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We generally neglect air resistance when we solve problems by ssuing ideal conditions.
We are ideal students..... so we assume ideal conditions :-).

Rolling friction is basically caused because of the deformation that occurs at the surface of contact. if the both the rolling body and the surface are perfectly rigid... then there will be no deformation... Hence there will be no rolling at all...
You may find the following link to be useful:
http://webphysics.davidson.edu/faculty/dmb/PY430/Friction/rolling.html

The freely falling explosive experiences zero acceleration in horizontal direction. So, there will be no change in the speed of the explosive in horizontal direction. So, the explosive continues to move with the same sped of plane. So, the explosive always lies right below the plane...

No that u have just few days in hand before ur exam, I suggest u the following things:
- What ever u already know, make sure that u r strong in that by practicing a lot in those topics. Any question from those topics should not go unattempted by u.
- Regarding the topics which u haven't covered, just revise the concepts. Take help of ur teacher or friends if u find difficult to manage time.
- Blank mind is a general problem for everyone, when they are not confident. Now, what u need to do is to believe that you are good in all the topics u know. Forget about the topics u dont know, as u are any way not going to do well in those topics. But as far as the known topics are concerned, be extremely careful and be well prepared.
- The most important and essential thing is practice. Keep practicing as much as u can. Attempt as many mock exams as u can. This reduces anxiety.

All the best for ur exam!!!

Well done Akhil !!!
The whole crunch of the problem is to get the common series...

There is no specific condition for using universal substitution. If the problem becomes very tough to solve using pure trigonometric concepts, and if u feel that converting the trigonometric equation into an arithmetic one helps, then u use the methods of universal substitution.

Let ax2.by2+............+c=0, represent the equation of a general conic section.

Let mx+ny+p=0 be the equation of a chord. Now, as pointed out by Judasrising, the homogeneousing technique results in a pair of straight lines, with origin as their point of intersection and this pair passes through the point of intersection of chord and the conic section.

Lets now look at the justification part:

- The resulting equation is homogeneous. So, (0,0) is a point on the resulting homogeneous equation.

- Let (x1,y1) and (x2,y2) be the points of intersection of the chord and the conic section. It can be easily seen that these two points also satisfy the homogeneous equation.

So, now we conclude that the resulted homogeneous equation is the equation of a curve, passing through origin and 2 points of intersection of the chord and the conic section. The only which we need to prove is that the equation represents a pair of straight lines.

Try getting a condition for the homogeneous equation to be pair of straight lines. the condition we get will be same as the condition for the chord to intersect the conic section. So, the homogeneous equation represents a pair of straight lines, when ever the given straight line intersects with the conic section.

Hence justified.

Algorithm for second problem:

the problem can be made simple by interpreting it in a different perspective.
Consider a, b, c as there different boxes. U have 4 red ball (red is the divisor 2 in the product), 5 green balls (green is 3) and 3 blue balls(blue is 5).

Now the problem is to place these 12 balls in 3 boxes, so that each box is filled with atleast one ball. Once again principle of inclusion and exclusion... So, the problem is same as the previous one..

Algorithm for the first question:

The students for whom there must be atleast 1 gift given are specified. So u dont need to select these 4 students. So, 6C4 doesnt come into picture.

The algorith to solve the problem is as follows:
- First distribute 9 gifts among 6 students without any restrictions
- then subtract the cases where atleast one of the 4 specified students dont get a gift. For this case, u need to use principle of inclusion and exclusion

Cheers

Is the man inside the sphere or out side??
Is the sphere at rest or in motion??
Is the man in the vertical line passing through the centre mass of sphere?

Why dont you include a figure to make the problem clear??

Iberis is correct...
Once u r confident if P&C, getting hold on Probablity is not a big deal. Probably u r feeling inconvenient with probability just because your approach could be wrong one. U dont need to go for any advanced book to get acquainted with probability.Just go ahead with ur 12th standard text book. Do it sincerely without overlloking even a word. U will be done with probability. There is no technique as such to get hold on probabilty. And remember, it is practice which makes u confident. So, solve as many problems as you can....

Cheers

Normal forced never perform work on rigid bodies. Here the sphere is rigid. So, even after the collision, KE is conserved, provided no non-conservative force acts on the body..

The basic definition of Elastic collision is that the energy is conserved. If u review the definition of Coefficient of restitution, u will find that it has nothing to do with energy.

But mathematically, it happens that coefficient of restitution is 1 for perfectly elastic collisions.
So, both the definitions are true for perfectly elastic collisions.
Now coming to the problem, I would like to stress on the exact definition of Coefficient of restitution.

e = (rel. vel. of seperation)/(rel. vel. of approach)
In this definition we talk about two relative velocities. Note that both the relative velocities MUST be taken in the direction of common normal. So, inthis particular proble, when u use the definition of Coefficient of restitution, u need to take the relative velocities along the common normal. ie. normal to the sphere at the point of contact. If u go by this method, u will get the same answer as u get by conservation of energy method.

The min no of tosses required for B's cash to go to zero is 20.
So, he will go bankrupt on 21st toss (if he looses on 21st toss)