Problem Solving Strategies

Ice Wizard

03:56

@Priya use V/I

Mayank

04:36

@JohnRennie Good Morning Sir

@JohnRennie I tried it.

John Rennie

@Mayank hi :-) Great picture. What was the angle?

120°

14 images?

Mayank

30 degree

12 images

@JohnRennie

Sorry for delayed response. I was in the middle of an exam.

John Rennie

Hmm 12 = 360/30 not 360/30 - 1

Mayank

Yes

There are many such cases.

John Rennie

04:52

What do you get for 120°?

(with the object not on the divisor?)

Mayank

Wait

@JohnRennie 2 Images

John Rennie

OK, that works then.

Mayank

Lemme send the image

@JohnRennie

John Rennie

OK, the formula works for that

Mayank

It doesn't work always.

Like in case of 60°

@JohnRennie How many images are there in 2nd photo?

John Rennie

05:06

Hmm, OK so it depends on the angle of view i.e. you can see 5 or 6 images depending on where you put your head.

I think I found that from the simulator. If you use the "observer view" and move the observer around you can change the number of images.

Mayank

So what answer should be given in exam?

@JohnRennie

John Rennie

Let's wait until you have some practice questions and we'll see what sort of questions the JEE actually asks

Mayank

^^

John Rennie

05:54

@Mayank I simulated the 60° mirrors on the optics simulator site, and I found that the number of images was very sensitive to the exact angle of the mirrors. For example this gives 5 images as it should:

But with only a tiny change to the mirror angle I get six images:

This looks very much like your picture i.e. the image farthest from the observer is split in two, so I wonder if what you are seeing is down to a small error in the mirror angle.

I've marked the "split" image in red.

Mayank

@JohnRennie Such a small degree can split image into two

:-O

John Rennie

Try it for yourself. Move your mirrors slightly and see how it affects the splitting of the image farthest from you.

Mayank

@JohnRennie So in 30° setup

The farthest image split

John Rennie

Let me try it with the simulator ...

Mayank

Ok

John Rennie

06:09

I can get the same to happen at 30°

Only a tiny movement of the mirror is needed to cause the splitting.

I suspect that n = 360/θ - 1 is correct, but only if the angle is exactly right.

Mayank

This must be the case

@JohnRennie

John Rennie

Yes

Mayank

That small mirror gap can't be removed whatever the angle is.

@JohnRennie

John Rennie

Real mirrors have a layer of glass on top of the reflecting surface, so you can't make the mirrors meet like you can in the simulator.

Mayank

Hmm

Safdar

06:19

This might be why this is happening, sorry for interupting

https://www.youtube.com/watch?v=ZAufb115KGU

This might be because when the angle changes slightly. it shifts from case 1 to case 3 (in video shared), so number of images goes to $\frac{360}{\phi} - 1 \to \mathrm{floor}\,\left(\frac{360}{\phi}\right)$ which increases the number of images by 1, moving it in the other direction would not change the number since the floor also decreases by 1.

John Rennie

I think n = 360/θ - 1 is correct, but it has been surprisingly interesting looking into this.

Mayank

Why some people say that Number of images(n)=360/Ø when n is odd and not placed at angle bisector?

John Rennie

Well that's θ = 360/3 and there are only two images as the formula says ...

Mayank

@JohnRennie So they are wrong.

John Rennie

1 hour ago, by Mayank

Your own experiment shows that.

Mayank

06:24

What happens at angle bisector?

Can you simulate that case?

@Safdar Why does case 3 formula work?

John Rennie

Still only two images

Mayank

Oh!!! They are wrong.

Many people are wrong to be precise.

Safdar

@Mayank I don't know the proof. I've not focused much on physics (not my forte)..

Mayank

You got a great percentile in physics in jee main.@Safdar

Safdar

@Mayank Correction: not that interested in physics. If you notice, my percentile for chem (what I actually focus on) was the lowest.

Mayank

06:31

@JohnRennie Thanks a lot. Now I can proceed to next topic peacefully.

John Rennie

:-)

Mayank

@Safdar These are just marks.

Safdar

@Mayank Exactly my point. thanks for noticing ;D

Mayank

:-) Best of luck for advance IIT-H student.

@Safdar

Mayank

06:44

@JohnRennie But I still didn't get the derivation of that formula.

John Rennie

I don't think there is a simple derivation ...

Mayank

Yes

I want to know who first made this formula and how.

Maybe by observing all cases?

John Rennie

I have no idea. I would have to Google for it.

Safdar

1

Q: How to prove the number of images of two mirrors inclined at $A$ is $360/A -1$

Consider two mirrors constructed as follows. How can we prove the number of images is $360/A -1$? How about the case in which $360/A$ is not an integer? Seriously I have no idea how to start solving it analytically rather than by using geometric construction with compass and straightedge.

geometry

Mayank

Why Maths.SE not PSE?

Safdar

06:53

@Mayank It's more or less a geometric construction.. Maths and physics have a lot in common. Quoting XKCD, one min..

Mayank

matrix multiplication, or a combination of complex multiplication and complex conjugation - 🤦

Mayank

07:37

@JohnRennie Is there a way to know if volume of wire remains constant or not when it is stretched while finding new resistance?

John Rennie

That doesn't mean the wire is being stretched, it means twice as much wire is being used.

Mayank

Doubled means not stretching

John Rennie

Doubled means using twice as much wire. i.e. if we started with one metre of wire then doubling means we add an extra metre.

Mayank

Take a look at this illustration

John Rennie

Ah, OK, that is poorly written. By doubled it means folding the wire in half.

i.e. if you had a wire of length one metre you fold it in half to get two wires of length half a metre.

Mayank

07:46

How would we know what the examiner mean ?

By doubling.

John Rennie

Hopefully in an exam the questions will be written more clearly.

From the JEE questions I've seen they are usually clear what they mean.

Mayank

Ok :-)

Mayank

09:33

@JohnRennie How to show that ∆R/R=2∆l/l?

R=rho*l/A

R=rho*l^2/v

Then

dR/dl=2I

John Rennie

We start with:

R = ρ L / A

The differentiate to get:

dR = ρ/A dL

OK so far?

Mayank

Ok

John Rennie

Now we want dR/R on the left side so we need to divide both sides by R. But we know R = ρL/A so we can divide the left side by R and the right side by ρL/A.

OK so far?

Mayank

Ok

John Rennie

That gives us dR/R = ρ/A x A/ρL x dL = dL/L

Mayank

09:41

Yes

John Rennie

So we get dR/R = dL/L, which is the correct result.

Mayank

:-O

A uniform wire is stretched uniformly by 0.5%. Its resistance change by

@JohnRennie

shelton Benjamin

Which computer software i can use to experimentally verify variation of z factor with increase in pressure of real and ideal gas?

John Rennie

@sheltonBenjamin no idea, sorry

@Mayank what's your answer to this?

Mayank

Taking a long way give 1%

But teacher used ∆R/R=2∆l/l

John Rennie

09:46

It does? You'll have to take me through how you derived this.

shelton Benjamin

My friend wanted an intuitive way of understanding this variation like he wanted to feel the phenomenon by experimentation can you please tell another way

Mayank

@JohnRennie I didn't derive it.

John Rennie

Are you assuming the volume stays constant, so the area decreases as the wire is stretched?

Mayank

Yes

John Rennie

@Mayank ah, OK, when I did the derivation above I assumed that the area stayed constant.

Mayank

09:50

Oh ok

John Rennie

If we are going to keep the volume constant we need to write A = V/L. OK so far?

Where the volume V is constant.

Then we get R = ρL/A = ρ L²/V

OK so far?

shelton Benjamin

two blocks of mass m1 and m2 are connected by ideal spring suddenly a sudden jerk is given to m1 which gives its velocity of V1 find the maximum compression in the spring during the subsequent motion

Mayank

@JohnRennie ok

shelton Benjamin

If I solve this question from a one frame of reference using work energy theorem why doesn't the method works

M1 frame

John Rennie

@Mayank so now when we differentiate we get:

shelton Benjamin

09:53

I think it is wrong time to ask question sorry

John Rennie

dR = ρ/V 2L dL

Yes?

Mayank

Yes

John Rennie

And we do the same trick as before. We divide the left side by R and the right side by ρ L²/V

(because R = ρ L²/V)

dR = ρ/V 2L V/(ρ L²) dL

Mayank

Got it

Thanks :-)

John Rennie

:-)

shelton Benjamin

09:57

Sir can you help me now

two blocks of mass m1 and m2 are connected by ideal spring suddenly a sudden jerk is given to m1 which gives its velocity of V1 find the maximum compression in the spring during the subsequent motion

John Rennie

@sheltonBenjamin the way you do questions like this is to work in the centre of mass frame

shelton Benjamin

If I solve this question from a one frame of reference using work energy theorem why doesn't the method works
M1 frame

How to do it from m1 frame of reference

John Rennie

The m1 frame of reference is a non-inertial frame.

It's going to be exceedingly difficult to do the calculation in that frame.

shelton Benjamin

I took the spring force kx divided by m1 and x m2 to get the pseudo force then using the work energy theorem why doesn't it work

Also included the work by the work by the spring force on m2 m2

John Rennie

I'm sorry, but hammering nails into my kneecaps and working in non-inertial frames are things I don't do.

3

shelton Benjamin

10:06

Does it mean my concept is wrong

I still want to get an idea of how can i do it from m1 frame of reference please sir tell me a source

Archer

10:23

@JohnRennie any idea on how to trace the history of a field?

John Rennie

@Archer the history of a field?

What do you mean? The history of a topic in physics like e.g. electrodynamics?

Annamalai Sriram

15:10

I have a doubt about this problem which is related to capacitors.

First part of Solution

Why the same charge flows from each capacitor ?

Can't we solve it using Equipotential method ? ( Vcom = sum of charges on capacitor/equivalent capacitance)

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Problem Solving Strategies