Hi @JohnRennie

Are you available to discuss my question from earlier?

@JohnRennie hello sir

Can we apply kirchoff's law on circuit that have 0 resistance but have inductor

Why does current not become infinite in AC circuit with inductor and 0 resistance

@HarjotDhillon Hi :-)

@Bumblebee Hi :-)

@HarjotDhillon With a resistor the current is limited because the resistor converts the energy supplied by the battery to heat. The battery increases the kinetic energy of the conduction electrons, but the resistor converts that KE to heat and slows the electrons back down again.

With an inductor the current is limited because the inductor converts the energy supplied by the battery to energy stored in the magnetic field of the inductor. As before, the battery accelerates the electrons but the inductor uses their KE to create a magnetic field and that slows the electrons down again.

The big difference between an inductor and a resistor is that with an indictor the energy is not lost. The inductor can transfer the energy stored in its magnetic field back into the energy of the electrons.

That's why in an AC circuit the power dissipated in an ideal inductor is zero.

@Bumblebee There isn't an exact answer to this. Typically a new idea starts off with only a few scientists believing it. If it's a good idea then over time more and more scientists believe it until it becomes an accepted fact.

Some ideas are accepted very quickly. For example special relativity was accepted quickly, but string theory is still not accepted even after 40 years.

hmmm ok

@j I was just wondering because usually in the Natural Sciences, it seems that something needs to be "sufficiently" proven in order to count as scientific knowledge. I'm just trying to gauge where that "sufficiently" is.

@Bumblebee The usual test is reproducibility i.e. can someone else do the same experiment as you and get the same results?

ah

so the replication of results

Yes

Hi :-)

hi :-)

The way I would do this is to write down the lens equation and rearrange it for the image distance 𝑣.

but this is mirror

Is it a convex mirror?

s cars have convex mirrors

So the focal length is 1m (¹⁄₂R).

Yes?

yes

I can never remember sign conventions, but I think we take 𝑓 to be positive because it's in the direction of the light. And 𝑢 is negative.

Does this make sense so far?

ya f is positive fo the convex and u is negative for all the cases

yes

ok

OK so far?

yes yes

Now differentiate both sides wrt time. On the left we get dv/dt, and that's the velocity of the image, which is what we want. Yes?

the velocity of the image is = -1/(x-1)^2

Careful. On the right we need to do d/dt[ x/(x-1) ]

yes

And dx/dt is just the velocity of the runner - call this V.

ok

And this is now routine. Yes?

but how do we got the V

V = dx/dt, and x is the distance to the runner, so V is just the speed of the runner.

And the question tells us V = 5 m/s

ok thanks a lot :-)

You're welcome :-)

This technique of differentiating both sides of an equation can be very useful.

You'll find other JEE questions where you need to use it.

yeah thanks a lot

A square wire of side 3.0 cm is placed 25 cm away from a concave mirror of focal length 10 cm. What is the area enclosed by the image of the wire? (The centre of the wire is on the axis of the mirror, with its two sudes normal to the axis). @john can u draw an diagram to show me how they have kept i jsut need the diagram and not the steps of solving

ok

I'm answering another question at the moment. I'll ping you as soon as I'm free.

ok

@jo

@john the first question i have asked no for that if we do for the 9m iam gettig an diffrent answer

@sudarsan Hi :-)

this question

@sudarsan I thought we had done this?

ya one min i have derived an formula for this iam just confused with the sign convention and can u just hold on for few mins i will send the small dervation

OK ...

byjus.com/question-answer/…

in this link it is there the anwer the answer is different from what i got

That's because they are calculating the average velocity over a second, not the exact velocity at 39m.

What we did was differentiate to find the exact velocity at v = 39m

is both are correct

according to the question

The question asks for the velocity at 39m

Not the average velocity between 39 and 34m

So your answer is correct and the Byjus answer is wrong.

Does this make sense?

mm

i agree but this answer was given in all of the website

OK, but I still think we are correct.

how

Our and the Byjus answer should be similar, but not exactly the same.

mm yes yes

i cna send you the pic of the derivation but i dont have the option for to upload

You mean your work?

yes

ya i just subtituted the F (focal length ) and made an formula

so the formula will be V(velocity of the image)=5d/dx[x/(x-f)]

ya i also got the smae

Where x is -39, -29, -19 and -9

ok then if we sub -9m in the (x-1)^2 in the denominator we get 100 on the denominaotr

so -1/20m

-1/20 ms^-1

Yes

but the answer byjus have given is 1/10 positive

i mean +1/10ms^-1 which is way different from our answer

At 39m our equation gives dv/dt = 0.003125 and Byjus gives 0.003571

So they are pretty close.

ya for teh -9m

we are getting different answer

For the smaller distances the Byjus method will be less accurate.

but why it is inaccurate

yes

At 9m we get dv/dt = 0.05m/s and at 4m it's 0.2m/s. Yes?

s

yes

So if you try to take the average velocity from 9 to 4m you'll get a value very different from the actual velocity at 9m.

It would make more sense to to wait 0.1s i.e. the runner moves from 9 to 8.5m

Or even 1 millisecond.

The point is the Byjus method is pretty useless for small distances.

mm ok ok

While our method works at all distances.

and our method is logical though

Yes

ok thanks

You're welcome :-)

@JohnRennie hello sir

Does negative power factor mean power is being supplied to our AC battery

The power factor cannot be negative.

But what about 91 degree

Oh wait now I get it

Our angle cannot be more than 90

Yes. And cosθ is positive everywhere in this range.

hi @ john have u cleared jee advanced have u seen the 2016 jee advanced question paper

Hi :-)

I'm retired. The last time I did a physics exam was in 1983 :-)

a thin rod of length f/3 is placed along the optical axis of concave mirror of focal length f such that its image is real and elongated just touches the rod what will be the magnification can u draw an diagram to illustrate it and solve this one @john

@sudarsan Hi :-)

Sorry for the slow reply - I was on the phone.

its okay:-)

I don't understand the question.

I don't understand what the relationship between the object and the image is.

ok one min

Did you type the question out exactly?

yes

the question is till the WHAT WILL BE MAGNIFICATION

after that its my request to do so:-)

I don't understand what:

and elongated just touches the rod

means.

Can you post an image of the question so I can see exactly what it says?

the image is elongated at it just touches the end of the object

currently iam not having that option i dont know why

can tell me where to click to get that option now

I think you cannot upload pictures until your reputation reaches a certain level, though I don't know what that level is.

i will upload the pic along with the question

i have uploaded

have u got the iamge

OK so the point C is at u = 2f so it reflects onto itself.

The point A is at u = ⁵⁄₃f and it reflects to the point A'.

i mean what i have uploaded iam just confused with the image made by the mirror and what i have sent

ok

And the magnification is the length A'C divided by the length AC. Yes?

but i have one doubt do we have the magnifaction for tha iamge which is elongated like this

The magnification is just the length of the image divided by the length of the object.

but i have studied only the (height of the iamge)/(heigt of the object)

ok ok

ok this far iam okay

That's true whether the rod is normal to the axis or parallel to the axis like here.

mm

So you just need to find where A' is, then the length of the image is A'C and the length of the object is AC.

Just divide the two lengths to get the magnification.

@JohnRennie Will you be around this evening?

ok to get the length of the image

At least I'm pretty sure that's what the question means.

we have to minus the

pa'-ap

@KavinIshwaran Maybe. My mother is making a special meal tonight and that may not be finished by 21:30.

@ka

@kavin are u from tamilnadu

@sudarsan The image starts at A' and ends at C, so its length is the length A'C.

@JohnRennie Ok :-)

ok

The point A is at u = -⁵⁄₃f

then the opposite way like ap-pa' right?

@sudarsan Yes.

So the position of A' is 2.5f

then the a'c is 0.5f

(I think that's right)

then 2f/0.5f

sorry

Yes, so the modulus of the magnification is ¹⁄₂f/¹⁄₃f = 1.5

mm

yeah thanks a lot

And the sign is negative because the image is inverted.

So m = -³⁄₂

I can see why it's confusing because we normally take the magnification normal to the optic axis.

But I think this is what the question means.

mm that's what i got confused

hi @john

Hi :-)

i have uploaded one question to you on the ratsauce

OK, what do you want to ask about it?

see that even its easy they have given the refractive index of water and glass they are asking for the Vg (velocity in the glass) now we know c =3*10^8 so the answer is in one step right like iam telling C(speed of lihgt)/Ug (refractive index of the glass)

which is equal to the Vg(veloctiy of the glass)

That's true, though I guess they mean you to not use your existing knowledge of the speed of light in a vacuum.

but iam getting an differnet ansewr

That's because if you calculate c from the data they give the result is 2.925 m/s not the actual value of 2.9979 m/s

ok thnks