Strategy for the last three months....!!

Are you wondering as the peak time January has approached and is about to end, yet there is a lot left to prepare all the subjects for IIT-JEE…!!

Well, to resolve and guide you better in such a situation, we bring you a few things that can help you too….

As the examination date is approaching near, the number of hours of study has even been increased to 16-17 hours a day by many of the students. But, to stay competent, it is not only enough to study a lot and work too hard. Rather, a smart work is advised at this time.

The first step to a smart preparation process is to make your fundamentals strong. You might be having a strong conceptual foundation by now to solve good questions and may even have the skill to solve many calculations. Due to the unpredictable pattern and  questions of the examination, it is better that all types of questions like objective, subjective, assertions-reason, multiple matching etc are practiced well. In mathematics, the questions might involve more than one concept. So, one has to be comfortable with the combination of several concepts in a single question. For example, Permutations and combinations and probability always can be mingled together. With these two topics, many questions can be made.

Cover the topics which helps you to score well  : The topics which score well can help you the best when conquered them first. The concepts of vector and scalar triple products are very important and scoring in Vector Algebra. In Conic sections, the formulae are very important. The complete knowledge about the fundamental properties regarding the nature of roots in Quadratic Equations is quite helpful. One of the most scoring topics of JEE mathematics is Progressions and Series. Differential and Integral Calculus’ are also very important. ‘Geometric interpretation of the derivative’ is one of the most crucial topics and hence should be focussed upon really well. Only the important topics have been listed here, but one should not limit his/her preparation to these topics only. It is better that the entire syllabus is revised before the exam to boost you up.

In physics, as per the general trend, Mechanics and Electricity and Magnetism are the topics carrying highest marks while both are equally important…!! Other topics which should be focused are Thermodynamics, Waves, Modern Physics, Optics, Fluid Mechanics etc are also important. Even though the topic Measurement and Errors is often neglected but quite a few questions are asked from this topic as well.

Start your preparation in a smart way : Solving previous years question papers helps you in a very good way. Practice and solving various mock tests makes you prepare to face the main examination hours. Adjust your sleeping hours to be more efficient during the examination hours. It’s the time now to change your old habits.

Revise and go through all the important points which you might have noted long before. Test your concepts and learn any special tricks if involved. Last but not the least, be confident, calm and relaxed to give your best.

HYPER CONJUGATION

·   When  sigma electrons are in conjugation with pi- bond then this conjugation is known to be ‘Hyperconjugation’.

·   Generally used to compare stability.

Required Conditions:

1.Compound should have at least one sp2 hybridized carbon atom of alkene or alkyl carbocation or alkyl free radical.

2. Alpha carbon with respect to sp2 hybrid carbon must be sp3 and it should have at least one alpha-Hydrogen.

If both the conditions are fulfilled then only  “Hyperconjugation” occurs otherwise not.

Ex 1:Lets consider an example of 

                  

                

          ·    Alpha-carbon is the carbon attached to the sp2 - hybrid carbon.

          ·     Alpha-Hydrogen is the Hydrogen attached to the alpha-carbon atom.

         Hence,

          This compound has three alpha-hydrogens

       Type 1: Pi-Conjugation

         1.

    

            Two alpha hydrogens in the above compound.

         2.

   

             Five alpha-hydrogens in the above compound.

         Type 2 : Positive Charge conjugation

              

         1.

                                                 

                 Three alpha-hydrogens in the above compound.

         2.   

                 Six alpha-hydrogens in the above compound.

           Type 3: Odd electron conjugation

           

           1.

                    Three alpha-hydrogens in the above compound.

          2.

 

                  Five alpha hydrogens in the above compound.

            Applications:
         
            Hyperconjugation is generally used to compare stability of alkene.
             No. of Hyperconjugative structures = 1 + No. of alpha hydrogens.

             No. of  alpha hydrogens is proportional to the No. of Hyperconjugative
             structures.

                            

                 2. Stability of carbocations

                         

               3.  Stability of free radicals

      

 Reverse Hyperconjugation:-

    
     Only possible if a halogen is in conjugation with pi- bonds.

  

    Three alpha chlorines are in conjugation with the pi bond.whenever 3 halogens are in conjugation with pi bond and hence it is known as reverse conjugation.

Are you a carrot, an egg or a coffee bean...??

A young woman went to her mother and told her about her life and how things were so hard for her. She did not know how she was going to make it and wanted to give up. She was tired of fighting and struggling.

It seemed that, as one problem was solved, a new one arose. Her mother took her to the kitchen. She filled three pots with water and placed each on a high fire. Soon the pots came to a boil. In the first, she placed carrots, in the second she placed eggs, and in the last she placed ground coffee beans.

She let them sit and boil, without saying a word. In about twenty minutes, she turned off the burners. She fished the carrots out and placed them in a bowl. She pulled the eggs out and placed them in a bowl. Then she ladled the coffee out and placed it in a bowl. Turning to her daughter, she asked, "Tell me, what do you see?"

"Carrots, eggs, and coffee," the young woman replied. The mother brought her closer and asked her to feel the carrots. She did and noted that they were soft. She then asked her to take an egg and break it. After pulling off the shell, she observed the hard-boiled egg. Finally, she asked her to sip the coffee. The daughter smiled as she tasted its rich aroma. The daughter then asked, "What does it mean, mother?"

Her mother explained that each of these objects had faced the same adversity - boiling water - but each reacted differently. The carrot went in strong, hard and unrelenting. However, after being subjected to the boiling water, it softened and became weak.

The egg had been fragile. Its thin outer shell had protected its liquid interior. But, after sitting through the boiling water, its inside became hardened! The ground coffee beans were unique, however. After they were in the boiling water, they had changed the water.

"Which are you?" the mother asked her daughter. "When adversity knocks on your door, how do you respond? Are you a carrot, an egg, or a coffee bean?" Think of this: Which am I? Am I the carrot that seems strong but, with pain and adversity, do I wilt and become soft and lose my strength? Am I the egg that starts with a malleable heart, but changes with the heat?  Does my shell look the same, but on the inside am I bitter and tough with a stiff spirit and a hardened heart? Or am I like the coffee bean which actually changes the hot water, the very circumstance that brings the pain? When the water gets hot, it releases the fragrance and flavour.

If you are like the bean, when things are at their worst, you get better and change the situation around you. When the hours are the darkest and trials are their greatest, do you elevate to another level? How do you handle adversity? Are you a carrot, an egg, or a coffee bean?

No matter what you feel like during your preparation process, at the end of it awaits a sweet result of your hardwork and talent.

Puzzles to challenge you..

Hello guys...!!
Looking for some interesting and challenging brain teasers..??
Here are few questions which  you can give a try..

1. Three friends met after a longtime at a restaurant. Their total bill was 15 rupees while each of them contribute 5 rupees. The waiter takes the money to the owner who is a friend of these three customers and the owner asks the waiter to return 5 rupees to them.
The waiter is not only poor at mathematics but also dishonest and instead of returning back 5 rupees, he took 2 rupees for himself and returned back one rupee to each of the three. Now, each of the three effectively paid 12 rupees and the waiter has 2 rupees with him and this total comes 14 rupees. Where has the other one rupee go from the original 15 rupees.?

2.   X*(4+(Y/100))=108.81

      find the value of (X+Y) by solving the above equation

3.  There are 100 soldiers in a circle in an order 1 to 100. The first soldier is given a sword. He kills next person (i.e.number 2)  and gives the sword to next to next (i.e. number 3). It is continued till only one person is left. Which number is survived at the last?

The answers to the above questions will be posted soon after 30 valid answers in comments.

Distribution of current and voltage in parallel and series

         Here is a brief explanation to know about the distribution of current and voltage in                    series and parallel connections.

 Distribution of current in parallel connection:

         When all the devices are connected using parallel connections, the circuit is referred to as a parallel circuit. In a parallel circuit, each device is placed in its own separate branch. The     presence of branch lines means that there are multiple pathways by which charge can traverse the external circuit. 

       When resistors are connected in parallel, the potential difference across them is equal and current gets divided in inverse ratio of their resistance. Because 

 where, i is the current, V is the voltage and R is the resistance.

And for parallel circuit V remain same so

For example

       so 

 Distribution of potential in series circuit : When resistors are connected in series, the  current through is same and potential is distributed in the ratio of their resistances  because 

                  And for same i, 

   

                  For example, 

                 so 

            

  

     Finding out equivalent emf of more than one dissimilar cells connected in parallel:

     

     Suppose n number of cells having emf’s  and their internal resistances are   are connected in parallel as shown. The complete network can be replaced by    a single source of emf    and internal resistance .

            

                 and

      

                  or

     For two cells in parallel.

Please note that positive terminals of all the cells are at one point and negative terminals at other common point. If some of the cells are oppositely connected then,

                      

                       

CONSTRAINT RELATIONS

In classical mechanics, a constraint is a relation between coordinates and momenta (and possibly higher derivatives of the coordinates). In other words, a constraint is a restriction on the freedom of movement of a system of particles.

The only difference between a constraint equation and eg a conservation equation is that a conservation equation is physics, but a constraint equation is geometry.

With example, a system of three pulleys at heights p q and r, we get physics equations (usually F = ma) for each pulley, but the "a" in F = ma is different for each pulley (in fact, it's p'' q'' and r'' respectively)
so we need a geometric equation relating p q and r …
usually this simply tells us the length of the string in terms of p q and r …
since we know that that length is constant, we can differentiate once (or twice) to get a neat "constraint equation"

The following are different examples of Constraint Relations. 

Two masses tied to a string going over a friction less pulley m1 > m2 as shown in Fig. 3.1.

If 'a' is the common acceleration of the masses and 'T' the tension in the string, then, for mass m1

                  

Similarly for mass m2, the net force which gives 

          

From Eqs. (i) and (ii) we get,

                                                        

                                               and 

Two masses in contact :

 Figure 3.2 shows two masses m1 and m2 in contact placed on a horizontal frictionless surface. A force F is applied as shown and as a result the masses move with acceleration, which is given by

                              

The force on m1 is F and the contact force on m2 is 

                                                             

                                

Three masses in contact:

 If three masses m1, m2 and m3 are placed in contact and force F is applied to mass m1,  the three masses move with acceleration (see Fig. 3.3)

            

                         

The force acting on mass m1 is F. The contact forces acting of masses m2 and m3 respectively are

                                

                          

                        

                                                               and