Problem Solving Strategies - 2018-03-31

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2:34 AM

i'll have a look at this within 2hrs

@JohnRennie did you have a look at that floating blade

?

3:03 AM

@MadhuchhandaMandal yah that's correct

2 hours later…

4:56 AM

hey Everyone! I want to simulate the light absorption in silicon nanowire on COMSOL...does anyone have experience in this? I had a few questions regarding this

5:20 AM

@Eliza Hi Eliza. The Problem Solving room is mainly meant for discussing specific exam or homework questions. The room for general physics discussions is The h bar.

Though the h bar is pretty quiet at this time of day so I doubt there's anyone around with experience of COMSOL.

@JohnRennie hi

Morning :-)

Sir ,how some scientists are able to measure speed of light ?

how it is possible to measure?

There are lots of ways to measure the speed of light. For example we can bounce light off the Moon and time how long it takes to travel there and back.

sir,in a solar eclipse how the moon is able to block the view of sun which is 50 times greater than its size?

5:28 AM

The Sun is a lot more than 50 times bigger than the Moon :-)

Yes sir I meant that

then how moon is able to block sun ,isn't it smaller?

The Moon is a lot closer. If you hold your hand up between your eye and the Sun your hand can block out the Sun. But your hand is a lot smaller than the Sun is.

why it is like that?

What matters is the angular size of the object. That is if you draw two lines to opposite sides of the object then the angular size is the angle between those lines.

I didn't understand ,sir

5:36 AM

user image

The angular size is the angle $\theta$

oh

user image

For small angles the angle is approximately just 2r/d

i.e. it's the ratio of the object's size to its distance away

okay sir i get it

It turns out that the ratio 2r/d is the same for both the Sun and the Moon. They both have an angular size of about half a degree.

That's why they look the same size, so the Moon can exactly cover the Sun.

oh

sir my next question

5:41 AM

There's no special reason for this. It's just chance.

Is there a big bang just before the big bang which created us?

To understand what physicists mean by the term Big Bang have a look at this:

206

Q: Did the Big Bang happen at a point?

TV documentaries invariably show the Big Bang as an exploding ball of fire expanding outwards. Did the Big Bang really explode outwards from a point like this? If not, what did happen?

general-relativity cosmology big-bang black-hole-thermodynamics

What are the other theories regarding origin of universe?

5:58 AM

There aren't any theories really. Just various speculative ideas that have nothing much in the way of experimental support. Right now we have to admit we have no idea of what actually happened at the Big Bang.

1 hour later…

7:10 AM

you mean that this theory is baseless

7:38 AM

@Akash.B I think what John means is that there is empirical evidence in support of the Big Bang but no one really knows what actually happened during it

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any one can help me with this?

ok nvm I got it

:/

Study the 1st book of 12th Physics. It usually has a high weightage in Mains if I remember correctly

Are the board exams over?

yah @Sid

actually these days i'm just mock tests only :)

If board exams of all subjects are over, then, you are more likely to get a tougher paoer

well let's see...

don

't scare me :(

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7:50 AM

Obviously. Our Mains paper was in the middle of the exams. So,we got an easier paper. If you have one week time after the exams, then, a hard paper is waiting for you

if we cut the string, what will be the angular acceleration of the rod?

@Sid hmmmm. nvm, it's too late to think of these things

if we take torque about CoM due to tension in string, we get a different angular acceleration. if we take torque about point B due to weight (mg), then we'll get different angular acceleration. but our teacher said that angular acceleration is the same from wherever you apply it. so, why isn't this the case here?

(take rod to be AB from left to right; not shown in diagram)

@GaurangTandon don't you have to consider the moment of the Tension T as well?

@Sid in case 2, moment due to tension is zero about B

and string was but at point A

(sorry i'll be back in 10mins)

Okay, I didn't know which string was to be cut. Either way, you simply use Torque=I times angular acceleration, no?

Also, if you are cutting the string, then you would have to take the pseudo force into account.

$3g/2l$ ?

8:04 AM

It is usually advisable to take Torque about COM do that you don't have to worry about pseudo force and stuff

@Tanuj yep that's right

@Sid yah I know but in this case it doesn't give the right answer :(

@GaurangTandon what is the right answer ?

@Tanuj this is the right answer

How would we even do it about com ?

why can't we? ang. accn must be same about all points

8:09 AM

@GaurangTandon yea , I'm asking how would we

That makes no sense. That's what I have learnt in the last 3 years

How do we not get the right answer?

@Tanuj T=mg/2 just before cutting at the point A.

oh ok, I got it

Tension at point B is probably changes after string at A is cut

but why?

yea , must be complicated.

sid also said we don't need to consider pseudo forces about com so idk what to do.

@JohnRennie can you help us out here?

we're stuck :(

Also @GaurangTandon where are you doing these from ?

8:11 AM

@Tanuj mock papers online

@GaurangTandon what website ?

@Tanuj al**n

;)

have you paid ?

no, i'm expected AIR 1, so they gave it to me for free.

of course I paid (-_-)

allen test series looks a lot different ! That is not allen i'm sure

8:13 AM

@GaurangTandon I keep forgetting how to do these rotating rod type questions, but as I recall you need to take into account the fact the rod moves downwards as well as rotating.

Isn't there one guy who has that allen series ? Madhu something

@Tanuj well idk; this is how it looks on my screen :/

you may be using the mobile app

@JohnRennie that's what I have tried and apparently the answer is wrong

@GaurangTandon okay cheers

@JohnRennie ah I didn't try that; lemme see again

@Sid what did you get?

8:17 AM

@GaurangTandon I didn't solve but it's obviously not 3g/2L

yep @Sid; @JohnRennie by that I got 3g/l instead :(

Apparently the torque due to the pseudo force has to be zero... But I am not quite sure why..

@Sid Did you not say torque due to pseudo force about com has not to be taken into account

@Tanuj that's because pseudo force acts on the COM (Or so I have been told).

But, if you are calculating torque about B, then, torque due to pseudo force exists.

@Sid okay , the answer is wrong then ? Nvm , I'll just leave it

8:32 AM

I don't know. It will bug me now for the rest of the day

yep same here

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@GaurangTandon @Tanuj halp I have my paper tomorrow and I'm freaking out

@AvnishKabaj i'll get back to you in half hour, probably with a solution

@GaurangTandon That would save me from cardiac arrest

@AvnishKabaj what paper?

Madhuchhanda Mandal

9:14 AM

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Can anyone explain me the solution?

They are using the formula of R.I= Sin (A+d)/2 ÷ Sin (A/2)

I dunno how you are supposed to find the exact value without a calculator but most probably the answer is either the highest value of the options or the lowest value given in the options

9:38 AM

@JohnRennie Suppose we have two unit +ve charged particles placed at a certain distance apart. Now a third negative charge -Q is placed midway. Now for what values of Q the three particles will be at equilibrium? (I know to balance the electrostatic forces, if the separation is given.)

Anonymous

@AvnishKabaj Is the answer to Q.48 (D)?

Anonymous

Everything's just like compound pendulum- If that's the answer.

@RaviPrakash Q would just be -1/4, no?

@RaviPrakash Suppose the two charges are at $+d$ and $-d$ with the third charge at the origin.

@JohnRennie Okay

Then?

9:42 AM

Now just calculate the force on one of the unit charges - the force on the other charge will be equal and opposite from the symmetry.

And then just set that force to zero and see what the resulting equation looks like.

yes, It would be interesting

@JohnR any ideas you have on that rotating rod problem? I can't sleep right now because of that question

@DivyankaS.Chaudhari yes

@Sid fiitjee workbook

@DivyankaS.Chaudhari Salooshan?

@Sid I had a quick go at it and couldn't get the correct answer. With this type of problem there's a right way to approach it, and when you know the correct approach the problem is straightforward. The trouble is that I can't remember the right approach ...

@AvnishKabaj oh, duh.

9:47 AM

@RaviPrakash are you OK with the problem now?

@JohnRennie |Q| = 1/4.

Anonymous

@AvnishKabaj Just write the torque equation, then torque= I*\aplha. Find \alpha. Then from that find \omega and then time period.

@RaviPrakash Correct! :-)

Anonymous

@AvnishKabaj Not sure about the smooth one. Smooth surfaces in rotation always confuse me.

But doesn't the system will be in equilibrium for all values of Q? @JohnRennie

9:50 AM

Ok

@DivyankaS.Chaudhari thanks

@RaviPrakash No. Suppose you make Q very small i.e. so it's effectively zero. Then the two charges of +1 repel each other and accelerate away. So they aren't in equilibrium.

@RaviPrakash for further practice, try to find the position of equilibrium if the two given charges are of same magnitude but are of opposite sign. You can take the third charge to be of any magnitude and of any sign.

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@JohnRennie see

That line doesn't mean that the system is in equilibrium

Yes, as I balance the forces on $ q $ it becomes a true statement

10:00 AM

@RaviPrakash t's good to develop some physical intuition about problems like this. If q can have any value then it can have the value zero. Yes?

Yes.

But if it has the value zero then we just have two positive charges, and that system cannot be in equilibrium.

I think what the book means is that the central charge is always in equilibrium because the forces on it from the other two charges are equal and opposite.

Exactly.

But the question asks abut the system of all three charges i.e. all three charges have to be in equilibrium not just the central one.

@JohnRennie the question in the book doesn't ask that

10:03 AM

@JohnRennie Yup

hi

@Sid I know

26 mins ago, by Ravi Prakash

@JohnRennie Suppose we have two unit +ve charged particles placed at a certain distance apart. Now a third negative charge -Q is placed midway. Now for what values of Q the three particles will be at equilibrium? (I know to balance the electrostatic forces, if the separation is given.)

:)

Did anyone have heard about upcoming earthquake in Delhi?

@Sid ah, OK, ISWYM. I thought we were still talking about Ravi's original question.

10:06 AM

@RaviPrakash what?

There is a social (maybe fake) viral message saying that

I don't think earthquakes can be predicted beforehand.

why universe is keep'in on expanding ?

@Sid that question ...

I've just had a closer look at it, and isn't it obvious?

hmm

10:09 AM

Or am I missing something?

@JohnRennie more likely that I am missing something

Let me draw a diagram ...

@JohnRennie sir?

@Akash.B because that's what was observed by scientists and is the only explanation to some observations

@Akash.B you asked that a while ago and I linked you to an answer on the PSE explaining it ...

10:11 AM

@Sid so space isn't really expanding

okay sir I hope you will give me an exact answer

@Akash.B just read the wiki article bro.

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he rod is rotating about the left end, so $I = \tfrac{1}{3}mL^2$

scientificamerican.com/article/…

okay then it is because of big bang

@JohnRennie okay

The torque is $mgL/2$, so we get: $$ mg\frac{L}{2}= \tfrac{1}{3}mL^2\alpha$$

10:16 AM

@JohnRennie but, wouldn't there be a pseudo force so that translational acceleration is 0? Then, what about torque due to that pseudo force?

We're working in an inertial frame so there are no pseudo forces

Huh...

How did I miss that for the last 2 hours?

10:30 AM

@Sid what's the question

@Sid True

That was a fake message

google.co.in/amp/zeenews.india.com/india/…

This confirms it.... Yeah!

@DivyankaS.Chaudhari In which state you're? currently ?

Anonymous

10:47 AM

@RaviPrakash Maharashtra. Lol why?

11:29 AM

hi

11:46 AM

@JohnRennie please have a second look once more, it would help clear our concepts :)

2 hours ago, by John Rennie

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oh missed that

lemme read

wait, I meant the other one

torque due to tension about CoM

that gives a different answer

@GaurangTandon link?

4 hours ago, by Gaurang Tandon

if we take torque about CoM due to tension in string, we get a different angular acceleration. if we take torque about point B due to weight (mg), then we'll get different angular acceleration. but our teacher said that angular acceleration is the same from wherever you apply it. so, why isn't this the case here?

@JohnRennie ^^

@GaurangTandon that would mean working in a non-inertial frame, which is going to complicate things ...

12:02 PM

oh, but how much? :(

Too much :-)

:P

alright I've totally dropped that idea

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@GaurangTandon just in case you still want it

In the com frame

yay

@AvnishKabaj what's that third equation, though, from the top?

12:18 PM

Subtituted the right most equation into equation number one then multiplied the entire equation by l/2 then added this new equation to equation number 2

totally made my day

and I'm sorry I couldn't get that solution for hollow cylinder yet

I've asked my teacher to help me out

but he hasn't replied yet, i've seen this question before

but can't seem to recall exactly

No problemo

1:16 PM

"Entropy change of surrounding in spontaneous adiabatic process is "

I think this should be zero (reversible adiabatic) or more than zero (irreversible adiabatic). any takers?

2 hours later…

2:58 PM

@GaurangTandon zero for reversible adiabatic

@Abcd yah that's also what I said

so thanks

i was just confirming

$\Delta S = nC_v \ln \left(\dfrac{T_2}{T_1}\right)+ nR \ln \left(\dfrac{V_2}{V_1} \right)$

@Abcd this is for system

question asks for surrounding

@GaurangTandon I am thinking.

$dS_{system}= dS_{produced}+ \dfrac{\delta q_{rev}}{T}$

The entropy of a closed system must increase in an irreversible
adiabatic process:

user image

Anonymous

3:16 PM

@AvnishKabaj Seeing some discussion, I thought that you didn't quite get how to reach the answer for your question. Here is the soln for Q.48

Anonymous

ibb.co/mGmEVS

@GaurangTandon Do you see how we have connected irreversible final states through reversible processes?

3 mins ago, by Abcd

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reading

We first consider the special case of an adiabatic irreversible process in a closed system.
This special case will lead to the desired general result. Let the system go from
state 1 to state 2 in an irreversible adiabatic process. The disconnected arrowheads
from 1 to 2 in Fig. 3.10 indicate the irreversibility and the fact that an irreversible
process cannot in general be plotted on a P-V diagram since it usually involves nonequilibrium
states.
To evaluate $S_2 - S_1= \Delta S_{syst}$, we connect states 1 and 2 by the following reversible

$S_3 = S_2$

@Abcd what's this? from where? and it doesn't seem to be a complete passage?

" temperature of a certain heat reservoir." where did this come from?

3:23 PM

We next either add or withdraw
enough heat q3→4 isothermally and reversibly at temperature $T_{hr}$ to make the entropy
of the system equal to S1. This brings the system to state 4 with $S_4 = S_1$.

@GaurangTandon Isothermal process = Heat reservoir = Const Temperature

${S_4- S_3 = \dfrac{q_{\ce{3 -> 4} }}{T_{hr}}}$

Since S is a state function, we have for cycle $\ce{1-> 2 ->3 -> 4}:$

$$0 = \int dS_{sys}= (S_2-S_1)+(S_3 - S_2)+ (S_4 - S_3)+ (S_1 - S_4)$$

Anonymous

@AvnishKabaj Please ignore previous link. New link- preview.ibb.co/nH2k4n/1522510163564_622808512.jpg

$\implies S_2 -S_1 = -\dfrac{q_{\ce{3->4}}}{T}$

Madhuchhanda Mandal

@jee_aspirants What do you do to avoid mistakes such as : 1-2 = 1 and such silly mistakes ?

@MadhuchhandaMandal let me know once you get it.

@MadhuchhandaMandal to be honest ,you can't actually. At the end of the day, you will surely do something like that in the exam hall.

I guess, you can just try to double-check your calculations...

3:34 PM

@Sid honestly , I would need a time limit of 5 hours for the exam. I can't finish the paper in 3 hours as is.

Madhuchhanda Mandal

I get some more time because I don't attempt chemistry :-D

@MadhuchhandaMandal lol really ? Why ?

Madhuchhanda Mandal

Still end up doing such mistakes :/

@Tanuj My target marks is 190 and I don't know much of Chemistry

Except physical

@MadhuchhandaMandal wow ! I wouldn't even expect 120 with no chemistry

what college are you in btw ?

Madhuchhanda Mandal

XD

3:37 PM

@MadhuchhandaMandal Look for 250, else, you will end up here or somewhere similar if you don't do advanced well...

Madhuchhanda Mandal

@Tanuj State Government college Kalyani Government Engineering College

@MadhuchhandaMandal okay , so you're looking to get into an NIT ?

Madhuchhanda Mandal

Yeah

@MadhuchhandaMandal me too. Hope you get there ! Gotta go .

Madhuchhanda Mandal

@Tanuj 7days to go !!!

3:40 PM

@MadhuchhandaMandal I know right !

cheers mate

Ha! I am glad I don't have these worries/excitement anymore..

Madhuchhanda Mandal

@Sid Hahaha !!!

IIT - G right?

Hm?

Anonymous

But these days never come again, such competition. It's pretty exciting. All the best everyone for the mains!

@DivyankaS.Chaudhari And these days are the ones which take away almost all motivation and hunger among undergrads

Madhuchhanda Mandal

3:46 PM

@Sid You are studying in?

Almost all my friends in college are not motivated because of the rigorous JEE Preparation they underwent.

Anonymous

I see, didn't know that. Yet to experience.

It's like the "exam is over, I can't study anymore attitude".

@MadhuchhandaMandal NITRKL.

Madhuchhanda Mandal

@Sid And that's eating the quality of Btech Grads

@MadhuchhandaMandal Yep. More than 70% of the students just go through the motion, hoping that a degree would get them a job.

@DivyankaS.Chaudhari You're appearing for JEE this year?

Anonymous

3:51 PM

Nope, next year.

Madhuchhanda Mandal

Being at RKL , You can imagine our situation... I am eager to study but no facility available.. Neither from the faculty's end nor are the peers interested

@DivyankaS.Chaudhari oh well, enjoy life. If you get the time away from JEE Preparation. :P

@MadhuchhandaMandal Don't expect anything from the faculty. Profs don't give a damn whether you know about the subject or not. If you want to read something, look it up on coursera or any other online site

Anonymous

Haha, I do. I do everything. I'm like jack of everything, master of nothing.

Madhuchhanda Mandal

I don't get my doubts cleared. If I ask them something either they avoid it or like give some made up explanation which they themselves know is wrong

@DivyankaS.Chaudhari that's actually good. You need to be decent in all the subjects to do well in JEE. Being very good in one subject and knowing nothing in the other won't help

@MadhuchhandaMandal Oh, well. I guess, our profs are a bit more helpful than that

Madhuchhanda Mandal

3:58 PM

Must be.

@Abcd i totally didn't get this sorry! :/ but it's so vague I wouldn't worry about asking on main

i'll rather say pass...

and move on

@DivyankaS.Chaudhari thanks

I got the solution for both tho

@GaurangTandon its given that way in Levine

very complicated language

yah :(

@GaurangTandon this same question is eating me since months

1

Q: How would one calculate the entropy change for an adiabatic irreversible process?

If two states $A$ and $ B$ are connected by a reversible path, they can never be connected by an irreversible path during an adiabatic process because: $\underbrace{\Delta U}_{\text{state function}} = \underbrace{\Delta W}_{\text{path function}}$ And $W_{\text{reversible}}\neq W_{\text{irrever...

Madhuchhanda Mandal

4:04 PM

@Abcd which one?

@MadhuchhandaMandal Entropy change of System and Surrounding in Adiabatic Irreversible Process

Anonymous

@AvnishKabaj @AvnishKabaj What's the second one's soln?

@GaurangTandon Levine is also defining an irreversible path between the two states

Anonymous

@Sid @Sid Yep!

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@DivyankaS.Chaudhari energy conservation since no friction

4:07 PM

@Abcd oh well :(

Anonymous

@AvnishKabaj Got it, Thanks!

@AvnishKabaj energy is also conserved in the first case because static friction does no work

Anonymous

That's right @GaurangTandon so , is there any flaw in my solution?

Anonymous

Because, then the answers would be different

i didn't read it yet

but you haven't applied energy consv., so why are you worried?

Anonymous

4:14 PM

Yeah, I have written the torque equation directly and solved.

Anonymous

No, I'm not worried, but your point is right as well. If we apply energy conservation, the answer would be same in both questions. But, it is not so.

i'm really confused

:/

Madhuchhanda Mandal

@Abcd Although Reversible adiabatic process won't exist between A and B but Infinite number of Reversible process would exist between the two points. And for entropy, process doesn't matter.So you can pick one of the processes and calculate it. Yay !! That's the summary of the discussion.

@GaurangTandon There's some issue with it

It was getting very difficult to write the energy

user image

Since, you guys are in-touch with mechanics, mind helping me out with this problem?

4:24 PM

@Sid if you don't mind, i'm just curious, why are you still solving these questions in college?

I know, we can draw FBDs, but does the application of P, affect the motion of A?

@GaurangTandon Part of Coursework, bro

oh!

A subject called, "Engineering Mechanics"

You will read that in College...

@AvnishKabaj answer is (2) for question 48?

@Sid hmmm ok

Hm? Someone pinged me here...

Madhuchhanda Mandal

4:31 PM

user image

@Sid Yeah The graphs..

I removed the previous one cause it contained an error

user image

Anonymous

@MadhuchhandaMandal Could you please mention the values of P where the graphs change? Because, I was trying to draw and got confused.

I just wanna verify the answer for this, my answer is coming 1, don't need the solution though

@GaurangTandon no

It's (D)

@AvnishKabaj :(

and for q49?

4:37 PM

I know this one was a toughie

@GaurangTandon (A)

Madhuchhanda Mandal

@DivyankaS.Chaudhari Okay wait

27 lb

(80*10.2*180/g) is the second value of P

(I don't know g in FPS system)

@MadhuchhandaMandal 32 ft/s/s

Anonymous

@MadhuchhandaMandal 4590?

Anonymous

4:52 PM

'_'

@DivyankaS.Chaudhari Okay Maharastra. Why lol?

Madhuchhanda Mandal

@DivyankaS.Chaudhari Yeah so that point is beyond 60lb. So we don't need to consider that point

(P/180) * (M80) = (W80 * Static friction coeff of 80)

And M80 = W80 / g

Anonymous

5:15 PM

@RaviPrakash Sorry, just got habit of writing that anywhere because you asked suddenly

Anonymous

@MadhuchhandaMandal I think I get it. I'll process this when my brain starts working.

Anonymous

Thanks tho!

Madhuchhanda Mandal

Welcome !!

user image

Is there any short process for these types of problem?

5:36 PM

@MadhuchhandaMandal yeah , "Tukka Tareka"

Madhuchhanda Mandal

(I mean without Calculating individual heat transfer and work done)

@Tanuj really?

5:50 PM

Hello I have a question that can unit of torque be joule?

@Madhu according to conventions, we use that efficience is net work done over heat supplied to system. ONLY take into account heat supplied.

@KingTut NO

Why sid

Joule = Newton metres = torque, is it wrong because joule is for energy only historically speaking

Tutankhamun , right ?

@KingTut Same dimensions doesn't mean same that you can interchange units.

That's like saying the unit of Power should be Volt-Amperes instead of Watt.

(Unit of Power can be Volt-Amperes but that's not for you guys to know right now)

Tutankhamen, Tutankhamun, Tutankhaten, I go by many names

5:56 PM

@KingTut Joule is also named after the Scientists and his work was based more on Energy and so that's what the International Standard is.

@KingTut sweet :)

Madhuchhanda Mandal

@KingTut Yeah .. How do we find by just looking a process that heat is supplied or rejected ? (Ofcourse without actually Calculating)

(And if that's not possible then it will take same time as individual calculations for each process)

@Madhuchh I think process is ABCDA but arrow is not proper, anyway work net is integral of pdv which is area of box, and its easy to predict where heat may be positive or negative

6:59 PM

@KingTut I know that....But in my 10th grade book it was specifically mentioned that don't use Joule for N-m.

Even I wondered why...

I think its only for consistency in physics atmosphere.

We intuitively associate joules with energy.

That's the reason I guess

[Question:]

A circular loop of string rotates about its axis on a frictionless horizontal plane at a uniform rate so that the tangential speed of any particle of the string is $v$. If a small transverse disurbance is produced at a point of the loop, with what speed relative to the string will the disturbance travel on the string

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Mar '1831

Problem Solving Strategies

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