@Pizza That's not even possible...

Hang on, in what units?

@Stuti $m/s^2$

@mechanist If you want a mathematical proof see:

@JohnRennie Hi !

Hi :-)

I have not had a chance to look at it yet.

Ok :-)

@mechanist If you want a physical proof consider a constant vector field E through the surface. Then the flux through dA is:

dΦ = E . dA

So the total flux is:

Φ = ∫ E . dA

Since E is constant we can take it outside the integral:

Φ = E . ∫ dA

But since E is constant every field line in E that enters the surface must exit it again so the total flux through the surface must be zero and so:

0 = E . ∫ dA

And since E is not zero this is only possible if ∫ dA = 0

@Pizza @Stuti if you use Stuti's method and set g = 9.81m/s² the answer is indeed 4.055.

@Pizza Are you happy you understand the method? If not we can go through it.

@JohnRennie Can you help me with a mathematics problem ?

I can try ...

I have two inequalities. x > 3a^2 , x > (9a^2)/t how can I find at which interval t lies in ?

Isn't it just 3 ≤ t < ∞

If negative values are allowed then t can have any negative value

I too thought it is, But they have given [1/3 , 1) U (1, 3]

@JohnRennie I thought they said 4055, my mistake.

I must be misunderstanding the question ...

Are you sure both inequalities are greater than?

Yes

But what does it mean to say x > a and x > b where a and b are different? Both can be satisfied only when x is greater than whichever of a and b is greater.

Yes

I don't understand the question. Sorry :-(

Its Ok :-)

@JohnRennie When we say an equipotential surface it does not mean the electric field is zero there right ?

Correct, though it does mean E must be normal to the surface.

Yes

@JohnRennie Suppose if we have:

A gaseous spherical cloud of gaseous particles moves in a circular orbit about the common center due to their mutual gravitational force. is it same as a two body problem when we consider a single particle of mass m and the other particles' mass combined to be M, the centripetal force is given by GMm/r^2 taking the whole of mass M is concentrated the center ?

Is it a spherical shell i.e. all particles are the same distance 𝑟 from the centre?

Not a shell it has a density as a function of r

Suppose you are a distance 𝑟 from the centre so there is some mass outside and some mass inside.

Yes

Does this make sense?

Yes

So the orbital velocity at 𝑟 would be the same as if all the M(r) was concentrated at the centre.

Is that what you are asking?

Yes

so when we take a particle m at radius r. it is part of the mass M(r) right

So can we treat it as a two body problem ?

Yes, though we can assume m ≪ M(r) i.e. take the limit m ⟶ 0

I remember you saying gravitational field is not like electric field and treating field as field from COM is valid only for spherical symmetry

In other cases some multipolar theorem is used you said (I am not sure about the name)

Correct, but we assume there are many very small individual masses, e.g. like a cloud of gas where m is one molecule, and then we can approximate the gas as a continuous fluid so it is spherically symmetrical.

I thought taking the mass from the body itself is not valid

@KavinIshwaran Consider the case of only three masses. Then we get the three body problem and that is not solvable.

Yes

We can only solve the system when there are so many masses that they can be averaged out to give a spherically symmetric continuous distribution.

In that case m ≈ 0 so taking it out of M(r) leaves M(r) unchanged.

Ok :-)

If m is comparable to M(r) our approximation breaks down and the system cannot be solved.

Yes

@KavinIshwaran I had a look at that problem with the two inductors, and I wonder if there is a simple way to solve it.

@JohnRennie What do you think about my approach ?

The work done against the EMF induced in the inductors has to be equal to the energy stored in the inductors.

Yes

Does that match the answer key?

@JohnRennie ¹⁄₂MI₁I₂ ?

Actually that term is a guess :-)

It seems plausible.

Oh :-)

The answer key says a numerical value

55 mJ

Hmm, I get close - I get 50mJ

hmm

Aha! I Googled it and for mutual inductors there is no factor of ¹⁄₂. The energy is MI₁I₂.

And that gives the correct result.

See here

Yes !

I get it :-)

Look like my method is one of the worst thing to do :-)

I tried your method and it would work but leads to a pair of simultaneous differential equations that I didn't know how to solve.

@JohnRennie for a LR DC circuit the expression for voltage is ke^-Lt/R

I thought by including the mutual inductance I thought I could modify the expression

@JohnRennie I don't understand the question :-(

@KavinIshwaran Suppose we give the particle a small horizontal velocity u. Then at first it moves in a circle, but at some point the velocity decreases to the point where the centripetal force no longer balances gravity and then the string goes slack.

OK so far?

Yes

Once the string goes slack the particle moves in a parabola, so it will rise up to a peak then fall back, and at the peak it is moving horizontally.

The question says this peak occurs as the particle crosses the line AB i.e. at x = L/8 if we take the the initial position as x = 0.

@JohnRennie Can you draw a diagram ?

Give me a moment ...

Ah, I think this can only happen when the mass has rotated to above O. So the diagram is (deliberately?) misleading.

Oh !

@JohnRennie thanks for that link, somehow I never managed to find it while searching.

@KavinIshwaran Like this

Does this make sense?

Ah !

Yes :-)

@JohnRennie Hello:) Are you available to discuss a problem now?

I'm not sure how to do this question.

@Bumblebee Hi :-)

The decay needs to conserve various conservation laws.

For example it has to conserve lepton number. Yes?

Yes

On the left the lepton number is zero, and on the right the anti-muon has a lepton number of -1. So Z must have a lepton number of +1.

Yes?

How do you know that it is an anti-muon?

μ is a muon. Yes?

Yes

Ohhhh

And a muon is basically just a heavier electron i.e. the muon has charge -1 and the anti-muon has charge +1 just like an electron and positron.

Oh, so it's an anti-muon.

I understand now

OK :-)

Can you take it from here?

No, i'm still not quite sure

I figured out that the mediating particle has to be a W+ boson

Well X must have a lepton number of +1. Yes?

Yes

And the options given are for a neutrino or an anti-neutrino. Yes?

And the neutrino has a lepton number of +1 while the anti-neutrino has a lepton number of -1. Yes?

Yes

So X must be a neutrino.

Got it. Thanks :)

OK :-)

@JohnRennie Hi! Are you free?

Hi, yes I'm free :-)

@JohnRennie What is the relation(relative-narrow/broader) between the slits of YDSE and double slit diffraction?

Do you mean how does changing the width of the slits in the YDSE affect the pattern?

Uhm I had done something related during JEE time. But don't recall now.

I'm not sure what you are asking ...

YDSE is double slit diffraction, so there is no difference

@JohnRennie Uhm like in YDSE the intensity of maxima remains constant while in double slit diffraction the intensity keeps decreasing

What are 𝑠 and 𝑆? Are you thinking of the experiment where we place a single slit before the double slits to produce a coherent light source?

@JohnRennie What is the difference between YDSE and Fraunhofer diffraction from double slit? Is it only that in YDSE , the wavefront coming on the slits are spherical while in diffraction one they are plane?

But YSDE is Fraunhofer diffraction.

The wavefronts from any slit are cylindrical (not spherical) but in the Fraunhofer limit we are far from the slits so the radius of the cylinders is large and consequently they look approximately like flat planes.

And the YDSE is done in the Fraunhofer limit.

I don't understand what you think the difference between YDSE and double slits is. They are the same thing aren't they?

Yeah I know that diffraction and interference is same phenomenon.

By YDSE I mean

In which this is observed

and by fraunhoffer diffraction I meant this setup

in which this is observed

Ah, OK, so this is a single slit placed before the double slits. And that single slit acts as the source of light for the double slits.

You've got the meaning of Fraunhofer diffraction a bit confused.

Fraunhofer diffraction is the case where the screen is so far from the slits that the light rays from the slit can be considered parallel. When we do the YDSE we are always working n this limit.

In the diagram with the lenses those lenses are being used to refract the light from the slits to make it parallel so it's achieving the same result as the screen being far from the slits. The lenses aren't essential - they are just an experimental convenience.

@Wolgwang @JohnRennie What will happen if I reduce the size of slits(b) in this setup?

The last graph in your second diagram is a result of the width of the two slits being comparable to the spacing between the slits, and that would be the same in both the experiments.

@Wolgwang A single slit produces a diffraction pattern that is a sinc() function. Yes?

Yeah

That's what this is.

The width of the central peak and the spacing between the minima are proportional to 1/b

So as b increases the central peak gets narrower.

OK so far?

@JohnRennie Actually I am unable to derive this

The radiant exitance was found by integrating the source over the half hemisphere enclosing the disk (the disk doesn't emit radiation "backwards"):

Any idea about how to find the radiant intensity?

@Wolgwang The derivation of the diffraction pattern for a single slit is somewhat messy. If you really need the details there are various good articles you can Google, otherwise I would just accept the result.

(Yeah I am not able to derive this from that)

(From Jenkins)

What we are taught to do is put $\beta=\frac12kb\sin\theta $

I am just unable to think clearly about graph of Intensity vs $\beta$ and then use this relation to do something for b.

There is a good description here

@Wolgwang @JohnRennie What will happen if I remove the lense (on the right) here? Will I get a YDSE pattern?

The pattern is the same in both experiments (assuming the slit widths and spacings are the same). The lenses are just a convenience that allows us to move the screen closer.

So how are we getting two different patterns? :'-( I thought that in the YDSE setup the slit is infinitely narrow so we can't integrate like in the single slit diffraction .

The patterns are the same if the slits widths and spacings are the same.

If you use a a very narrow slit in one experiment and a wide slit in the other then that will cause the difference between them.

I don't think the YDSE specifically says the slits have to be very narrow, though we often do it that way to make it simpler.

Oh I thought this from the following paragraph

@JohnRennie Can you explain the paragraph when you get time?

Reading now ...

@Wolgwang This is talking about the definitions of the terms interference and diffraction and whether they mean the same thing or not.

(tl;dr they mean the same thing)

At the start of the paragraph it suggests the definitions that interference happens when two or more light sources interact, and diffraction happens when there is only a single light source.

OK so far?

Yes

But what the paragraph says is that this is a meaningless distinction.

Suppose we take a single slit and we put a very thin line down the middle. This line divides the slit into two, so now we have two slits. Yes?

Yes

But if we make the line very thin it will have no effect on the pattern, or at least such a small effect that we cannot see it.

So the diffraction pattern from the original single slit and the interference pattern from that same slit with a very thin line down the middle are the same.

Does this make sense so far?

Yeah

So really there is no difference between diffraction and interference.

We can imagine limiting cases

@Wolgwang So I can say that this is just a special case of of Fraunhoffer diffraction in which the parameters like slit width, distance etc are taken such as the factor sin^2beta/beta^2 =1, right?

e.g. two infinitely narrow slits are just two distinct sources.

@Wolgwang Yes

Okay.

Thanks a tons

:-)

OK :-)

@KavinIshwaran was this problem solved? I see a solution in the old chats but I'm not really able to comprehend it well enough. Can you solve it if you got it?

@Swan Hi !

I can explain if you want

@KavinIshwaran Go for it man

@Swan Hi !

When the point mass reaches the bottom, lets say it has attained velocity v

and the velocity attained by the cylinder be V

Now conserving the energy gives a relation, gR = v^2/2 + 2V^2

Now, when we consider the system of cylinder and the point mass, when the point mass is at the bottom total impulse given to the point mass is same as the impulse given by the mass to the cylinder. that is Mv

and the cylinder has a velocity V, total momentum change is 2MV. This is given also by the friction on the floor. so 2MV = J_f + Mv

@KavinIshwaran I think this is the part I'm having problems with. I would need to revise impulse. Thanks a ton :)

Ok :-)

@JohnRennie Hi !

@JohnRennie and @Stuti yes sorry , i thought in reality I thought he wrote $7 x 5$ but in reality it was $7.5$, anyway yes, it's correct thank you very much :)

@KavinIshwaran Hi :-)

@JohnRennie Sorry I wanted to discuss one question but I have to go now :-(

@JohnRennie Thank you for the clarifications :-)

Bye :-)

@JohnRennie Hello sir

Hi :-)

Are you free right now?

If it's quick ... it's getting late.

sarthaks.com/3266484/…

This question it easy but iam getting the wrong ans i tried dividing it into 3 capacitor about line passing through d/2 distance

That's what I would do.

Iam getting the wrong ans we can discuss it later if you are running late

Yes

Let's do it tomorrow then.

Ok np:) bye

Bye :-)

@JohnRennie

A block of mass $m_1 = 4kg$, resting on a frictionless horizontal surface, is attached to a spring, of spring constant $k = 242 \frac{N}{m}$, in its rest state . A second block of mass $m_2 = 5kg$ moves on the plane with speed $v_2$ and collides elastically with the first body in the direction of the spring axis. After the collision, the block attached to the spring compresses the latter by a maximum length $\Delta = 39cm$. Determine the initial velocity of the second block $v_2$.

could you help me solve this exercise?

However, I don't want you to give me the solution right away, I would like to understand how to tackle these types of exercises

The problem is that I have never done physics and now I find myself in difficulty

@JohnRennie, I have read that Cramer's rule cannot distinguish between infinite solutions and an inconsistent system. Why is that true? Plus, even the full method of finding solutions ie $Ax=B$ then $det(A)*x=adj(A) \cdot B$ fails when A is a matrix of all 1's and $B=\begin{pmatrix}1\\1\\0\\\end{pmatrix}$ as $det(A)=0$ and the rhs is 0 too, suggesting infinite solutions, but it's actually an inconsistent system.

Is there even a way to know when we will have infinite solutions?