Problem Solving Strategies

2:10 AM

@AvnishKabaj Maybe it was to check whether students can adapt to new things.. in various decimal types also they gave new values for constants..

Anonymous

@Abcd Don't know. But it doesn't matter.

Anonymous

@Jasmine You'll never know unless you join. It depends from centre to centre. Our centre isn't thaat good.

Anonymous

@samjoe I think that's a brilliant idea to stump those who don't read everything :P

samjoe

2:25 AM

Yeah exactly, in one question I completely missed that they had given $R=8$. I proceeded with my calculation and spent some time, then I saw that value. I think it was done to make calculations simpler

Many coachings have released answer keys. I was just looking over at allen's : link

Abcd

5:18 AM

Based on the compounds of group 15 elements , the correct statement(s) is (are)
(A) Bi2
O5 is more basic than N2
O5
(B) NF3 is more covalent than BiF3
(C) PH3 boils at lower temperature than NH3
(D) The N–N single bond is stronger than the P–P single bond

This is directly from NCERT @samjoe

John Rennie

5:34 AM

Morning :-)

Ugh, chemistry! :-)

Abcd

@JohnRennie Hey, I have one or two questions. Are you free?

starunique2016

yeah directly

Sid

@JohnRennie Don't you have a PhD in Chemistry? :P

samjoe

@Abcd yeah

Abcd

@samjoe But chemistry of your paper looks tough. I dont know why

There are so many "many reactions in one" type questions.

samjoe

5:49 AM

To me whole paper was tough just because of decimal type. Other than that paper was as easy as last year

starunique2016

inorganic is i think always from ncert

in some cases jd lee has to be preffered

John Rennie

@Abcd yes I'm around for a half hour or so ...

Sid

@samjoe Did you do better than last year? In terms of marks?

Abcd

@JohnRennie I am done with this question and have got the right answer. But I did it using 4 equations (of KVL and charge conservation), is there any symmetry hidden here or are there any other tricks to solve it?

Sid

And that Physics paper is a disgrace for IIT Exams...

John Rennie

5:58 AM

@Abcd What are you trying to calculate? $V_{ab}$ ?

samjoe

@Sid I cant check the exact marks, I only checked the options from answer key based on memory. I think I have lost quite some marks in decimal types, although i spent most time on them.

Abcd

@JohnRennie yes, I did.

@Sid @samjoe I think they are purposely setting easy papers now to demolish the coaching monopoly which they want to .

samjoe

@Sid Yeah they some questions were not even mains level. For example the unit of electric field, magnetic field relation. So many of that kind

John Rennie

@Abcd Let me draw a quick diagram ...

Abcd

@JohnRennie Okay

All the Kota teachers are always like "the tougher the paper, the better it is for our students"

samjoe

6:01 AM

@Abcd That is actually not good, be it coaching students or non coaching, those who prepare will suffer. Easy papers means those with mediocre level of preparation also get decent marks.

All that matters then is speed.

John Rennie

@Abcd Isn't it just this?

Sid

@JohnRennie I think Abcd is asking if there's some shortcut rather than the whole solution of the problem

Abcd

@JohnRennie I did it using the fact that $V_{ab}= V_{C_4}$

@Sid yes exactly.

I dont think there's any shortcut then...

John Rennie

I don't think there's a shortcut, but you don't need to use KVL

Abcd

@JohnRennie So, how will you do it?

samjoe

6:08 AM

Yeah just use series parallel. Also that $V\propto C^{-1}$

John Rennie

Just calculate the combined capacitance of C2, C3 and C4 - call this C234 - then you have C1 and C234 in series so you can calculate V234.

And that's the voltage across C3 and C4, so now you can calculate V4.

Abcd

@samjoe How to use $V\propto C^{-1}$

John Rennie

@Abcd suppose you have two capacitors C1 and C2 in series.

Abcd

@JohnRennie okay

Q will be same on both

Oh just V_1C1 = V_2C2

Sorry.

John Rennie

@Abcd :-)

Abcd

6:12 AM

samjoe

@Abcd in JR's diagram, if $V$ across $C_3C_4$ is $V$ then $V$ across $C_3$ is $\frac{VC_4}{C_4+C_3}$, same that you wrote

John Rennie

@Abcd Cripes, I got this wrong last time I tried it ...

Abcd

@JohnRennie The capacitors have dimensions $l\times l$, Initially, l/2 length of the dielectric of length l is present "inside" the capacitor. Find the time period of oscillation and the heat generated in the process of 1 oscillation.

@JohnRennie Who had posted this problem earlier?

John Rennie

You need to work out the energy as a function of the displacement $x$ of the block of dielectric from the centre.

samjoe

@Abcd I think we did it using $F = -dU/dx$

Abcd

6:14 AM

@JohnRennie I have it.

John Rennie

If you work out this energy then differentiate it you get the force pulling the dielectric towards the centre

Abcd

I have that too

John Rennie

And you'll end up with a simple harmonic motion.

Abcd

No its not simple harmonic.

@JohnRennie Force on the slab is:

Sid

@JohnRennie When's the next most probable year of a Royal Wedding in UK?

samjoe

6:15 AM

Yeah its not simple harmonic, we had to remember this force actually

Abcd

$$F= \dfrac{- \epsilon_o b \epsilon^2 (k-1)}{2d}$$

Where $\epsilon$ is the emf of battery

and b is the width of the plate

John Rennie

Just a constant force then i.e. independent of the position of the slab?

Abcd

The dimension of the plate are $l\times b$*

@JohnRennie yes

@samjoe do you know that this force exists because of edge effect?

John Rennie

Ah so the time to reach the centre is just $$ \frac{\ell}{2} = \tfrac{1}{2}\frac{F}{m} t^2 $$

Abcd

@JohnRennie I am done with the calculation of time period T = 4t

John Rennie

6:18 AM

Multiply that by 4 to get the period

Abcd

how to calculate heat generated?

samjoe

@Abcd Know nothing about that

Abcd

@samjoe You never thought about how this horizontal force exists

samjoe

@JohnRennie If its constant force then how will motion be periodic, i cant understand that

John Rennie

@samjoe the force changes direction at the mid point

samjoe

6:19 AM

@Abcd That is not edge effect exactly, i think its polarisation

@samjoe its both.

@JohnRennie oh yeah

@JohnRennie how?

Its pulling it in :P

@Abcd the potential is proportonal to $|x|$ not $x$.

Abcd

6:22 AM

@samjoe the direction change is not obvious from the force equation.

@JohnRennie How does potential come here?

John Rennie

@Abcd potential energy I mean i.e. $U(x)$

samjoe

3 mins ago, by Abcd

@samjoe You never thought about how this horizontal force exists

Abcd

@JohnRennie so?

@samjoe ?

samjoe

I was not at all talking about force equation. You can see that plates are attracting the dielectric.

Due to polarisation

@Abcd @John What is edge effect exactly and what role it plays here?

Abcd

@samjoe Look: because of edge effect, field lines are curved at the edges, as a result, there vertical components cancel and there horizontal components add to provide horizontal force

John Rennie

6:25 AM

The force is $-dU/dx$ i.e. the gradient of the line.

samjoe

Nothing about the polarisation? Even with your argument we can see that force is inward always.

@abcd ^

Abcd

@samjoe what do you mean

6 mins ago, by John Rennie

@samjoe the force changes direction at the mid point

@JohnRennie You mean the force changes direction when the entire slab is inside the capacitor?

John Rennie

Yes

samjoe

@Abcd by inward I mean that its always towards $x=0$

Abcd

@samjoe yes nothing about polarisation

John Rennie

6:28 AM

@Abcd Well, it's zero at the precise moment the slab is exactly between the plates. But it changes sign either side of that centre point.

I'm out of time. Back in a bit ...

Abcd

@samjoe How is potential energy zero at midpoint?

samjoe

@Abcd how can that be, there is polarisation that should also contribute

Abcd

@samjoe I think Potential energy is maximum when the entire slab is inside

Hey @Farcher :D!

It even makes logical sense that most PE should be there when entire slab is inside, coz of most no. of induced charges and their contribution to PE

Anyway, yes, the force changes direction at the centre.

samjoe

But that doesn't go well with F = -dU/dx

Abcd

How to calculate the heat generated?

@samjoe What? Why?

samjoe

6:39 AM

F should be negative for $x>0$ right

Abcd

@samjoe x is the amount of slab inserted inside

$U = \dfrac{\epsilon_o b[l+(k-1)x]\epsilon^2 }{2d}$, where x is the length of slab inside.

Its maximum when full slab is inside.

samjoe

Oh, it was never mentioned so I took it as something else

Abcd

@samjoe Is this potential energy function even continuous?

samjoe

I am giving a bitsat mock, ill try after some time

John Rennie

@Abcd $U(x)$ is continuous, but $dU/dx$ is discontinuous at $x=0$

Abcd

6:46 AM

@JohnRennie exactly, what does it imply?

John Rennie

@Abcd What are the axes of your graph?

Abcd

@JohnRennie Y axis is potential energy and x axis the value of x which is maximum when the blue line has its maximum (i.e. when the whole slab is inside)

John Rennie

$x$ is the displacement from the centre point i.e. $x=0$ when the slab is exactly between the plates?

Abcd

@JohnRennie This is x^

John Rennie

That's not a helpful way to define $x$ ...

Abcd

6:51 AM

Is this the graph of PE^

x is the length of slab inside.

John Rennie

1 min ago, by John Rennie

That's not a helpful way to define $x$ ...

Abcd

@JohnRennie my teacher did it that way :/

John Rennie

You should always try to take advantage of symmetry

Abcd

hmm

What do I do now @JohnRennie?

How to calculate the heat?

John Rennie

If you define $x$ as the distance of the centre of the slab from the centre of the plates then the system is obviously the same for $+x$ and $-x$ as they're related by a reflection about the centre of the plates.

Abcd

6:57 AM

@JohnRennie yes

John Rennie

@Abcd you can calculate the capacitance as a function of $x$ (however you define $x$) so you can calculate the energy stored on the capacitor $\tfrac{1}{2}CV^2$.

Abcd

17 mins ago, by Abcd

$U = \dfrac{\epsilon_o b[l+(k-1)x]\epsilon^2 }{2d}$, where x is the length of slab inside.

Should we start from scratch @JohnRennie, defining x the way you want to

John Rennie

No, just carry on.

Abcd

Okay so:

Work of battery force = $\Delta U + \Delta KE + \Delta Q_{heat}$

$v^2 = Fl/m$

$\Delta KE = \dfrac 1 2 mv^2 = \dfrac{\epsilon _o bl \epsilon^2(k-1)}{4d}$

$\Delta q_{charge}= ((C_f-C_i)\epsilon (k-1)\epsilon_0 lb)/(2d) $

I am not getting the right answer on paper.

I am getting heat = 0

which is wrong obviously.

John Rennie

I must admit I would have guessed the heat was zero just on general principles.

Abcd

7:04 AM

@JohnRennie battery is doing work ...

John Rennie

If the heat is not zero that means energy is being lost as heat, and the oscillation would be damped.

Abcd

current would be flowing

and that would generate heat

22 secs ago, by Abcd

@JohnRennie battery is doing work ...

John Rennie

The battery does work for half the cycle, but for the other half of the cycle the battery has work done on it.

The net work done by the battery is zero

Abcd

but every time current flows

heat is generated

John Rennie

Right, but we're assuming an ideal battery and wire so $R=0$

The heat is $I^2R$ so if $R=0$ the heat is zero.

Abcd

7:06 AM

@JohnRennie no, my teacher said that $R\ne 0$

he said "you dont need R for this question"

John Rennie

In that case I don't understand the question

Abcd

1

Q: Heat produced when dielectric inserted in a capacitor

When a capacitor is connected to battery, it stores $\frac{C V^2}{2}$, while battery supplied $CV^2$ energy. Therefore, $\frac{C V^2}{2}$ energy gets lost as heat. When a capacitor is already charged and a dielectric is inserted in this charged capacitor (which is still connected to the battery),...

I found this right now^

Jasmine

7:54 AM

@Abcd was it taught to you? what is point 4 I can't understand. will it come later in the chapter?

Can anyone help me understand this

John Rennie

@Jasmine this is related ...

5

A: What is the force between two charged objects when the space between them is partially filled by a dielectric medium?

A dielectric effectively behaves as if it was thicker than it is. If the dielectric constant is $K$ and the thickness of the dielectric is $t$, then for calculating the force it behaves as if the thickness was $t\sqrt{K}$. To see this let's take the example we know about where the dielectric fil...

Jasmine

@JohnRennie such a simple explanation thanks

Abcd

Jasmine

Now I know what they meant by equivalent distance

Abcd

@JohnRennie How to find Va- Vb

Principle of Superposition gives 0

But answer is-8 V @JohnRennie

@Jasmine @Sid Any idea?

John Rennie

8:08 AM

8V?

Abcd

@JohnRennie $-8 \pu{~V}$

John Rennie

That has to be wrong

The circuit is symmetric about the horizontal plane i.e. reflecting in the horizontal plane doesn't change it.

That means $V_a = V_b$

Sid

@Abcd that's wrong

Abcd

@Sid Okay, this is from HCV and it looks like a misprint

@JohnRennie Replace the lower battery by 24 V one. My new edition of the book has a misprint, the older editions have the correct data.

John Rennie

OK so superposition gives $V_{ab} = 16 - 8 = 8$ (give or take a sign)

You've got the hang of superposition then? I'm surprised it isn't taught at JEE level as it's so useful.

Abcd

8:14 AM

@JohnRennie It can be solved using equations of KVL too...

John Rennie

@Abcd Superposition is easier though :-)

Abcd

@JohnRennie I'd just like to see how you did it using superposition.

Sid

@JohnRennie when there are no current sources. I dislike current sources

John Rennie

@Sid true ...

Akash. B

8:36 AM

@JohnRennie sir, can I share my view about quantum theory

John Rennie

@Akash.B that isn't really what this room is for ...

Sid

@Akash.B maybe post it on the H bar?

Akash. B

@Sid any other rooms

Avnish Kabaj

The h Bar

General chat for Physics SE (physics.stackexchange.com). For M...

6

Akash. B

8:56 AM

@AvnishKabaj they use to avoid me because I don't know anything

John Rennie

@Akash.B you give the impression that you just want to chat, and physics is just an excuse to chat. There's nothing wrong with that, but this room isn't for general chat. The h bar tends to be looser, but most of the people that hang out there are pretty expert so they usually want to chat about quite advanced stuff.

Abcd

9:25 AM

@JohnRennie I am unable to use symmetry arguments in questions like these:

Why should both ones and both threes have the same magnitude of charge each. It's just not obvious.

@JohnRennie Are you there?

John Rennie

I'm trying to think of a simple argument for this ...

There is symmetry because the circuit is centrosymmetric

samjoe

transverse

transverse symmetry

Abcd

@JohnRennie didn't get you

samjoe

Rotate it by 180 and you get same circuit.

Abcd

Okay,so?

John Rennie

9:32 AM

@samjoe ah, good point. I was thinking about inversion symmetry but yes it has a simple rotational symmetry.

samjoe

So charges are same on both 1 micro Farad have same charge and similarly for 3 micro Farad

@John I didnt understand how we use superposition in that earlier circular circuit example.. how we can add 24 V battery

Abcd

@samjoe one is negative side and other is positive side

John Rennie

@Abcd If you swap the voltages at the two ends you don't change the magnitudes of all the charges, you just change their signs.

Abcd

@JohnRennie @samjoe Also what if both ones were at top and both threes at bottom

samjoe

wheatstone :D

Abcd

9:38 AM

@samjoe will charges received by both ones be same

John Rennie

@Abcd that's now symmetric about the vertical plane

samjoe

yeah...

Hmm symmetry

samjoe

10:02 AM

@Sid do you know how to use circuitlab.com ? I cannot seem to simulate my circuit due to some error, which it is not showing

hmm nice tool

samjoe

1:21 PM

@koolman sup dude what are you doing these days

Sid

1:33 PM

@samjoe Haven't used it, no.

starunique2016

1:45 PM

Abcd

Can you tell me the date of the next aits?

samjoe

2:01 PM

@Sid Hm I think I found it I was finding voltage across capacitor in a circuit

Koolman

3:53 PM

@samjoe Nothing just enjoying holidays

Abcd

4:04 PM

@samjoe how is circuitlab.com? What do you use it for? Is it going to be of any help to me?

samjoe

@Abcd actually I am still figuring out how it works. It looks like it can solve electric circuits. I have tried few circuits, still to know how to make it work properly. Interface is really friendly though.

@Koolman Yeah but how are you enjoying.

Here in delhi its hot as hell. Temperatures remain 40-44 throughout noon, evening.

Sid

@samjoe Yeah, Delhi is weird. For some reason, you always have the urge to drink more and more water when you are there

Abcd

4:21 PM

@samjoe or @Sid can you now explain the symmetry of that question?

Sid

@Abcd which question?

Abcd

7 hours ago, by Abcd

@Sid this

Sid

That has Rotational symmetry, I think.

So, current across both 1 and both 3 will be same respectively

Abcd

@Sid But please explain why both 1s and both 3s have same charge

@Sid why

Nehal Samee

@Sid So , we can consider same charge $q_1$ for the capacitor $1 \mu F$ ...?

Sid

4:24 PM

I think you can solve normally by KVL and find that the charges will be same

I am trying to think of a good reason as to how to explain this...

Abcd

@Sid Okay... please let me know when you've got it.

Sid

Ah, Okay. Mark the capacitors as $C_1, C_2, C_3 and C_4$

Abcd

@Sid marked

Nehal Samee

@Abcd does this help m.youtube.com/watch?v=owT4WjwfY-A ? I haven't seen this although ...

Abcd

@Sid Are you there?

Sid

4:30 PM

@Abcd I am drawing you a diagram, just hold on, please

(Sorry, for the horrible diagram). Now, C1 and C4 have the same value and C2 and C3 have the same value

Rotate the whole thing by 180 degrees/

What do you get?

Hmm... This would be clearer to you if you draw it in a star-shaped thing..

Abcd

@Sid You mean DeltaStar conversion?

Sid

No. You don't need that for this..

Abcd

@Sid we get the same thing.

@Sid ?

@Jasmine ?

Sid

Look at this @Abcd . If you rotate it by 180 degrees, then ,C4 comes in place of C1, C2 comes in place of C3. Yes?

Jasmine

By mistake

Abcd

4:43 PM

@Sid Yes, then?

Sid

So, voltage source remaining constant, the current/charge that flows is now dependent on the equivalent capacitance of that network. Yes?

Abcd

@Sid okay, then?

Sid

But, we know C1 and C4 are same. And C2 and C3 are same. So, the equivalent capacitance remains the same. Yes?

Abcd

hmm

Sid

And since, C1 and C4 are same, the charge division that happened in the first case will be the same in the second case.

So, C1 and C4 have same charge going through them.

Similar is the case with C2 and C3.

Jasmine

4:47 PM

@Sid is this also called wheatstone bridge?

Sid

@Jasmine Yes. But, this is unbalanced Wheatstone Bridge.

Nehal Samee

@Sid Can we use ∆→Y transformation here ?

Jasmine

@Sid and balance if the ratio across them is same

Sid

@Jasmine Yup.

@NehalSamee Yup, you can do that

Although, I dislike doing Delta-star transformation for capacitors because I do it for resistors. :P

@Abcd Did you understand? Or do I have to explain more?

Abcd

@Sid not getting, just because it becomes the same thing how can we say both will have same charges

Sid

4:52 PM

@Abcd Okay. Let's think about the case, I drew first. The original one. Let charge across C1 be $q_1$ and that across C3 be $q_3$

Similarly, the charges across C2 and C4 being $q_2$ and $q_4$ respectively.

Then, if you solve the equations, you get some values of charges, yes?

Now ,the total charge in the system will be $q_1 + q_2$

Yes?

@Abcd Is this clear enough so far?

Abcd

22 mins ago, by Sid

@Sid yes

Sid

Okay, after rotation, Equivalent capacitance remains same. So, total charge will be same. yes?

Abcd

@Sid hmm

Koolman

@samjoe movies , games , friends ,internet etc

Sid

@Abcd Do you see why Equivalent capacitance will be same?

Abcd

5:00 PM

@Sid coz we are getting back the circuit we started with

Sid

Yup! So, Total charge will be same. That is, $q_1+q_2$

After Rotation, we have C4 instead of C1 in that position. But, the values are same. So, charge across C4 and C1 are same.

starunique2016

who will win?

tell me

csk or srh

Abcd

@starunique2016 srh i guess

@starunique2016 Are you jee 2019 aspirant from FIITJEE

starunique2016

nopes

yes

i was asking you the dates of aits

Jasmine

I am the biggest ever fan of DHONI

Abcd

5:06 PM

@starunique2016 they are not released

starunique2016

csk too lucky

i hate dhoni

Jasmine

I quite forgot about today's match

If you enculcate serenity from CAPTAIN COOL THEN YOU WILL SCORE BETTER IN JEE PAKKA

starunique2016

then i like dhoni

i am his biggest fan

stuff...stuff ...stuff

Abcd

@starunique2016 just out of curiosity are you a girl or a boy?

starunique2016

boy

Abcd

5:09 PM

ok

Jasmine

@Abcd I always wanted to know @Abcd are you a girl or a boy?

Abcd

@Jasmine boy

Jasmine

Ok

For finding equivalent capacitors do you use field method also

I mean it's not shortcut but do you have an idea

@starunique2016 cool

starunique2016

jasmine you must be a girl ,,,otherwise it would be very akward,,,

Jasmine

@starunique2016 ?

starunique2016

5:12 PM

csk wins

any onw up of coding...do the first one

jasmine just joking dont mind

Jasmine

Ok

I dint have any csc or ip as subject

starunique2016

Abcd

Sid

Damn. CSK and Last-Over Heists. Still a better love story than Twilight

starunique2016

anyone for the code...

Sid

@starunique2016 So, what do you want to know about that? As to how to write that code?

starunique2016

5:19 PM

can you please write the code.

or you may tell me the idea...i am very poor at coding

Sid

I am not writing the code. But, all you have to do is read the number and the location. And then, check whether the location is less than the size of the array. if yes, just insert the number in the location and push the numbers after it to the next positions. If no, Just append it at the end

Swapnil Das

9:06 PM

Anyone prepping for JEE should try out this site embibe.com

I just gave my test on Limits (Mains level). Got 83/120 (1 hour test)

The h Bar

Transcript for

Problem Solving Strategies