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05:35
@JohnRennie good morning. Are you free
@Zerix morning :-)
@JohnRennie are you free for giving hints
Yes ..
Okay
@JohnRennie does speed of sound increase with pressure?
How does this work
I've answered a question on that on the main site. Let me have a quick look for it.
3
A: What causes an increase in sound speed in a medium?

John RennieThe speed of sound is given by the Newton-Laplace equation: $$ v = \sqrt{\frac{K}{\rho}} $$ where $K$ is the bulk modulus (i.e. a measure of stiffness) and $\rho$ is the density. The physical interpretation of this is fairly obvious. Stiffer substances recoil faster from a displacement so incre...

05:48
@JohnRennie Stiffer substances recoil faster from a displacement so increasing the stiffness increases the speed of sound
Didn't understand
Is this intuition
Suppose you have a spring with a force constant $k$ and you compress it a distance $dx$ then the force is $F=kdx$. A stiffer spring has a higher $k$ so it produces a greater force for the same displacement.
If you have some mass attached to the spring then the greater force means a greater acceleration so the stiffer spring accelerates object attached to it faster.
Okay. Got it
As you increase the air pressure you are in effect making the air stiffer.
Okay. Got it
That's nice way to understand the equation
I've just logged into work and we have all sorts of problems this morning. Probably due to storms we're having in the UK at the moment. I may be tied up for a bit sorting this out.
05:56
Okay
That would be your first storm of 2019 xD
 
2 hours later…
07:27
@JohnRennie
 
2 hours later…
09:31
@JohnRennie hi. Are you free now
@Zerix hi, yes, panic over. All my servers are working again :-)
@JohnRennie for disintegrating, Q value should be positive
Is this correct
Disintegrating? Q value? What area of physics are we talking about?
Nuclear physics
@JohnRennie let me post a question
It would make more sense
Ah you mean the energy released ina nuclear decay?
09:40
Q11
The Q value comes negative
How is disintegration possible?
The binding energy of a nucleus, or indeed any bound state is negative, i.e. if you start with an isolated neutron and proton and bring them together then when they form a bound state they emit energy.
This means the mass of a bound state is less than the total mass of its components.
OK so far?
@JohnRennie okay.
But how is it negative
Nucleus has less energy than proton and neutron
If you look at the masses given in the question you'll see that the mass of the deuteron is 1876MeV while if you add together the messages of the proton and neutron they come to 1879MeV. i.e. the deuteron is 3MeV lighter than it's components.
So less energy means more stability
@JohnRennie yes
Well, suppose you want to split the deuteron back into a proton and neutron, that means you need to supply 3MeV of energy to do this. It's kind of like ionising an atom. If you want to split a hydrogen atom into a proton and electron you have to supply 13.6eV of energy.
09:47
Okay
So the answer to the question should now be obvious.
@JohnRennie (d) ?
Correct!
The deuteron has to absorb a 3Mev photon to supply the energy needed to split it apart. Just like a hydrogen atom has to absorb a 13.6eV photon to ionise it.
@JohnRennie So Q value here is like excitation energy
It's a little more complicated than that.
Deuterium is stable. It won't fall apart by itself. That's because the deuteron has less energy that the decay products.
09:51
Okay
But consider tritium. Thus is also a bound state, so the tritium nucleus weighs less than a proton and two neutrons. So it should be stable with respect to decay into a proton and three neutrons. Yes?
Yes
But tritium is unstable. It decays with a half life of about 12 years. So we have something weird here. Can you suggest why it is unstable even though it's a bound state?
@JohnRennie no idea
Maybe more neutrons
Let me draw a diagram
09:57
Tritium is a bound state so when you form it from a proton and two neutrons that releases energy. So to make tritium decay back to a proton and two neutrons you would have to supply energy.
That makes sense
But tritium doesn't decay into a proton and two neutrons. There is another bound state 3He that is even lower energy than tritium is.
So tritium will spontaneously decay into 3He (and an electron)
And this releases energy.
i.e. you would have to supply energy to convert 3He back into tritium.
OK so far?
Okay. Is that beta decay
Yes. In tritium decay one of the neutrons converts to a proton and an electron. The electron is released as a beta particle, and the remaining two protons and neutron form a lower energy state than the single proton and two neutrons in tritium.
So as a whole the process releases energy. That energy is released as the energy of the electron.
The point of all this is that when we talk about a Q value we are always refering to a spontaneous decay like tritium to 3He and the Q value is the amount of energy released in this decay.
10:04
Okay
@JohnRennie so it can be both negative and positive
Got it
We wouldn't normally use the term Q value for the energy we have to supply to split up a nucleus. Q value always refers to energy released and it's always positive.
That's just terminology. I'm not sure what term is used to describe the energy we have to supply to split up a nucleus.
Q12
Wait. I think I miscalculated
@JohnRennie not getting answer. What to do here
@Zerix I'm not sure ...
10:20
I found a formula here. Is this what we have to do
The Q value is the initial mass minus the final mass i.e. it's the amount of mass that got turned into energy. In this case we are told the Q value is 50MeV.
Yes. Okay
So how to relate this with KE
The KE of the alpha particle is 48MeV, so there's another 2MeV unaccounted for ...
Ah, of course!
The missing KE is the kinetic energy of the daughter nucleus.
@JohnRennie is this like conservation of energy
If you work in the COM frame the total momentum has to be zero, so the alpha particle goes one way and the nucleus goes the other.
From the info given you can calculate the momentum of the alpha particle, so the momentum of the daughter nucleus must be equal and opposite. You know the KE of the duaghter is 2MeV so you calculate its mass.
NB energy and momentum are linked by $p = \sqrt{2mE}$
10:28
Okay .lemme try then
@Zerix it's really simple. You don't need a complicated calculation.
@JohnRennie
@blue_eyed_... hi
@JohnRennie, A coil of wire wrapped on a hollow tube and a light bulb are connected in series to an ac source. What happens to the brightness of bulb when an iron rod is inserted in the tube?
@blue_eyed_... well, what is the iron rod going to do to the inductance of the coil?
10:37
@JohnRennie woah. Finally got it. So I take COM method is the best bet in these cases
At first I was actually calculating the momrntum xD
@JohnRennie, Increase ?
@Zerix yes. It's easy to forget that both the emitted particle and the nucleus recoil after the decay. I must admit I forgot in this case, which is why I wasn't sure how to do it at first. I take it you got the answer (A)?
Yes
@blue_eyed_... yes, the inductance will increase. If you draw a phasor diagram it should be obvious that increasing the inductance is going to increase the overall impedance.
@JohnRennie what to do in case of recoil. The next question includes it too.
10:40
@Zerix work in the COM and remember that momentum is conserved
Okay
@Zerix in Q13 you are told the velocity of the alpha particle is $u$, so the momentum is $p_\alpha = m_\alpha u$. Yes?
Yes
And the momentum of the daughter nucleus must be equal and opposite so $p_d = m_d v_d = -m_\alpha u$
Yes
10:43
And you know what the mass of the daughter $m_d$ is ...
Yes
so the answer is ... ?
@Zerix really? If you add $m_d$ times C to $m_\alpha u$ to get the total momentum does it add up to zero?
Lol. It's B
Silly old me
10:48
:-)
11:12
Last question
In (a) energy is supplied
In (c) energy is emitted
How to see in (b) and (d) ?
@JohnRennie pls leave this
OK ...
I am getting too careless
11:34
@JohnRennie what is the difference between binding energy and mass energy
@Zerix I would have to see the context. Is this from a question?
The fact that energy is to be supplied or energy is emmitted depens on the mass energy rather than binding energy .
@JohnRennie let me post question
Q22
The binding energy of two deutron comes to be 1.1Mev
For helium it's 1.75Mev
Mass energy of two hydrogen is nearly sane to that of helium
Do you mean q23?
How do we decide energy is emmitted
Q23. Sorry
You do this with a cycle. Let me draw a diagram ...
Is that obvious? Or do I need to explain what I've done there?
Oops, just spotted an error. 1.1MeV is the binding energy per nucleon, so the D binding energy is 2.2MeV. Replace 1.1MeV on the diagram by 2.2MeV.
Likewise replace 7MeV by 28MeV
@Zerix there
11:49
It makes sense
@JohnRennie but what is difference between binding energy and Q energy here... Both are same?
The Q value is the energy released when two deuterium nuclei fuse to form a helium nucleus. Yes?
Yes
The binding energy is: for D the energy released when p +n fuse, and for He when 2p + 2n fuse.
So they are equal......
(the binding energy per nucleon is the above energy divided by the number of nucleons)
11:54
Or same in a sense
Well the answer to your question is 23.6eV
And that isn't equal to any of the binding energies ...
I am getting confused
For first deutrium to p +n.binding energy is equal to Q value?
Similarly for the second?
The term Q value specifically means the energy emitted in a nuclear process. In this case the process in question is 2D -> He.
The Q value for this process is calculated from the binding energies of the D and He nuclei
@JohnRennie why not calculate individual Q value of the steps and add
4.4+28
That is in effect what we did. But I don't think I would describe those energies as the Q value. That's just not what the term is usually used for,
12:00
@JohnRennie okay. I got what we did here
@JohnRennie could you tell me what is happening in Q22
What is meant by probability there
If you take any particular radioactive nucleus then the probabiity it decays in one half life is 1/2. Yes?
Yes
I never knew there is a chance for it to not decay too
The question is asking: suppose we are watching one particular nucleus and we watch it for five half lives and it doesn't decay. Then for this particular nucleus what is the probabulity that it will decay in the next (sixth) half life?
@JohnRennie 1/2?
Yes.
The point is that just because a nucleus survives five half lives that doesn't make it special. Just by pure chance we expect $1/2^5$ nuclei will survive five half lives.
It's like if you flip a coin five times and it comes up heads every time. Then you flip it one more time what is the probability that it will come up heads again?
12:09
1/2.independent events
Yes. Though you'd be surprised how many people think that isn't true :-)
So is radiavtivity is just probability?
@JohnRennie I can imagine xD
@Zerix yes
Woah. That's new
@JohnRennie gtg now. See you Tomorrow
12:13
@JohnRennie Hi
@Dante hi :-)
small doubt in thermodynamics!
Two moles of ideal helium gas are in a rubber balloon at 30C. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35C. The amount of heat required in raising the temperature is nearly (take R=8.31 J/mol K)
They've used nCpdT
I think the point about the balloon is that it exerts a constant pressure on the gas. So this is a constant pressure process.
12:17
There is no energy required in expansion. Then what's wrong in using (f/2)R(T2-T2) ?
@JohnRennie Constant pressure or zero pressure?
It can't be zero pressure ..
Do they mean work done will be zero in expansion
It's a really poorly worded question, but when I saw it my immediate assumption was constant pressure.
2
They mentioned it can be assumed it requires no energy in expansion.
@JohnRennie Hmm, I see. I thought the same but it was JEE advanced 2012 question, therefore it was better to confirm.
 
3 hours later…
15:41
@Zerix I'm around, but eating lunch!
@JohnRennie I was just scrolling hbar and saw blue's message. Happy birthday!
@Zerix Thanks :-)
2
16:50
@JohnRennie Many Happy returns of the day!! :-D
2
17:29
@JohnRennie Happy birthday sir ! :)
2
17:41
@JohnRennie Happy birthday sir! :)
2
18:10
@JohnRennie Also, a small conceptual doubt- What exactly is an instanteous axis of rotation? I always used it in problem solving as the point where perpendiculars drawn to the velocities of any two points of a rigid body meet, but this is obviously not the actual meaning of the thing
@JohnRennie Also, on a separate note, why does the need for the Coriolis force arise when we are analysing,say, a coin rotating about an axis while sitting on that coin? it seems to me that centrifugal force is good enough for a pseudo force?
@YUSUFHASAN I suggest you to watch HCV's video on coriolis force. It explains why centrifugal force isn't enough.
I have forgotten that though but I recall that the explanation was great.
(not in syllabus as well)
@Abcd Ok...is it on YouTube?
@YUSUFHASAN yes
@Abcd Yeah ik it is not in syllabus...But experts at my coaching are guessing that this and a few other topics in physics might be given as a new comprehension
@Abcd Thanks BTW!
@YUSUFHASAN it has come in 2016 I guess and jee doesnt repeat.
18:17
@Abcd Yeah...But it maybe interesting to know
@YUSUFHASAN Not sure we can follow our interests in last 2 months :) .... But you are 99.7 percentile person so maybe you can manage ;)
@Abcd That was my chemistry percentile...I am not even going to tell you the other two… And how they botched up the net percentile :(
@Abcd Suffices to say...The kind of mistakes I made in the other two...You would judge me really bad...And then have a good laugh
oh let me see my chem percentile
@YUSUFHASAN No I avoid judging anyone :)
@Abcd Yeah? Well...It is really amazing that I know about frontier MO theory...and then in the paper I forgot the basic integral dx/underoot((x^2+ a^2)) in tension
@Abcd Now you see? :)
@YUSUFHASAN no formulas going blank in the head is a cmn mistake and one is so stressed that one doesn't feel like deriving it in exam hall.
18:27
@Abcd Yeah, that maybe true...But I am a real legend at physics and maths….Perhaps I am not as talented as you, good at everything @Abcd XD
am good at nothing
Oh yeah? What's your overall percentile? @Abcd
@YUSUFHASAN 99.125..., yours?
@Abcd I managed to bring my percentile just at the edge of the 99 percentile decimals inspite of my chemistry percentile...That's how "good" I am at the rest of the two..And I am not sharing the entire number...It just sucks
@YUSUFHASAN I feel dis percentile system sucks, such high numbers like 99 out of 100 mean nothing (in terms of rank) in reality.
18:33
@Abcd Yeah, I also feel that this percentile system is biased....The marks I am getting...I should have had a much better rank with these marks in previous years...I was shocked at the number when it was converted
And the damn idiots never disclosed the actual marks
Happy Birthday John Rennie
πŸŒπŸŽ‚πŸŽ‚πŸŽ‰πŸŽ‰
belated*
@YUSUFHASAN that makes no sense
@Abcd Not for UK
@AvnishKabaj Smart!
It's belated in India
18:35
@AvnishKabaj What doesn't make sense?
@Abcd ;)
@YUSUFHASAN percentile being biased
@AvnishKabaj I mean...I know the technicalities...I have seen the details countless times..But then, all the 10 toppers from my coaching(who are expecting ranks in 3k in advanced) were present on 9th and 10th, probably because they registered early, idk. So, it obviously affected my percentile, and I still don't know my marks exactly for my own satisfaction
In general, 9th Jan people have got higher percentiles that other shifts with same marks.
but that was because 9th jan was v.lengthy
And many toppers from various institutes were there on 9th and 10th...Fiitjee, Akash,VMC etc. I feel that the student pool wasn't that well mixed, although I don't claim any =thing
14 secs ago, by Abcd
but that was because 9th jan was v.lengthy
People are saying 9th Jan is easy, but they didnt examine the lengthiness of the papers
18:41
You must have the wrong rank list
270 = 1k in mains
My friend for 282 and was 700
In other slots kids with 300-310
I atleast know that these 10 guys are geniuses...And they should have been more jumbled up
Haven't even made it under 1k
@AvnishKabaj Exactly my point...Mains papers vary subtly in difficulty, but still, 300-310 was definitely 1k in the previous years
@YUSUFHASAN Same here, most of my batch toppers gave on 12th Jan.
shuffling wasn't uniform
but how can they shuffle?
they have no means to shuffle
they dont know if you are topper of ur coaching or not.
I am pretty sure that the shuffling was uniform
18:45
How?
When you're dealing with large numbers
Well idk
Man
Maybe it was maybe it wasn't
@Abcd They can randomly shuffle roll numbers in between slots...All the roll numbers were given serialwise as per registration...And they were sent in the same order... Otherwise Iw ouldn't have seen all these magical guys on my shift all together
dude they have no data to shuffle uniformly with equal number of toppers and average students in each slot
@YUSUFHASAN yes random shuffling. Not on the basis of intelligence or something
They just had 10th class data and that's pretty useless.
Abh to Kar me Kya karoge
Mains/advance prep
Go go go
Maybe for 2nd JEE mains they'll use 1st JEE mains percentiles for shuffling
@AvnishKabaj "kar me kya karoge" means?
18:49
@Abcd Yeah, and they should also release the marks...I atleast want to see my marks along with my percentile even if for self satisfaction
Ofcourse I have checked...But still...Something which looks a little good besides the lousy percentile

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