just saw that sheshannsh did some question as recently as 2 days, I thought he always locked himself up in his room and studied all day. surpise to see toppers using internet [Irony]

@ADG What's the answer? I couldn't do it

maybe 2^[sth]

Options:

I'm not seeing it (now started working on it)...

1,2,3,4,5,none of these

@AndreaL. MathJax -------------->

@ADG Nonono, I said I don't want to see the solution because I started working on it ;)

OK

got it

Ok wait i'll try again too

I thought it is a JEE standard ques so maybe 3 mins are enoguh?

$2^5$

@AndreaL. 6 min over?

@Arpan correcto.

:D

another tricky one:

I got the recursion now...

$11I_5-10I_4=2^{(5)}$

Great :D

Next!

$$\sum_{\rm k=1}^{\rm \infty} \frac1{\displaystyle\rm \sum_{\rm l=1}^{\rm k}({\rm{3l^2-3l+1}})}$$

This one: is it true that$$\sum_{n=0}^\infty\frac{n}{2^n}=\sum_{n=0}^\infty\frac{1}{2^n}$$?

Is it 3 raised to l?

One by one!

@Arpan edited.

Sry @Arpan, I may have overloaded the situation :/

@AndreaL. yes the difference of hand sides is zero

Let's do ADG's first :D

LHs and RHs

it's too easy @AndreaL. since exponential function is [spoiler]

For this question you can have as much time as you wish

@ADG You got it @ADG! (by the way, the proof it's easy, but I must say it's nice)

@Arpan should I pin the question at hand so it's easy?

@ADG Are you sure there is a +1 in the denominator?

@Arpan a+1? where is it, see this---------------------->

$3l^2-3l \color{red}{+1}$?

yes it is

Oops messed up :P

$$\sum_{k=1}^\infty \frac{1}{k^3}$$ .-.

you got it

What do i do with this .-.

apery's constant.

@ADG Not familiar with it

Ok tell:

$$\sum\frac1{k^2}$$

$\frac{\pi^2}{6}$

how?

oh regular mse user.

Read somewhere :p

intersting isn't it.

Very

Ooh I got one (btw the last exercise was 1/k^3 at argument, after simplification)

I have a real solvable one

it's geometry

but do it by hand don't use any software.

Okay

Line L:3x+4y=15

Circle C1:x^2+y^2=25

These intersect at A and B

Circle C2: with diameter A and B

A circle C3, touches C1 internally and C2 externally, it's radius is?

3x+4y=15?

yes

Hint: [Should I give you?]

No wait

@AndreaL. are you gone?

Mh... I'm trying to figure out where are the tangent points of C3

I have even evaluated C1 and C2

Me too

Almost there

Well it's easier than it looks you just have to solve a quadratic equation.

Hold on man

I'm

Are the points of connection between C1, C2 and C3 unique?

draw a rough sketch, the problem is stated enoughly.

Radius = 1?

nope

i.sstatic.net/PoFq1.png Here I don't see a unique C3 circle... maybe I'm overlooking something

oops sorry

took the wrong circle

Radius =2

@AndreaL. Oh that. You have to find the maxima

A-HA!

@Arpan maxima you have to find forgot to state

Yes getting 2 still

So now it becomes $d=4$, or $R=2$ ;)

Yep

correct. you people found 5,0 and 7/5,12/5?

and then?

actually -7/5, 24/5

ya I do remember from today's test

radius of C2 is 4

then what'd you did @AndreaL.

Oh you're not asking me.

I got 7/5, 24/5

Mh... let me check again

@Arpan I already asked you but you didn't told

anyways anyone tell how'd you did then next?

Radius of C2 is 4

then

For maxima, centre of C3 should lie on line joining centre of C1 and C2

@Arpan why?

the real question is this

[OK I know]

done

more?

yeah sure

The same line should connect C1,C2,C3, but I'm slow today, please bear with me :/

Gas

from $a\to b\to c\to d$

ideal monoatomic

at a $P_0,V_0$

at d $e^3P_0,V_0/e$

a to b and c to d are adiabats and b to c isotherms

find work done in b to c

@Arpan ?

@AndreaL. ?

Yes?

done?

wait bro

Question is wrong?

why?

oh sorry i read wrong

a to d is?

a->b adiabat, b->c isotherm, c->d adiabat

a zigzag process [on Pv diagram]

show diagram it is not making sense

Nono it is

@AndreaL. xD

ok

i'll just skip this one :D

Nobody said it was a closed process

i know

b->c is compression or expansion?

hooh

hihi*

@Arpan expansion

something like this

All are expansions?

@Mann wwhere weeeere yyyou

Doing question paper , thinking why do i still make stypid mistakes.

I meant that stypid xD

@Mann I also make stupid mistakes D:

I know dem feels .-. , adg no know dem feels

He scholar :(

scholar? ntse and kvpy only.

Don't we have to use adiabatic equation of state for a-b and c-d then isothermal work done in bc?

ONLY

@Mann ok then tell the answer

Trying .-.

the question is atleast 12 mins old

I just came D:

ok you came on 24 and it's 27, 3 min atleats for you too

Is temperature at b given?

@Arpan $T_0$

LOL :D

Answer is $2nrT_0$?

no idea .-.

Or maybe one

correct now teach me how.

@ADG Do consider giving all data in the question.

sob

now tell me how did you solved it?

xD

PV^gamma

and PV

?

Just go on assuming variables and start from a and build to d.

Yes using those two equations

Here $L=RT\ln V_c/V_b$

Then using the pressure and volume at d, you'll get a good relation

LOLing at messages around 11:49 here

@AndreaL. no W=RtlnVc/Vb

@Arpan done

@ADG She wrote the same thing .-.

she?

He, sry ;)

vc / vb = e^1/2 :D

he sorry :po

*sorry :p

or wait

i am close to it

@Mann this reminds me of math+*ot

I know a "she" andrea too :p

> LOLing at messages around 11:49 here

@ADG Ok man, easy

I am getting $\frac{1}{2}nR T_b$ ._>

K

@ADG Give more questions na

@Mann dimensionally incorrect rejected

@Mann $\ln(e^2)$

oo?

I had e^1/2 d:

D:

a LOL question:

No :o

You messed!

Naa, I don't wanna do it again xD

Chodo, next :D

Friction is a force that aids us daily, in fact so much that we don't even pause to appreciate its importance, we would not be able to wear pants or jeans without friction, we would have to live like naked in the jungle.

Okaaaaaay.

What was relation in adiabatic equation for v and t?

The shirt is supported at our shoulders but if we stand up the weight of our jeans is to be like curved surface o a vertical cyclinder

I cant really remember these equation .-.

Aren't you having ROFL's?

OFC

The force of friction acting between this curved surface and jeans balance the weight of the jeans.

Dafak did i read D:

$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$

To understand this mathematically, let us consider a vertical man whose waist is a rigid cyclinder having circumference of 90 cm.

Ah thank you lord :D

@AndreaL.

Didn't want to deduce it again lol

Hahah, thank you

He wears a jeans of mass 550gm [only] using an elastic massless belt which can be assumed to be on elastic string of force constant 500N/m and circumference 85 cm when not extended. The coefficient of friction between waist and jeans is 0.5

Whta is the fricton force acting between the man's waist and jeans:

Options:

7.5N,12.5N,25N,50N

Dahell .-.

@Mann what?

That question made lol :D

me*

me 2 that's why:

More questions :D

> [past]a LOL question:

I am sad they didn't took a woman and jean. That would be more interesting

$$V_b=V_c\left(\frac{P_a}{P_d}\right)^{1/(\gamma-1)}\left(\frac{V_a}{V_d}\right)‌​^{\gamma/(\gamma-1)}$$

??

@AndreaL. maybe that's just rearrangement

bandar kahinka ! :D

here comes the caterpillar

Nice one :D

Not today .-.

Yes! XD

From here $\gamma = 1+1/R$

o?

we already know that $\gamma=C_p/C_v=5/3$

stop it. plz

Good Golly @Mann :D

@AndreaL. what is golly?

Here's a good question

Here's an explanation:

?

Ready?

?

Ready?

I'm writing it (i can't type faster)

I'm posting a question

damn too small :p

Ok here

The term "good golly" comes from an usage of this song here, whether its meaning it's close to our "perbacco, caspita" in Italy, or an exclamation to express surprise (in a good way). The only english equivalent that I know is "oh my"

A is a Solid cylinder, B is hollow cylinder

Mass of Plank is equal to cylinders that is $m$.

Friction is sufficient that there is no slipping anywhere, find time period of small oscillations of the Plank.

@Arpan so should I sit on that trolley?

@ADG If it helps you to solve

Doing?

yup

@Mann?

don't siturb, now I'll have to do it from starting

Out of the blue $T=2\pi\sqrt{m/k}$

@AndreaL. Haha, nice guess ;) But no xD

done

@AndreaL. You mean $2\pi\sqrt{\frac{m}{k}}$?

Yep, typo. The dimensions were screaming

@AndreaL. Still no xD

$\omega=k/[\frac 56+\frac 7{10}-1]m$

Which is.....?

15k/8m

Nice :D

$$T=2\pi\sqrt{8m/15k}$$

@ADG Actually no

@ADG Typo?

Answer is $$2\pi\sqrt{\frac{15m}{8k}}$$

haan haan thike, paka mat

@ADG wtf

what's happening?

@ADG You're being rude

@Arpan OK sorry

I heard you got banned once

@Arpan sometimes things don't come well online, likhe aur sune me antar aata hai. Sorry

chalega yaar

@Arpan `` ;( "" (<---Cry face)

bana nhi dhangse @Mann help indeed

@Arpan well do you wish to talk on the ban thing?

not really, no

maybe chatting would improve my social relationship buildings and remove my bad personality traits?

apolizing would help maybe and prevention of rudeness too?

JEE Admit cards are here!

jeeadv15.iitb.ac.in/jeeAdv/stud_login.jsp

@ADG

@Mann

@user223679

Pin the link!

Helloo everyone.

@AndreaL. @cirpis @Rammus

@Sawarnik Hi!

@AndreaL. Hi!

Hahah, well, today I chose to give it a try ;)

@AndreaL. Actually, it looks quite an Indianized chatroom.

But anyway, I am really having problems finding the motivation to study chem :(

Too devoid of logic, it is.

@Sawarnik I noticed that, it's interesting ;)

:D

The motivation will follow through your interest, say, you can even reflect an aspect of another field (i.e. logic or math) and try to apply or represent its meaning it as much as possibile with chemistry. It worked very well with me now that I'm studying Thermodynamics I and II. As example, I tried to demonstrate each Maxwell Relation by hand, using the math I like

@AndreaL. Thermo may be fine, but right now in chem I am being thrown facts that have no reasoning, which is demotivating.

@Sawarnik Analitic Chemistry?

(the most counterintuitive branch ;))

@AndreaL. heh .. i don't even know what that is, i.e i don't know the basics of chem right now. Hopefully it would change.

:D

Being thrown facts that have no apparent reasoning is in fact demotivating. But a solution would be signing every unclear aspect and then ask for clarification and/or further proofs. If you have a good teacher/professor he wouldn't be mad at you for that

@Arpan Thanks, but I have downloaded the admit cards already and sorry if you're feeling bad of my rudeness. I will surely improve on that.

My lates fad is bismuth:

enlarge it

or see here en.wikipedia.org/wiki/Bismuth

@Sawarnik How do I always manage to miss you?

I was owe'd .-.

The ban hammer striked me .-.

@ADG @Arpan

@ParthKohli Exactly what I was thinking :/