@JohnRennie Hi :)

@JohnRennie sir, when you are free just have a look at my question. There are a quite a lot of models that describe why refraction happens. like fermat's principle of least time, huygens principle and marching soldiers analogy. but non seem to make sense. what is the exact reason that light bends ?

@KavinIshwaran Light bends because the light wave has to be continuous at the interface i.e. the magnitude of the electric field cannot change discontinuously at the interface.

can you explain ?

It's a long explanation and I'm doing something else at the moment.

no problem !

pls explain when you are free !

@JohnRennie Hi :-)

Are you free?

Hi, yes I'm free :-)

> A block A of mass 4kg is placed on another block B of mass 5kg, and the block B rests on a smooth horizontal table. If the minimum force that can be applied on A so that both the blocks move together is 12N, the maximum force that can be applied on B for the blocks to move together will be :

@JohnRennie What's wrong?

Should that be:

> If the maximum force that can be applied on A so that both the blocks move together is 12N

Otherwise the question doesn't make sense.

@JohnRennie Hi :)

@Wolgwang I get 15N as well.

@JohnRennie Let me check.

@KavinIshwaran Hi

@JohnRennie Hope I am not disturbing you :)

@KavinIshwaran I get lots of questions from students about coursework, and my priority has to be those questions because they're important for the JEE. I'm happy to do more general discussions, but only if they don't interfere with me answering coursework questions.

@JohnRennie no problem !

@JohnRennie can you explain this ?

Not now. Probably not this morning as I have lots of questions waiting for me this morning.

Oh thats fine ! we will keep this tomorrow or whenever you are free

@JohnRennie Yes, It is maximum

@JohnRennie What is happening in this solution?

I'm busy answering another question at the moment. I'll ping you when I'm free.

Ohk :)

It's saying that the frictional force is 12N, but I don't understand how they come to that conclusion.

If the frictional force is 12N then 27N is the answer because we apply 27N to B and 15N is used accelerating B leaving the 12N friction to accelerate A at the same rate.

But how they conclude that the frictional force is 12N from the starting point:

> The minimum force that can be applied to block A so that both the blocks move together is 12 N.

completely escapes me.

@JohnRennie Thanks :-) These websites sometime write anything.

@JohnRennie Hi , can you please help me with it

on the phone ...

@JohnRennie ah ok no problem , just have a look whenever you're free :)

@AshishAhuja free for a second?

yes sure

come to my room

@YashAgrawal i guess we need some kind of symmetry argument

I think rotational symmetry along with assigning potentials should work, let me try it out.

@satan29 maybe , I tried using it to assign potentials to the points but I couldn't get any further ;/

@YashAgrawal answer is (D) 11?

@AshishAhuja yes

From the symmetry and the constraints on charge we can assign these charges to the capacitors.

(I haven't assigned the signs, just the magnitudes)

I'm guessing a few Kirchoff's later we can find the values of all the charges.

@JohnRennie Sorry but I am not properly getting how you assigned these charges

with refer to JR's diagram, here's what I did: imagine attaching a battery across 1, 4 of emf E. Let potential of point 4, i.e V_4 = 0, then V_1 = E. Assume V_3 = x, V_5 = y, then due to rotational symmetry V_2 = E - y, V_6 = E - x. Now we need to solve for for the two variables x, y which we can do by using the fact that sum of charges on any 3/4 adjacent plates of neighbouring capacitors = 0.

@YashAgrawal We are connecting our circuit to points 1 and 4, so charge can only flow into or out of the system at points 1 and 4. Yes?

yes

So at point 1 a charge Q₁ flows onto one of the capacitors and Q₂ flows ono the other. We don't know what these charges are, but that doesn't matter.

OK so far?

yes

If a charge Q₁ + Q₂ has flowed in from point 1 an identical charge Q₁ + Q₂ must have flowed out from point 4. Yes?

yes

And the circuit is rotationally symmetric i.e. if we rotate it 180° it stays the same. All that happens is the direction of the voltage is swapped. Yes?

right

So the charges on C₁₂ and C₄₅ will have the same magnitude because they are exchanged by the rotation of 180° (C₁₂ means the capacitor between points 1 and 2).

And you can pair up all the capacitors expect for the middle one.

In fact the middle one has to have a zero charge because the rotation sends it to itself.

I need to go soon ...

So we need to hurry this up.

yes

@JohnRennie sir you may write it all in a go

That's how I can label the two rightmost capacitors the same as the two leftmost capacitors.

ah ok

Now look at point 2. No charge can flow in or out of this point, so the net charge on all the capacitors connected to this point must sum to zero. Yes?

yes clear

got how C(23) has Q1 + Q3

Cool :-)

And that's it. We now have the charges on al the capactors except for the middle one, and the middle charge has to be zero.

Main confusion was being in that swapping of charge in end capacitors

OK. Are you happy with it now?

@JohnRennie yes

I'll leave you to grind through the Kirchoff calculations. I need to go.

@JohnRennie yes , I was not able to see that rotational symmetry initially

@JohnRennie ok , bye :)

Bye :-)

When a knife penetrates into cardboard travelling at some distance then knife travelling from some distance will gain some velocity due to gravity before entering cardboard. So what will be the acceleration during penetration. Doesn’t the Newton third law give acceleration?

@cOnnectOrTR12 Assuming the knife is in freefall ?

@cOnnectOrTR12 "When a knife penetrates into cardboard travelling at some distance then knife travelling from some distance will gain" I dont undertand what it means though

@AshishAhuja yeah got it , but I wonder that middle one would have then an 2V potential difference , contradicting the statement of JR that middle one (b/w 3 and 6) have no charge

@KavinIshwaran yes free fall. A knife of some mass is dropped. It drops from some distance and penetrates some distance inside

ok

@cOnnectOrTR12 then

We use Newton third law to calculate v that is before entering cardboard. That will be initial velocity

@cOnnectOrTR12 that depends upon the distance

So to calculate a we use that eqn again with v as u

@KavinIshwaran yeah yeah

This a is the acceleration of knife while penetration

But my acceleration is wrong

@cOnnectOrTR12 acceleration at the instant before penetrating the knife ?

@KavinIshwaran no. That will be -g. Acceleration during penetration?

@cOnnectOrTR12 after it starts to penetrate the knife no longer accelerates

it will decelerate

Yeah that

I calculated a using Newton third law. But that is not the acceleration during penetration

@cOnnectOrTR12 ok, if i am ryt, you are asking the deceleration after it startes to penetrate

I mean third eqn

Yeah!

But the information is not enough to calculate the deceleration

@cOnnectOrTR12 do you have the exact question ? can you send a pic ?

Distance is given. How deep it penetrates

@cOnnectOrTR12 send a pic of the question

An open knife edge of mass 200g is dropped from height 5 m on a cardboard. If the knife penetrates distance 2m into the cardboard, the average resistance offered by the cardboard to the knife edge is—

@cOnnectOrTR12 ok for this you dont need newton's third law

you wont get an answer

Third eqn!

@cOnnectOrTR12 i guess you are taking about equal and opposite reaction ? am i ryt ?

That too. Yeah. We need to find force for which we need a. Mass is given.

ok wait a min

😊

@cOnnectOrTR12 sorry got sidetracked

v^2 = 2as ryt ?

since u = zero

so a = 10 and s = 5 so v = root100 = 10 ryt ?

where v is the velocity before the knife strikes the cardboard

it is given that it travels 2 m

Yeah! a=-g and s=-5

@KavinIshwaran ok, yeah

no, s = 5 only, and a = 10 not -10 since the acceleration is along the direction of gravity

before striking the cardboard

@cOnnectOrTR12 now, v is the initial velocity before it comes to rest ? yes ?

@KavinIshwaran how? we are using upwards is +ve. So!

@cOnnectOrTR12 thats when you are accelerating upward, but in this case we are accelerating downward.... clear ?

@KavinIshwaran but using upward is +ve sign convention. Acceleration due to gravity should be -g

s is going downwards. So s=-5

@cOnnectOrTR12 that often depends upon the reference we take, but now we just need to take g as 10

@cOnnectOrTR12 s is not going downwards. the knife is... the s is just the distance between the knife and the cardboard

Yeah! And I took standard reference as upwards is +ve.

@KavinIshwaran I mean body.

Proceed!

v will be u. v=0. Then?

then again take the third equation

Yeah

since v = 0

a=25 I remember

But it’s wrong

-(10^2) = 2a2

@cOnnectOrTR12 wait a min

@cOnnectOrTR12 what is the answer the book says ?

7 N

I guess they are asking the force, not deceleration

@cOnnectOrTR12 ahhh, yes as i expected

YeH

no problem, the a = 25... till here its clear ?

Yes 👍🏻

-25

initially then the weight of the knife is considered as a force ? yes ?

Weight is force at all times

just a min I am getting a call

I didn’t call you. Who’s calling

in phone call

😊

ok

shall we continue ?

Bilkul

lets so the net force acting on the knife is 0.2*25 = 5n ryt ?

Yeah

weight of the knife is 0.2*10 = 2n, since it is against the cardboard, it is -2n, yes ?

Yeah -2N.

so x - 2N = 5N, where x = the resistance/ force offered by the knife, do a simple math, your answer is 7N ? got it ?

@cOnnectOrTR12 any doubt ?

@KavinIshwaran But when we started calculating v then we put a=-g , How is it possible without a defined weight it has reached a certain v. Again when we calculated a how is it possible it has reached some distance without some weight? So I think there is some weight element included in the formula. No need to find the weight separately

@cOnnectOrTR12 weight is a force yes ?

Yes

F = m*a m = mass a = the body's acceleration ? yes

Yeah

but we cannot say the otherway around, like if the object is accelerating at 2 m/s and have a mass 1kg, its not like when it hits an object the force it exerts on it is 2N

Then ?

if the object is accelerating at 2 m/s^2 and and have a mass 1kg, the force acting on that body is 2N, not like when it collides it exerts a force 2N

But from start to finish there is one force acting and that is gravity.

no

after it starts to penetrate, the frictional force acts on knife

Leave that!

ok

Then there is only one. Due to this gravity it accelerates. And due to this acceleration there is a generation of another force. So there are actually two forces. Weight and the force due to acceleration

hmm... its a wrong interpretation

@YashAgrawal if the middle capacitor has no charge, then V_3 = V_6, which seems... unlikely.

@AshishAhuja yeah , I am getting difference between them to be 2V

@YashAgrawal ..how are you getting 2V? Did you assume the EMF of the battery to be a certain value?

@AshishAhuja getting the potential "difference"

But yes I did assumed battery potential to be 10V

@AshishAhuja what you got as x and y ?

@YashAgrawal ah ok. I haven't done the calculation but yeah it seems wrong to me that the central capacitor will have no charge.. maybe JR made a mistake in a hurry.

@YashAgrawal x = 2E/5 and y = E/3

So point 3 and 6 would have potential 2E/5 and 3E/5 respectively

Difference being E/5 (= 2 for E = 10V )

isn't V_6 = E - x = 2E/3?

E - x = E - 2E/5 = 3E/5 ??

oops yes my bad

yeah yeah it does come out to be E/5

Yes

anyway, to my understanding the rotational symmetry here simply means that V_12 = V_45 and V_16 = V_43 (magnitudes). I don't think it implies that V_36 = 0

@AshishAhuja V_12 = V_45 and V_16 = V_43, can you explain its reasoning mathematically , I think I understood it more as a fact

well here's my understanding: if you rotate the diagram 180 degrees about an axis perpendicular to the plane of the diagram, you get the same circuit diagram. This to me intuitively convinces me that V_12 = V_45 and V_16 = V_43

if you're asking for a proper mathematical proof, I've never seen one but one could probably do it by assigning variable potentials to each of the junctions and solving (would be like 4 variables lol)

I agree that 12 and 45 overlaps , but how does that intuitively convinces that there potential difference would be same ?

this is a wheatstone bridge.. let's just assume that instead of the galvanometer we have a resistance R (sorry for this lol but I don't know how to draw a neat circuit diagram for what I want to show)

yes

Let use take some values of the resistors such that R_1 = R_x = A and R_3 = R_2 = B and we know that the bridge resistance is R

and now we have connected this setup to a battery as shown in the diagram

ok

are you convinced that the current through R_1 = current through R_x and similarly current through R_2 = current through R_3?

(this is a simpler case of rotational symmetry)

yes convinced about current

we can assume potential of two junctions to be x and y and then apply nodal equations to get symmetry result

I had seen it before if that's what you wanna tell

now instead of these resistors we could also have capacitors, i.e C_1 = C_x = A and C_3 = C_2 = B and instead of the galvanometer we have some random capacitance, do you agree that charge of 1 = charge of x and charge on 3 = charge on 2?

@AshishAhuja yeah

had to, no other way out of charge distribution

ok, so looks like you are convinced of this because you have solved it by using potentials/nodal method/whatever. Now, if you notice rotating this diagram by 180 also gives the same diagram, so this also has rotational symmetry, right?

yes agreed

ok so similarly, the original question you have, when you rotate the diagram by 180, the two capacitors which overlap (essentially they are indistinguishable) will have the same potential across them.

just like in this simpler case the two indistinguishable capacitors have the same potential, which you agree with.

I think I got what you wanted to emphasize , since charge distribution would be symmetrical so by Q = CV , potential distribution also have to be symmetrical

since Capacitance is same

yes, that is true.

so as a thumb rule can we say that anyhow rotating , flipping circuits we concide with orginal circuits and then pd b/w those conciding points would be same ?

tbh with most forms of symmetry I have absolutely no clue how to actually prove them, its more or less just an idea which gets stuck in your head. I have no clue how to prove this rotational symmetry either, but without using this the question will become too lengthy.

@YashAgrawal yes, I would say so. Works even with resistors.

@AshishAhuja yes , in resistors it's actually easier to believe because current having only magnitude we can connect our battery reverse to prove symmetry arguments

@AshishAhuja thanks for helping out :)

@KavinIshwaran why?