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06:50
@AniruddhaDeb How many stereoisomers are there in the middle compound?
 
3 hours later…
09:50
@Safdar no stereoisomers (0)
@AniruddhaDeb So 4 isomers in second case and that is it?
10:02
@Safdar the function P(x) has only 3 roots in that way...
in which some are repeated
@RahulVerma Oh.. i was thinking about the remaining two roots..
@Safdar it happened with me also :P
 
2 hours later…
11:59
@Safdar compound on right has 2 stereoisomers and that's it
also anyone here giving BITSAT?
@AniruddhaDeb I am.
12:15
@AniruddhaDeb so am I
16 evening
@AniruddhaDeb and me also
@RahulVerma Did you guys figure this out :
2 hours ago, by Safdar
user image
@HrishabhNayal Yup, a $\in R$, and b is a perfect square.
@HrishabhNayal Since, there are repeated roots: the roots can be $r, r\omega, r\omega^2,r\omega, r\omega^2$ or $r, r\omega, r\omega^2,r,r$ (since the two roots left have to be conjugate to each other)
So we get r=2 using $\displaystyle{\prod \alpha =32}$
12:33
Btw, why I'm seeing $'s instead of mathjax??
@RahulVerma try refreshing..
room topic changed to Random discussions: A group just to unwind, lazy chats and no stressful discussions. Only two other rules - No arguments, let's be nice. For chrome users: chrome.google.com/webstore/detail/se-chat-mathjax/… For mathjax rendering: math.ucla.edu/~robjohn/math/mathjax.html For a detailed Mathjax guide: math.meta.stackexchange.com/questions/5020/… [[jee]]
@Safdar no change. Ah, I see. I'm using firefox, instead of chrome
yo sorry I went off for a bit
@Safdar I still don't see what you are tryna say
Product of roots = 32..
12:40
ok that I get but...
how does this solve the problem?
@HrishabhNayal you get all the roots since all you don't know is $r$
three of the 5 roots are $r.r\omega,r\omega^2$
The other two have to be conjugate to each other so it can only be $r\omega,r\omega^2$ or $r.r$
@Safdar ok
yeah thanks
did you guys figure this out too?
in Room for Safdar and Rahul Verma, 28 mins ago, by Rahul Verma
@Safdar how to find molar heat cap in ^
for a polytropic process $PV^x=constant$ $$C=C_v + \frac{R}{1-x}$$
since the air is in the bubble : $$P=\frac{4T}{r}$$
$P$ is the pressure of gas $T$ is surface tension and $r$ is radius of soap bubble
@HrishabhNayal $P_0 $ is zero since it is in vaccum?
@Safdar that is what I think, yes
This means $$Pr=constant$$
@HrishabhNayal So, you get $PV^{1/3} = c$. Nice..
12:51
@Safdar yup
So you get $C=4R$
that is if my calculations are correct :D
nvm.
@Safdar when is your bitsat?
@HrishabhNayal Day 3 shift 2
@Safdar ok. any other exam you giving?
Advanced.. that is it..
12:56
ok
@Safdar btw where were these ques from?
> Assume that the heat capacity of the soap film is much greater than that of the gas in the bubble.
Nice way to say isothermal conditions.
@HrishabhNayal FIITJEE AITS..
@Safdar ok
@HrishabhNayal ioc.ee/~kalda/ipho/kaz-T1.pdf This was an IPhO question :O
@Safdar hol up..
thats tyrannical
how are we supposed to do that in ~3 min?
@HrishabhNayal Welcome to AITS..
13:05
@HrishabhNayal would you like to have that pdf, from which scrshot was taken?
@Safdar no thank you. I'll see my way out XD
Solution.
@RahulVerma If I say yes,I might not be able to sleep tonight .....so yes !
@Safdar almost same I think
I still can't get the second part tho...
@HrishabhNayal they used a solid angle..
@Safdar you guys got another way?
13:16
@HrishabhNayal nope.. none considering the wall thickness.. Atleast not me.
@RahulVerma cool thanks!
@HrishabhNayal you might revise that statement soon..
2
@Safdar lmao
13:26
@Safdar very true XD
@HrishabhNayal +3 points..
@Safdar for what?
@HrishabhNayal Solution of one part(+1) and courage to ask for a AITS paper($+\infty$)..
@Safdar haha lol
in JEE Maths Zone, 43 mins ago, by Tapi
Let $f $ be a real-valued differentiable function defined on the real line R such that its derivative $f'$ is zero at exactly two distinct real numbers α and β. Then,
(A) α and β are points of local maxima of the function f.
(B) α and β are points of local minima of the function f.
(C) one must be a point of local maximum and the other must be a point of local minimum of f.
(D) given data is insufficient to conclude about either of them being local extrema points.

I thought the answer was C, but some answer key says it's D. Can someone explain why?
what do you guys make of this?
> Some Answer key
?
13:30
@Safdar maybe some say D some don't
@HrishabhNayal then we need to first check validity of keys before comparing..
I've been thinking for this about a while now
@Safdar check this : $f(x) = x^3 $ if $x \lt 1$ and $(x-1)^3 +1$ for $x \geq 1$
$f(x)$ is not differentiable though
@HrishabhNayal take something much more simpler. $f'(x)=x^2(x-1)$
I've been trying to conjure a function with two saddle points
@Safdar ah, so either or both of the roots are repeated... Nice!
13:38
@Safdar hmm.. $0$ is a saddle point
So D is right then?
yep..
https://drive.google.com/drive/u/0/folders/1mD_0QzPygyXhfKQPCIVx3NBPDtkE7run

Solutions (Don't look until you've attempted once)

@AniruddhaDeb @RahulVerma @HrishabhNayal @RahulVerma @satan29
@Safdar where did you get them?
@HrishabhNayal Old faculty where I study.
Cool
in JEE Chemistry Club, 9 hours ago, by Hrishabh Nayal
Can someone tell me how oxidation state of metal affects metal - carbon bond and carbon - oxygen bond in metal carbonyls?
@HrishabhNayal More negative charge on central atom, weaker the bonding.. so shorter $\ce{C-O}$ bond
Also there is only physics BTW..
13:46
@Safdar something is better than nothing :-)
14:06
What is potential due to a charge q inside a sphere or radius R?
...the charge is at dist r from the centre
@RahulVerma $\displaystyle\frac{kQ}{x}$ where x is distance.. Wdym actually?
@Safdar a hollow conducting sphere has a fixed charge q inside it at dist r < R. That was the condition given.
@RahulVerma so at centre kq/r..
depends on where you wanna find it for..
@Safdar hmm... forgot that. I want to find potential at surface of sphere
@RahulVerma Is this the second question?
14:13
@Safdar yeah, a part if it. They said it is kQ/3R, where 3R was radius. But, I didn't understood how?
14:39
Nvm, I got it now
The field effects of inner charge is cancelled by inner induced charge to make stable config. and so net potential is due to outer induced charge only!
 
3 hours later…
17:34
@Safdar solutions for AITS? I already downloaded those tho
@HrishabhNayal I got the first option, left the SHM one. 1 mark it was :P
@AniruddhaDeb Could you share the download link?
You need to log on to reg.fiitjee.com
then 'past question papers' tab, it's been uploaded there.
@AniruddhaDeb It doesn't work on my device.. Was it a .rar file?
@Safdar I didn't know that, interesting. You know the cylinder-wedge numeric type Q in paper 2 was from SS Krotov, which is basically a collection of moscow physics olympiad questions.
@Safdar yep it's a rar file. You'll need to unrar it on a computer
It doesn't allow me to do anything..
17:39
You have winrar or some other program on your pc for compressed files?
@AniruddhaDeb It is a macOSX.. it has an application which recognises everything but .rar files.
Oh yeah ive encountered this when I had a mac. I changed my archive manager to 'the unarchiver' as a result (I think you can get it on mac app store). For a short term fix, use an online unarchiving site
@AniruddhaDeb lemme try that then..
something like (extract.me)[extract.me] or (unrar.online)[unrar.online]
@AniruddhaDeb got it.. thanks..
17:44
Np :)
https://drive.google.com/drive/folders/1KBEr78a3Vq6wLVORFwBs2rn19qOqA5uV?usp=sharing

AITS Q and solutions

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