Problem Solving Strategies

4:03 PM

when you have a Euler graph where two of the edges are odd but one is 1 while the other is 3 can you visit all the edges

this is my euler graph [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]] I can visit all the vertaces but not all the edges is this problramatic?

if I start from an odd vertex i can visit all the edges

but I can't visit all the edges form an even vertex

I was under the impression that I can visit all the edges from any vertex but I might be wrong.

Abcd

4:33 PM

Dude please post this stuff in chemistry room. @Dante

@sammygerbil Are you there for few physics problems?

sammy gerbil

@Abcd hello

Abcd

@sammygerbil Before that can you transfer the above 3 messages to chemistry room? You can use room owner powers to do that.

Rick

is this a chemistry room?

Abcd

Nah

JEE Chemistry Club

For Mathjax see: chemistry.meta.stackexchange.com/questions/89/…...

Rick

sorry I thought it was problem solving

Abcd

4:35 PM

This is^

Rick

my bad

Dante

@Abcd Yeah, sorry there was no one there, that's why I put it here.

Abcd

@Dante If someone pings, you automatically get the notification "you are pinged in XYZ room where you arent present"

Dante

Oh, right.

sammy gerbil

→ 5 messages moved to JEE/High School Chemistry Problems

Abcd

4:43 PM

@sammygerbil First 28 then 27

sammy gerbil

@Abcd What do you think for #28? Is your answer 4? What is your reason?

Esha Manideep

@Abcd think of an inelastic collision, that is clearly against your answer ;)

Abcd

@sammygerbil my answer is 4 because work done by internal forces is always 0

and

$\Delta KE= \text{Work}$

Avnish Kabaj

Ke is not conserved during collision

Like total energy is always conserved

Abcd

@sammygerbil Do you see the error?

Avnish Kabaj

4:48 PM

But even in an elastic collision ke energy isn't conserved .While the collision is taking place some of it changes into potential energy

Is the answer 2

sammy gerbil

@Abcd Work done by internal forces is not always zero. Where did you get this from?

Abcd

@AvnishKabaj hmm

@sammygerbil i read it somewhere. Cant recall where.

@sammygerbil But internal forces are always in equal opposite pairs right?

so shouldnt work be 0?

sammy gerbil

@Abcd Yes they are always in pairs but that does not mean that total work is zero.

Abcd

@sammygerbil equal and opposite $\implies \text{Zero Work}$

sammy gerbil

@Abcd Yes but the forces act on different objects.

Work done on object A equals force on A times distance moved by A. Work done on B equals force on B times distance moved by B.

+ve work is done on A and B. Total work is not zero.

Work is not a vector.

Abcd

4:55 PM

@sammygerbil ohh :( . Thanks. Concept cleared.

@sammygerbil 27 now.

Wait.

And what about linear momentum thing?

sammy gerbil

@Abcd Total linear momentum can only be changed by an external force. It is a vector.

If there is an explosion, so that parts A and B fly apart, their linear momenta are always equal and opposite.

Avnish is correct. Answer is 2.

Abcd

@sammygerbil How can you claim total linear momentum can only be changed by external force?

Does that follow from equal and opp reasoning?

Avnish Kabaj

Linear momentum is force×time

Work = f×d

Abcd

$\vec F = \dfrac {d\vec p}{dt}$

sammy gerbil

@Abcd Yes. The forces are always equal and opposite, and they act for exactly same amount of time. So impulses are equal and opposite. Impulse = change in momentum. So change in momentum is equal and opposite for both parts.

Abcd

5:00 PM

@sammygerbil Thanks. 27 now.

@sammygerbil I feel info is missing from 27.

Avnish Kabaj

Why does the negative half cycle of the modulating signal get clipped

sammy gerbil

@Abcd First imagine the two cars to be stationary. Then work out speed of 3rd car relative to other two.

Abcd

But if I do random nonsense, I get (4). Which is the correct answer.

@sammygerbil third car's speed wrt any frame is not given.

sammy gerbil

@Abcd Of course not. That is what you have to find!

Abcd

@sammygerbil just 5 minutes time is given! Nothing else, for the third car.

sammy gerbil

5:03 PM

Don't know what "random nonsense" means. What is your calculation?

Abcd

@sammygerbil I did $(30+x)\times \dfrac{4}{60} = 5$

And got x = 45

But I didnt mark it in exam :(

Because I just did this randomly from the given information.

sammy gerbil

@Abcd Yes that is correct.

Avnish Kabaj

@sammygerbil @Abcd

3 mins ago, by Avnish Kabaj

Why does the negative half cycle of the modulating signal get clipped

sammy gerbil

Because relative velocity is 30+x kph.

Abcd

@sammygerbil does "they " in second statement refer to 3rd car and set of 2 cars?

@sammygerbil ^^^

@AvnishKabaj yet to revise this chap.

Avnish Kabaj

5:05 PM

Ok

Abcd

completing waves and modern archive today

Avnish Kabaj

Okok

sammy gerbil

@Abcd I found the question difficult to understand also. The two cars are moving to right at same speed of 30 kph, 5 km apart. The 3rd car is coming towards them moving left. It passes lst car then 2nd car. Time between passings is 4 mins.

Abcd

@sammygerbil how can it pass the two cars separately?

sammy gerbil

@Abcd They are 5km apart, one behind the other on the same road.

They are not side by side. All 3 cars are on the same road, moving in same straight line. 1st and 2nd cars moving right, 3rd car moving left.

Abcd

5:10 PM

@sammygerbil I thought distance between their doors was 5 km :(

@sammygerbil ya i thought they are side by side

@sammygerbil then leave it, its easy.

@sammygerbil new question:

sammy gerbil

If you imagine 1st and 2nd car at rest, 3rd car moves 5km in 4 mins which is 75 kph. From that you have to subtract 30 kph which is speed of 1st and 2nd cars. But it is same as your calculation, which is correct.

@Abcd Some of these problems the hardest part is understanding what the question means!

Next?

Abcd

Intensity of light from a source is : $\dfrac{500 }{\pi} \pu{W/m^2}$. Find the amplitude of the electric field in the wave.

@sammygerbil I tried various things for this^

Like:

$E_o^2 \propto I $

for wave on string intensity is given by:

$I = \rho V 2\pi^2 f^2 $

But couldnt get the electric analgoues of $\rho, V$

I tried this formula also:

$E = \dfrac{hc}{\lambda}$

Couldnt arrange anything into a cool relation between intensity and field.

sammy gerbil

$I=\frac12 \epsilon_0 cE^2$

Abcd

???

That was for intensity of field in space between capacitor

$u = \dfrac 12 k \epsilon_o E^2$

Where u is energy density (not intensity ?)

@sammygerbil Oh i see you have edited.

@sammygerbil how did you get that relation?

sammy gerbil

2

Q: Relation between intensity of light and amplitude of electric field?

A question in my textbook involve finding the electric field amplitude at a point in space given the intensity of light. It uses the following equation to solve it: - $\frac{1}{2}I=\frac{1}{2}\epsilon_{0}|E_{rms}|^2c$ But where did this equation come from? I am unable to find an explanation for...

Abcd

5:18 PM

$E = Bc$

$\dfrac{1}{\sqrt{\mu_o \epsilon_o}} = c$

:/

sammy gerbil

Also here : en.wikipedia.org/wiki/Intensity_(physics)

Abcd

@sammygerbil no answer

sammy gerbil

@Abcd ??

Abcd

@sammygerbil do you know how to derive it using the techniques i tried?

@sammygerbil i mean that question had no answer.

sammy gerbil

@Abcd True. But it gives correct formula. (Factor of 1/2 in front is a typo.)

Comments suggest how it can be derived.

You don't need to derive it just memorise and use it.

Abcd

5:23 PM

Kk

@sammygerbil next question

sammy gerbil

An alternative form is $I=\frac{E^2}{Z_0}$ where $Z_0=120\pi$ Ohms is impedance of free space.

Abcd

oh that would give the answer more neatly

@sammygerbil Attempt is:

$A \cos (k x)= 5\sqrt 3$ where $x= 10 cm$

Similarly:

$A \cos (2kx) = 5 \sqrt 3$

$\cos 2kx = \cos kx$

$2\cos^2 kx - 1 = \cos kx$

$\cos kx = -1/2 \text{ or } 1$

$A= 5 \sqrt 3$

or

$A = 10 \sqrt 3$

= None of these.

@sammygerbil Whats wrong with the attempt

@sammygerbil please reply

sammy gerbil

@Abcd The loop has a node at both ends so I would use sine instead of cosine.

Abcd

Ok :(

sammy gerbil

Suppose the ends of the loop are A and D and the 1st point is B and the 2nd point is C. Then AB=BC=CD=10cm.

The phase at D is $\pi$ so the phase at B is $\pi/3$.

We have $A\sin(\pi/3)=\frac{\sqrt3}{2}A=5\sqrt3$.

Abcd

5:39 PM

@sammygerbil how phase at D is pi

sammy gerbil

@Abcd The "loop" is a half wavelength of the standing wave with amplitude $y=A\sin(kx)$.

$y=0$ at A and at D.

Abcd

@sammygerbil didnt get

Loop has to be full.

For two points to have same amplitude

in standing wavr

e*

sammy gerbil

Look at a standing wave eg arborsci.com/cool/big-standing-wave-small-effort

Each "loop" is half a wavelength.

Abcd

Ohkay

but my method would also work

@sammygerbil Last 3 questions for the hour:

sammy gerbil

@Abcd Don't think so. Try drawing a diagram. Apply your formula to that.

Abcd

5:50 PM

@sammygerbil First one:

I did it by putting $\phi = 0$ in all options and rejecting option 1 because thats the formula for normal case. What is the formal method to solve it?

No idea for next 2.

@sammygerbil i meant my method with sin not cos.

sammy gerbil

@Abcd #1. Yes that's good thinking. We can reject option 2 because that says intensity is zero when phase difference is zero, obviously wrong.

Abcd

@sammygerbil ya i know i rejected it

@sammygerbil i want to know the formal method

I got my answer as C easily

sammy gerbil

@Abcd You said reject option 1. But we can reject that also because if intensity from each slit is different the combined intensity will never be zero.

@Abcd How did you get C?

You mean option 3?

I agree, option 3.

You wanted to learn "trick" methods so that you can get the answer quickly. Formal method will take too long in an exam.

Abcd

6:06 PM

@sammygerbil yeah, more used to a b c d than 1 2 3 4 in options

@sammygerbil am curious still

@sammygerbil cool

sammy gerbil

@Abcd Formal method would be to write amplitude of interference pattern as $A=A_1+A_2\cos\phi$. Then total intensity is $A^2$.

Abcd

@sammygerbil why A2 cos phi

sammy gerbil

@Abcd Because there is a phase difference of $\phi$ between the waves. Doesn't matter whether you write cos with A2 or with A1.

If you prefer you could write $A=A_1\cos\phi_1+A_2\cos\phi_2$ and $\phi=\phi_2-\phi_1$.

Abcd

@sammygerbil didn't get

Why exactly A1 + A2 cos phi?

Shouldn't we be doing vector addition of A1 and A2 (phasor method)

sammy gerbil

@Abcd Yes if you want. It amounts to the same thing. Adding two oscillations with a phase difference between them. Try it using whichever method you want.

2nd question : Use conservation of angular momentum about point of contact ; $mv_0r=I\omega=\frac32 mr^2\omega$.

3rd question : Calculate the circumference $C$ of the drop which is squashed between the plates. The force on each plate is approx $\gamma C$.

I am not sure how the angle of contact can be used in this calculation. Also the volume of the drop. Seems to be too much information here.

16. Physics | Surface Tension | Force Between Two Parallel Plates Due to Liquid Between Them

►

Abcd

6:42 PM

@sammygerbil Sorry but I really dont get how you quickly got $A1+ A2\cos \phi$.

Could you give concrete explanation?

How to make the intensity function $I(\phi)$?

sammy gerbil

@Abcd At some point the amplitudes from slits 1 and 2 are A1 and A2. There is a phase difference of phi between them. They have same frequency so phase difference does not change with time, it depends only on position.

If you do not like this explanation use one which you do like.

eg phasor addition.

Sorry will be AFK for 20 mins.

sammy gerbil

7:19 PM

@Abcd Back.

Abcd

@sammygerbil OK, doing optical instruments atm.

@sammygerbil are you OK with a simple concept doubt?

sammy gerbil

ok

Abcd

least distance of vision.

8 cm for children

200 cm for old people

Now, the definition is "nearest stuff we can see"

So shouldnt it be more for children?

because they can see larger distances?

Avnish Kabaj

Nearest distance is the nearest point which the eye can focus on

Abcd

Oh damn.

misunderstood the definition.

Avnish Kabaj

7:26 PM

:P

Abcd

thank you!

Avnish Kabaj

Wait

See this question it will help you a lot

2

Q: Magnification vs Magnifying Power

I've read it a lot of times. But I've not been able to get around magnifying power and magnification of a simple microscope and the difference between them. Can someone explain?

geometric-optics optical-materials

Abcd

thanks.

sammy gerbil

@Abcd No. In children the lens material is more flexible and muscles are stronger because they are newer. So the eye is more "adaptable". It can adjust more easily to focus on closer objects - and further ones also.

Abcd

@sammygerbil i was interpreting the definition of least distance of distinct vision like "child can see distinctly objects till 8 cm only" and "elders can see till 200 cm" which was wrong

sammy gerbil

7:32 PM

@Abcd That should be 20cm not 200cm.

Abcd

@sammygerbil lol ... i think 200 is maximum possible distance ...

200 is so much wrt the eye!!

@sammygerbil 200 cm is written in my book for 60 years of age...but I cant find any online sources to confirm it.

but nothing to worry about

useless thing

@sammygerbil one more confusion.

sammy gerbil

@Abcd yes?

Abcd

$m= v(\frac 1 v - \frac 1 f)= (1- \frac v f)$

Now for simple microscope, why will we put $v= D$ where D is least distance of distinct vision?

@sammygerbil for first formula i have used $m = v/u$ along with lens formula/

Here D appears to be what I thought initially. :(

16 mins ago, by Abcd

@sammygerbil i was interpreting the definition of least distance of distinct vision like "child can see distinctly objects till 8 cm only" and "elders can see till 200 cm" which was wrong

This^.

@sammygerbil Also, sorry for concept questions. Please tell whenever you arent OK with it frankly. Wont ask then.

sammy gerbil

7:47 PM

@Abcd To view something comfortably it should be at the least distance of distinct vision. Objects which are closer appear bigger so they can be seen more clearly, but the closer they are the more difficult it is for the eye to focus on them. A distance of 20-25 cm gives the best compromise. So that's where you want the image to be.

Abcd

@sammygerbil D= 25 cm means object should be at least (or at most?) 25 cm (that is externally 25 cm away from the eye?) away from us so that we can see it properly?

sammy gerbil

@Abcd The closest distance to focus without discomfort is D=20cm. Any closer you can probably see it by squinting but you will quickly tire out your eyes doing this.

Abcd

@sammygerbil but something at a distance of 2 cm is also very clear to me ...

if not 2, then 5 ....

like if I keep a pen 5 cm away i can see it.

sammy gerbil

You could not read a whole chapter of a book at 5 cm from your eyes. For most adults the comfortable distance is 20-25cm. They can focus on things closer for short periods, a few seconds, but the eye will quickly tire and become uncomfortable.

Abcd

@sammygerbil BTW, confirmed with one more book. 200 cm = 2 m is indeed the correct value for adults..

sammy gerbil

7:59 PM

Older adults have trouble focussing even at 20 cm, so they often need reading glasses.

@Abcd What does the book say about it exactly?

Abcd

@sammygerbil Now my question is why is there no "maximum distance of distinct vision " ? ... Like we cant see any far off things after like 20 m or something...

@sammygerbil both books^

sammy gerbil

@Abcd We can see things which are 20m away. I think what you mean is that we cannot read a book which is 20m away. This has to do with resolution. See quora.com/What-is-the-resolution-of-the-human-eye-in-megapixels

@Abcd Yes both books say 25cm is standard for adult, but at 60 it can be 1-2m.

Abcd

@sammygerbil but cant see after 500 m or 1 km at least!

or can we?

sammy gerbil

@Abcd Please define what you mean by "cannot see after 1km". You can see stars which are hundreds and thousands of light years away.

Abcd

@sammygerbil ohh...

yes you are right

i was confusing it with "cant read a page or cant see a human 100 m away"

JD_PM

8:10 PM

@sammygerbil Night professor. I am stuck with maths. Do you have some free time?

sammy gerbil

@JD_PM yes

JD_PM

Thanks. OK let me contextualize a little bit. I have to get the pressure of a gas as a function of its density. This is the integral I have to solve. imgur.com/a/mDhUUnQ Do not worry about the terminology. $\epsilon_f$ is the fermi energy.

Now I will explain what I have done

And in what exactly I am stuck with.

We know to what a specific part of this integral equals to: imgur.com/a/dxn1Ipd

So $\frac{N}{V}$ is: imgur.com/a/k0DfHIT

Of course we also know what $g(\epsilon)d\epsilon$ is: imgur.com/a/IHxGwuo

Now I should be able to plug everything into the pressure's integral and get this result, but I do not why I do not get it: imgur.com/a/xlh9g1r

@sammygerbil please if you need more details let me know

sammy gerbil

8:35 PM

@JD_PM Sorry I don't want to work on this problem. Please ask John Rennie about it.

JD_PM

@sammygerbil Okey professor, good night.

sammy gerbil

@JD_PM good night,

JEE Chemistry Club

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