@JohnRennie are you receiving my pings?

@ayc hi, yes, I got the pings. However I'm only around in the mornings (UK time).

The answer given is (a) and (d) but I don't understand it. 1. Is plane polarised the same as linearly polarised?

Plane polarised means the electric field vector always points in the same direction. With circular polarisation the direction of the field rotates about the direction of travel.

And yes, linear polarization is another way of saying plane polarization.

Here we see that there are two mutually perpendicular electrical field vectors. However I do realised that to be circularly or elliptically polarized the two E vectors should be 90 degrees out of phase... But here, it is not so. Both are in same phase. So, "the electric field vector always points in the same direction" . Hence this is plane or linearly polarised... Is this a valid explanation?

Yes

Thanks ;-)

I found the diagrams here very useful-hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polclas.html#c1 . Thanks for your help ;-)

@tatan there's a misprint in the question that I'm sure you've spotted. Option (a) should presumably be $E_2\mathbf i - E_1\mathbf j$.

However for part (a), I don't see why this is correct. for E1 along along positive i, the B vec is along +ve j and for E2 along j, the B vec should be along -ve i . So , (a) shoudn't be correct right?

@tatan options (a) and (b) are the same so there's obviously a mistake in the question. $\mathbf E \cdot \mathbf B$ should be equal to zero.

Shouldn't it be the opposite? I mean ExB should be give the direction of wave propagation right?

E and B are perpendicular so E.B=0

I mean to say, shouldn't it be E_1 j - E_2 i ?

E_1 j - E_2 i = - (E_2 i - E_1 j)

Ah, I see what your saying

I got by reasoning from ExB gives the direction of wave propagation and its +ve k here ...

Um, I'd have to draw a diagram to be sure which B should point

ok

Either way, the question has a mistake

yep... that's for sure. Its disheartening to see mistakes in books that are being circulated nationally and get reprinted every year without corrections ;-(

@Johnrennie hi

@Johnrennie are you free now?

Yes, I'm around for a while

@JohnRennie do you recall poiseuille's law?

@JohnRennie if no viscosity had been there then should the height different occur?

Do you mean the Hagen–Poiseuille equation?

@JohnRennie yep.

@JohnRennie $V =\frac{ \pi pa^4}{8\eta l}$

The HP equation applies when the flow is dominated by viscous forces. If you make the viscosity zero then the flow will be dominated by inertial forces and the Darcy Weisbach equation will apply instead.

Actually, no, hang on, the DW equation applies to turbulent flow. For inertial dominated flow you'd end up with something like Torricelli's law.

@JohnRennie i see . Now suppose you have a 20 cm^3 of kerosine ( density .8g/cm^3) and 30 cm^3 of water underneath . A bottle . If you hole the bottom then what will be the velocity of water. Now the energy at top layer of water has same pressure due to kerosine as the bottom of water so only the weight of water only acts on the bottom of water. So does that mean kerosine has no effect on the water's velocity.

I'm guessing this is inertial dominated.

@JohnRennie in this case if i take many layers of water the velocity would be different. So whats the mistake in concept im suffering ?

In that case the velocity of the water is given by equating the work done by the pressure pushing the water out and the change in kinetic energy of the water flowing out through the hole.

So what matters is the pressure at the hole.

And since the weight of the kerosine does increase the pressure the kerosine does matter.

@JohnRennie what had you said if you took Bernoulli equation. The top has no pressure , is there potential energy at top kerosine ?

@JohnRennie From when to when?

@ayc I'm here from around 05:00 to 13:00 UK time. I am around later in the day but I don't want to get into anything complicated later in the day.

@JohnRennie im kinda confused with potential energy and pressure energy .

Ohk........

@Johnrennie would you stay 13:00 uk time today?

@Nobodyrecognizeable what matters is the pressure at the hole. For a single fluid this is directly related to the potential energy change because pressure = $\rho g h$. In this case, where there are two layers, I would stick to just using the pressure.

@Nobodyrecognizeable I'm around until about 13:00. It depends on factors that are sometimes outside my control.

@JohnRennie ok . Thanks. Can i ask few more sums i would have to take screenshots of them.

Yes, you can certainly ask ...

@JohnRennie here goes the first one.

@JohnRennie the torque created must be counteracted by the table's mass.

I'll just be a few minutes. Something has come up at work.

@JohnRennie ok no problem if you come back please ping me.

@Nobodyrecognizeable because the table is circular its top sticks out beyond the legs. That means a weight placed right on the rim produces a torque trying to topple the table.

@Johnrennie the torque should be balanced in new or old center of mass?

Here M is the mass of the table and m is the mass placed on the rim.

@JohnRennie ok.

@JohnRennie the cm would be $md/(m+M)$

For worst scenario d should be 1 m.

Now im asking for which center of mass the torque should be accounted.

This is a straightforward torque balance problem. Take moments about the position of the legs, so the anticlockwise torque is $mgd$ and the clockwise torque is $Mg(r-d)$. Equate the two and solve for $m$.

@JohnRennie why $Mg (r-d)$ ?

Torque of body works from the center of mass.

@JohnRennie great figure all clear now . Lemme calculate.

That's what the toppling looks like.

@JohnRennie yep. Great figures btw. Im getting a relation : 20(1-d) = md

Which d should i account for?

Calculating the distance $d$ is simple geometry ...

@JohnRennie d= $\frac{20}{20+m}$

@Johnrennie should i take another leg and equalize the moments ?

That gives me $Mg(r-d) =mg(2r-d)$

I need to work now for a bit

@JohnRennie fine . If you have time again ping me back

@JohnRennie by solving above two equations i got m=20.

@

$r= r/\sqrt 2 +d$

@JohnRennie 47 kg . Option c is correct.

@Nobodyrecognizeable maybe you should review this question and make sure you understand what is happening here and why the calculation works as it does. This is the sort of question that you should find straightforward.

@JohnRennie Are you there

@Abcd morning :-)

@JohnRennie A block of mass m is resting on a wedge of angle $\theta$. With what acceleration should the wedge move so that the mass $m$ falls freely?

If the mass falls freely then it obeys the equation $\Delta y = \tfrac{1}{2}gt^2$. That means the wedge must accelerate sideways to make the height of the wedge decrease at the same rate.

@JohnRennie okay, then?

Let's draw a diagram ...

You want the horizontal acceleration of the wedge, $a_x$, to be such that $a_y = g$. Yes?

@JohnRennie ?

@JohnRennie $a_x$ will be same for wedge and block?

@JohnRennie yes

Like that ...

@JohnRennie ....In this question:i.sstatic.net/4oMa2.jpg In the (a) why is the potential difference across C2 Va-Vb?....It should be Vb-Va right?....could you explain?

@ayc let me finish Abcd's question ...

@JohnRennie ...its alright..I can wait..just ping me once you are done!

@Abcd and the question says the block falls freely so it accelerates downwards at an acceleration of $g$.

In effect it's like you are pulling out the wedge from under the block at a rate that matches the free fall of the block.

So if you consider the point on the wedge directly under the block then that point must accelerate downwards at $g$ so it doesn't impede the block's fall.

@JohnRennie ya makes sense.

@JohnRennie but the point under the block cant accelerate downwards anyway.

Hi guys, anyone can help me with Chernoff bound?

@Abcd what matters is the height of the wedge above the table immediately under the block. When I say point on the wedge I mean this height.

@Theantomc this room is meant for high school level problems. You may ask your question on maths main chat room or h bar.

@JohnRennie OK, then?

@Abcd thanks :)

@Abcd we need $d^2h/dt^2 = g$ for the block to fall freely. Yes?

one second. trying.

$\tan \theta = \dfrac h x$

$\dfrac{dx}{dt} = a$

$x = h \tan \theta$

$\tan \theta \dfrac{dh}{dt} = a$

$\tan \theta d^2 h/ dt^2 = da/dt$

$\tan \theta g = \dfrac{da}{dt}$

$da = g\tan \theta dt$

@JohnRennie ^^ ?

Start with the obvious relationship $h = x\tan\theta$. OK so far?

yes

Differentiate twice wrt time. The angle is constant so we get:

$$ \frac{d^2h}{dt^2} = \frac{d^2x}{dt^2} \tan\theta $$

hmm

Is that an English hmm or an Indian hmm? :-)

indian

OK, and we 've already decided that we want $d^2h/dt^2 = g$ so we get:

$$ \frac{d^2x}{dt^2} = \frac{g}{\tan\theta} $$

Oh my mistake above is I put $dx/dt = a$

Yes @JohnRennie $a = g\cot\theta$

Got it thanks.

Cool :-)

@ayc I need to work for a few minutes, but you're next in the queue :-)

@JohnRennie Ohk..but just ping me..I'll be waiting

@ayc back

@JohnRennie ..Yeah please answer that question

@ayc Appearing for JEE '19?

yes

So remind me what we're looking at and what you're asking (I've got the picture and I'm looking at it now)

@ayc how's mains prep?

@JohnRennie In (a) why is the potential difference across C2 Va-Vb?....It should be Vb-Va right?....could you explain?

@Abcd Physics and Math are fine and chem isn't good

@ayc are you from FIITJEE

yes

@Abcd you too?..which branch?

lol almost everyone here is from FIITJEE

@ayc yes am also from FIITJEE

@ayc what's your score and rank in AITS?

@ayc It just depends on the direction you're measuring the potential change. It doesn't matter what direction you take as long as you are consistent. In this case I'd guess the working assumes that $V_B > V_A$ so they are taking $V_B - V_A$ as the potential increase as you move left to right.

In that case $V_A - V_B$ is the potential change across $C_2$ as you move from left to right.

@Abcd My last one:3500(around

@Abcd Yours?

@ayc last one was 2000 PT 2 Advanced and before that all were around 1000.

@Abcd ohk

@JohnRennie Is ayc's question answered?

@Abcd @ayc is your question answered?

@JohnRennie In the fig:4.43 they say that Va-Vb is the potential difference across c1,c2,c3.If they take Va-Vb or Vb-Va as potential increase as you move from left to right then dont you think that in either case C2's potential difference has to differ from the others?..I still cant understand how can it have the same as that of the others

@JohnRennie Please dont mind the last message...I got it

Cool :-)

@Abcd your turn

@JohnRennie I have one more question..its gonna take time to upload pic...anser @Abcd first

@JohnRennie why $v_1 \ne v_2$ ?

I can't make out what's happening at the top of that picture ...

@JohnRennie block is connected to pulley

@JohnRennie pulleyis connected to incline

Ah ...

@JohnRennie I feel $v_1 = v_2$ because the string begin utilised by $v_1$ has to come through $v_2$

Like so?

@JohnRennie yes

If we call the length of the string $L$ then $AB + BC + CD = L$

waitt

@JohnRennie First I want the answer to this^

@Abcd that's what I'm about to answer

$v_2 = dAB/dt$ and $v_1 = dCD/dt$. Yes?

@JohnRennie ok

@JohnRennie yes

So $v_1 = v_2$ only when $dBC/dt = 0$ i.e. only when the length BC is constant.

@JohnRennie Isn't $\dfrac {d BC}{dt} = d{CD}/dt = d{AB}/dt$ because its the same string and same string cant have different velocities.

$AB + BC + CD = L$ so:

$$ \frac{dAB}{dt} + \frac{dBC}{dt} + \frac{dCD}{dt} = 0 $$

@JohnRennie i know you will differentiate it and stuff but I dont get why above argument is wrong

The change in the distances between the points B and C is not equal to the velocity of the string

It's equal to the sum of the velocity of the string and the velocity of the upper pulley

Actually, no, it's just equal to the velocity of the upper pulley

@JohnRennie why

The point B is fixed while the point C is moving horizontally with velocity $v_2$.

ok

The distance BC is equal to $x/\sin\theta$

yes

So $dBC/dt = d(x/\sin\theta)/dt$ which is going to get a bit messy because $\theta$ is also a function of $x$ not a constant.

It certainly isn't going to be $dx/dt$ i.e. it isn't going to be equal to $v_2$

@Abcd is the answer c?

@ayc wrong

@Abcd ..ohk

What happens if we take $\theta$ constant ...

@JohnRennie we cant

We would get $dBC/dt \sin\theta = v_2$

@Abcd ...My bad I think it should be d for sure..is it?

Hmm

It's got to be (d) but I've made an algebra error somewhere ...

Because when $\theta = 0$ BC is momentarily constant so at that point $v_1 = v_2$

And (d) is the only equation where this is the case.

@Abcd ..This is what I think:Let change in length of string CD be x1.Let wedge move by a distance x2.Chnage in length of AB will be x2.Chnage in length of Bc will be x2sintheta.......total change is zero...X1-X2-X2sintheta=0.......X1=X2(1+sintheta)...hence ...V1=V2(1+sintheta)

@Abcd ..I forgot to mention signs.....cahnge in length of CD=+x1.....change in length of AB will be (-x2) and cahnge in lngth of Bc will be (-x2 sintheta).

@JohnRennie Can I ask my question,now?

@ayc I guess so. Abcd has gone quiet ...

@JohnRennie i.sstatic.net/bbCmk.jpg

In that question ..How do i DETERMINE THE AMOUNT OF HEAT DISSIPATED IN CHARGING THE CAPACITORS

@ayc Where does it ask about the energy dissipated in charging the capacitors?

@JohnRennie, good morning

Are you free after ayc

@JohnRennie I just wanna know!

@ayc the problem with calculating this directly is that with zero resistance you get an infinite current for zero time. This means you can't simply calculate the dissipation.

You have to give the wires some resistance, write down the formula for the energy dissipated in that resistance then take the limit as the resistance goes to zero.

All very messy, which is why we never do it that way.

@harambe yes, and I think I've answered ayc's question so go ahead

@JohnRennie So when you take the limit answer wont be zero right?

@ayc correct

@JohnRennie imgur.com/a/sXuKtwx

Q26 (b) part

By sidways jerk, how can I calculate the time.. Intislly I thought to divide the length by velocity but it's not working

If you've done part (a) you have an equation for how the velocity varies along the rope

Yes

V=√gx

$$ v = \frac{dx}{dt} $$

Rearranging gives:

@JohnRennie I have once solved a question where A capacitor ,initially uncharged,was connected with a battery of emf v and the heat dissipated in charging the capacitors turned out to be (1/2)C(v^2) which was also equal to potential energy stored in the capacitor.....In our case we can find the equivalent capaciance and calculate the energy stored to be 108 micro joules....Now here also can I say that heat developed will be 108 microjoules only.......

$$ dt = \frac{dx}{v(x)} $$

@JohnRennie got it

I got the hint

Cool :-)

@ayc yes, that should work

The answer matches finally

@harambe BOOM! :-)

@JohnRennie...I have one last question...Just give me 5min

@harambe it is always worth remembering that you can do an integral either way. If you have the velocity as a function of time you'd do $\int dx = \int v(t)dt$, and if you have it as a function of distance you can do $\int dx/v(x) = \int dt$. This can be a very useful trick.

Yeah... This method slips my mind usually

@JohnRennie This is how I understand the charge distribution based on what you've taught me recntly:We know that electrons are emitted at negative end of E1.These electrons repel other elecrons in the path E1 to B to C5 thus leaving negative charge on one plate of C5.Now this negative charge accelerates electrons towards A leaving positive cahrge on the other plate of C5 ..Is this understanding correct?

Basically yes. Electrons from the negative terminal of E1 leave the battery and hit C5 where they pile up to form the charge labelled -Q1.

This negative charge pushes electrons off the other side of C5 so that side gets and equal and opposite charge +Q1.

I think I would use superposition to get the voltages on the capacitors rather than Kirchoff, though whatever you're used to is probably best.

@JohnRennie hi . How did you get that relation with d the legs should be placed at the circumference.

@JohnRennie Could you explain the charge distribution in another problem I have ..I simply can't think at all....Ill post the pic..give me 2min

@Nobodyrecognizeable the question says:

It specifically says the legs are placed on the circumference.

Illustration 4.20.....(b)

@JohnRennie circumference is the outermost point of table is n't it?

@JohnRennie ....Thats the question

@Nobodyrecognizeable yes i.e. the legs are at the edge of the table and arrange symmetrically i.e. equal spacing.

@ayc looks straightforward. My instinct is always to use superposition when I see two or more batteries but Kirchoff's laws will work too.

@JohnRennie could you explain the charge flow?....as inuitively as we understood in the last problem?

@JohnRennie but you say $ d =r(1-1/\sqrt 2)$ then how is the distance of the legs from center of table same as radius of table.

@Nobodyrecognizeable Huh? If the legs are on the edge of the table then the distance from the centre of the table to the legs is automatically equal to the radius.

@ayc I think it's dangerous to start using the simple intuitive arguments for complicated circuits. I advise against it.

Just use Kirchoff or superposition.

@JohnRennie when you are free, please ping me

@JohnRennie i'm (might be) misunderstanding r-d isn't it the distance between the table and a leg.

@JohnRennie Ohk!....Btw...Thank you for your time!

@Nobodyrecognizeable those are the distances

@harambe free

@JohnRennie where are the legs?

@Nobodyrecognizeable At the corners of the square

@JohnRennie imgur.com/a/pAjP6eq

Q33

Don't know how to proceed

I have just written their wave equation with amplitude A1 and A2

OK so let's take our time zero to be the moment the first wave is transmitted, so for the first wave we get $A_1 = A\sin(\omega t - kx)$.

@JohnRennie fine . Thanks professor. Have a nice day goodbye.

Okay

The second wave is transmitted a time $T$ later, so its equation is $A_2 = A\sin(\omega(t-T) - kx)$

Do we take their amplitude same?

It's not given in the question

The amplitudes are only going to matter for the last part of the question, and that part of the question specifically says the amplitudes are both 2mm.

Okay

@JohnRennie one more doubt

Part (a) asks for the phase difference. The phase is the argument to the sine function i.e. the bit inside the brackets. So the phase of the first wave is $\omega t - kx$ and the phase of the second wave is $\omega(t-T) - kx$. Just subtract the two to get the phase difference.

Is T time period here

Because in the derivation we discussed, it had T as time period

No, $T$ is the difference in transmission times i.e. 0.015 seconds. I wrote it as $T$ to keep the working general.

I agree it is a poor choice of symbol because $T$ is often used for the period.

Oh well :-)

@JohnRennie you did (t-T) because the second wave lags the first wave by time T?

Let me draw a diagram

Okay. Diagram helps a lot

The first wave is $A\sin(\omega t)$ - ignoring the $x$ part for now, take $x=0$

The second wave (blue) is transmitted 0.015s later so it starts a distance 0.015 along the time axis. Yes?

Yes

So to move the blue curve left to coincide with the red curve we need to subtract 0.015 off its times ...

that is, if we define $t' = t - 0.015$ we'll get the equation of the second curve to be $A\sin(\omega t')$

So you have substracted so that the two waves coincide at one time?

What I'm saying is that if we subtract 0.015 seconds the waves coincide, and if they coincide they're described by the same equation.

But you can look at this more simply ...

But that would mean they have no phase difference

We know the first wave is zero at time zero. Yes?

Yes

And we know the second wave is zero at time $t = 0.015$. Yes?

Yes

And if I write the second wave as $\sin(\omega(t - 0.015))$ is it zero at time $t = 0.015$?

Yes

So subtracting 0.015 works.

But if the waves councide, then how does phase difference work

The waves coincide if I define a new variable $t' = t - 0.015$ and write the second wave as $\sin(\omega t')$.

Because the two waves are then $\sin(\omega t)$ and $\sin(\omega t')$

Yea

So it's phase difference in time t reference but not in t'

I've succeeded only in confusing you, haven't I?

Kinda lol

I understood that waves coincidence if we define a new variable t'

OK, forget all that stuff about shifting waves. Just note that if we subtract 0.015 it works i.e. it correctly tells us that the first zero is at $t = 0.015$.

Yes

Actually, with this post you're basically there ...

We know the first wave is $\sin(\omega t)$.

And if the waves coincide with the new variable $t'$ that means in terms of this new variable the second wave must be $\sin(\omega t')$.

OK so far?

@JohnRennie I have understood the other wave starts at the=0.015 at the same place where the other wave started. By this time, the other wave has traveled some distance and there is phase difference. Can you comment if this is wrong

It's wrong

@JohnRennie okay. I had a feeling this was wrong

We are looking at both waves at the same point in space i.e. at the same $x$. That is, we are sitting at constant $x$ and comparing how the two waves change with time.

We are free to choose any value of $x$ to make our comparison, and it's obviously convenient to choose $x=0$.

Okay

Waves are confusing because they change with both t and x. But a lot of the time we are only changing one variable. So for example in part (a) we are taking constant x and looking at the two waves as a fuunction of time.

In part (b) we are going to take constant time and look to see how the two waves vary with x.

By constant x you mean x=0.. That means displacement of the wave at x simple Harmonically

You can choose any value of x to keep constant. The two waves have the same frequency and the same wavelength, so their phase difference does not depend on x. That means we are free to choose any value of x we want to find out what the phase difference is.

Okay

And choosing $x=0$ is convenient. When we get on to part (b) choosing $t=0$ is going to be equally convenient.

Okay. So at (a) , we coincided the t wo waves at x=0

The equations you wrote earlier

In (a) we are sitting at $x=0$ watching the two waves go by.

Okay

The first wave is $A\sin(\omega t -kx)$ so if we are sitting at $x=0$ and we are seeing how the amplitude changes with time at $x=0$ we get $A_1(t) = A\sin(\omega t)$.

Okay

And I think I've convinced you that the second wave is $A\sin(\omega(t-0.015) - kx)$ ...

Yes for x=0

So if we sit at $x=0$ we get $A_2(t) = A\sin(\omega(t - 0.015))$

Yes

(a) asks us for the difference in the phase. Can you now see how we get this?

Yes

So it is ... ?

-0. 015 omega

Yes. And you should now be able to guess what the answer to (b) is ...

Yes 4π

@harambe well, $4k$

Simplified the expression by putting values and it came 4π though yes 4k

For part (b) the second wave is going to be $A_2 = A\sin(\omega t - k(x +4cm))$

(I think I mean "x + 4cm" not "x - 4cm")

Yes it should be x+4

Cool, you're getting the hang of this.

Honestly waves confuse everybody because it's having the wave depend on both t and x at the same time that is confusing.

I am just proceeding how I proceeded at (a) . Took (a) as reference

@harambe yes, that's exactly the way to do it.

Yes.. I can officially say I hate Waves.

Next part is easy. Hopefully I can do that lol

Once you pass that critical point in understanding waves they get really easy. But it's hard work getting there.

There isn't really any way round this except to keep at it. Draw lots of diagrams!

@JohnRennie about resonance. The key is that wave has to travell twice so that it can add constrictively. But the next line directly says that length should be multiple of lamda. I am still confused why

The wave repeats after lamda so that's bwhy?

For constructive interference you want the phase difference to be a multiple of $2\pi$. Yes?

Yes

As you move along a wave the phase changes by $2\pi$ once every wavelength.

Yes

So if your reflected wave differs from the second wave by an integral number of wavelengths that guarantees the two phases differ by an integral multiple of $2\pi$.

Got it

@sammygerbil good afternoon. Are you free for a question

@harambe Yes. Has John Rennie finished for the day?

I believe yes. He is unavailable after 6pm IST.

ok.

@sammygerbil imgur.com/a/pRmai3l

Q39

Do I have to calculate resonamt frequency seperately in left and right wires

Also would this mean one end of the wire is open

@harambe There is only one wire, which is stretched between the two pulleys. Both ends of the wire and also the middle are nodes (zero displacement).

Okayl

Like that. The weights are just there to determine the tension.

Okay. Got it

I was thinking the wave will start from the point of attachment of masses... A silly thought

Hello, I have a problem. Can someone help me with it?

The diagram I had drawn for it was: