4:11 AM
@Hercules hi :-)
You can't approximate a sheet to a sphere!
@ronakjain
@Protein hi :-)
Hindi?
Hello bro
@Protein yes I can talk in hindi
Toh kya tumne is question Ko solve kar Liya hai ?
Bhai pehala Banda Mila jo SE Hindi bota ðŸ˜€hai
4:16 AM
@Protein are yar India mein rhte hai , islie hindi to aati hai
Han pehele Kiya tha
Toh kya tumhe iska answer pta hai m Voh mujhe check karna the ki mera answer Sahi hai yah nahin
@Protein are kya hua ? Kuch to response do
Yar trying to solve
Itna teda hai
@Protein toh mein answer bhejta Hun
Iska answer match karna ki Sahi hai ya nahi
@Protein are yar thoda quick response diya kar
@Protein aur vaise yeh question tumhare notes mein kaise hai yeh toh Hercules ne post Kiya that
4:45 AM
@Hercules Calculate the angle subtended by the sheet, then the proportion of the flux that passes through the sheet is that angle divided by 2 pi.
4:56 AM
@JohnRennie is the answer I posted above for the problem of Hercules is correct ?
5:34 AM
@ronakjain I haven't done the calculation ...
@JohnRennie ok are you free now to look at my problem ?
@ronakjain yes, I think the other question is about done.
6:10 AM
@ronakjain How did you reach this answer?
@HrishabhNayal is my answer correct ?
@ronakjain I am working on it. I just want to know your method :-)
@HrishabhNayal firstly consider the wire to be at a distance r from the centre of rectangular plate
Than consider an elementary rectangular part of it at a distance y from centre of it and width dy.
@ronakjain Yes I can follow. I used the same method
The distance of the rectangular elementary part will be sqrt(y^2 + r^2)
And by this you can find the electric field at the position of elementary rectangle
6:15 AM
@ronakjain And area is $L \ dx$
@HrishabhNayal you also know the area.
Now angle between electric field and area vector will be theta
@HrishabhNayal please don't use mathjax. Use simple letters.
@ronakjain Ok
@HrishabhNayal now Cos(theta) = r/sqrt(r^2+y^2)
And the the flux of elementary rectangle = ECos(theta)dA
@ronakjain Now we multiply and integrate right?
And integrating it from
y = -w/2 to y = w/2 and you will get the result
Yes. This is my method. Were you trying a different one
6:20 AM
@ronakjain Nah
@HrishabhNayal ok
@HrishabhNayal do you know the answer to the problem ?
@ronakjain I get essentially the same answer as you. And I think I remember this from one of NTA's mock test. This should be correct
@HrishabhNayal do you know about the official site where u can get the previous year AIEEE papers with solution
@ronakjain No not official but they should be on some of the coaching institutes website. I think
@HrishabhNayal actually I was looking for a problem . The solution to which was confusing . The answer I got on three website s were all different
6:25 AM
@ronakjain If it is a physics one You can post it here and see what others have to say.
@HrishabhNayal ok wait a minutes.....
The problem of AIEEE - 2005
I think answer should be 4I but the Allen solution say it is 2I and other coaching says it is I.
@HrishabhNayal let me check
@HrishabhNayal so from this what do you think ? Is my answer correct ?
@ronakjain wait there is more
@JohnRennie can you help me with a problem?
6:38 AM
@HrishabhNayal I have also derived the expression for electric field and intensity and found that intensity should be proportional to square of slit width
@ronakjain Yeah it should be 4I
@HrishabhNayal yeah
how are you guys able to insert images?
i can't
anyone?
@AbramIvanov open the page in desktop mode and you will be able to upload images
@AbramIvanov I think there is a 100 rep requirement meta.stackexchange.com/questions/124078/…
6:45 AM
i see
@AbramIvanov Workaround . Ask a question (but don't post it). Include the picture you want. Grab the imgur link generated. Copy and paste into chat. Delete the draft question
genious minds here
i fell dumb
@AbramIvanov Not my original idea though. :-)
yaaah
@AbramIvanov what do you need to find? Acceleration of particles? Tension is strings? Something else?
6:55 AM
@HrishabhNayal both
my answer is $\frac g5$
i will have to go now, be back soon
@HrishabhNayal can you tell me how to use mathjax in mobile. I have to face a lot of problems while posting mathematical terms
4

We'll be using robjohn's bookmarklet to make mathjax work on your mobile device. As a matter of fact he's pretty much written a step by step guide for doing so. Copy the text from the box below. Create a bookmark to this (or any) page. Replace the contents of the bookmark URL with ...

Watch this if you cannot follow
@HrishabhNayal let me try ...
@ronakjain I have to go now see ya
7:00 AM
@AbramIvanov try asking @JohnRennie or @satan29 . I am a bit rusty on mechanics currently. :-)
@JohnRennie Hi.
Can i ask my question now?
@HrishabhNayal hi
7:18 AM
@Ohw hi, yes, what do you want to ask?
help me with this pls
@JohnRennie hi.
How can i calculate density of the $object$ which is a hollow
Cube ,side length 1m and made of steel .
how thick is it?
Negligible thickness.
@Ohw you mean the average density of the whole object?
7:29 AM
Yes.
Just mass/volume then. If the cube has side of 1m the volume is 1 m³ so the density is numerically equal to the mass of the steel in the cube.
I don't think that's the correct way.
If i do this what will the difference between a hollow and a solid cube.
@Ohw If you have two steel cubes of the same size, one hollow and one solid, the solid cube will have much more steel in it and therefore will weigh much more.
@AbramIvanov are the pulleys massless?
If the cube is filled with water will i just add the density of the previously calculated denity and the density of the water?
@JohnRennie yes sir
7:37 AM
@Ohw the density is always the total mass divided by the total volume. If the cube is filled with water the total mass is the mass of the steel walls plus the mass of the water inside the cube.
@AbramIvanov I get a weird FBD then:
do you also get T=0?
@JohnRennie same as mine
There is a non-zero net tension on the pulleys so they would have an infinite acceleration.
but this seems wrong
@JohnRennie Sir i have another check my answer question.
That means the string would immediately go slack i.e. the tension would be zero.
7:39 AM
@JohnRennie i find it hard to digest tho
is it possible in real?
@JohnRennie You missed tension on the second mass.
Ah yes, I did. But there is still an unbalanced force on the left pulley.
@Ohw what's the question.
-3

No current will flow from a,b and C resistors since all the current will now be flown from the wire without any resistance. This wire made the circuit short.

This is what my diagram looks like after rearranging the components and i cannot spot my mistake.
Forget the text "without wire".
Give me a moment to have a play at rearranging the circuit ...
@Ohw this is the original circuit:
Okay
7:53 AM
And if I redraw it to make the layout more obvious I get:
which is a simple circuit.
I don't understand why my diagram is incorrect.
Your diagram isn't incorrect. It's the same as mine ...
If the current starts from positive terminal of the battery why won't all current moves through red wire?
(In my diagram.)
7:58 AM
I've marked the red wire on my rearranged drawing. Do we agree this is correct?
@JohnRennie I don't agree that your diagram is correct.
OK ...
No disrespect.
Would you like to correct it and show how you think it should be?
@JohnRennie Hello sir :-)
8:00 AM
@Jasmine hi :-)
I have shown it the way i can possibly think.
@Ohw but our diagrams are the same, just laid out differently.
Probably after Ohw
Please explain "If the current starts from positive terminal of the battery why won't all current moves through red wire" with reference to my diagram.
Won't current take the path of least resistance?
8:05 AM
The current from the battery I0 splits at the red wire. Some goes via the red wire and some goes through a. Yes?
there is a circuit with only 1 resistance. and we connect it a wire over it will the current be divided?
I'm not sure what you are asking ...
You mean like that?
The current is not divided here?
I've marked the potentials at the two points in the circuit, Va and Vb.
Okay.
8:12 AM
The potential difference across the resistor is Va - Vb, so the current through the resistor is Ir = (Va - Vb)/R
Yes?
Yes.
But if you connect the points with an ideal wire of zero resistance the potential difference across an ideal wire must always be zero. So that must mean Va = Vb, and therefore that the current through the resistor is zero.
So all the current flows through the wire and none through the resistor.
Then why isn't this happening in my diagram too?
(Original question)
Because the circuit in the question isn't the same as your simple circuit above.
The current flows in this circuit are more complicated.
You haven't got just one wire in parallel with one resistor.
After reaching the junction bde , the current will flow in backwards direction?
8:21 AM
That's what the current flows look like
Okay.
Are we done now. Can I move on to Jasmine's question?
I just have one more problem in this question
OK ... ?
8:25 AM
@ManasDogra I have a bit of a queue at the moment I'm afraid ...
@JohnRennie ok ok whenever you are free...
@ManasDogra this is the most elegant answer I've ever seen:
132

This question was prompted by Can matter really fall through an event horizon?. Notoriously, if you calculate the Schwarzschild coordinate time for anything, matter or light, to reach the event horizon the result is infinite. This implies that the universe ages by an infinite time before someone ...

@Ohw yes ... ?
I have labeled the junction as numbers
OK, so the red wire makes V1 = V4
8:28 AM
In your diagram , i think you have connected 1 and 3 points
Starting of A and starting of B are connected and is the only confusion i have from your diagram .
Thank you so much. It clears it all.
If you compare the voltages across all the resistors you'll see they are the same in my and your diagrams.
@Jasmine are you still around?
@JohnRennie Yup
8:33 AM
A simple doubt
What do you want to ask?
There are 3 small balls colliding with bigger ball
What is the definition of the term impulsive in this context?
@Jasmine is there a misprint in the book?
@Ohw I don't know
8:37 AM
I would have guessed impulsive just means a force that acts for a very short time, and in that case I would have said all three collisions produced an impulsive force ...
@JohnRennie The general impulsive force
@JohnRennie yes, I feel that as well
@JohnRennie Tension only changes in the A and C.
@Ohw but that's not the definition of an impulsive force
That's what the text says "when is tension impulsive"
Oh wait, sorry, the book is saying whether the tension T is impulsive or not.
8:39 AM
Yes, I agree. Both the angled collisions would be impulsive.
@Jasmine but all three collisions would change the momentum of the ball.
I don't think we can answer this. It's just poor writing in your book and only the author knows what they actually meant to say.
@JohnRennie but what about Normal contact force
Which one? The tension in the string T?
I guess in case A it is the Normal which is entirely responsible for change in momentum
@Jasmine In the b case normal force is horizontal, there is no change of momentum in y direction just after the collision.
8:43 AM
B is quite clear
Just the reverse of A happes in C.
All three collisions produce a force that has a horizontal component and will therefore change the momentum of the ball.
Only the vertical directon of normal force reverses in A and C.
@Ohw In case of C wouldn't the string become Slack ? how can tension be impulsive in that case ?
Not immediately, and obviously the collision will depend on the mass of big ball.
8:49 AM
@Ohw Will tension in case of C in any way be responsible for changing the momentum of ball ?
The definition of impulsive force is basically a Large force acting for a short period of time however if there is no force during collision how can we call it impulsive as in case C the tension would become 0
Momentum of ball is changed by normal contact force in all cases
Due to this value of tension changes.
@Jasmine I think the author means to increase the momentum rather than change by tension force.
@JohnRennie So did anyone calculate quantitatively the T in your question after that?
will tension we impulsive if the small ball hits the big ball head on from bellow ?
*below
@Ohw but that wont be the definition of impulsive force
8:57 AM
@AdvilSell If the string becomes slack value of tension changes from T to 0 , since the value of Tension changes there will be change in momentum due to this
@JohnRennie what do you think
@Ohw But tension doesn't takes any momentum from the ball to become zero , it becomes zero because the string becomes slack , if it was a rod instead of a string then tension would be impulsive , I think , What do you say ?
9:22 AM
sorry for commenting on your discussion force of normal reaction will be impulsive in all the case because it does impart the momentum in all the cases but the tension will in case A because as the ball hit the bigger ball there would be some tightness in the string
which makes the ball just slightly jump above due to tension and i feel the second angled one too produce the same thing
@JohnRennie hi morning sir i got my bag back
hey!
hi..
@Jasmine iwould say that The Tension constraints the ball to move perpendicularly to the string (which arises from the fact the string is inextensible). In light of this, It seems clear that T will be impulsive for collision with A
Since the collision with A imparts a velocity along the horizontal(Vx) and also along the downward vertical (Vy). The tension becomes impulsive to "kill" this downward momentum mVy. (i.e J-Mvy)=0
The collision with A is the only one which imparts a downward momentum to the ball
9:39 AM
How can tension be "non-impulsive" for collision B? There seems to be a sudden surge in tension due to the centrifugal force and hence...
@GuruVishnu i guess youll have to consider the definition of impulsive rigorously
in case of B, yes the tension changes suddenly, but it remains at the value mv^2/r for a considerable period of time
In case Of A, tension exerts a downward force so that Mvy is destroyed, and then it stops acting in the downward direction.
i.e, it acted for a very, very short period of time
the book is probably defining impulsive to be a force that acts for a very,very short period time. and in light of this, I would not call B impulsive
Guys i have a bit of a soft question
9:57 AM
Fine. Again a local terminology misinterpretation. If B is also impulsive then, in the same way, C would also be impulsive as it suddenly changes from mg to 0. For A, F-t graph: _/\_, but for B and C it looks something like _/---
right
@GuruVishnu Hi!
Hi!
although for C, the string slacs, and so the tension cant be impulsve
Yes. Both B and C are similar and that'll be the only difference. The minimum is zero for C whereas it's mg for B.
By the minimum, I meant either of the two horizontal regions in the F-t plot.
10:01 AM
Can you please explain how can an adiabatic expansion be irreversible? I mean the heat exchange b/w system and surrounding is zero, so entropy should not increase?
I guess that's from Cengage; just guessing from the font face
@GuruVishnu yes, thats definitely cengage lol
@JohnRennie ah! Sir. I didn't see you in H bar so I thought that you were offline. Can you please have a look at the above question.
@JohanLiebert what is your definition of irreversible?
@satan29 it can't be reversed in time?
Though I haven't seen a formal definition.
10:06 AM
btw, I would also like to tell you, that the entropy change of the system in an irreversible adiabatic process is non-zero :)
@satan29 I can't seem to get it. I think this might be due to my inability to understand the equation dS=dQ/T. Shouldn't dQ be 0?
@JohanLiebert Some time ago, I learnt from an answer on Chemistry.SE that all reactions are in equilibrium and there's nothing called 'irreversible' reaction. Maybe, is this what you were looking for?
the correct equation is ds=dq_{rev}/T
i.e heat exchange for a reversible path
the issue with irreversible paths is that the variables P and T arent very well defined
AFAIK the temperauture isnt spatially uniform throughout your system
so we make use of the fact that enropy is a state function
and so the entropy change depends only on the initial and final state variables, and not on the path taken
so i suppose i go from (p1,v1,T1,) to (p2,v2,T2) irreversibly.
the entropy change would be the same if
i go from (p1,v1,T1) to (p2,v2,T2) reversibly
10:22 AM
so for a reversible process, dq_rev=du+(nRT/V)dv. therefore,ds= dq_{rev}/T= du/T + nR*dV/V
@JohanLiebert the equation:
du=nCvdT. using this, we get ds=nCvdT/T +nR*dV/V
$$dS = \frac{dQ}{T}$$
applies only to reversible processes. For an irreversible process we the equation should be:
which we can integrate to get delta S= nCvln(T2/T1) + nRln(V2/V1 )
$$dS \ge \frac{dQ}{T}$$
@JohanLiebert so $dS$ can be positive for an irreversible process even when $dQ = 0$
10:25 AM
btw @GuruVishnu thank you so much for that transcript a couple of days ago!
No problem! But did you mean the surface tension thing?
yes
@satan29 Yes , that's what I was saying

1 hour later…
11:44 AM
@JohnRennie thanks sir.
12:07 PM
@satan29 I have a objection with a word kill
You have used above instead you should use that every body want to remain at lowest energy so to try do that tension forces arise to balance the force which creates the changes in energy!