Problem Solving Strategies

harambe

12:09 AM

@sammygerbil you still awake

sammy gerbil

@harambe hello early riser!

harambe

Needed some help with Doppler effect

@sammygerbil imgur.com/a/yyhSQJe

In Q 66 is it asking for me to calculate the frequency change before and after the bullet passes

In Q68,would not remaining in the straight line with source affect the frequency

sammy gerbil

@harambe Q66 is cut off.

harambe

Q67 *

sammy gerbil

@harambe Q 68. I don't understand what you mean.

@harambe Q 67. Yes it is asking for the difference between before and after the bullet passes.

harambe

12:17 AM

@sammygerbil Yeah. It came out slightly confused to me

So in Q68, I would use the formulae of apparent frequency for both trsins and add it?

sammy gerbil

@harambe It asks for the frequencies heard, not the sum or difference. Use the formula to get the frequency heard from each train. You need to find the velocity of the train in the direction of the observer.

harambe

Okay. Got it

@sammygerbil to be equidistant from the two trains he should be 1.2 km apart. How can he be equidistant with 500 m?

sammy gerbil

@harambe He is 500m from the track. Not on the track. Draw a diagram.

harambe

12:34 AM

I thought the trains we're not in same track.Interpreted it wrong lol

harambe

8:39 AM

@JohnRennie hello. Good morning

John Rennie

@harambe morning :-)

harambe

I had a doubt about nuclear forces. Why is force stronger if spin of nucleons are parallel

John Rennie

That's a JEE question???!!!

harambe

Given in my textbook

Is it worth knowing at my level?

John Rennie

I don't think it's possible to answer that usefully at JEE level because you need to know about a property called isospin.

Anonymous

8:47 AM

@harambe I don't think so :P Which textbook are you using?

John Rennie

Isospin

In nuclear physics and particle physics, isospin (I) is a quantum number related to the strong interaction. More specifically, isospin symmetry is a subset of the flavour symmetry seen more broadly in the interactions of baryons and mesons. The name of the concept contains the term spin because its quantum mechanical description is mathematically similar to that of angular momentum (in particular, in the way it couples; for example, a proton-neutron pair can be coupled either in a state of total isospin 1 or in one of 0.). Unlike angular momentum, however, it is a dimensionless quantity, and is...

harambe

HCV

John Rennie

1

Q: Does the strong force increase or decrease with aligned spins?

The deuterium exists only with the proton and neutron of aligned spin, which suggests that the residual strong force is greated with aligned spins, i.e. the binding energy is greater if the spins are aligned. On the other hand the mass of $\Delta^{+}$ is greater than the mass of proton $p$, even...

That explains it, though I suspect the answer will be largely incomprehensible

Anonymous

@harambe HCV theory questions are usually answerable using just the text given in the corresponding chapter. Pretty sure you just need to remember it as a fact (that nuclear force is spin dependent and that's it) and don't require a detailed explanation.

Anonymous

@EshaManideep Neamen and Boylestad are good books

harambe

8:51 AM

Yea. It's all too for me

Esha Manideep

@Blue Thanks, let me have a look, what sid mentioned yesterday were also nice :)

Anonymous

@EshaManideep Why do you need it though?

Anonymous

For the transistors chapter?

Esha Manideep

@Blue Yep, transistors mainly

Anonymous

I wouldn't spend too much time on it, if I were preparing for JEE (it's not in Advanced, and even in Mains they will at max ask only one question from the transistors chapter). Pradeep Kshetrapal and Physics Galaxy lectures should suffice. Read NCERT well. :)

Esha Manideep

8:55 AM

@Blue Even I am not spending, I will finish it off today in 2 hours maybe ;)

harambe

9:14 AM

@JohnRennie imgur.com/a/bhXAU7D

I didn't understand the significance of this neutral atom in this derivation.. Why are they using it for here

Never mind. I got this

Nobody recognizeable

@JohnRennie you there ?

Nobody recognizeable

10:54 AM

@JohnRennie are you here ?

John Rennie

@Nobodyrecognizeable hi

Nobody recognizeable

@JohnRennie 11th.

@JohnRennie i suspect $c'= 2\pi R (1-\frac{R^2\omega^2}{c^2})$

John Rennie

That's a famous problem called the Ehrenfest paradox.

Ehrenfest paradox

The Ehrenfest paradox concerns the rotation of a "rigid" disc in the theory of relativity. In its original formulation as presented by Paul Ehrenfest 1909 in relation to the concept of Born rigidity within special relativity, it discusses an ideally rigid cylinder that is made to rotate about its axis of symmetry. The radius R as seen in the laboratory frame is always perpendicular to its motion and should therefore be equal to its value R0 when stationary. However, the circumference (2πR) should appear Lorentz-contracted to a smaller value than at rest, by the usual factor γ. This leads to the...

Nobody recognizeable

@JohnRennie if the coordinates aren't euclidean then what are they (lorenz coordinates)?

John Rennie

Born coordinates

In relativistic physics, the Born coordinate chart is a coordinate chart for (part of) Minkowski spacetime, the flat spacetime of special relativity. It is often used to analyze the physical experience of observers who ride on a ring or disk rigidly rotating at relativistic speeds, so called Langevin observers. This chart is often attributed to Max Born, due to his 1909 work on the relativistic physics of a rotating body. For overview of the application of accelerations in flat spacetime, see Acceleration (special relativity) and proper reference frame (flat spacetime). From experience by inertial...

But note that the spacetime is still flat. This is still special relativity not general relativity. It is the coordinates that are curved not the spacetime.

Nobody recognizeable

11:04 AM

@JohnRennie coordinates curved not spacetime?? Any related question on main site?

John Rennie

The Rindler coordinates are another example:

Rindler coordinates

In relativistic physics, the coordinates of a hyperbolically accelerated reference frame constitute an important and useful coordinate chart representing part of flat Minkowski spacetime. In special relativity, a uniformly accelerating particle undergoes hyperbolic motion, for which a uniformly accelerating frame of reference in which it is at rest can be chosen as its proper reference frame. The phenomena in this hyperbolically accelerated frame can be compared to effects arising in a homogeneous gravitational field. For general overview of accelerations in flat spacetime, see Acceleration (special...

But to understand what it meant by curved coordinates you really to need to have a sound grasp of differential geometry (which is the maths that underlies general realtivity).

Nobody recognizeable

@JohnRennie ok. I will have that in the bucket list.

John Rennie

Without this I doubt the analysis of the Ehrenfest paradox will be easy for you to understand.

Nobody recognizeable

@JohnRennie yep. But sometimes questions need to show you a lot left to learn . Anyway are you staying for 1:30 today? If so ill have to discuss a doubt.

John Rennie

No, I'm only here for another hour or so, then I won't be back until Monday morning.

Nobody recognizeable

11:26 AM

@JohnRennie if my clock ticks slowly will i age less? Though clock will say that but what about me will i see aging effects?

@JohnRennie this is Sunday , professor. You are generally here in mornings . So monday morning is normal you get here at 5 o clock uk generally.

John Rennie

@Nobodyrecognizeable Yes, I'll be around from about 5 a.m. on Monday.

Nobody recognizeable

@JohnRennie yep you keep forgetting me. Anyway the previous one is the question.

John Rennie

When we talk about clocks in SR we don't mean some specific mechanical device. The clock records the proper time of the observer holding it, so the time recorded by the clock is the same as the time experienced by the observer. You will age by the time shown on the clock.

Nobody recognizeable

@JohnRennie i see. Thanks professor. Gotta go . Have a nice day professor, goodbye.

John Rennie

Bye

ayc

1:03 PM

I have a problem!....anyone ?......just ping me

Esha Manideep

@ayc Please tell the problem

ayc

@EshaManideep I can't understand how they gave those potential values?..those (x-20),(-45) and others

Esha Manideep

@ayc He thought that the centre's potential is x, then KVL does it all

ayc

@EshaManideep I know,but...wait I'll upload the answer!

@EshaManideep Once check the answer!Could you explain briefly how they got (x-20)-10

Esha Manideep

1:20 PM

@ayc It's the potential difference across the 1 Ohm resistor

ayc

@EshaManideep I know that.I want to know how they came to the value:(x-20)-10 as the potential across the 1 ohm resistor

Esha Manideep

@ayc On one side of the resistor potential is 10 and the other side it is x-20

Anonymous

@ayc Potential at point A: xV. Potential at right side of the 1 ohm resistor is thus (x-20)V, since there's a 20 drop due to the cell. Potential at the left side of the 1 ohm resistor is 10V. You have the potentials at both ends of the resistor and now you can write the current through it as (x-20)-10.

Esha Manideep

@Blue He got the first part, and second one he didn't ( I guess so).... happens sometimes :P

ayc

@EshaManideep @Blue Alright!..What was I thinking!?

Anonymous

1:29 PM

@ayc Is it clear now? :)

ayc

@Blue clear and sounds funny:the question I asked

Anonymous

Happens. Move on :P

ayc

@Blue @EshaManideep Thank you for your time!

Abcd

1:53 PM

@sammygerbil please ping me ven you are free

ayc

4:22 PM

anybody free?..I have a problem.

Esha Manideep

@ayc Yep, somehow I feel totally lazy to do my own work today :P

ayc

@EshaManideep Why is the upper limit taken as infinite?...........Why the number of solutions is equal to coefficient of t^5 .....It should be t^6 right?

Esha Manideep

@ayc Yeah,even am not sure about t^5. But as far as that infinite is concerned please give a 2 min thought about it. You should be able to figure out by yourself

ayc

@EshaManideep Take your time..no rush ..I'll be back in an hour or maybe sooner

Anonymous

4:43 PM

@ayc That doesn't seem right to me. You need to find the coefficient of $t^6$ from $(t+t^2+...)^4$.

Anonymous

$t^0$ shouldn't be there in that sum as a dice doesn't have a $0$ face.

Anonymous

Ideally, you'd just need to find the coefficient of $t^6$ from $(t+t^2+...+t^6)^4$. Using an infinite sum just makes it simpler.

ayc

5:06 PM

@sammygerbil Could you look at my question?

@Blue Lets forget t^6.Why infinite sum?

sammy gerbil

@ayc I think you can use the Stars and Bars Theorem.

ayc

@sammygerbil My book has stuff like this:It first talked about stars and bars theorem and then it used the multinomial theorem.After reading both of them I felt that stars and bars theorem is only a special case and multinomial theorem is general.In the multinomial theorem there are cases and the first case is the one in the picture.In the Illustration the answer (6+4-1)C(4-1) looks similar to what you would get if you used Stars and bars theorem.

@sammygerbil The book solved it taking upper limit as infinite.I dont know what that means......could you tell me How do I relate this case with stars and bars theorem

Anonymous

@ayc You can use a finite sum. It doesn't matter. You'll get the same answer.

Anonymous

You basically need to find the coefficient of $t^6$ in $(t+t^2+...+t^6)^4$

Anonymous

Coefficient of $t^6$ in $(t+t^2+...+t^6)^4$ = Coefficient of $t^6$ in $(t+t^2+...)^4$

Anonymous

5:21 PM

The latter has a nice form

ayc

@Blue Are you sure its t^6 ?...becoz I don't think this book(Cengage mathematics) has mistakes..I study from these books only and I haven't found even a single mistake until now.

Anonymous

1

Q: Number of ways to get 17 by rolling dice 4 times (with combinatorial argument)

A dice is rolled four times. In how many ways we can get total of 17? We can solve this by finding coefficient of $x^{17}$ in the ordinary enumerator $(x+x^2+x^3+x^4+x^5+x^6)^4$. But I feel this requires little bit more effort. We can directly solve similar problem like: In how many wa...

combinatorics generating-functions

ayc

@Blue Yeah.I have checked that question already

Anonymous

@ayc Find out the correct answer using the formula true blue anil gave. Find the coefficient of $t^5$ and also find the coefficient of $t^6$. Check which one matches. Pretty sure you need $t^6$, as you need a sum total of $6$.

Anonymous

If you don't find it intuitive yet, think of it like this:

Anonymous

5:31 PM

You have $4$ bins consisting of all permissible power of $t$ ($t^0$ is not permissible because each $x_i$ needs to be at least $1$)

Anonymous

Also, $x_i$ can at max be $6$

Anonymous

So you have four bins, each consisting of powers of $t$ ranging from $1$ to $6$

Anonymous

Now you need to pick up $t^i$'s from the four bins such that $t^{\sum i}=t^6$

Anonymous

And in this method, you're basically finding the number of ways it's possible to do that

ayc

@Blue Ohk!..........I have to go now.......Thank you for your time!

sammy gerbil

5:47 PM

@ayc Am I looking at the right question 6.122? Simplest answer is that no die can show less than 1 so start with all 4 dice showing 1. Total so far is 4. We need a total of 6 so there are 2 more spots to distribute. Each spot can be placed on 4 dice so answer is 4x4 = 16.

I don't understand how they get 84.

Have I misunderstood the question?

sammy gerbil

6:05 PM

@ayc Stars and Bars solution : The total required is 6 so place 6 stars in a row. There are 4 dice so we need to place 3 bars in the gaps between the stars. No die can show less than 1 so bars cannot be placed in same gap. There are 5 gaps so the 3 bars can be placed in 5x4x3/3x2x1 = 10 ways. (Enumerating all the permutations, there are indeed 10. My first solution must be wrong.)

ayc

7:27 PM

@sammygerbil Yes you are looking at the right question.Actually once you put one dot on each the rest two dots can be put in 4c2+4c1 ways..which is 10....You've understood the question in the way I did .So I can't tell if we got it right or wrong.

@sammygerbil Even I don't understand how they got 84...Maybe we could understand better if we look at the way they did....I have no idea what they did...could you look more closely in the way they did..do you understand their steps?

sammy gerbil

8:22 PM

@ayc Blue seems to understand the generating function method. Best to ask him about that method.

@Abcd Sorry I have guests on Sundays so usually not available until the evening.

Abcd

8:35 PM

@sammygerbil Are you free now?

sammy gerbil

@Abcd yes

Abcd

@sammygerbil Could you please tell me the IST equivalent of evening in UK

@ayc Please upload straight pics

sammy gerbil

@Abcd UK is 5.5 hours behind IST. Time is now 20:40 (Sunday evening) here.

Abcd

@sammygerbil quite symmetric. IST is GMT + 5.5

@sammygerbil so Greenwich meridian must be passing through UK

I dont remember where Greenwich is

sammy gerbil

@Abcd Yes. Greenwich Observatory is in London.

Greenwich is a district of London.

Beside the Thames River. Long maritime history. Home of National Maritime Museum and Cutty Sark sailing ship.

Abcd

8:43 PM

@sammygerbil Did you get the solution to yesterday's left problems?

sammy gerbil

@Abcd No I did not look at the one about two cars and the wind and a siren.

Abcd

@sammygerbil pulley one?

sammy gerbil

@Abcd Can't find the link.

Abcd

23 hours ago, by Abcd

@sammygerbil please see my attempt also ^^^

sammy gerbil

@Abcd Didn't we solve this one? (No pulley here!) Answer B?

Abcd

8:49 PM

@sammygerbil i told you that B is my answer but thats the wrong answer

sammy gerbil

@Abcd What is the right answer? I don't see what we did wrong.

Abcd

23 hours ago, by Abcd

So $\omega^2 = 5k/m$

23 hours ago, by Abcd

So I got option D as the answer

This is a sleepy typo^

I meant option B there

@sammygerbil are you OK with my method as well?

sammy gerbil

Sorry I didn't check your method. You got same answer as me so I assumed the method is correct. Do you have a doubt about it?

Abcd

@sammygerbil Answer is C

@sammygerbil I have never got right answers using wrong methods but you never know ...

sammy gerbil

@Abcd I have an idea!

Abcd

8:52 PM

@sammygerbil ?

sammy gerbil

The question says the block touches the springs. It is not attached so the springs only act in compression.

In the diagram there are small gaps between the springs. They are trying to fool us!

Abcd

@sammygerbil ohh

@sammygerbil for ayc's problem are you damn sure its 10? Because even I get 10 using my stars and bars with constraints method.

sammy gerbil

@Abcd We all get 10. I wrote out all the permutations and there are 10.

Abcd

@sammygerbil how to proceed with the problem

sammy gerbil

@Abcd What do you mean?

Abcd

8:58 PM

@sammygerbil textbook is surely wrong then. I think it would be really helpful if @Blue could point out the error in the textbook's method.

@sammygerbil wait

Energy conservation?

Push it to right by x.

Then conserve energy?

then differentiate?

wont work

sammy gerbil

@Abcd Simpler to calculate half-period on left side and right side, then add.

Abcd

@sammygerbil how can we add half periods that way

sammy gerbil

Because when mass is on the left it is not affected by the spring on the right, and vice versa.

Abcd

i think the point is that time period has nothing to do with amplitude

sammy gerbil

Yes. I haven't used amplitude.

Abcd

9:04 PM

and I am fretting over the amplitudes on both sides.

OK Got it

sammy gerbil

There is a lot of unnecessary information in the problem.

They are trying to fool us again!

(Confuse us.)

Abcd

hmm tricky

some from today's exam

I had guessed $\lambda \propto \dfrac 1 {Z^2}$ in exam.

But that couldnt give the right answer.

What is screening constant?

In my attempt I assumed I/2 goes through P

So answer should be:

$\left(\dfrac{1+ \sqrt {1/2}}{1- \sqrt {1/2}}\right)^2$

doesnt match any option.

sammy gerbil

@Abcd Don't know. Moseley's Law came to mind but I don't see screening constant there. en.wikipedia.org/wiki/Moseley%27s_law

Abcd

Official solution is very long and scary.

I mean moseley's law doesnt give that graph

it gives a staight line

and questions are quite direct from it

sammy gerbil

@Abcd That is correct.

Abcd

9:12 PM

@sammygerbil why not giving the right answer then

sammy gerbil

@Abcd Don't know about the ration, that may be wrong. How did you arrive at this answer?

Abcd

ohhhh

$\sqrt \nu = a(Z-b)$

I remember my teacher telling us that $b$ is generally one.

And we know that:

$a= \sqrt{\dfrac{3Rc}{4}} = \text{constant}$

sammy gerbil

@Abcd Is $a$ the screening constant?

Abcd

So yes the problem is a moseley's law problem!!

@sammygerbil no b

sammy gerbil

Screening means that inner electrons shield an outer electron from the nucleus. So effective nuclear charge is lower for outer electrons.

Abcd

9:17 PM

@sammygerbil Amplitude is $\sqrt I$

So sum is:

$\sqrt{I}+ \sqrt{I/2}$

And difference is $\sqrt{I}- \sqrt{I/2}$

Squaring and dividing gives desired ratio

sammy gerbil

@Abcd Yes I think I understand. That looks reasonable.

Abcd

@sammygerbil y not correct answer then

@sammygerbil should I show you official solution

sammy gerbil

@Abcd Yes.

Abcd

sammy gerbil

9:32 PM

@Abcd That looks a difficult solution to do in 3 minutes.

Which option do they say is correct?

C (5)?

Abcd

@sammygerbil this was JEE Main Mock test. Questions are supposed to be solved in 2 minutes in it :P . 3 minutes are for JEE advanced questions.

@sammygerbil yes

sammy gerbil

@Abcd That does not seem logical. At lower level you get less time?

Are JEE Advanced questions really more difficult?

Abcd

@sammygerbil yes.

@sammygerbil JEE advanced has 20+20+20 PCM questions for 3 hours. So 3 minutes per question. Also there are two Papers separated by a 2 hour break. (so total 6 hours and 120 questions)

@sammygerbil JEE Main has 30+30+30 PCM questions for 3 hours. SO 2 minutes per question. There is only 1 paper.

@sammygerbil well its not actual exam so people (exam setters) can keep difficult stuff in paper i guess....

sammy gerbil

I think there must be an easier solution.

Abcd

yes....

sammy gerbil

9:42 PM

I am a little too tired to concentrate on it just now. Any other questions?

Abcd

@sammygerbil no am doing maths

sammy gerbil

@Abcd ok. Goodnight then.

sammy gerbil

11:30 PM

@ayc I am convinced 10 is the correct answer but still cannot see the mistake in my first attempt. I ought to have divided by 2, because the order in which the 2 extra spots are placed is irrelevant. But that gives 8 not 10. Can you explain why 4c2+4c1?

sammy gerbil

11:48 PM

@Abcd Q27 YDSE with Polarizer : Divide the electric field of light falling onto the slits into amplitudes parallel (p) and perpendicular (s) to the polarizer. Only p gets through the polariser and emerges from slit 2 whereas p+s emerges from slit 1. Max amplitude in the interference pattern is p+s+p (slits 1 & 2 in phasee) while min amplitude is p+s-p = s.

Now p and s are orthogonal, so for the max amplitude we have a p-component of 2 units and an s-component of 1 unit, which add to as vectors to give a magnitude of $\sqrt5$.

The min amplitude has only s-component so magnitude is 1.

Intensity is proportional to square of amplitude so ratio of max to min intensity is 5:1.

Transcript for

Problem Solving Strategies