@Akari You're getting the right answer -_-

$u^2+v^2= V^2$

@PolarBear You could energy conservation also here. $mg(\omega a)=\dfrac{V^2}{R}$

Where $V=\dfrac{1}{2} Bl^2 \omega$

It's not $l$, it's $r$ in the question.

@AdvilSell Getting the inductor one?

Doesn't look like simple LC oscillations to me.

@Dante No

Hi!

Could someone help me with this?

Multiple answer type.

@Dante Is the answer D

A,B

@Dante Sorry. Use lenz's law and get A, B

I didn't get you.

@Dante I mean if there are electrons in the surrounding region and induced electric field exerts force on them then induced current will flow which will tend to oppose the change in magnetic field

Oh, now I get it.

Thanks

But I didn't...magnetic field due to an infinite wire varies radially right? So why aren't concentric circles formed ,the tangents to which represent the electric field?

You have to look at magnetic field lines, not how they vary

Field lines are circular around the wire

So...how would the electric field lines look like?

Parallel (or anti parallel) to the wire.

Why? Why not tangential to those circles?

See, imagine the wire is vertical and current if flowing upwards with increasing magnitude,

Okay.. then?

Now, if the electrons in space around wire start moving upwards (along the wire), they generate a magnetic field in the form of same circular loops but opposite direction.

This is what lenz law tells us.

Now since the electrons are moving upwards along the wire, induced electric field also must be so.

If u took the direction of current as upwards . Shouldn't the electrons be moving downwards? @Dante

No, direction of electrons is opposite to that of current na, if they move downwards, current in the space is flowing upwards, so it's contributing to magnetic field against lenz law.

@PolarBear @YUSUFHASAN Did you understand the triangular frame question?

Jan main

Self inductance one

@Dante That question I asked JR sir on this group...he said that it is very difficult to comment on the self-inductance of a triangular system without calculation,and he wasn't sure it could be done as simply as it was shown in mains... He forwarded a few papers for that as well(scroll up for them) and the relations were quite complex there

Oh, I saw some coaching institute's answers, looked like bs. I think area is not the factor we have to consider there, electric field outside solenoid must be zero, only length should matter imo.

Yeah.. I saw the actual relation for that .. seemed quite complex.. it had factored in area,number of turns and whatnot

Oh, lol, screw it.

Okay.. so basically...the electrons moving in the wire are for electric field..but magnetic field is still circular,right?@Dante

@Dante self inductance increases by?

@Dante What are you equating energy dissipated in resistor to?

rate of change of potential energy of block 'mgv'

@PolarBear I think 3, will have to search that question.

@YUSUFHASAN "the electrons moving in the wire are for electric field" What?

@Dante The answer was 3

Yeah, length of frame increases by 3. How does area matter here?

I don't get it.

@Dante why are you equating it to V²/R

V²/R is the heat dissipated in the resistor

Yeah, potential energy of mass m is converted into heat

@Dante Nothing.. I was referring to direction of the electric field (in a really dumb way) .. Thanks!

The current flows due to magnetic field and not due to mass

What makes the current keep goin?

I got the answer also btw.

Rotating thing

Hmmm

Nice, thanks!

I still have the same question pending though. What is the magnetic torque.

@PolarBear The solution is correct

I got the answer

It's T/4 only as half the charge is the maximum that goes to B

@Avka How T/4?

I understand that only half charge on A flows

I think it this way, capacitor A initially ----q | | -q ---- 0 | | 0----(Capacitor B)

Now as q/2 flows

------q/2 | | -q/2 ---- -q/2 | | q/2 -----

So this way at T/4 B acquires charge q/2

Correct?

It's like simultaneously when charge q/2 flows from A, -q/2 flows from B

Isn't really that intuitive

That was the whole thing

T/4 because it will be a function of Cos 0 at π/2

@Avka Yeah, I get it. Thanks!

Thanks a lot.

@Mr.Xcoder you preparing for physics/junior science/informatics olympiads?

I guessed. Nevermind :)

Who are you?

@Mr.Xcoder are you from India going for OCSC?

dp/dt? Meaning? As in Newton's second law?

If you don't get that, it's probably not meant to be.

Is Mr.Xcoder from India/US?

hey @JohnRennie quick question when you're available.

@kylecampbell hi

@JohnRennie it's the same question as yesterday, I can't figure out why an alternative method for a hinged rod problem doesn't work. pretty sure I must be setting it up incorrectly.

doesn't yield the same result as conservation of energy

imgur.com/Zjv6bu4 here's the picture, you want to find the angular velocity when the rod hangs vertically. $m=4kg$ and $L=0.3m.$ I realize conservation of energy is probably the easiest way, but I'd like to know why you can't find the work done by gravity in terms of an integral of torque wrt angular displacement.

in principle, I don't see why not. specifically, the integral is $\int_{40}^{0} \tau d\theta$ where $\tau = mg\frac{L}{2}\sin(\theta).$ by the work energy theorem, you could set that equal to $\frac{1}{2}I{\omega}^2,$ assuming the rod started from rest.

Both methods should work

So what's wrong with my integral then?

my bounds are in degrees

Assuming $\theta$ is the angle to the vertical the integral is from $\theta=40°$ to $\theta=180°$, not from $40$ to $0$.

I thought $\theta$ was the angle between the force vector and the position vector, and since they're parallel, their cross product is 0 when the rod hangs vertically?

wait a minute... it would be 180 then.

That's $\theta$ isn't it?

yeah, I see

hello sir @JohnRennie

@user8718165 hi

Let me suggest a different way of looking at the problem. I just need to draw a diagram ...

@JohnRennie Ok sir

On the left we have the beaker before we immerse the ball. The weight of the fluid is $Mg$.

@JohnRennie ok sir got this much

But there's another way to look at the weight i.e. it's the force exerted by the fluid on the base of the beaker, and that force is the pressure of the fluid at the base multiplied by the area of the base. OK so far?

@JohnRennie ok sir...lets assume the beaker is massless..

Yes, we'll ignore the weight of the beaker to keep things simple.

OK. The pressure at the base is $P = \rho g h$, and the area of the base is $A$, so the force is $F = \rho g h A$.

And $hA$ is just the volume, so $\rho h A$ is the mass $M$. So the force is $F = Mg$. And that's the same as we had before. So the force is the same whether we just write it as $Mg$ or whether we use pressure.

ok sir

@user8718165 is this OK so far, because next we'll use the pressure idea to explain what happens when we submerge the ball?

@JohnRennie yes...got it upto here

OK, so now we submerge a ball of volume $V$. This displaces an equal volume of water upwards to the level of the water rises. It rises by some distance $x$ shown on the right diagram, where $xA = V$. So the depth of the water is now $h + x$.

@JohnRennie yes sir...got it

That means the pressure at the base has increased to $\rho g (h + x)$ and the force is now $F = \rho g (h + x) A$. And this force is just the weight, so the weight has increased as well.

The increase in the weight is $\Delta F = \rho g (h + x)A - \rho g h A = \rho g x A$

OK so far?

@JohnRennie ok sir

And we have already worked out the $xA$ is equal to the volume of the ball $V$, so we find that when we submerge a ball of volume $V$ the weight goes up by:

$$ \Delta F = \rho g V $$

So it doesn't depend on the mass of the ball, only on its volume.

@JohnRennie ok sir got it...this means any object(more dense, less dense, equal density as water) when submerged fully will increase the weight right...?

@user8718165 Yes.

@JohnRennie here, $\rho$ is the density of the submerged object, and $V$ its volume?

No, $\rho$ is the density of the water, not the object.

But $V$ is the volume of the object.

@JohnRennie oh oh sorry...got it

our calculation involves only the mass and density of water but only the vol. of object is reqd. because we need to know how much water the object has displaced right sir?

Yes. If you wanted to write the force as a function of the mass $m$ and density $\rho_o$ of the object you can do because $V = m/\rho_o$.

But I think it's simpler to just stick to using the volume of the object.

Relevant conceptual question: suppose you were holding a rather heavy lead ball (for some reason) on your boat in a lake. Then, you decide to throw it overboard. Does the water level of the lake stay the same as before, rise, or fall?

@kylecampbell Been there, done that :-)

@JohnRennie the water level shouldn't go up because previously my boat had displaced the amount of water required to keep the ball in place but now the ball itself goes into the water and displaces water...is it correct?

@KushalT. No, I'm not from India.

@KushalT. Yeah, physics.

@JohnRennie I was wrong... your answer says the water level should fall.

@user8718165 when the rock is in the boat it has to be supported by displacement of water i.e. it displaces a volume of water with a weight equal to its own weight $V_w$.

When the rock is in the lake it displaces a volume of water equal to its own volume $V$.

And $V_w > V$ i.e. it displaces more water when it is in the boat.

@JohnRennie if the volumes were equal, the stone would float??

If the volumes were equal the means the density of the rock has to be the same as the density of water. Yes?

yes sir

In that case the rock would have neutral buoyancy so it would neither float nor sink.

@JohnRennie submerge fully and remain there without submerging more?

@user8718165 it would just stay at whatever height in the water you placed it. There would be no upward or downwards force on it.

@JohnRennie oh ok sir

If you lowered the rock gently into the water then you're correct that it would descend into the water until the top of the rock was level with the top of the water. Then it would just stay there.

@JohnRennie if I just threw it there with force then due to my force the object will go down as no buoyant force will be there?

@user8718165 yes, if you threw it the rock would go downwards until the viscous drag from the water brought it to a halt. Then it would just stay at whatever depth it stopped it.

@JohnRennie ok sir got it...thanks a lot for helping me

The three SHMs will be given by $A\sin(x) + A\sin(x + π/6) + A\sin(x + π/3)$

Adding them we get

$A(\sqrt{3} + 1)\sin(x + π/6)$

Yet, according to the answer given sentence (ii) is correct. How?

@PolarBear won't the SHMs be $A\sin(x) + A\sin(x + π/2) + A\sin(x + π)$ ??

Uh oh, silly mistake.

No, they will differ by $π$

Ah yes! Silly mistake.

@Akari $\lambda/2 = π ≠ π/2$ (?)

@PolarBear draw a phasor diagram

@JohnRennie I did. They differ by $\lambda/2$ that's $π$. Right?

So, this? $A\sin(x) + A\sin(x + π) + A\sin(x + 2π)$

I think that's a misprint. I think it should be If each differs in phase from the next by $\pi/2$.

I automatically read it as $\pi/2$ and only realised it was $\lambda/2$ when I looked more closely.

@JohnRennie Oh, that's right I guess because I wondered uh what lambda?

yeah, $\frac{\lambda}{2}$ didn't make much sense in the context

Then the first and third wave sum to zero leaving only the second.

Yeah, so second's true. Got that. Thanks!

@JohnRennie Here in the solution how do they find the torque applied by magnetic field I the third line?

The context is that a disc is rotating, the one of radius r in the question and magnetic field B is perpendicular to it. A resistor is connected to it's rim and the circuit completes at the disc's centre.

I haven't worked through it, but at a distance $r$ from the centre the Lorentz force on an element $dr$ will be $BIdr$ so the torque will be $BIrdr$. Integrate that from $r=0$ to $r=R$ to get the total torque.

In fact that's correct isn't it, because we get $\tau = \tfrac{1}{2}BIR^2$ and substituting for $I$ gives the expression in the third line.

@JohnRennie Oh, correct. I forgot the fact that the current flows from the centre to the rim so I was thinking Lorentz force as $BI× 2πx$ which is incorrect.

@JohnRennie That seems correct.

Thanks for your help!

@Akari hi

@Akari There are some obvious restrictions on the total energy $E$. It can't be less than zero because for that to happen it would have to have a negative kinetic energy and that is impossible. Yes?

@JohnRennie hi

@JohnRennie yes

@Akari and the total energy cannot be greater than $V_0$ because if that was the case the particle would escape from the potential well and just head off to infinity instead of oscillating. Yes?

@JohnRennie won't $E - V_o = E_{kinetic}$ ?

@Akari yes

And E_{kinetic} is greater than or equal to 0, so I get $E >= V_o$ ?

Suppose the particle is at the bottom of the well. At the bottom of the well $V=0$ so the total energy is equal to the kinetic energy $E = T$. Yes?

@JohnRennie yes

OK suppose $E = V_0/2$.

Assume $E - V_0/2$ as kinetic energy?

Now start moving sideways in the well so $V$ increases above zero. Since total energy is conserved $E$ must stay constant. That means as $V$ increases $T$ must decrease.

@JohnRennie yes

@Akari oops, that was a typo. I meant $E = V_0/2$

@JohnRennie okay.

@Akari So once we have got halfway up the well, i.e. at the point where $V = V_0/2$, that means the kinetic energy has to fall to zero. Yes?

Yes.

That's the turning point i.e. the point where the oscillating particle stops and turns back to go the other way.

But wasn't it because we assumed E to be less than $V_0$ at the first place?

So are we sort of considering cases and checking if oscillations will take place?

@Akari yes, this is the point I'm making. If $E < V_0$ then the particle can't get out of the well because its KE falls to zero before it reaches the top of the well. So for $E < V_0$ the particle oscillates inside the well.

But suppose $E > V_0$. Then when the particle reaches the top of the well its KE $T = E - V_0$ will still be greater than zero.

Ohkay I understood it. Thanks a lot for your help. But what was wrong in whatever I was doing?

That it didn't prove that the particle will oscillate?

You posted:

What I would say is $E - V = T$ not $E - V_0 = T$

Ohkayy. Thanks a lot for your help.

The problem with your equation is that $V_0$ is a constant. What you equation tells us is the kinetic energy only when $V = V_0$.

Yeah right. I got it now.

Thank you.

Cool :-)

@JohnRennie we've considered that the whole place has the density $\rho$. Sorry missed it then. Just wanted to know if the weight of the ball had any impact?

because it doesn't have the same density $\rho$.

@user8718165 we concluded that only the volume of the ball matters. Yes?

@JohnRennie yes...I returned to the prob now...but in the region $A(x+h)$ the density in the region of the ball is not $\rho$ while in the remaining region it is $\rho$. So how do we consider the whole region to have density $\rho$?

In my equation $\rho$ is the density of the water. I am not considering the density of the ball (or the rod) at all.

@JohnRennie the mass of the region of $A (h+x)$ is $\rho A (h+x)$ ?

@user8718165 no. We don't know what the mass is because it's the mass of the water plus the mass of the ball.

@JohnRennie how is the increase in the weight $\rho xAg$? this amount of water was still present before but now it is at a higher height. Does more height with the same water increase weight?

@user8718165 because the pressure at the bottom is $P = \rho g (h + x)$, and the force on the base is the pressure times the area of the base.

@JohnRennie (sorry if I sound silly or incorrect) why are we always considering that the whole volume is covered with just water? the ball also occupies some volume so the force on the bottom of the vessel should also increase by mg where m is mass of the ball

The ball displaces the volume of water that was there before it was submerged.

The ball is supported partly by the water and partly by the rod. Yes?

@JohnRennie Hey

@JohnRennie yes... got it and does the same amount of water which was before in the vessel when kept at a higher height increase pressure at the bottom(due to displacement by the ball)?

Yes

@TheEastWind hi

@JohnRennie thank you so much

you're related to IT too right?

@TheEastWind yes

is google known to mess up the location of devices which were used to sign in to my account?

i mean i got a mail that my google account was accessed from an unknown device

it was a linux OS on a chrome browser(acc to google)

i did access my google account in an incognito tab using my phones mobile data

@TheEastWind Google gets the location from the IP address that it sees you connecting from. Typically it will get the location of your Internet company's main office.

For example Google thinks I'm in London even though I'm 200 miles away in a different city. That's because the Internet company I use is in London.

So that means that a chrome browser on a linux OS is actually what my ISP uses right?

@TheEastWind presumably you were using a laptop or PC with Linux installed?

Nope

thats what bothers me

i dont have single device with linux on it

Guys can someone please help me?

Let me try firing up an incognito tabe and see what happens ...

@RaphX be with you in a moment ...

Ok..

@TheEastWind I opened a new incognito window and signed into my Google account. I haven't had an e-mail yet, but it can take a while. I'll keep an eye open for the mail and let you know.

OK thanks a lot!!

@RaphX hi

The way I approach this type of problem is to call the unknown velocities $v_a$, $v_b$ and $v_d$ for Akash, Binoy and Deepak, and write down what I know about tjhem.

We know Binay started first, then Ashok started after Binay had run 6m, then Deepak started after Ashok had run 10m.

@RaphX OK so far?

yeah

Let's draw a quick diagram. Give me a moment to draw it ...

@JohnRennie Hi. I'm next :p

@RaphX OK, I've marked everything we are told about the race on this diagram.

Binay starts first at time $t_0$

How did you draw that diagram so quickly?

Then at time $t_1$ Ashok starts when Binay has run 6m, then at time $t_2$ Deepak starts when Ashok has run 10m.

@RaphX would you like to have a look at the diagram and check that it does show what the question tells us?

@kylecampbell practice :-)

I use Google Draw and I've got pretty quick at using it.

mad skills

I didnt understand the last case

Why is Ashok behind Binay?

@RaphX the question tells us:

> Akash started the race when Binoy had alread covered 6 m, and was 15 m short of the finishing line when Deepak finished the race

So when Deepak is at the finish line Ashok is 15m behind him. Yes?

Yeah got it

And it also tells us Binay was 10m behind Deepak at the finish. So we get the distances I've drawn for the time $t_3$ that Deepak reaches the finish line.

Now, the question asks:

> What was the ratio of speed of Akash and Binoy?

With questions like this you often have to just try a few things to see how to get the answer. I must admit I didn't spot it straight away. I realised how to solve the problem only while I was drawing the diagram.

@RaphX shall I explain what I did, or do you want to have another go?

Yeah, please explain

@RaphX Look at the diagram to see how far Ashok and Binay ran between the times $t_1$ and $t_3$. From the information we are given we can work out the distances the two people ran. Yes?

Got it thanks a lot! @JohnRennie

@RaphX Cool :-)

I always recommend drawing diagrams as things are usually easier to see ona diagram.

@Dante your turn :-)

Thanks again , I will remember that! @JohnRennie

@JohnRennie Could you tell me how steam distillation works ?

@Dante have you already attempted to read about it and run into problems?

I read bit of theory given in NCERT, it's too brief, didn't try any problem.

Give me a moment to dig out my old physical chemistry book ...

Okay

@Dante on the phone ...

Yeah, np.

@JohnRennie I think I changed my mind, I have my JEE main tomorrow morning and have got loads to revise, the stuff I asked above isn't very important for main, shall we discuss it tomorrow afternoon?

Sorry for the trouble!

@Dante no problem. I'm around tomorrow. Good luck with the exam!

Thanks!

(I can't find steam distillation in my book anyway :-)

Oh, ok, I didn't even google it thinking I'll have to spend time searching for it, will google it tomorrow before asking.

@JohnRennie Hi

Shouldn't this be zero?

@Dante remember that velocity is a vector

If $v_0$ is the original velocity and $v_1$ the velocity 60° later then the change is $v_1 - v_0$ where you need to use vector addition.

I think I did that, used coordinate system. Took initial velocity as $10 i$

After turning 60 degrees, it was found to be $5j,5\sqrt3i$

So the change was $(5\sqrt{3}-10, 5)$

Umm, wait, I think I took $10j$ initially

Then 60° later the velocity is $(-5\sqrt{3}, 5)$

Yeah

It's just the third side of an equilateral triangle isn't it?

Oh, got it.

I had done some calc mistake ig. Btw, take a look at this

@Dante what's the answer ? (10) ?

I have this question:

Here, $v = 4 × g = 40$. Now, by Doppler effect, $f_{app} = \dfrac{320 × 1000}{320 + 40} = \dfrac{8000}{9} ≈ 888$

@AdvilSell Yeah

@Dante okay :D

Answer's $894.75$

Any idea how they came up with this?

Oh, I am sorry. I thought you were done with your question Dante.

Just few seconds

@Dante you can get this by simplifying the magnitude of the difference of two equal vectors(of magnitude v here)

By writing $\cos \theta = 1-2\sin^2 \frac{\theta}{2}$

Yes, that's what the formula seems to end up with, but how did they start it exactly?

Is that for circular motion with sweeping of angle \theta?

Umm.. Start with what?

@PolarBear Yes

@Dante this?

Yes

@Dante It's very simple, just derive it.

Yeah, trying

It's known too.

Oh sorry, I didn't see that you asked how they derived this in first place

Yes, derivation is what I'm looking for.

@JohnRennie good morning

@Scáthach hi

Figured it out. Thanks everyone!

I need to go. I'll be around later or failing that tomorrow morning.

resonance.ac.in/answer-key-solutions/JEE-Main/2019/…

What point is 8th question trying to make?

How are we supposed to calculate it?

@AdvilSell @Avka @PolarBear @Akari ?

@Dante $\frac {P}{Q} = \frac{R}{X}$

for balanced bridge

@Dante reso site isn't loading

@Akari How do they expect us to calculate $\sqrt{400*405}$ accurately is what I'm asking.

For second case, $\frac {Q}{P} = \frac{R}{X}$

multiply both

@Dante use the differentiation technique to estimate the sqrt(405)

@Avka Page refresh karo, 2-3 baar

Also use the values of R as 400 and 405 in first and second case respectively

Binomial approximation

Saw it

Same thing as what advill said

oh yes, thanks!

@Dante Oops I didn't see this message.

np

Okay , Quick Question : what is $i^{i}$ ?

$e^{-\pi/2}$

@Dante correct !!

Is april maths tougher than jan maths?

@Dante idts

All the best @Scáthach!

@AdvilSell Hmm

best of luck @Scáthach !!

@Dante Same to you bro

@AdvilSell thanks.How did yours go

@Scáthach Mine is tmrw. Morning

@AdvilSell All the best to you too bro.Hope everyone does best

All the best @Scáthach

@Avka Thanks.How did yours go

@Scáthach 12 2nd

@Avka same .all the best

mr xcoderx can you explain your olympiad preparation

Also which country

@JohnRennie hi sir !!

@Opartunity hi

@Broly yes

@KushalT. What exactly do you want me to explain? How I train for them?

@KushalT. Romania

@Mr.Xcoder If that's the case, then I solve the problems from past editions of the olympiads, from locals (in the beginning of the year) all the way to the nationals. I also have a couple of textbooks that I try to solve fully (a couple of examples: Anatolie Hristev and Florea Uliu (both Romanian authors), the "Kvant" magazine (Russian - some of the problems can be found online in english as well - Quantum Magazine)). And lastly, I solve questions from IPHOs and APHOs and selection exams

Why is D a correct option?

Why is acceleration constant for a time interval? Won't it be instantaneous(not necessarily constant), and then go to 0?

@JohnRennie you there sir?