Problem Solving Strategies

PCMSE

04:37

@JohnRennie Good morning sir, can you please see below this message?

John Rennie

@PCMSE Hi :-)

PCMSE

Hello sir :-)

John Rennie

2 days ago, by PCMSE

Sir facing slight confusion, using our standard convention v=-2u and applying mirror formula 1/f=1/v+1/u, we get u=f/2 and v=f. Whenever you are done with his doubt.

@PCMSE This?

PCMSE

Yes sir

John Rennie

What sign convention are you using? Cartesian?

PCMSE

04:44

Assuming positive towards the direction where light is incident

Say towards right

John Rennie

I was just Googling for the sign convention. I learned a different sign convention when I was a student and I alwats have trouble remembering the Cartesian convention.

What I found is (assume light is travelling from left to right towards mirror):

The distance to the object is negative because it's measured from the mirror leftwards to the object.

For a concave mirror the focal length of the mirror is negative because it's measured leftwards from the mirror.

Does this sound correct so far?

PCMSE

Yes sir :-)

John Rennie

Hmm, there seems to be Cartesian and a New Cartesian sign convention for mirrors, and they are different ...

What equation have you been taught for mirrors? 1/u + 1/v = 1/f or 1/u + 1/f = 1/v ?

PCMSE

1st one sir

John Rennie

04:59

OK, that's the New Cartesian convention I think.

I think the sign convention is clear now, so what is your question?

PCMSE

Yes sir, so when we plug in f, v, u in the equation without any numerical value and no prior knowledge as to where the image will form, do we have to keep the variables as it is or plug as - f, - v, - u?

Also m=-v/u=2, so v=-2u

John Rennie

Let's call the modulus of the focal length F. Then for a concave lens f = -F and for a convex lens f = +F. Yes?

PCMSE

Yes

John Rennie

So we just plug the symbol f into the equation, and the actual value of f can be negative or positive.

And the same applies to u and v.

We write 1/u + 1/v = 1/f and the values of u, v and f can be positive or negative.

So if I've understood your question correctly you keep the variables as it is.

Does this help, or have I misunderstood what you are asking?

PCMSE

If we keep the variables as it is, then f=2u,which is not matching

John Rennie

05:09

For what problem?

PCMSE

@JohnRennie The tagged one sir, above it.

John Rennie

2 days ago, by pi-π

@JohnRennie A concave mirror forms on a screen an image of twice the linear dimension of the object; object and screen are then moved until the image is 3 times the size of the object. If the shift of the screen is 20 cm determine the shift of the object and the focal length of the mirror.

This?

PCMSE

Yes sir

John Rennie

Let's write the modulus of the focal length as F, then because it's a concave lens the focal length is negative i.e. f = -F. And the object and the image are both real i.e. both to the left of the lens, so u and v are both going to be negative numbers. The magnification is M = -v/u and this is also going to be a negative number i.e. the image is inverted. OK so far?

PCMSE

Yes

John Rennie

05:18

Note that the question just says an image of twice the linear dimension of the object i.e. it says |M| = 2. It doesn't say if M is -2 or +2, though we know it's going to be -2 because real images in a convex lens are always inverted. Yes?

PCMSE

Yes

John Rennie

So -2 = -v/u and this gives v = +2u

2 days ago, by PCMSE

Sir facing slight confusion, using our standard convention v=-2u and applying mirror formula 1/f=1/v+1/u, we get u=f/2 and v=f. Whenever you are done with his doubt.

I guess you took M = +2, and that gives you v = -2u, which is wrong.

PCMSE

Is it better to say -2=-(-v)/(-u)?

John Rennie

I wouldn't do that. u and v and variables that can be positive or negative.

PCMSE

Okay :) got it. Got my mistake @JohnRennie

John Rennie

05:30

In this case you know that the values in u and v are going to turn out to be negative numbers, but that does not mean you should write them as -u and -v. Just write them as u and v and the equation M = -v/u will always work.

@PCMSE If it's any consolation, everyone gets confused about sign conventions! :-)

PCMSE

Yes sir, thank you very much :-) finally cleared

Bye sir :-)

Robin Singh

@JohnRennie Hello sir :-)

John Rennie

@RobinSingh Hi :-)

Robin Singh

spoiler alert for anyone who hasn't attempted jee 2020 paper yet

Sir, I solved this one in the train's frame but if I try it in ground frame then I am getting wrong answer

John Rennie

OK, so this is a Bernoulli equation problem?

Robin Singh

05:36

Yes sir

John Rennie

For gases Bernoulli's equation is complicated because gases are compressible. I guess we have to make the assumption that the air is not compressed i.e. treat it as if it were an incompressible fluid. Is that what you did?

Robin Singh

Yes sir, I did the same thing

Because they also said assume laminar and steady flow

John Rennie

So in the tunnel frame the air flows backwards past the train at a speed v₀ = -¹⁄₃v₁

Yes?

Robin Singh

Yes sir

John Rennie

So in the train frame the air is moving at -⁴⁄₃v₁

Hmm, that means the train and the tunnel are going to measure different pressures ...

Robin Singh

05:45

Yes sir, this is the problem :(

John Rennie

I'm not sure how to do this problem. With Bernoulli problems it's just the fluid moving i.e. the surfaces next to the fluid are all stationary (or all moving with the same velocity). In this case the surfaces (the surface of the train and the surface of the tunnel) are moving relative to each other so you would have a velocity gradient in the layer of air between them.

\that means you'll get a pressure gradient i.e. the pressure in the layer of air is not constant but depends on the distance from the two surfaces.

satan 29

is the answer 9? (although not sure about what to do about different frames measuring different pressures)

Robin Singh

Yes sir ans is 9

Maybe I'll ask it in the main site after adv

satan 29

ask now

Robin Singh

Oh ok then

satan 29

05:50

you might regret it later : )

Robin Singh

Let's hoard some rep xD

@satan29 They won't ask similar questions in consecutive years xD

Jk

satan 29

i remember having to leave this question due to time pressure :/

Lalit Tolani

@RobinSingh You may ask now , as it will help us also as future aspirants

satan 29

paper 2 was honestly a calamity

Robin Singh

@JohnRennie Does this mean that the pressure right next to the train is what we get when we use Bernoulli equation in the train's frame?

@LalitTolani Yeah but some people might want to attempt in the last month before jee

@satan29 ikr like all inpho lvl questions

John Rennie

06:02

@RobinSingh I assume so ...

Robin Singh

Then i guess that's what the paper setter might be asking and if we solve it in ground frame we won't get N as integer and the paper instructions say that N should be an integer

EVO

06:40

18

Q: Bernoulli's equation and reference frames

So I was thinking about this while driving home the other day. I've never been quite clear on why when you drive with the windows down air rushes into your car. I thought this might be explained by Bernoulli's equation for incompressible flow, but I ran into what seems to be a contradiction. If ...

@RobinSingh

Lalit Tolani

07:27

@JohnRennie Hello sir, free??

John Rennie

@LalitTolani I'm busy for a moment I'm afraid. I can ping you when I'm free if you want.

Lalit Tolani

@JohnRennie Ok sir no problem , please ping when you are free

John Rennie

08:12

@LalitTolani Hi :-)

Lalit Tolani

@JohnRennie Hello sir

John Rennie

What do you want to ask?

Lalit Tolani

There is an infinite sheet on left

A conductor on right

John Rennie

So the sheet induces a charge separation on the sphere?

Lalit Tolani

When the induced charges cancel the external electric field , electrons inside experience no net force, yes??

John Rennie

08:14

Yes

Lalit Tolani

What will happen now if we connect them with a conducting wire as shown?

Will charge continously flow in a closed loop?

John Rennie

Obviously not otherwise we'd have a perpetual motion device :-)

Lalit Tolani

What is it?

John Rennie

Inside the sphere the fields of the sheet and sphere sum to give zero.

Lalit Tolani

Yes

@JohnRennie Only free electrons will flow

John Rennie

08:17

Outside the sphere the fields of the sheet and sphere sum, but they do not sum to zero because they vary differently with position. So outside the sphere we have some field that is a complicated shape. OK so far?

Lalit Tolani

Yes , very much complicated!!

John Rennie

And your connecting wire is a conductor that is in this (complicated) external field, so the external field induces a charge polarisation in the connecting wire. i.e. both the sphere and the connecting wire get polarised.

OK so far?

Lalit Tolani

Yes

John Rennie

So when you connect the wire to the sphere it's not like shorting a component in a circuit. The wire has been polarised and develops a field as well.

Lalit Tolani

what do you mean by shorting a component in circuit??

John Rennie

08:20

The field of the wire will be complicated, but we know that inside the wire the field has to be zero just like inside the sphere, so inside the wire the external field and the field induced in the wire cancel so they sum to zero.

@LalitTolani if you haven't studied circuits ignore this comparison.

Lalit Tolani

So overall field inside wire due to induced charge of wire , induced chare of sphere and due to sheet will be zero

John Rennie

Yes. It must be zero because the wire is a conductor.

Lalit Tolani

Similarly for conducting sphere overall field inside sphere due to induced charge of wire , induced chare of sphere and due to sheet will be zero

John Rennie

Yes

Lalit Tolani

No net charges will flow after some time , everything will look steady though there will be still some thermal motion

John Rennie

08:24

Yes

Lalit Tolani

The doubt arose when my teacher was explaining that when we place a conductor in external electric field there is an instantaneous current

To get a steady current we must connect a wire at both the ends

Thankyou so much sir :-)

John Rennie

Yes

Lalit Tolani

Are you free to discuss more?

John Rennie

You do not get a current flowing in a loop round the sphere and wire.

Yes

Lalit Tolani

What should be done so that we can get steady current?

In sphere

Somehow we must neutralise those induced charges

John Rennie

08:28

You cannot do this with a static electric field.

You need either an external electric field that changes with time, in which case you have a form of transformer.

Or an external magnetic field that changes with time, in which case you have a form of generator.

Lalit Tolani

Ok , I need to learn more before I could understand that

John Rennie

A static electric field is a conservative field. Yes?

Lalit Tolani

Yes

John Rennie

So if you move an electron round a loop back to its starting point the net energy change has to be zero.

Lalit Tolani

Yes

John Rennie

08:31

And that means the potential of the starting and ending point has to be the same i.e. the net potential difference as we move round the loop must be zero.

Lalit Tolani

Yes

John Rennie

So no current flows because there is no potential difference to make it flow.

Lalit Tolani

Yes , that's a perfect argument :-)

No current can flow in a closed loop

John Rennie

You won't have learned this yet, but when you have changing electric or magnetic fields the fields turn out to be non-conservative.

Lalit Tolani

Hmm , I need to learn more for that

John Rennie

08:33

So in this case you can get a current flowing round a loop. You'll learn more about this in time (probably far more than you want to :-).

Lalit Tolani

Yes

I will ping you in another room

Robin Singh

09:55

@EVO Ahh I did suspect this

Thanks!

Adil Mohammed

15:36

@JohnRennie are you free?

caramalizedTomato

16:26

Even I too tried the same thing, but what I think the problem is bernoulli's equation has a KE part, and that can have different values depending upon the frame, so that does alter the pressure, what changing the frame does is what we would see in a wind tunnel with the air blowing in front of the train, now blowing air and static air do have different pressures.
So if you account for that you get the correct answer

Transcript for

Problem Solving Strategies