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5:04 AM
hello
if we are in an inertial frame can me ever tell if the frame is moving or at rest ( you cant see outside)????
 
5:16 AM
@JohnRennie hello
 
@AdvilSell hi Advil. No, it is a basic principle in physics that there is no absolute velocity, so you cannot say if an inertial frame is at rest or not.
 
ok thanks sir
 
5:31 AM
@JohnRennie good morning. Are you free in half an hour
 
@Zerix yes, I'm around for several hours
 
6:15 AM
@JohnRennie What's the difference between deque and circular queue?
@JohnRennie ??
 
@Abcd hi
What is deque?
 
@JohnRennie double ended queue
 
give me a moment while I Google that ...
OK, so a deque is just a queue where items can be added or removed from both ends.
deque and circular queues are very different. I'm not sure where to start saying what the difference is ..
 
@JohnRennie hi
 
@Abcd maybe it would be best to explain what the types of queue are ...
@Zerix hi. Currently talking computers, though it shouldn't be long.
 
6:31 AM
Sure
 
@Abcd hello?
@Zerix do you want to go ahead while we're waiting for Abcd?
 
@JohnRennie sure. I just want some hints from you
 
@JohnRennie there
@JohnRennie Should I send a pic of that from my book or have your figured out the difference thru Google?
 
@Abcd I know what the queues are, but I'm not sure how to answer your question what's the difference because they are so different.
@Abcd are you really asking how they work or why we'd use them?
 
@JohnRennie No, my book doesn't make clear how they are distinct things. If I read my book I feel cricular queue = dequeue
@JohnRennie So Please tell what exactly are circular queues and deques
 
6:39 AM
The idea of a circular queue is that if the queue fills up new items added to the queue overwrite old items. Circular queues tend to be used when we want to keep only the latest additions to the queue and we don't care if old items get lost.
 
??
@JohnRennie Can you tell thru diagram
 
Suppose we have an array of length 10 that we're using as the queue. If n is the number of items in the queue we add new items using queue[n++] = newitem. Yes?
 
@JohnRennie We add using queue[rear++] = newItem in case of linear queue
 
OK. The point is that when n reaches 10 the queue is full and no new items can be added. Yes?
 
ya
 
6:46 AM
With a circular queue we'd do something like:
queue[n] = newitem;
n = (n+1) % 10;
So when n reaches 10 it gets reset to zero again and the next new item would go into queue[0] and overwrite what was originally there.
The idea is that if the queue gets full we only keep the most recently added 10 items and any older items are lost.
It's called a circular queue because it is as though the queue were arranged in a circle so when we reach the last place in the queue we go back to the front again.
Random picture from the Internet:
 
@JohnRennie then?
 
@Abcd there may be applications where you only want to keep the last N items (for some value of N) and you don't care if older items get lost.
For example suppose you're streaming live audio from the Internet on your phone and you pause the playback.
While you pause the playback your phone will be downloading the live stream and temporarily storing it in a buffer so when you unpause you can resume listening from where you paused. Yes?
 
yes
 
But your phone only has a limited amount of memory, so if you pause for too long the buffer will get full.
 
k, then?
 
6:58 AM
So what the streaming app will probably do is overwrite the oldest data. That would mean when you unpause you will have lost a chunk of the audio, but at least it means you'll still hear the latest part of the audio.
 
yes i c.
 
This is done with a circular buffer. When the rear position reaches the end of the buffer it wraps round and overwrites the data that was originally at the head of the queue.
 
@JohnRennie HOW is FRONT = REAR + 1 a condition for full circular queue?
 
When the circular queue is full it looks like this (I'll update my diagram above - give me a moment):
 
$0 (!=) 7+1$
 
7:03 AM
0 = 7+1 mod 8
mod just means the modulus operator. It's % in Java.
 
@JohnRennie But book says front = rear + 1
 
Circular queues use modular arithmetic
front = (rear + 1) % queuesize
where queuesize is the number of items the queue can hold - 8 in this case.
 
@JohnRennie yes but book doesnt mention %queuesize
 
Maybe it just assumes you know that ...
@Abcd Now you realise the queue uses modular arithmetic is it now clear how it works?
 
@JohnRennie Not perfectly, watching some videos. You may see harambe's question till then.
 
7:12 AM
@Zerix you there?
 
@JohnRennie eating lunch. I will get back in half an hour
 
@Zerix OK, ping me when you're ready
2
 
7:40 AM
@JohnRennie hi
 
@Zerix hi
 
 
Your key equation is $I_c = \beta I_b$. Yes?
 
The voltage is going to be reverse in phase
Yes
 
So what is $I_b$? (Just use Ohm's law)
 
7:42 AM
Bias voltage?
 
In questions like this you can generally assume the biasing has been taken care of and you don't need to worry about it.
 
Okay
 
In fact note that the question specifically says an input signal in active mode
 
Depending on the voltage sign
@JohnRennie 10^{-6}/6 A
I think this will be Ib
 
Yes, $I_b = 0.1666$ uA.
So $I_c = $ ... ?
 
7:48 AM
8.33X10^{-5)}A
 
I think that's a factor of ten out ...
I think it's 8.333uA
 
@JohnRennie yes
 
OK. So what is the voltage drop across the collector resistor?
 
@JohnRennie 0.9996A
 
Well, it's 1V. You have a rounding error giving you 0.9996V.
 
7:52 AM
@JohnRennie do we have to round off?
 
The base current is 1/6 uA. Yes?
 
Yes
 
So the collector current is 50/6 uA.
 
Yes
 
And you're multiplying this by 120000 ohms
That gives exactly 1V.
 
7:55 AM
Oh. Guess I converted into decimals that's why it came complicated
 
Yes
 
@JohnRennie so D?
 
You are quite correct that because we normally take the output to be $V_{ce}$ the output will be inverted, so yes it's (4).
See, it's easy once you get the hang of it :-)
 
How to analyse PnP transistor
 
A PNP transistor is kind of like an NPN transistor but upside down. A typical PNP circuit looks like this:
The equation is still $I_c = \beta I_b$.
But all the current directions are reversed compared to an NPN transistor.
 
8:06 AM
Okay
 
For an NPN transistor the $I-b$ current flows into the base and out the emitter, and $I_c$ flows into the collector and out the emitter, so $I_e = I_c + I_b$. Yes?
 
So just with reverse current sign?
@JohnRennie yes
 
@Zerix you flip the battery over so 0V is at the top and +V at the bottom, and that reverses all the currents.
 
@JohnRennie yes
 
@Zerix if you look at the PNP transistor the current flows are reversed so current flows into the emitter. A small part of that current flows out the base and the rest flows out the collector. So we still have $I_e = I_b + I_c$.
The reason I mention this is that the question is asking you about the relative sizes of $I_e$, $I_b$ and $I_c$.
So take answer (A). Can $I_c$ be the greatest current?
 
8:12 AM
No
 
Correct. Now answer (B). Can $I_e$ be the greatest current?
 
Yes
 
Correct. And answer (C) is asking in $I_e$ can be from $0.02 \times I_b$ to $0.05 \times I_b$. Can that be true?
 
Nope
Last is incorrect too
 
See how easy this is? :-)
 
8:34 AM
Still struggling here
The ratio of voltage gain is same a current gain
Ignoring biasing
I can get value of Ice in terms of beta
 
I don't understand the question. If it doesn't give us the value of $\beta$ I don't see how it's possible to answer.
I also don't understand what it means by the arrow labelled $R_o$
Ah, wait, that's a common base circuit ...
Not common emitter.
 
Common base.. That's new
 
To be honest I don't know how common base circuits work. They are almost never used so I've never had to learn them.
I can go away and Google it if you want, but it will take me a while ...
 
@JohnRennie better to. Leave this
 
Well you could get a common base question in your exams ...
 
8:46 AM
@JohnRennie how did you know it's common base
I don't see the base being grounded
 
Read this. It's a pretty good introduction.
 
First solution
 
Yes, but the solution doesn't mean anything to me since I don't remember how the common base circuit works.
 
@JohnRennie how to proceed if imagine this was a common base emmitter
At least I can learn the steps
 
I don't know. I'm currently reading the article I linked in an attempt to understand how the circuit works.
 
8:52 AM
@Blue @Abcd @AvnishKabaj @Dante is common base emmitter in syllabus? Didn't see in NCERT
 
Not common base emitter, just common base.
 
Lol
@JohnRennie leaving this first now..... I think I will proceed for this after confirmation
 
@JohnRennie difference between vaporisation and boiling
Should I. Post question
 
@Zerix Yes
 
8:59 AM
Q7
How to proceed here
I thought water boils..
How exactly is working being done
 
Yes the water is boiling. The question is using the term vapourised to mean boiled
 
@JohnRennie when is vaporize used
 
Work is being done because the volume of 18mL of water is ... erm ... 18mL.
 
Breaking bonds.... Molecules losing energy?
 
And the volume of the steam created from that 18mL (approx. 18 grams) of water is ... ?
Don't get hung up about vapourisation and boiling. It isn't relevant here.
 
9:02 AM
@JohnRennie shouldn't volume be 18 ml still?
 
@Zerix what is the pressure of the steam formed when the water boils?
(hint: it's in an open pan)
 
1atm
 
Correct. And 18g is one mole of water. Yes?
 
Yes
 
And the question says assume the steam behaves like an ideal gas, so the ideal gas equation applies: $PV = nRT$
 
9:05 AM
Okay
So work done Patm{Vf-Vi}
 
So, what is the volume of the 18g of steam created when 18mL of the water boils?
 
373R?
 
@Zerix yes. $dW = PdV$ and $P$ is a constant 1 atm so this integrates to $\Delta W = P \Delta V$
@Zerix you need to be consistent about the units. If we take the usual value for $R$ of 8.31 then you need $P$ to be in Pascals i.e. 101325 Pa.
These numbers should be seared into your soul! :-)
 
0.0821 L atm/K?
Will this work
@JohnRennie how to calculate final volume
 
The final volume is just given by $V = nRT/P$
n = 1, R = 8.314, T = 373 and P = 101325. That gives V = 0.0306 cubic metres.
The initial volume is 18mL i.e. $18 \times 10^{-6}$ cubic metres.
Ah, you spotted that? :-)
 
9:13 AM
@JohnRennie So the initial condition and final condition is applied when the gas/ water achieve equilibrium
We can't apply gas law between the process?
 
You're overcomplicating this. The question says we start with 18mL of water and end with 0.0306 m^3 of steam. And the change happens at a constant 1 atm of pressure.
 
So external pressure does work
It's so complicated
 
Suppose the water is in a piston, and as it boils it pushes the piston out. The piston is pushing against a pressure of 1atm so when its volume changes by $\Delta V$ it does work of $P\Delta V$. Like I say, you are overcomplicating this.
 
Q14
Is the gas expanding from 1 Litre to 10 Litre against 5 atm first and then 1 atm or individually by 5 atm completely and 1 atm completely?
@JohnRennie wait. I think I got this
@JohnRennie hello?
 
9:38 AM
@Zerix hi, I was waiting for you because you say you knew how to do it ...
 
@JohnRennie Q15
Is it asking about isothermal irreversible work?
 
I think the minimum possible work done on the gas is the reversible work.
Did you sort Q14?
 
Yes
For Q14, I used PV= Constant
Calculated the volume at each the steps
And used PV as work
It came out -14
Expansion= Negative
 
@Zerix yes, because the whole process is isothermal that must mean the work done is equal to the heat absorbed (so the internal energy $U$ remains unchanged).
 
Yes. Gotta try Q15 than
@JohnRennie I got the numerical value. About sign, it will get positive because compression means positive energy
 
9:48 AM
I can never remember whether work done on the gas is positive or whether work done by the gas is positive. An to make things worse chemists and physicists use different conventions for this.
 
Lol. I feel like when I start doing thermo questions of physics, I will be confused and lot xD
 
It's just a matter of doing enough questions that you get the hang of what the answers should look like. Then when you come to the exam you'll know roughly what the answer should look like even if you don't know exactly what it is.
 
10:14 AM
@JohnRennie hello
Can you help me with Q18
H= U+ PV
∆H=∆U + ∆VP
∆U= nCv∆T
∆U= Cv( 400/R-300/R)
V∆P= 300 J
Am I right here
 
There's a bit missing off the right end of the question ...
 
 
Well, it's adiabatic so dQ = 0. That means dU = dW (or possibly -dW)
 
I don't think that's gonna help here though
It's asking about enthalpy
So I was more concerned on using H= U+ PV
 
Well it's constant pressure so $\Delta H = \Delta U + P \Delta V$
 
10:24 AM
Yep. Looks like I was right but still getting wrong answer....
 
What is the answer?
 
500 J
I get 6.51 J.....
Wait I get 450
@JohnRennie I get 450 J
Can you verify your answer
 
Well $P\Delta V \approx 300$
 
Yes
 
I must admit I don't understand the question.
I don't understand what is going on during the irreversible compression.
Does it mean the gas pressure starts off at 0.25 atm and increases to 1 atm as the gas is compressed?
 
10:34 AM
@JohnRennie 0.25 atm.... Is it given In the question?
I assumed 1 ATM throughout process
@JohnRennie Q19 is same like this and pressure is given
I think I am leaving this question then and try my luck on 19
@JohnRennie volume is not given in Q19. How I am supposed to calculate temperature
 
You know the initial P, n and T so you can calculate V from the equation of state.
 
But what about finally Via
Ignore me
 
The problem with these questions is that I don't understand what the question means by the irreversible process. You'd need to get one of the other JEE students to go through exactly what the question means.
 
All these questions are based on such graphs
Something like that.
 
I just need a few minutes to do something ...
 
10:49 AM
Sure. I wasn't asking btw xD
 
11:09 AM
@JohnRennie can you can help me with phase change problem.... I suck at them
 
@Zerix I'm in the middle of something, but if you would like to post the question I'll have a quick look
 
11:26 AM
Q33
Don't know how to start
 
Work = $P\Delta V$
You're told that $P$ is 1atm
And you're told the volume change ...
 
Is ideal gas valid in phase change?
I get confused here
 
Work is always $\int PdV$
That's not just for ideal gases
 
Okay. I will remember this
 
It's just force times distance
I answered a question explaining this somewhere ...
 
11:29 AM
Can you post it here
 
4
Q: How much work is needed to compress a certain volume of gas?

Abdelrahman EsmatI want to know the formula (and what does the symbols stand for) for how much work is needed to compress a certain volume of gas?

 
11:44 AM
@JohnRennie last doubt
 
Yes?
 
Can you verify work done in Q34..... It is not coming correct even though We solved similar question earlier.....
@Dante is common base in syllabus? I didn't see in ncert
 
@Zerix You do this in two steps. First vapourise the two moles (i.e. 38mL) of water to get 38mL of steam at 100°C. The internal energy increases by twice the molar heat of vapourisation. OK so far?
 
Ok
 
Now let the 36mL of steam expand isothermally to the final volume. The internal energy change for this second step is zero. Yes?
 
11:50 AM
Yes
 
Because the temperature is constant at 100°C
 
Figured
 
This isn't what actually happens of course, but inernal energy is a state function so as long as we start and finish in the correct states we can choose any path to move between the two states and the internal energy change will be the same.
 
Okay
 
So the change in internal energy is just $2 \times 40.66$ kJ/mol
 
11:54 AM
@JohnRennie what about work done in vaporisation
 
We're doing the change in two steps. First we vapourise the water without changing the volume, so the work done is zero because $\Delta V = 0$. Yes?
 
Yes
But why constant volume
 
In the second step we let the steam expand isothermally to the final volume. But in an isothermal expansion the work done is equal to the heat flow i.e. $dW = dQ$, so $dU = 0$.
@Zerix remember that I said we can choose any path between the initial and final states, because the internal energy change doesn't depend on the path?
 
@JohnRennie yes. Okay
Got it
 
So I'm choosing a path in which the first step is constant volume. I'm choosing this path specifically because the work is zero and that makes the calculation of $\Delta U$ easy.
 
11:59 AM
@JohnRennie that gives 81.32 kJ
 
Yes. Are you now going to tell me that's wrong? :-)
 
But the answer is 75.12 kJ
 
Ah ...
 
Where are we wrong
Because I think the answer is correct
I gtg now
@JohnRennie see you tomorrow
 
Bye
@Zerix ah got it!
 
12:09 PM
@Zerix No
They've given syllabus on official website.
 
12:25 PM
@JohnRennie
 
@Abcd hi
 
@JohnRennie What does highlighted part mean?
 
It means you write an interface with the methods required to manage a queue. Stuff like AddToQueue(), RemoveFromQueue(), QueueSize() or whatever.
 
@JohnRennie why to create interface for this? how is it useful?
 
Then you write some classes that implement the interface to give some concrete examples.
 
12:31 PM
1 min ago, by Abcd
@JohnRennie why to create interface for this? how is it useful?
 
Well the various forms of queue will share a lot of methods and member variables. Writing an interface means all the shared stuff is defined in just one place. Otherwise you'd be repeating the shared code in all the different classes.
 
Oh okay.
@JohnRennie If String str is defined in interface then we need not redefine it in class implementing it?
 
@Abcd Yes
You can't put the code for methods in an interface, but you can define the variables there.
I need to go now. Back tomorrow.
 
 
3 hours later…
3:30 PM
@Zerix it's not
 
 
3 hours later…
6:10 PM
@JohnRennie Hi, are you there?
This is the paragraph
This is the question
 
@YUSUFHASAN hi Yusef. It's the end of the day for me now. You need to catch me in the morning. I'm around from about 05:00 UTC.
 
@JohnRennie Okay thank you!
@JohnRennie BTW you got my name a little wrong.. It's Yusuf,not Yusef... Not that I mind :)
 
Ops, sorry :-(
 
It doesn't matter XD.. People have even called me "Yousuf".. So by those standards, your's is an honest mistake ;)
 
6:44 PM
@Don'tbeax-tripledot we're you able to solve the rotation question? I couldn't match the answer. Sorry
 
6:54 PM
@Zerix Not yet, but I didn't even get the chance to retry it. I've been busy in the last few days
 
@Don'tbeax-tripledot I will ask JR tomorrow. Hope you don't mind
 
7:11 PM
@Zerix Sure, I don't mind at all.
 

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