@yuvrajsingh morning :-)

@JohnRennie morning.

@yuvrajsingh hi

How are you feeling this morning?

Good, since I couldn't, t get sleep in night, so I wake early in the morning. @JohnRennie

I guess try to relax a bit to get ready for the exam.

Yes sure,, you will be amazed the whole night library was open

Ye sure.

@JohnRennie

Wow! They obviously expect the students to work all night :-)

Yeh, here at IIT, selected the top brains in India, so everyone want to compete.

Many of the student are scared of this competition.

We have total 33 exams, including test and viva, and quiz.

@JohnRennie

Today it is just starting, I can see the buches of book, the hectic engineering books around the students.

The workload at the colleges seems very heavy. But do you get the chance to make friends and socialise as well?

Seriously it depends if you ares studious boy, then you get every chance to, to learn more and more,.

And if you are a student who just joined IIT for fun, he can involve in sports meets etc, here we have gym Khanna and activities club, I clouding robotics and lot other.

Student who have similar thinking can join the respective group.

OK, so there is lots to do apart from just working.

I used to go canoeing at college. I found it a really good way to relax.

What is it,.

@JohnRennie

Let me see if I can find a link ...

cucanoe.co.uk

Aah, it is similar to river rafting, here I and my friends usually go for this in weekends, here it is famous because we near river Ganga.

@JohnRennie

@yuvrajsingh that sounds like really good fun :-)

Seriously, we are planning for this after the exam.

Good :-)

Let me tell you a story.

It's good to do something that is physically tiring after you've been sat in your room studying for too long.

Or incident.

@yuvrajsingh yes?

It was about haunted house.

When I was in class 10.

Our school taken us to bangarh a haunted Kingdom, and a place.

I think there is wiki article on it. @JohnRennie

> Bangarh is the historical place situated in Gangarampur, West Bengal, India. Bangarh was the ancient city which was the administrative centre of Kotivarsha Vishaya (territorial division), itself part of the wider administrative unit of Pundravardhana Bhukti, which had Mahasthangarh as its capital[1] in the period of Chandras, Varmans and Senas. After the Senas were defeated by the Muslims under Bakhtiar Khilji, Devkot was established as their capital were Bakhtiar died

Is that the place?

No it is in rajasthan alwar

It is bhangarh @JohnRennie

Bhangarh Fort?

Yes.

Is it haunted?

It is one of the scary haunted house, in India, usually palace closes for tourist after 6P.M

So many person, who gone inside the fort after 6 P. M Were lost.

I think there are so many videos available on net.

When did you go there? After dark?

Come on there is police outside so we can, t go, there is temple just outside the main town.

I think our pics is available on our school Web page, ST. XAVIER school jaipur.

stxaviersschooljaipur.com

The most haunted fort in India :-)

@JohnRennie hi......

@JohnRennie hi.

I have so many question.

@yuvrajsingh hi :-)

Hi sir there was a question,

Yes?

I have photon, I drop it from height of 1km, under earth gravitation field change in frequency of photon will be.

Gosh, that's a hard question! That's the gravitational red shift which you don't normally study until you learn general relativity.

Did they seriously ask you that?

Yes sir,.

There was more questions I have one more

@yuvrajsingh you can do it like this:

Yes sir.

The photon starts out with energy $E = h\nu$. Convert it to a particle with mass given by $mc^2 = E = h\nu$. I know you can't really do this, but it's a thought experiment.

Yes.

Now let the particle fall 1km and its energy increases by the potential energy $mgh$ so the total energy is $E + mgh$. Finally convert this back into a photon with a new frequency $\nu'$ given by $h\nu' = E + mgh$.

And that gives you the new frequency $\nu'$.

Sorry to ask what will be it's initial energy.

It mean when I just photon is dropped.

The initial energy is just $E =h\nu$

Ooh. Sorry. But sir what value of mass i take here.

The mass is given by $E = mc^2 = h\nu$

So $m = h\nu/c^2$

Ooo. Again I am stupid.

I have one more question.

@yuvrajsingh Not at all. The scientist who came up with this idea was ... care to guess?

OK.

It was Albert Einstein!!

Yes.

I have one more question.

Yes?

It was imagination not in exam.

I I drop a apple from window of the train.

What will be the trajectory of apple.

And why it is.

In whose frame? The ground frame or the train frame?

Trajectory would be same in all frames I think.

@JohnRennie

No. In the train frame the apple starts out motionless i.e. when it drop it the apple is momentarily at rest with respect to your hand. So ignoring air resistance it will fall straight down.

Just like if you dropped it when you were sitting in the carriage.

What about ground frame.

In the ground frame it starts with a horizontal velocity equal to the train, then accelerates downwards at $g$. So the trajectory is a parabola.

In opposite direction.

yes, a $y = -x^2$ type parabola.

OK.

we can calculate this if you want.

Sir today I am tired I have so many, which think we're out of the syllabus, can I catch tommorow, sir please.

Yes sure at least I want to complete this.

yes I'll be around tomorrow.

@JohnRennie

@yuvrajsingh you want to go through the calculation?

Yes if you agree.

Take the origin to be the point where the apple is released. At this point its velocity is $\mathbf v = (v_t, 0)$ where $v_t$ it the speed of the train. So the horizontal position is given by $x = v_t t$. OK so far?

@JohnRennie

Ok

The vertical position is just $y = -\tfrac12 g t^2$. It's negative because the appls falls down with acceleration $g$.

And from our first equation $t = x/v_t$ so substituting for $t$ in the equation for $y$ gives:

$$ y = - \tfrac12 g \frac{x^2}{v_t^2} $$

$$ y = - \frac{g}{2v_t^2} x^2 $$

Got it.

Let say I take good sleep now, and meet you in morning. :)

@yuvrajsingh OK. Good night :-)