@ScientistSmithYT for general questions I'd probably ask in the h bar chat room.

Strictly speaking this room is for JEE problems, though we chat about other stuff too.

@YUSUFHASAN hi, how's it going? Are you at college now?

Hello sir Goodmorning@JohnRennie

@user8718165 morning :-)

@JohnRennie I've got some questions sir

@user8718165 yes ... ?

@JohnRennie Sir is it necessary that the North and the South Pole of a magnet have exactly equal strengths?

It depends what you mean. Magnetic field lines are always loops i.e. they can never start of end anywhere. That means every field line that comes out of the North pole must go back into the South pole.

@JohnRennie That means the strengths are equal... And it makes sense too

But you could make a magnet from a wedge shape so the metal had different widths at the North and South poles. That means the field strength i.e. the number of field lines per unit area, would be different at the two ends.

@JohnRennie That means if the shape of a magnet is exactly symmetrical...the strengths are equal....right

@user8718165 yes

It's because no magnetic charges exist i.e. there is no magnetic equivalent to a positive or negative electric charge.

Field lines can only start or end on a charge.

@JohnRennie sooo......nice..... Thank you sir

@JohnRennie sir another question

@user8718165 yes ... ?

@JohnRennie a small doubt.

If charge moves in direction of electric field then PE decreases right?

@JohnRennie Life is going on sir :) , and I am almost at college,as a few rounds of counselling are still left. Say,you wouldn't have heard about the IISERs,have you?

@YUSUFHASAN I haven't, sorry.

@Aladdin yes, if you move a charge $Q$ a distance $x$ in the direction of the field lines (i.e. from positive towards negative) then the PE decreases by $\Delta U = QEx$

Ahh okay

@JohnRennie That's all right, they are much more newer than the IITs(established only in 2006),but yes,it is expected they will catch up in reputation soon :-) . Anyway,in the curriculum at my college,I will be required to study physics for 4 semesters before deciding my major,so I will prolly be back soon enough with a few more doubts :)

@JohnRennie suppose...a glass is filled with water just full to the brim (the water isn't forming a hump)...at the surface pressure is atmospheric pressure....right?

@YUSUFHASAN you're always welcome to ask. I'll be here :-)

@user8718165 yes

@JohnRennie now lets carefully place a lid...such that the water just touches the lid...and no water falls off

@YUSUFHASAN it's nice to have that flexibility. It means you get a chance to figure out what you enjoy before committing to the full four years. Cambridge works like that. You do lots of different subjects in the first year and only specialise from the second year onwards.

@user8718165 OK ...

@JohnRennie now what's the pressure at the surface of the water?

@JohnRennie hii sir

@user8718165 there isn't a simple answer to that because the process of putting the lid on hasn't been precisely defined.

The pressure on the top of the lid is 1 atm. If you take the lid to be infinitely stiff then none of that pressure is transferred through to the liquid.

@JohnRennie yes...so the pressure on the surface of the water is 0...right

But what happened at the instant the lid touched the water? In real life the lid will touch some point on the water surface first and the water surface will curve where it touches the lid. That means the surface tension will create some pressure inside the water.

As the lid completes its journey down the air-water interface will move out towards the rim of the glass and the curvature and therefore the pressure will change as it moves.

@JohnRennie okay...sir...got it....is that pressure measurable?

@user8718165 yes, you can put a pressure gauge in the glass.

There's nothing especially weird about what's going on, it's just that exactly what happens in the final microseconds before the lid touches the glass depends on tiny details of the lid and water surface shape and how they meet.

@JohnRennie now I get it...I was completely puzzled when HC Verma said that putting a lid doesn't change the pressure at the surface of the water...its equal to atm

@user8718165 Verma shouldn't really have said that as it's kind of a special case.

oops

@JohnRennie sir...can I show you?I might have misinterpreted stuff :-)

@JohnRennie thats okay :-)

@user8718165 I'll have a look if you want to post the page ...

@JohnRennie actually I saw it a week ago...I still remember some bits n parts XD

@user8718165 I'd just let it go.

:51051862 I'll have to watch it later ...

@JohnRennie as you wish sir:-) just have a look once...when you get time

@JohnRennie hi. You here?

@Aladdin hi

I had done 18 but I am having problems with 17

My answer is not matching for 17

I guess the answer is D because the angle should be greater than 180 yet I can't get the exact value for this

I would guess the answer is (A) ...

It's given D though

Ah, oops, it wants the lowest tension. Yes, it will be (D). The answer (A) would be the highest tension not the lowest.

The tension in the string is proportional to $v^2$ so we need to find the point where $v^2$ is lowest. Yes?

Yes

The PE and KE are related by $PE + KE = E_{total}$ where the total energy is constant. That means the lowest point for the KE is the highest point for the PE. OK so far?

Tension depends on electrostatic and gravitational potential too

@JohnRennie yes

@Aladdin hmm, true ...

Give me moment to calculate where the PE is highest. I'm curious to see how it relates to the answer.

Okay

The answer given comes from just considering the centripetal force.

Okay. The centripetal here depends on both gravity and electrostatic

i.e. find the point where the PE is a minimum.

Our answer here is for the scenario where both electrostatic and rension contribute to centripetal force. How does centripetal force being lowest means any one of them is lowest

I'll have to have a think about this to get it clear in my head, but I'm fairly certain that the tension does just depend on $v^2$. That's because $v^2$ also depends on gravity and the electric field so when we calculate $v^2$ we are automatically including the gravitational and electric forces.

Yeah $ v^2$ depends on both gravity and electrostatic

So all we have to do is find the angle where $v^2$ is a minimum

And that's the angle where the PE is a maximum

P. E is $V = mg\ell(1-\cos\theta ) + qEl\sin\theta$

@Aladdin I would take the PE at the centre of the circle to be zero. Then the PE is negative to the lower right and positive to the upper left.

The vertical distance downwards the centre is $y = \ell\cos\theta$ and the horizontal distance right is $x = \ell\sin\theta$

And the PE relative to the centre is $PE = -mgy - mgx$

Okay

That gives us: $PE = -mg\ell\cos\theta - qE\sin\theta$

So $dPE/d\theta = mg\ell\sin\theta - qE\ell\cos\theta$ and the maxima and minima are at $dPE/d\theta = 0$

Electrostatic should be positive though... We moved opposite electric field right?

I may have messed up the signs. I always find it confusing getting the sign right. Usually I charge ahead and fix up the sign at the end.

Our equation is going to give us:

$$\tan\theta = -\frac{qE}{mg} $$

I get the same from my expression too

So yes I got a sign wrong. It's obvious I got a sign wrong because the minimum PE has to be at $0 <= \theta <= \pi/2$ and $\tan\theta >= 0$ in that region.

We get two solutions for $\theta$ because $\tan\theta$ is periodic, so we get a solution at $\theta = \arctan(qE/mg)$ and another at $\theta = \pi + \arctan(qE/mg)$. The first is the minimum and the second is the maximum.

Okauy. That would make sense

You could calculate the second derivative and show that it's negative at one angle and positive at the other if you felt like it :-)

@JohnRennie hello. I had some doubts again

@Aladdin hi

sarthaks.com/366064/…

@JohnRennie I didn't understand the answer... What is area enclosed area of circle here....

Gauss' law tells us that the flux through a surface is equal to the charge inside the surface divided by $\epsilon$. So the flux through our sphere is equal to the charge inside the sphere divided by $\epsilon$.

Yea but charge inside the sphere should depend on infinite charge sheet

They are using π$r^2$ though

We are told the charge density of the sheet is $\sigma$, so the charge inside the sphere is $Q = \sigma A$ where $A$ is the area of the sheet that is inside the sphere.

Yes

The part of the sheet that is inside the sphere is a circle of radius $r$ where $r^2 + x^2 = R^2$. Yes?

Didn't get this

Does that help?

That's a side view so we see the infinite sheet edge on.

I don't see the circle

Can you outline it

Let me draw what you'd see if you were looking from the right:

The intersection of a sphere and a plane is a circle.

Okay it clicks now

Black circle is intersection

Wow this picture makes it. More clearer

That's from a Stack Exchange question

@JohnRennie that's beautiful sir

@user8718165 not my drawing :-)

@JohnRennie okay :-)

@JohnRennie hello

@Aladdin hi

In the first question, there is no mention of gravitational energy

I thought the change in gravitational and electrostatic will be equal to kinetic here

My expression includes gravitational energy too but it's not in any option

I suspect the plane of the paper is meant to be horizontal i.e. the rod and particle are lying on a frictionless horizontal surface.

Hmm that would make sense

So it's going to be A isn't it?

Yes

@JohnRennie do you know how to do question 21

I am not good with field lines and stuffs

A field line shows the direction of the force on a positive charge.

First of all the setup is symmetric about the +Q charge so the field lines must be symmetric about the charge +Q. That immediately eliminates one of the diagrams. Yes?

Yes

Now consider this location for our positive test charge:

Which direction is the force on our positive test charge going to be?

Right

Yes, and remember that the field lines show the direction of the force. So the field lines at the position of the red dot must be horizontal.

Okay

And that rules out two more of the diagrams.

So D?

Yes, D.

Wow. So that's what we do in these questions

Hello@JohnRennie

@user8718165 hi

@JohnRennie Sir one question

@user8718165 yes ... ?

@JohnRennie Is it true that when a glass of water (not necessarily filled) is left open to atmosphere... The water is compressed however so slightly :-/

If the depth of the water in the glass is $h$ then the pressure at the bottom of the glass is $P = \rho g h$. Yes?

@JohnRennie okay sir

If you subject any object to a pressure $P$ then that object will be slightly compressed by the pressure. The compression is given by:

inverse of bulk modulus

$$ \frac{dV}{V} = \frac{1}{K} P $$

where $K$ is the bulk modulus.

@user8718165 aha, you were ahead of me :-)

@JohnRennie but that eqn is ahead of me... sir :-)

So in principle the water at the bottom of the glass is compressed by $\rho g h/K$

@JohnRennie sir K is temperature...right

But for water the bulk modulus is about $2.15 \times 10^9$, so it isn't compressed by very much :-)

@user8718165 no. $K$ doesn't mean Kelvin. It's just the symbol commonly used for the bulk modulus.

@JohnRennie I remember having seen it somewhere in my book...but I was puzzled...so I wanted to ask you...

@JohnRennie so that means under atmospheric pressure the loss in volume is very less....

@user8718165 if we rearrange the equation to $dV = PV/K$ we can substitute some numbers.

Take one cubic metre of water, so $V = 1$. The bulk modulus of water is $2.15 \times 10^9$ Pa and one atmosphere is about $10^5$ Pa.

That means the compression of our one metre cube of water by one atmosphere is:

$$ dV = 10^5/2.15 \times 10^9 \approx 4.7 \times 10^{-5} m^3$$

$dV = 47 cm^3$

@JohnRennie yes sir....got it

@JohnRennie that's too much volume...is it?

I'm not sure if 47 cc is a lot or not. We tend to think of water not compressing at all.

@JohnRennie okay sir...got it

@JohnRennie thank you so much :-)

@user8718165 :-)

@JohnRennie finally I'm able to get an intuition...the the final height of the column will be 99.9953 cm...that's pretty pretty negligible sir ;-)

The change in the side of the cube will be the cube root of $4.7 \times 10^{-5}$.

Duh, no, it will be one third of $4.7 \times 10^{-5} \approx 1.5 \times 10^{-5}$

@user8718165 do you want me to explain how I got that?

@JohnRennie sir its already very late now...and I'm troubling you....I want to know so you can tell me tomorrow sir :-)

OK :-)

@JohnRennie goodbye sir

@user8718165 Goodnight. See you tomorrow.