speed will the left sphere collide with the right one. If the collision be assumed elestic, each sphere gains equal speeds (equal to haf the impacting speed of the left sphere) in opposite directions, but the angular velocity of the left sphere remains at the value it was before collision.

Now, its a simple problem of the spheres being given a vertain configuration of linear and angular velocities and finding their speeds at pure rollning is a standard task.

Please check whether or not the distance between them is given

Manipal> Anna > the other two

Manipal is a bit costly, but on the whole, its reputation amongst private colleges is staggering

Repled at http://www.goiit.com/posts/list/0/geometry-how-we-find-the-orthocenter-if-three-vertex-are-given-73792.htm#363828

Replied at http://www.goiit.com/posts/list/0/geometry-how-we-find-the-orthocenter-if-three-vertex-are-given-73792.htm#363828

Take any two vertices, write the equation of the line joining them. Next wite the equation of a line passing through the third vertex and perpendicular to the previous line.  This gives you the perpendicular from the third vertex

Rapeat the steps for another pair of vertices to get the perpendicular from another third vertex. You get the equation for another perpendicular from a different vertex.

Solve the equations of the perpendiculars to get the coordinates of the orthocenter

The question level is simple. Just revise your JEE notes and NCERT books. also see some previous year papers

In Ranchi, as a coaching institute, FIITJEE is the best.

Howvere, in Ranchi, there are a lot of excellent private tutors who teach better than the FITJEE teachers.

Physics: Navendu Shekhar Mishra

Mathematics: S Kar

Chemistry: Hari Om Pandey

Dear Suresh,

Life's tough ... hmm ... well no new revelation. Specially for you. Well, if you want to study in some college, then you'll have to convince your parents that they must allow you to go to coachings. If they don't agree then ask them to hire private tutors for you.

I know that its very hard for you, but then buddy, you gotta try. Understand what you need, try to make your parents see that. If they don't see it, rethink your options.

Draw the FBD. Let the length of the thread be L and the separation between the masses be x. Also assume the angle made by the string to the vertical to be . Then

equations of motion

T-mgcos-(kq²/x²)sin=ma_n

-mgsin+(kq²/x²)cos = ma_t

where a_n and a_t are the normal and tangential acceleration components. Now at the moment of release v=0, so a_n = (v²/R) = 0

Also note that I have taken positive tangential direction away from the other charge.

So net force alon the string, ma_n = 0

Net force perpendicular to the string, ma_t = -mgsin+(kq²/x²)cos

Substitute the values, the trigonometric ratios can be found from the diagram

You can find the tension from the first equaion.

At the initial moment a = a_t = -gsin+(kq²/mx²)cos

This is a very standard type of problem solved by conservation of linear and angular momentum. refre to HC Verma solved examples in Rotational chapter.

Let a' be the acceleration of the sphere's cm wrt the wedge. It will be directed down the plane. Let a be the acceleration of the wedge. Let be the angular acceleration  of  the sphere. Now as the sphere is observed in the reference frame of the wedge, there is a pseudo force ma on the sphere directed opposite to the acceleration of the wedge.

If  N be the normal force between the sphere and the wedge,

N=mg cos - ma sin and Nsin = ma

This gives a = gsincos/(1+sin²)

Now, for no slipping a' = R

a cos + g sin - (f/m) = (5f/2m)

Substitute the value of a and find f

Given the pattern of JEE nowadays, its of utmost importance that you complete the syllabus if you want to get through. In an objective paper, you have to score the maximum and that you can't do if you don't know the entire syllabus. So your first aim should be covering the entire syllabus, that is being reasonably proficient in all the things. After that you can start working on areas you find easy and work harder for those you find hard. But syllabus coverage is the most important priority

About how to study ... well, just study ... take track of all the time you waste, and convert it to study time.

No the cutoff doesn't go down.