@Bohemianrelativist uh.. library?

@napstablook I died laughing at this comment

@JohnRennie hi

Hi :-)

I wanna ask something about laptops

My professors says that I should switch to linux from windows

But some of my seniors say that linux is slow and u may face issues while attending online classes..... so I'm a bit confused

People have a tendency to think that what they are used to using is best.

So you'll find Windows users telling you Windows is best and Linux users telling you Linux is best. Neither of them are to be trusted! :-)

@JohnRennie correct but I believe you'll give a fair advice; )

Both Windows and Linux work fine. Which you use is a matter of personal preference. If you might ever find yourself working in IT then it's good to have some knowledge of Linux because it is widely used in India, but otherwise there is no particular reason to choose one over the other.

Are you using Windows at the moment?

Yup

What are the main apps you use?

MS Office?

Widely used is Microsoft teams and I will have to install compilers for different programming languages .

And yeah ms office and word will come in use for college projects...

If you rely on Microsoft software then that's a good reason to stick with Windows. There is a MS Teams app for Linux, but it's always going to be best supported on Windows.

And if you're going to be programming then MS Visual Studio is by far the best development environment there is, and it's only available on Windows.

There are equivalents for MS Office on Linux, but I don't think they are as nice to use.

Ohk . So I'm going with windows

So it sounds to me as if you should stick with Windows. You can run some Linux apps in Windows 10 anyway, and in Windows 11 you can run pretty much all Linux apps if you ever need to.

Installing both is a bad option right? It might slow down the laptop?

Installing both is fine. I know lots of people who have done that and it works fine and does not slow down the laptop. But given that Windows has a lot of Linux support built in you probably don't need to install Linux.

Ohk then

Thanks ...

You're welcome :-)

← 2 messages moved from The h Bar

@Ishwaran Hi

@JohnRennie hello !

@Ishwaran Hi :-)

In the above Question the only hint they give is that the object is placed at the focis of Lens 1

You need the light coming from the final (concave) lens to be parallel, and that means the (virtual) image for the concave lens has to be at its focal point. Yes?

Yes !

So you need the final virtual image to be here.

As you say, the object is at the focal point of the first concave lens, so the light coming from the first lens will be parallel.

Like this.

@Ishwaran OK so far?

The object is placed at O

Lets take

Yes, and O is at the focal point of the first convex lens.

Yes

So the light coming out of the first lens will be parallel.

Can be ?

If you have an object at the focal point then the image is formed at infinity. Yes?

Yes

And if the image is at infinity then the light rays going to the image are parallel.

(for a point object)

Absolutely thats what I came to ask that what if we take the object that placed on O :

So image will again form at infinity

But

In a different angle ?

You mean if the object isn't a point, i.e. some of the object is off the principal axis?

Yes

Then you are quite correct that the light rays to the image won't be parallel. They are only parallel for a point object. However I take the question to mean the object is a point object. It wouldn't make sense otherwise.

Like this

@JohnRennie Then lets proceed with it

So do you agree with my diagram so far?

Yes

So the light rays entering the second lens (f = 40cm) are parallel, and that means the second lens will form an image at its focal point i.e. 40cm to the right of the lens. Yes?

Yes

And this is where i stuck

So y has to be 30cm. Yes?

Yes !

The problem is after coming out of the first lens the light rays are parallel

The x value should not affect the overall result

But in the options have a definite value for y

x*

I agree. The value of x doesn't matter. I think it's a mistake in the question.

What does the solution say?

The answer key comes tomorrow....

I think (2) and (3) are both valid answers.

I too think so...

The options must have an error otherwise the question

Thanks sir for sparing some time ! :)

@Ishwaran You're always welcome to ask questions here :-)

@napstablook close too early to go.

@JohnRennie can i ask a doubt?

@AdilMohammed Hi :-)

Yes, what's the question?

Its the torque acting on a parallelogram...

there are two ways to do it right, one is Force acting on the sides times the perpendicular distance

and the other one m×B

To be on the same page, lets consider a Parallelogram with side x,y carrying current I

theres a magnetic field along b

A parallelogram, or a rectangle?

parallelogram

I need to see a diagram as I'm not sure what you are describing.

yess i am drawing it

@JohnRennie which application do you use for drawing?

I use Google Draw

is the method i wrote beneath the picture correct for finding the torque?

I must admit that it hadn't occurred to me to do it that way, but yes it seems valid. Does μ = IA apply to all shapes of current loop or only to circles?

I had that same thought

The forces on the two horizontal wires are ±BxI sinθ, out of the page at the bottom and into the page at the top. Yes?

indeed

So wouldn't that make the torque τ = BxI sinθ × y sinθ

yes

(assuming the axis of rotation is through the middle of the parallelogram)

So the two methods differ by a factor of sinθ

@JohnRennie do you mean the axis of rotation passes through the midpoint of the upper and lower sides?

Does the question say where the axis or rotation is?

I was assuming this would be the axis.

@JohnRennie nope

@JohnRennie oh yes you are correct

If the parallelogram is just free, i.e. the axis isn't fixed, then it would rotate about an axis that isn't horizontal. That's because there would be some torque about a vertical axis as well.

That would make the calculation complicated.

ohhh

In that case I guess your first approach would be the best one.

the force times perpendicular distance one?

Writing the torque as B × μ, where μ = AI

To be honest I don't know the answer to this. It isn't a calculation I've done before.

@JohnRennie oh okay, so the formula applies for all shapes. it's only that it gives the net torque?

@JohnRennie haha i am gonna try doing this experiment at the college lab, when I get into it, (if they allow that is) and tell you the results of my findings

OK :-)

thanks for your time sir

Sorry I couldn't help.

its allright, atleast now i will have something interesting to do at college😂

@JohnRennie Hi! Are you going to sit on the armchair now? :)

Yes, it's about time to log off and relax for the rest of the evening :-)

Ok, I'll talk to you tomorrow then. Bye