Problem Solving Strategies

2:03 AM

@PhyMan nope

Anonymous

2:28 AM

@Yashas Got the answer now?

Anonymous

I get $$\frac{\pi}{4\sqrt{2}}\ln(\frac{\sqrt{2}+1}{\sqrt{2}-1})$$

did not bother doing math becaz brain was dead

m doing optics today

Anonymous

I'm revising carbonyl compounds

Anonymous

:P

I haven't started chem revision yet :/

Anonymous

2:31 AM

Are you doing practice papers ?

Anonymous

I'm only redoing the papers from 2016 to 2009

I am reading the textbook

not reading

glimpsing

Anonymous

I see

when I feel that I have forgotten something, I read it again thoroughly

Anonymous

I forget everything if I don't revise for a month

Anonymous

2:33 AM

My brain has very low retention power :P

my brain has negligible retention power

Anonymous

lol

Anonymous

Mine too

Anonymous

Yesterday my father came to know someone got 360/360...and said that I am useless :P Why do the newspapers have to publish those ranks!!!!!

Anonymous

Argghhhh

Anonymous

2:36 AM

:P

Anonymous

Anyway, lets go back to study :)

3:34 AM

@blue @Yashas

Anonymous

Yes?

Anonymous

What?

that's the answer!

Anonymous

We did that question few days ago iirc

:o

Mononitration in each step

3:35 AM

At that time we don't have the solution

We had a solution

we just had a confusion with the nitration

Anonymous

@Yashas Obviously

Anonymous

Want to try an organic question? I'm giving one

ok

Anonymous

3:37 AM

Predict the product of this reaction ^

Gilman's reagent :o

SN2

is the last atom Cl?

two possible cases

Anonymous

@Yashas Yes

CH3 - C(=CH2) - CH2 - CH3

oops

fixed

I think that is wrong

there must be some magic happening

nobody wud give such a direct question

nucleophile can also attack double bonded carbon

Am I correct @blue

3:40 AM

aha

Anonymous

@Koolman

Anonymous

Yes, nucleophile can attack double bond

I knew something extra was there

Anonymous

But I want to know which product is major

Anonymous

and why

3:41 AM

then double bond will shift and LG will remove

@blue the one which @Yashas told

I think my answer is not the major product

Mine could be a product but I think it is minor

Anonymous

@Yashas Your product is fine. Now say it will be major or minor and why

if my product is formed, the carboanion is going to have difficult attacking the double bond

Anonymous

Carbonyl??

Anonymous

Where is the carbonyl ?

3:43 AM

*double bond

@blue i think it will be major , as both ions will be close to each other

Anonymous

@Koolman Which ions?

LG and cabocation

6 mins ago, by blue

Anonymous

@Koolman There is no carbocation formed in the mechanism...

3:45 AM

@blue If the reaction goes through SN1, you will get a carbocation.

Anonymous

@Yashas The reaction will not go through SN1...and even if it does that will be 1 percent of the total

Anonymous

It is always SN2 in such cases

Anonymous

The nucleophile is a strong one and the leaving group is attached to 1 degree carbon even though there is a conjugated double bond

If the nucleophile attacks the double bond first

then you are going to lose the planarity

SN2 becomes difficult

The delta sign tells that we can ignore the kinetics

Anonymous

@Yashas If nucleophile attacks double bond then a chain mechanism is followed, where Cl leaves...no carbocation formed

3:47 AM

SNai?

Anonymous

@Yashas It's just 1,4 addition like you do in alkenes...

Why does the Cl leave?

there is no conjugation for 1-4 addition to happen

how is the CH2-Cl carbon in resonance with the double bond

Anonymous

It is an indirect type of SN2.

@Yashas double shift between C and C(14 ) and LG leaves

Anonymous

^

3:49 AM

hmm

I should revise organic :p

SN2 prime

SN2'

Anonymous

The main reason why this occurs is the hindrance due to the large group attached to the alpha carbon

I'm doing physics and math

Anonymous

It is much easier to attack the double bond

kept chem for final weeks

Anonymous

3:51 AM

For the nucleophile

Anonymous

@Yashas It is better to do all together, imo :P

:|

so the major product is second product

Anonymous

@Koolman Yes, because of hindrance. SN2' product is major

second product?

where are the products? 0o

Anonymous

3:52 AM

@Yashas Notice that there is a labelled carbon atom

Anonymous

You'll get two products

:o

Anonymous

One due to direct attack on C-Cl

but where is the product?

3:53 AM

Anonymous

@Yashas You wrote one of the products toh

@Yashas these two are the products

Anonymous

And for the other one follow Koolman's mechanism

Organic is easy revise :P

Anonymous

Organic looks easy

Anonymous

3:54 AM

:P

Revising organic is like: reading... AHAHAHA...I remember...reading.... AHAHAHAHAH I remember....

4

get all problems correct for the next 2 days

then start getting bad lol

I just need to solve 1000 problems

and I'll be set

wth they have done in this

4:15 AM

How was all of your jee mains rank?

Anonymous

@Koolman The upper curve is for the liquid composition

Anonymous

The lower curve shows vapour composition

Anonymous

At each point go traverse towards right keeping P constant

Anonymous

You'll get vapour composition from liquid composition

@blue @Yashas @Koolman How was your jee mains rank?

Anonymous

4:22 AM

@Programmer I got around 3k rank. (225 marks) @Koolman got around 206 marks and @Yashas got 200+ or 220+ (he has not specified)

Nice

Wow!!!!

@blue how they got liquid composition of Z2 at P1

Anonymous

@Programmer You appeared for JEE mains?

Going for IIT for sure?

Anonymous

@Programmer Trying to..

4:23 AM

This is jee mains

I will appear this year

2018

jee adavance is left

Best of luck all of you for advanced

Anonymous

@Koolman See the point where the horizontal line at P1 cuts the lower graph.

@blue then whats the role of of solution Z2

return below 0.4

4:26 AM

Can I share a physics problem ?

My answer is not matching

Anonymous

@Koolman See the upper graph gives composition of the solution at pressure P1 and lower gives the composition of the vapour at P1.

Anonymous

You'll need both of the graphs for solving the question

At P3 the Xa should be 0.75

Anonymous

@Programmer Sure. But everyone here is quite busy. We might not be able to answer immediately.

Anonymous

But yes, go ahead and ask

4:28 AM

@blue No problem. I am sharing

and @blue at P2 Xa is 2

Anonymous

@Koolman Right.

Anonymous

@Koolman 0.4

That means the solution given is wrong

@blue yeah sorry 0.4

Anonymous

4:30 AM

What does Z1 and Z2 mean!!

Anonymous

Doesn't make sense

they are solution

@blue yes

Anonymous

@Programmer Please write the problem you are facing

@blue @Yashas @Koolman @MadhuchhandaMandal Please try the problem

question is bad

Anonymous

4:31 AM

Along with your attempt

Anonymous

We don't do others homework

Anonymous

You need to show your attempt

Anonymous

We will help spot the mistake

Anonymous

@Programmer

My answer is not matching with the answer given

Anonymous

4:31 AM

@Programmer Please upload your solution then

Anonymous

@Koolman Yeah :/

Ok sure

@blue i.stack.imgur.com/PF0Bk.jpg

Anonymous

@Programmer You didn't take friction into account.

Ok!!!!! But please help. How should I take that?

Anonymous

@Programmer First draw a free body diagram. Friction acts opposite to motion of chain along the incline. f=(mu) mg cos(theta)

4:41 AM

Ok. Thanks I am trying

So won't it be:

(5 lambda H g)/4*(1+mu cos theta)

Anonymous

@Programmer Weight of the chain should be taken along incline

Anonymous

Use sin(theta)

Anonymous

@Programmer Haven't you learned in class 11 how to draw free body diagrams?

Ok so (5 lambda H g)/4 *(sin theta + mu cos theta)

Anonymous

You seem to be having problem with the basic concepts.

Anonymous

4:45 AM

@Programmer Looks fine

@blue yeah I did some solving but not that much seriously

Anonymous

I didn't check your calculations though

Anonymous

The concept is fine now

Anonymous

Check your calculations

Anonymous

You should get the correct answer

4:47 AM

(25 lambda H g) / 16

But still not matching

Anonymous

Hmm, you didn't take the mass of the hanging part

Anonymous

F+hanging part pulls the rope

Ok (in this case mg only I think?)

Anonymous

@Programmer Yes. Calculate m of the hanging part properly

(33 lambda H g)/16

Total

But no match

Anonymous

4:59 AM

@Programmer I think you made some calculation error.

Anonymous

Did you use F+mg=weight on incline+friction?

@blue Do you think his approach is correct?

Anonymous

If you wish you can explain him the concepts

Hmm.. Ok nice language from now

@blue Do you think his approach is correct?

Anonymous

@MadhuchhandaMandal No.

Anonymous

5:02 AM

I pointed out some of his errors

Anonymous

I need to go now

Anonymous

Maybe you can point out the rest of his errors

Anonymous

Probably he made a mistake in writing the force balance equation

Anonymous

5:28 AM

@Programmer I urge you to show me your complete solution. I'll help you find the mistake.

Otherwise you would get bitteb

*bitten

Has he done any solutions to see us??

Huh

Anonymous

@Yashas Could you please tell Madhuchanda to stop? I'll report him to a moderator if he continues being rude.

→ 5 messages moved to Trash

→ 6 messages moved to Trash

→ 7 messages moved to Trash

Report about me @blue

→ 3 messages moved to Trash

Anonymous

5:31 AM

@MadhuchhandaMandal Fine.

If you are so sensitive to a word like "mad"

Let us change topic and move on.

Bye all of you

I won't talk as long @MadhuchhandaMandal is active

Anonymous

@MadhuchhandaMandal You are welcome to use this chat room provided you speak to people politely. You scared off a new user today.

Anonymous

5:35 AM

@Programmer I don't think passive aggression and immaturity helps.

Ok... Lets Change the topic

Anonymous

Yes, go on

So I am rechecking my calculations wait

Weight of inclined part = lam * length of inclined part * sin theta

Anonymous

@Programmer Right.

*g

Length is H/sin 37?

= (5H/4)

Anonymous

5:39 AM

@Programmer Yes

So weight of inclined part = ((5H/4)*Lambda*g * sin 37) ?

Anonymous

fine

= (5H/4 * Lambda * g * 4/5) = H * lambda * g

Now weight of inclined part = (H/2 * lambda * g)

?

Anonymous

@Programmer That is weight of hanging part?

Oh sorry... Weight of hanging part

Anonymous

5:44 AM

Friction?

Weight of hanging part=(H/2 * lambda * g). Correct?

Ok. Now friction

Anonymous

Ok

(3/4)*(Weight of inclined part)*(cos theta)?

Anonymous

yes

So friction = (3/4)*(H* lambda * g) * (3/5) Right?

Anonymous

5:47 AM

yes

So, Friction =(9/20) * H * Lambda * g

So total force = total weight+friction

Anonymous

No F+weight of hanging part=weight on incline + friction

Ok... I get that

F = (9/20)*(lambda * H * g)+(H * lambda * g) -(H * lambda * g)/2

Anonymous

Yes looks fine

But solving, F = (19/20 * H * lambda * g)

Anonymous

5:53 AM

What is the answer given?

Anonymous

I can't see any conceptual error

C

((5/2)*Hlambdag)

Anonymous

I can't find anything wrong :/

Anonymous

Maybe someone else on the main site can help

Anonymous

Which book is this?

5:55 AM

Its previous year question paper

Fiitjee

AITS

Anonymous

Hmm, I'll try and let you know if I solve it

Thank you very much

This chatroom is awesome

I misjudged previously

Sorry for those comments

Anonymous

I think we need to use the value of $v$ given

Anonymous

The constant velocity is misleading

Anonymous

I think it is the instantaneous velocity

Anonymous

5:58 AM

Anyway, I got to go now

Anonymous

Cya!

Ok. Please let me know if you can solve ir

it

Anonymous

Sure, I'll try