@YusufHasan Thank you :-)

@JohnRennie When I resolve the velocities I see the object moving in a helix

But when I resolve the magnetic field, I am seeing the object moving in a circular path. Am I missing anything ?

@KavinIshwaran Hi

Hi

How are you getting a circular path ?

@KavinIshwaran Hi :-)

@RonaldBecker I am visualizing the Magnetic force direction

@JohnRennie Hi !

We can resolve the velocity into a component parallel to the field and a component perpendicular to the field.

Yes. Then I see the particle moving in a helix

But If I resolve the magnetic field I see the particle moving in a tilted circular path

Then the parallel component tells us how long the particle takes to travel the distance 𝓁 because it is not affected by the magnetic field.

And the perpendicular component determines the radius of the the helix and the period of the circular motion.

Yes

@KavinIshwaran It's only the component perpendicular to the field that matters. You can ignore the component parallel to the field.

Yes. Suppose if we don't resolve the velocity, but if we resolve the field only, the component of field perpendicular to velocity will make the object move in a circle right ?

Hmm, it had not occurred to me to resolve the field into two components. I'd have to think about that approach.

In both the ways we must see the object moving in a helix right. But in the second case I am seeing the particle moving in a tilted circular path... I think I am missing something

@JohnRennie Hello sir

Hi :-)

Are you free right now?

Yes

Do you know what fraunhofer diffraction is ?

Yes. When you do diffraction problems they generally assume the distance to the screen is large compared to the distance between the slits. This is called the Fraunhofer limit. It simplifies the calculations in many ways.

What is use of lens in it ?

Whether you need a lens or not depends on how the apparatus is set up. I don't think there is a general answer.

@JohnRennie Hi !

Hi :-)

Did you find anything wrong in my approach ?

@JohnRennie Is used to make rays focus or make it parallel to screen

I had a quick go at visualising the geometry involved but got tangled up and didn't get anywhere. It isn't priority to be honest since resolving the components of the velocity is the obvious way to do this.

@RonaldBecker I'd have to look at a particular example.

@JohnRennie Ok then :-)

@JohnRennie does fringe width also decreases in diffraction?

I'm not sure what you're asking. The fringes are formed at specific angles so the width between the fringes depends on the distance to the screen.

I mean we know that intensity of fringes decreases as we move away from the center

Ah, OK.

But here is has shown that length of fringes also reduce

In general we get some equation line nλ = d sinθ so we get an equation for the sine of the diffraction angle.

If θ is small then to a good approximation the spacing between the fringes and the width of the fringes is constant.

This is the approximation we generally use in JEE questions.

But for large θ the sine function causes the fringe spacing to change. Exactly how it changes will depend on the experimental setup.

Hey sir I had problem in this question if you can help me out

@RonaldBecker So yes at large angles the spacing and width does change.

@AayushSethia Hi :-)

What is the question?

Just a sec

@JohnRennie Ah this figure shows fringes for large angle

blob:web.whatsapp.com/397eb072-766d-4659-a3e7-26301ea7180e

@RonaldBecker Yes

Ok thank you sir

Sorry just a minute

@AayushSethia OK :-)

How do I upload pictures

I am a fellow Jee aspirant

I think you can't upload pictures until your rep reaches some minimum limit. Can you upload the picture to imgur.com and post the link here?

A horizontally oriented tube AB of length l rotates with a constant angular velocity ω about a stationary vertical axis OO’ passing through the end A (Fig. 1.86). The tube is filled with an ideal fluid. The end A of the tube is open, the closed end B has a very small orifice. Find the velocity of the fluid relative to the tube as a function of the column height h.

I hope you can find the picture online

This is a variant of deriving Torricelli's law. Are you familiar with Torricelli's law and how it is derived?

For this question I have seen people applying Bernoulli equation at point B but although isn't point B inside the tube

I think at point B and A the vel. should be same so how can Bernoulli be applied ?

@AayushSethia Like this?

byjus.com/jee/ie-irodov-solutions-part-1-hydrodynamics

On this site they have given a pictorial representation

It is Q10.

Yes

The problem with using Bernoulli's law is that in Bernoulli's law we generally use the gravitational acceleration 𝑔 and that's constant. Here we have the centripetal acceleration rω² and that changes with distance from the pivot.

But the same principle applies.

I have found out the pressure difference between point a and b

But the problem is with vet.

velocity

If we move some mass of fluid 𝑚 from the inner surface of the liquid to the point B its PE changes due to the work done, and then if that mass squirts out from B the change in PE has to equal the change in KE. Yes?

Yes

ok

Suppose we have some small mass of fluid 𝑚 shown by the red rectangle at a distance 𝑥 from the pivot, then the centripetal force is F = mxω²

Yes?

yes

Yes?

@JohnRennie I think I found where the problem arises when we resolve the field.

Yes ... ?

As the angle changes during circular motion the initially parallel magnetic field component is no longer parallel and it will influence the trajectory I guess

But I will stick with resolving the velocity :-)

OK :-)

It will become too complicated if we do that way ?

I find it really hard to visualise how things move in 3D unless the trajectory is really simple.

So I always choose the simple way to do a problem :-)

But the am I right with the conclusion I said above ?

It tests If I really understood the ideas :-)

I'd have to think about it.

Ok :-) Please tell me if that idea is wrong.

@JohnRennie How to do this ?

Reading now ...

The phase of the supply is fixed and is the same for both branches.

Yes

So if we consider the LR branch we can use a phasor diagram to find the phase of the current relative to the AC source. Call this θ₁.

Then we can use a phasor diagram to find the phase of the RC branch relative to the AC source. Call this θ₂.

OK so far?

Yes

And the question asks for the difference between the phases of the two branches, and that is just θ₁ - θ₂.

Yes?

I'm not sure why they give the voltage of the source as that is not necessary for the answer.

I get it :-)

OK :-)

@JohnRennie hello sir

@HarjotDhillon Hi :-)

Why molar specific heat don't have dimension of mol^(-1) in its dimensional formula

A mole is dimensionless because it's just a number.

It's just Avagadro's number.

But then why Faraday constant have dimensions [M⁰L⁰TAmol-1]

There's a difference between units and dimensions. we need to make sure we are talking about the same thing.

Dimensions are what we use in dimensional analysis i.e. [L], [M] and [T].

A mole has no dimensions.

But there is a unit of a mole.

@HarjotDhillon Hmm, OK, I don't think the mol should be there ...

Thank you sir :-)

@JohnRennie Hi !

Can you draw phasor diagrams for that question ?

Yes

We are treating the branch as a LR series circuit ?

So θ = 60°

@KavinIshwaran Well it is an LR series circuit!

Yes

But I was confused if we can take an individual branch and calculate its phase and then subtract them

I am used to constructing phasors and then calculate phase differences in them

Both branches have the same voltage across them. So whatever happens in one branch does not affect the other branch (assuming the power source is ideal). Yes?

Yes

So in branch 1 we find the current lags the voltage by 60°

@JohnRennie Can you draw me a phasor diagram for this ? I am not used to parallel combinations.

But we re not doing any calculation of a parallel circuit!

What I'm saying is that if you remove branch 2 it does not affect the current in branch 1.

Yes?

Yes

So we remove branch 2 and then calculate the current in branch 1 using the same series phasor diagram that we are used to i.e. just the standard diagram with R on the x axis and +ωL on the y axis.

OK so far?

Yes

Now we go back to the original circuit and remove branch 1 leaving just branch 2. Again this does not affect the current in branch 2 so we can now use a standard phasor diagram with R₂ on the x axis and -1/ωC on the y axis.

Yes?

Yes

And from out first diagram we get θ = +60° i.e. 60° above the horizontal.

Yes

And from the second diagram we get θ = ... erm ... I haven't calculated it yet. Minus something.

tan(θ) = 1000/root3

So in the second branch θ ≈ -90

It's tan⁻¹ of a large number so it's about -90° to a good approximation.

Yes?

Ah yes

I wonder if there's a mistake in the question as that seems a weird angle.

The answer is given to be 150 degrees

So I think it is fine :-)

But one of the options is 150° and that's the answer we get.

@KavinIshwaran Yes.

@JohnRennie Thank you for the clarifications :-)

You're welcome :-)

Hello @KavinIshwaran !

@khaxan Hi !

Does lonepair-sigma-lonepair count as a conjugation?

What is the compound ?

just a second

Position 3 , remove hydrogen

OG question: compare acid strengths in decreasing order

It doesn't come under conjugation

Okk, thank you !

I guess you are taking carbanion to have a lone pair and it forms a conjugated system ?

Yep, that's what I was thinking

it does not come under conjugation.

Yes

Ok :-)