Ya, because you can deduce from the molecular formula.

A double bond is equivalent to one degree of unsaturation and a triple bond is equivalent to two degrees of unsaturation.

A cyclic ring is equivalent to one degree of unsaturation

So from the molecular formula, you can degree of unsaturation and thus determine the possible structures having the formula.

E.g.

Suppose a compound has 4 degrees of unsaturation.

So the molecule is likely to have one of the following in its structure:

1)Two triple bonds

2)Four double bonds

3)Four rings

4)A triple bond,two double bonds

5)Two double bonds,two rings

6)Three rings, one double bond

7)Three double bonds, one ring

8)One triple bond,two rings.

Therefore you can determine the various structural isomers.

The degree of unsaturation of a compound is half the number of hydrogen atoms which have to be added to its molecular formula so as to obtain its corresponding alkane.

Eg Ethene C₂H₄

2 H atoms needed to make it Ethane

Hence Degree of Unsaturation=2/2=1

Similarly Ethyne has two degrees of unsaturation

In order to find the degree of unsaturation of compounds containing Chlorine,Nitrogen,Oxygen,etc.

1)For a compound containing chlorine

Replace the chlorine atom  in the formula by H atom and proceed as above using the new formula obtained.

2)Oxygen

Simply remove the atom from the formula and proceed as given above, applying the above method to the new formula

3)Nitrogen

Remove the N atom and ONE H atom from the formula and then apply the given algo. to the new molecular formula.

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AIEEE 6230

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Joined UICT Chemical Engineering.

Sorry double post. Something really wrong with the goiit server.

Well this is a very looong question.

Firstly, find the centre of this curve.

5x^2+5y^2+6xy-8=0=f(x,y) (Say)

For that you solve the equations

df/dy=0

and df/dx=0 simutlaneously

(Hope you know this. To find the centre of any second degree curve, differentiate the relation partially w.r.t x and then partially w.r.t. y

and set the two derivatives to zero. solve the two equations to obtain centre coordinates.)

Here, you can see that

df/dx=10x+6y=0

and df/dy=10y+6x=0s

so centre is clearly, origin (0,0).

Now consider a  variable line passing through this centre (origin) making an angle  A  with positive x axis

So the equation of this line is y=xtanA

Now consider a point on this line at a distance "r" from the centre(origin), and which also lies on the curve

As the point lies on this line  and also on the curve,

its X and Y coordinates may be written as x=rcosA and y=rsinA(Clearly.)

Further as the point lies on this curve, the point must satisfy the equation of the curve

so putting x=rcosA and y=rsinA in the eqn of the curve,

5r²cos²A+5r²sin²A+6r²sinA.cosA-8=0

therefore, r²(5sin²A+5cos²A+6sinA.cosA)-8=0

i.e. r²(5+3sin2A)=8

r²=8/(5+3sin2A)

Now r=sqrt(8/5+3sin2A)

now, r will be maximum when the denominator is minimum

The minimum value of the denominator is for sin2A=-1

i.e. r_max=sqrt{8/[5+3(-1)]}

=sqrt(8/2)=sqrt(4)=2

minimum value of r will be when the denominator is maxmimum

that happens for sin2A naximum i.e. for sin2A=1

i.e. r_min=sqrt(8/{5+3})

=1

You can find the points by using this data.

sin2A=-1

implies A=-pi/4

so the line is y=-x

and the points for max. distance are [-sqrt(2),sqrt(2)] and [sqrt(2) , -sqrt(/2)]

Similarly for the minimum distance,

y=x is the line since A=pi/4

so, the two points are [-sqrt(1/2),-sqrt(1/2)]and [sqrt(1/2),sqrt(1/2)] nearest to the origin.

 Basically the equation represents an ellipse with major axis along the line y=-x and minor axis along the line y=x

Length of major and minor axes are 4 and 2 respectively. The centre of this ellipse is the origin.

No, see there are only n-1 a's inside parentheses. So including those n-1 a's and the remaining a outside the brackets, I took aⁿ

let r be the common ratio of the GP

therefore,

T1=a

T2=ar

....

....

Tn=ar^n-1=b

now, P=T1T2......(Tn)

=a(ar)(ar²)(ar³).......(ar^n-1)

=aⁿ(r.r².....r^n-1)

=aⁿ(r^{1+2+........+n-1})

=aⁿ(r)^n(n-1)/2

Therefore, P²=a²ⁿ(r)^n(n-1)

since b=ar^n-1,

P²= aⁿ(ar^n-1)ⁿ

i.e. P²=aⁿ(b)ⁿ

P²=(ab)ⁿ

Proved.

Dudes, Dudettes, please don't embarrass yourselves by opening your mouth about things you aren't aware of :|

Firstly, it's BIT Mesra and not BITS Mesra.

Karna,

BIT Mesra is an excellent institute, and it is comparable to NIT Tiruchirapalli, I think I can say this because I had enquired about this institute. For the past few years, it has been ranked among the top 20 engineering colleges of the country. I would put the top NITs and BIT Mesra on the same plank. Of course, there is a lot of debate on this topic but it's a very good college. Anyway the BIG CHEESE is an IITian and very knowledgeable as regards colleges, and his opinion would have been more important to me than the others, if I were you.

As for those who say they don't believe the rankings, Karna is talking about the Outlook rankings and not about the farcical India Today rankings.

 Only BIT Mesra has a rather weird admission criterion(donno about this year.) Last year they had fixed a certain no. of seats for EACH STATE.

Akshay aka Tantri_The_Mantri_ aka Morocco Mole aka Red Flag

December 27,2007 13:06:49 IST

Chemical Engineering

University Institute of Chemical Technology, Mumbai \m/

I love this place :D

Who's talking about leaving here? Only the frequency of the visits will decrease. I have another week to go before my college begins, so I'll spend some time here before leaving :)

Sincere advice to all younger guys and those planning to sit for JEE in 2010. Please don't get addicted to the internet.  goiit.com is incredible, but only as long as you use it constructively. Otherwise it will become quite the opposite, I warn you.

But I'm really going to miss some wonderful people I met on this site: Nugorama, Mirka, bladeX, Siddhant,karna, Killer Kiran,rc12, taran, Sujit  and many many more. I've learnt a lot from these guys, and I hope I managed  to teach them something in return :P

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1) the initial part relates to angles and their measurements which isn't tough at all.

2) Trigonometic identities questions are repetitive but there are some really good ones. It didn't take me very long to solve those.

3)Trigonometric equations problems are a bit tricky

4) Inverse functions sums and those trigonometric series sums are not very difficult.

5) Properties of Triangle : I never managed to make head or tail of this nonsense. I didn't study it and only solved a few sums. Several were very very tough.

6) Miscellaneous examples : Didn't  try. Never got time.

My suggestion  is since you are going to  give JEE in 2010, it would be better to solve only selected problems because there isn't much time.

If you spend so much time solving misc. examples, when will you study things like calculus and coordinate? Ask a sir or senior to mark a few problems to solve. In anycase, trigonometry doesn't have that much weightage in JEE.