goIIT: IIT JEE AIEEE CET WBJEE VITJEE BITSAT IIIT AICET Engineering Entrance Experts, Student Discussion Community, IIT JEE Preparation Online

i voted 6 times

1) Inertial mass. This is mainly defined by Newton's law, the all-too-famous F = ma, which states that when a force F is applied to an object, it will accelerate proportionally, and that constant of proportion is the mass of that object. In very concrete terms, to determine the inertial mass, you apply a force of F Newtons to an object, measure the acceleration in m/s2, and F/a will give you the inertial mass m in kilograms.

2) Gravitational mass. This is defined by the force of gravitation, which states that there is a gravitational force between any pair of objects, which is given by

F = G m1 m2/r2

where G is the universal gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them. This, in effect defines the gravitational mass of an object.

As it turns out, these two masses are equal to each other as far as we can measure. Also, the equivalence of these two masses is why all objects fall at the same rate on earth.

thank u

thanx

xcellent work akku

the name of the boy is SONU

isnt it

excellent work dude

hey dude!

the situation is as given in figure.

speed of car A in m/s = 42/3600 x 1000 = 35/3 m/s
speed of car B in m/s = 60/3600 x 1000 = 50/3 m/s

let time taken to overtake is t seconds.

now dist covered by car A in t seconds = 35t/3 m
and dist covered by car B in t seconds = 50t/3 m

as it is clear from figure that in overtaking car B has moved 10m extra than car A.

so 35t/3 + 10 = 50t/3

15t = 30
t = 2 seconds.

road dist travelled by car A during overtaking = 70/3 m
and road dist travelled by car A during overtaking = 100/3 m

pls rate if crrect.........

Arithmetic

Progressions

An arithmetic progression is a sequence in which each term (except the first term) is obtained from the previous term by adding a constant known as the common difference. An arithmetic series is formed by the addition of the terms in an arithmetic progression.

Let the first term on an A. P. be a and common difference d. Then,

General form of an A. P.:

a, a + d, a + 2d, a + 3d, ...

nth term of an A. P.:

a + (n - 1) d

Sum of first n terms of an A. P.:

n/2 [2a + (n - 1) d] or

n/2 [ first term + last term]

This idea was from the mathematician Carl Friedrich Gauss, who, as a young boy, stunned his teacher by adding up 1 + 2 + 3 + ... + 99 + 100 within a few minutes. Here's how he did it:

He counted 101 terms in the series, of which 50 is the middle term. He also realised that adding the first and last numbers, 1 and 100, gives, 101; and adding the second and second last numbers, 2 and 99, gives 101, as well as 3 + 98 = 101 and so on. Thus he concluded that there are 50 sets of 101 and the middle term is 50. So the sum of the series is:

50 (1 + 100) + 50 = 5050.

This can be rewritten as:

100/2 (1 + 100) + 50 = 5050 or

101/2 (1 + 100) = 5050

Arithmetic mean. Given x, y and z are consecutive terms of an A. P., then

y - x = z - y

2y = x + z

y is known as the arithmetic mean.

Properties of A. P. (summary of the above points mentioned)

Given a sequence u₁, u₂, u₃, ... u_n-1, u_n, u_n+1, ...

1. u_nis in the form a + (n - 1)d.

2. u_n - u_n-1 is a constant (common difference).

3. u_n+1- u_n = u_n - u_n-1

Geometric Progressions

A geometric progression is a sequence in which each term (except the first term) is derived from the preceding term by the multiplication of a non-zero constant, which is the common ratio. A geometric series is formed by the addition of the terms in a geometric progression.

Examples:

1) 3, 6, 9, 12, ... first term 3, common ratio 3

2) 4, -8, 16, -32, ... first term 4, common ratio -2

Let the first term be a and common ratio be r.

General form of a G. P.:

a, ar, ar², ar³, ...

nth term of a G. P.=

ar^n-1

Sum to first n terms of a G. P.:

Geometric mean. When x, y and z are consecutive numbers in a G. P.,

y² = xz

y is the geometric mean

Properties of a G. P. (summary of the abovementioned points)

1. nth term is in the form ar^n-1, where a and r are constant

2.

is constant for all n (common ratio).

3.

Sum to Infinity

The sum to infinity is a finite value the sum of the first n terms of a geometric series tends to when n tends to infinite. Sum to infinity only exists when a series is convergent.

Sum to infinity is given by the expression:

and only exists if :

Why?

The sum of first n terms is given by:

As

. (only if

)

Hence

.

Thus the sum to infinity is given by

If

,

, and the sum to infinity will not exist.

Harmonic Progressions

A Harmonic Progression (HP) is is a series of terms where the reciprocals of the terms are in Arithmetic Progression (AP).

The general form of an HP is
1/a, 1/(a+d), 1/(a+2d), 1/(a+3d), .....

The nth term of a Harmonic Progression is given by
tn=1/(nth term of the corresponding AP)

In the following Harmonic Progression

The Harmonic Mean (HM) of two numbers a and b is

The Harmonic Mean of n non-zero numbers
is

Relation between AM, GM

& HM

that is, AM, GM, HM are in Geometric Progression.

For two positive numbers,
AM ? GM ? HM equality holding for equal numbers.

For n non-zero positive numbers, AM ? GM ? HM

Permutations        and combinations

When        we talk of permutations and combinations in everyday talk we often use the        two terms interchangeably. In mathematics, however, the two each have very        specific meanings, and this distinction often causes problems. 

In        brief, the permutation of a number of objects is the number of different        ways they can be ordered; i.e. which is first, second, third, etc. If you        wish to choose some objects from a larger number of objects, the way you        position the chosen objects is also important. With combinations, on the        other hand, one does not consider the order in which objects were chosen        or placed, just which objects were chosen. We could summarise permutations        and combinations (very simplistically) as 

Permutations        - position important (although choice may also be important)        
       Combinations - chosen        important, 
       which may help you to remember which is which.

Pictures on a wall

Suppose        you have to put some pictures on the wall, and suppose you only have two        pictures: A and B.

You        could hang them

Order        1                      or

Order        2                    

Not        much of a choice, but it leads on to the difference between permutations        and combinations.

THE        IMPORTANT DIFFERENCE

As        mentioned above, there is an important difference between permutations and        combinations. In this case, for permutations the order of events is important:        order 1 is different from order 2. For combinations, however, it does not        matter which picture was hung first. In this example there are two permutations        (A, B ? B, A), but only one combination (A, B = B, A).

Another        way that you may find useful to help you remember is to consider a combination        lock. On combination locks you have to turn dials with numbers on so a particular        number is given, e.g. '1, 2, 3, 4'. But they do not unlock when if the order        is changed (e.g. 2, 1, 3, 4). In this case the order is important.  So combination        locks should not be called combination locks but 'permutation' locks.

  
       A        permutation lock! 

Now        we know the difference between combinations and permutations let us consider        a more complicated picture-hanging problem. First we will look at permutations.

Two        company, three's a crowd!

This        time you have three pictures, called, not surprisingly, A, B and C. How        many different permutations are there for you to hang your works of art?        A worrying problem indeed! Let's hang them up. 

When        we hang the first picture we can choose from all three.

or

When        we get to the next picture we have only two to choose from

or

Finally,        for the last picture there isn't a choice ? it's the one that remains.        So we get the six permutations

Now        this can be worked through, but what if we had 10 pictures. How many permutations        would that be? Time for some maths to make it easier.

For        the first picture you had a choice of three. For the second picture you        had a choice of two and then you had the last picture, with Hobson's choice        (no choice!). Written algebraically this is 3 × 2 × 1. For large numbers        this is very time consuming to write out, which is why mathematicians use        factorial notation.  

A        bit harder

What        happens if we did have 10 pictures and wanted to choose our three most favourite        to hang up? How many permutations would we have then?

For        the first picture on the wall we could choose from all ten. For the next        we could choose from the remaining nine, and for the third choice we could        pick one from eight. 10 × 9 × 8 = 720.

For        this choice of pictures we have 720 permutations, but this isn't 10!.

10!        = 10 × 9 × 8 × 7 × 6 × 5  × 4 × 3 × 2 × 1 = 3628800

We        used the first three terms (10 × 9 × 8) but not the rest (7 × 6 × 5 × 4        × 3 × 2 × 1).  But what we can do is work out 10! and then divide by the        other bits:

The        final term is much easier than it it looks at first. It relates the number        of items to choose from to the number of choices we made (hanging 3 pictures        out of 10).

Not        surprisingly mathematicians have some shorthand for this too:

 where        n is the number of different objects and r of them are to        be arranged.

Combinations

So        far, we have considered hanging pictures on a wall where the order they        are hung is important. But what if the order is not important? Consider        the hanging of the three pictures above, we had six permutations. But the        three pictures were the same in the each permutation, they were all pictures        A, B and C, however they were placed. They were only one combination. We        had 3! permutations from three pictures and one combination. If you like,        these three pictures made one set, and the same is true for each group of        three pictures. 

Consider        the problem of choosing three pictures from a set of ten. If we took the        first three pictures (1, 2 and 3) we could arrange these in six different        ways (permutations), as we did with the three pictures A, B and C. Now,        if we take the next possible set of three: 2, 3 and 4, we can arrange these        in six different permutations. Likewise for all the other sets of three        pictures; each set of three can be arranged in six different ways, i.e.        each set of three pictures will have six permutations. We know from the        example above that there are 720 permutations when choosing three pictures        from ten. And each set of three chosen can be arranged in 3! different ways.        So the number of sets of three (i.e. the number of combinations) is 720        ÷ 3! = 120.

.

In        the same way that permutations have shorthand, combinations have similar        shorthand. All we have to do is divide the number of permutations by the        number of permutation in each set. So, the right-hand side of the following        equation is the same as the equation for the number of permutations except        for an additional r! term in the divisor (which corrects for the        number of permutations of each set). Note, also, that the P (for        permutation) is replaced by C (for combination).

.

If        you have a scientific calculator you should see these labelled (on some        calculators they are separate keys, on others they are second-function keys).        In general, you enter the number of items to choose from (n) then        the nCr or nPr button        and then the number of items to choose (r).

Circular permutations 

They are ordered arrangements of objects in a circle. While order is still considered, a circular permutation is not considered to be distinct from another unless at least one object is preceded or succeeded by a different object in both permutations. In the set of objects, one object can be fixed, and the other objects can be arranged in different permutations. Thus, the number of permutations of n distinct objects that are arranged in a circle is (n-1) !.

EX. There are (6 - 1)! = 5! = 120 ways to seat 6 persons at a circular dinner table.

Suppose you have 2 objects with masses M and m that exert a force on each other as shown. 

A summary of the mass, force & acceleration is given in the table below

                  Quantity       Mass M       mass m     
       Mass       M       m     
       Force       -F       F     
       Acceleration       aM         = -F/M       am         = F/m     
    

Newton's laws of action/reaction have been used.  Calculating quantities such as distances between them looks complex, but there is a nifty tool.  View the system in the reference frame of one of the objects (say mass M).

Now mass M is stationary, but mass m is still free to move under the force F, with one catch.  When one views a situation in the frame of an object with acceleration a, all objects have as there acceleration the acceleration due to the forces on it minus this acceleration a. For the case of mass m

a_m´ = F/m - a_M = F·(1/m + 1/M) = F/(m·M/(m + M))

In other words, mass m behaves as if it instead has a mass

m = m·M/(m + M)

This is called the reduced mass.

I have tried my best to expliain the concept!

pls rate is crrect....................

i think u shud go 4 morrison & boyd.
its the best book 4 organic

it will disintegrate by

-decay.

look in ₁₃Al²⁹ it has 2 extra neutrons than in ₁₃Al²⁷ which is stable.
so in order to become stable its neutron will disintegrate into a

-particle and a proton to reduce the energy caused due to the higher no. of neutrons.

the actual reaction is shown below

₁₃Al²⁹ -------------------> ₁₄Si²⁹ + ₀e^-1

the reaction process may continue depending on the stability of ₁₄Si²⁹.

hope u have understood

ok mr obaid u r forgiven but on the condition u must not repeat the same thing again.

kool article.
did u write dat urself