@KavinIshwaran yes I would like to know that

@HarjotDhillon Hi

@KavinIshwaran hello

@HarjotDhillon One minute

So this would be the intermediate after aldol reaction. yes ?

Yes

Can you see what is happening ?

Yes positive charge shift from secondary to tertiary

So next step double bond formation

Are you talking about acid catalysed or base catalysed aldol condensation?

base...

In base catalysed there is no carbonation formation

But in our example there is no two alpha hydrgens...

So carbocation will form during the aqueous work up (consider the OH group in the carbocation position), this will produce carbocation and rearrangement will happen.

But there are already many hydroxide ions present in solution so I don't think its going to be favoured by equilibrium unless our carbonation react further very quickly.

Aldol addition is kinetic controlled. but the condensation reaction is thermodynamically controlled. so I would say the condition is enough for the rearrangement to happen

Ok then there would be double bond formation

That is what I would answer

Normally it won't happen. I am considering the whole process into two, one is aldol addition in basic condition, and the condensation in aqueous condition at increased temperature

Oh is a bad leaving group in a basic condition. the base catalyzed mechanism you linked there undergoes E1cb elimination. but there is another alpha hydrogen in the place. but in our example it isn't.

Anyhow. It doesn't matter if the reaction happens in base or acid medium. Know that your example can undergo condensation

Tbh if this question again come in exam I would choose statement 1 to be true. If it was considered true in 2022 then there is no way they can mark it wrong now. I don't want to go from whole process of claiming and challenging.

Ok

Thanks for your help :-)

Like the linked q and a in Chemistry chat, there are exceptions like that. still a five membered ring can undergo condensation considering ring expansion

@HarjotDhillon Welcome :-)

@HarjotDhillon Are you writing NEET this week ?

Yes

On 5th may

All the very best :-)

Thank :-)

How dont know how the exam compares to JEE. But it is truly a challenge considering the amount of immense competition it has.

I*

@JohnRennie Hi !

Hi :-)

Did you find a solution?

We were discussing this question before. And I got the answer key

Looks like B and C are right

I can have a go at it using the ideas we discussed, but not right now as I have to go out soon.

But I found a solution for this question. and I couldn't able to understand it.

@JohnRennie Ok :-) we will discuss later

Post the link to the solution so I can have a look at it.

It looks vague

Hi @JohnRennie

@sanya Hi :-)

You've pinged at a bad time. I'm going out now. I'll be back around 8:30 p.m. Indian time.

Okay:) @JohnRennie

? Do you see any fualts in it? Otherwise, can you point me to the correct Q&A site to ask this question?

@tush I suspect this is just a typo, but it should be:

$$ \Phi = \sum_{\lambda_1}^{\lambda_2} \Phi_i \delta\lambda $$

But you've basically written ∫Φ(λ) dλ so why not just do the numerical integration?

@JohnRennie Correct, it is a typo

@JohnRennie I want to first understand what I do before letting the computer do the dirty job :)

@JohnRennie What if the binning is not constant?

If the binning is not constant instead of a fixed δλ between each reading you will get a variable δλᵢ

I see. Thanks!

In that case just sum Φ(λ)δλᵢ

Hello. I have some questions about this problem:

"A platform of M kg moves with a velocity of 2 m/s on straight horizontal rails without friction. On the platform, and at rest relative to it, an elastic spring of negligible mass and a spring constant of 1000 N/m is arranged in the direction of motion, initially compressed by 20 cm, and two bodies, attached to the spring, one at each end, of masses m1 and m2 kg, respectively.

At a certain moment, the spring is released, causing both bodies to start moving on the platform in opposite directions, being shot off. Assuming no frictional forces exist, calculate the velocity of the three objects."

First of all, is not clear if the spring is attached to the platform or if it isnt.

If it isnt, I guess we can solve the problem by doing $m_1v_1 + m_2 v_2 = 0$ and $ 1/2 k x^2 = KE_1 + KE_2$

But if it is attached, the speed of the platform its going to change. Naming it $v_p$ we have: $m_1v_1+m_2v_2 = Mv_p$ and $1/2 k x^2 = KE_1+KE_2 + KE_p$

@JohnRennie please ping me when you're free :)

@sanya I'm free now :-)

But I miss one equation, since i have three variables ($v_1,v_2$ and $v_p$) but two equations

How could I get the third equation?

(I am not sure if it is even solvable)

Posting the ques....

Position vector of q is î+j(hat)+2k(hat) and position vector of 4q is 2î+j(hat) +3k(hat). Find the position vector of neutral point and find the charge Q that could be placed here so that q is in equilibrium

The image is whats done and I got the positon vector of the neutral point after this I need the magnitude of distance between the neutral point and q for the second apart am I supposed to find the magnitude twice and see which is smaller and then use that or is there a faster way to do this?

@sanya You have the charge q at (1, 1, 2) and 4q at (2, 1, 3) and you find the neutral point. Let's call this (x, y, z).

OK so far?

Yes

Then the vector from the neutral point to q is (x-1, y-1, z-2). Yes?

Yes

Yes but the position vector of the neutral point has √q1 and √q2 in the equation so wont I have to take out the magnitude twice once with positivel sq root and once with negative

Oh I see.

I think there's only one neutral point ...

The neutral point lies on the line in between q and 4q

And it's one third of the way along that line from q.

Yes?

Yes because when I tried to find out the magnitude of position vector of Q one of it was √-something so we dont take that..

The vector from q to 4q is (1, 0, 1) so the vector from q to the neutral point is oe third of this i.e. (¹⁄₃, 0, ¹⁄₃)

@JohnRennie How'd you see that ?

Call the distance from q to 4q 3s

Okay

Then on third of the way along this is 𝑠

Okay

So the distance from q is 𝑠 and the distance from 4q is 2𝑠.

Then the force from q is kq/s² and the force from 4q is k4q/(2s)²

Yes?

Force or field?

@sanya Force on unit charge, so yes it's the field since the force on a unit charge is equal to the field.

The point is that k4q/(2s)² = kq/s²

So the forces from the two charges are the same so this is the neutral point.

Ah ok

@JohnRennie Then I dont have to do this

I'm no sure how you ended up with a √q in the equation. But there is only one neutral point so you only need to do one calculation.

@Odestheory12 If I understand the question correctly the spring pushes the two masses away from each other and they shoot off the ends of the cart. Since there is no friction between the masses and the cart the velocity of the cart is unaffected.

Yes?

But that is if the spring is not attached, right?

I guess the spring could be attached at its mid point and the outcome would be the same.

But I'd guess the spring is not attached anywhere.

So the two masses fly off the ends of the spring leaving the spring lying on the cart.

Yeah, if the spring is not attached I can solve it with conservation of momentum and energy, but I was interested in the case of the spring attached

Do you mean the spring attached to the two masses, or the spring attached to the cart?

attached to the cart

At it's mid point?

Yeah

Then that would make no difference since the mid point of the spring doesn't move relative to the cart.

You are right, the reaction force is independent of the mass

@JohnRennie if you do the calculation I get $$ \frac {\vec I (2√q_1 + √q_2) + \vec j ( √q_1 + √q_2) + \vec k (3√-q_1+2√q_2)} {√q_1 + q_2} $$

Not sure if I made a mistake with the tex the formatted version is not coming through :(

Then its simply $m_1 v_1 + m_2 v_2 = 0$ and $\frac{1}{2} k x^2 = KE_1 + KE_2 $ right?

It's fine here. If you're on a phone I don't think MathJax works.

@JohnRennie Oh I see why then

@Odestheory12 If you work in the rest frame of the cart, then in this frame the two masses are initially stationary.

So after they have moved apart their total KE has to be equal to the energy stored in the spring and their momenta have to be equal and opposite.

Yeah, exactly ^^

@Odestheory12 So yes, this gives the velocities of the masses in the rest frame of the cart.

Thanks, much appreciated.

Then you need to add the velocity of the cart to get the velocity in the ground frame.

@Odestheory12 You're welcome :-)

@sanya We can go through the calculation if you want ...

Yes please

Also note that its not -q in the k vector bracket and in the denominator its √q2 not only q2 :( Sorry im using it for the first time should have been careful

@sanya Is this the method you used?

Yes

Sorry for the delay in messages my wifi is unstable

So if call the q charge 1 then x₁ = (1, 1, 2) and x₂ = (2, 1, 3)

Yes

4|x₁ - n|² = |x₂ - n|²

It feels as though this method is going to get very messy ...

@JohnRennie yes

@JohnRennie it was fairly ok I didnt substitute though I just used the general equation..so maybe

How did you get square roots of charges in the final result when there aren't any charges in the equation we've got so far?

I have to say I don't see how this method works ...

I think it's kind of the same as what I did but more complicated.

@JohnRennie if we go by this we take the square root of the first equation and then divide it with the second equation

Oh I see. But here the charge q cancels out of the first equation doesn't it?

q₁ = q and q₂ = 4q so we can divide both sides by q to get just:

1/|x₁ - x₃|² = 4/|x₂ - x₃|²

Oh I was trying to get the magntiude with q1 and q2 so I thought of substituting at last but I see that would make it complicated

And we only take the positive root because magnitudes are always positive.

Yes got it

OK :-)

Sorry feels like wasted your time , this was simple . Thankyou

You absolutely have not wasted my time!

For you to see how we old timers simplify problems like this is very important!

You are always welcome to ask questions like this :-)

I see :)) thankyou again

@JohnRennie hello sir

@JohnRennie we discussed a problem where we had to prove if a charge was in stable or unstable equilibrium some days back... I just got the answer sheet for it back and it states that it is in stable equilibrium in the line but if we consider all directions (3d space) it is in unstable equilibrium...can you tell me why's that? Its not like theres a charge placed in any other dimension in whose field the charge can move, in any other dimension I would assume df/dx=0,hence neutral eqbm

@JohnRennie it was this conversation