Problem Solving Strategies

Scáthach

4:14 AM

@JohnRennie good morning. Are you free

John Rennie

@Scáthach morning :-)

Yes I'm free.

Scáthach

@JohnRennie An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to 220 V(rms) , 50 hz Ac supply,, the series inductor needed to work for it is close to

John Rennie

@Scáthach this is like that resistor and inductor in series question that you did yesterday

The arc lamp needs a current of 10A, so with a 220V supply you can work out what impedance is needed to get a current of 10A. Yes?

Scáthach

Yes

John Rennie

And you know the arc lamp has a resistance of 8 ohms because it draws 10A from an 80V supply.

And for a resistor and inductor in series $Z^2 = R^2 + L^2$

Scáthach

4:27 AM

@JohnRennie got it

@JohnRennie dc equivalent is ac rms

John Rennie

Yes

Scáthach

I get confused in this statement sometimes

John Rennie

The AC rms voltage and/or current is produces the same power as the DC equivalent.

Scáthach

So in this question why you took irms to be 10 then

John Rennie

So if you have some resistor R and you're calculating the power dissipated $I^2R$ then for example 10A DC and 10A rms AC dissipate the same power.

@Scáthach yes

In most cases you can just treat DC and AC rms as equivalent.

Scáthach

4:31 AM

@JohnRennie okay. Got it

ayc

4:48 AM

@JohnRennie Reagrding Light:Is maxwell's theory our current level of understanding of light ? Do we've much more in detail concepts ?

and ofcourse Max planck theory is there...anymore than those?.

John Rennie

@ayc light is ultimately described by quantum electrodynamics, which is a quantum field theory. Maxwell's equations emerge as the classical limit of quantum electrodynamics.

ayc

@JohnRennie Most theories have a basic platfrom upon which it is built for ex: special relativity is built upon two postulates....Do we have any such postulates in QED .or its just mathematics?

John Rennie

In classical electrodynamics (i.e. Maxwell's equations) there is a field called the electromagnetic four-potential. This isn't directly observable, but from it we can obtain the electric and magnetic fields.

What quantum field theory says is that this field should be quantised i.e. it should have states like the quantum states of a particle. So we write the EM four-potential as a sum of these states. The states are called Fock states.

The result is a theory that explains things classical EM cannot. For example it explains how an electron-positron pair can be created from a photon. This is an easy thing to observe experimentally, but Maxwell's equations cannot predict or explain it.

Se QED is actually well motivated. It's certainly not just a mathematician's fantasy. But it is much, much more complicated than regular quantum mechanics like you would learn in the first year if an undergrad physics degree.

Scáthach

@JohnRennie hi

John Rennie

@Scáthach hi

Scáthach

5:01 AM

@JohnRennie how to calculate time period lost by pendulum due to change in lebgth

John Rennie

@Scáthach the pendulum period is $T = 2\pi\sqrt{L/g}$, so if you change $L$ the period changes.

ayc

@JohnRennie I don't understand any of the terms,but I get the story..all theories develop in a certain way:First we give a theory and later we find that its not complete, then we add some complicated mathematics and then finish it i.e we prove that our intial theory is just a special case of our current,modernized,theory !..right?

John Rennie

@ayc basically yes. All physical theories are approximations that work fine within a limited region. Outside that region we need to find a more complicated theory that includes the original theory as a subset.

e.g. Newton's mechanics works fine for speeds much less than $c$. At high speeds we need special relativity, which has Newtonian mechanics as a low speed subset.

Then for high masses we need general relativity, which has special relativity as a low mass subset.

Scáthach

5:40 AM

@JohnRennie hi

John Rennie

@Scáthach hi

Scáthach

@JohnRennie hi

imgur.com/a/R1fZB2H

I know the charge present between A snd B

But still what to do to get electric field zero between them

John Rennie

Use Gauss' law

Consider a sphere with some radius $r$ between $a$ and $b$ i.e. $a \le r \le b$

Scáthach

Yes

John Rennie

Calculate the charge inside the sphere. This will be the charge at the centre plus the charge in the region between $a$ and $r$. The flux through the sphere is the field times the surfce area i.e. $E 4\pi r^2$ and this is equal to $Q/\epsilon$.

Scáthach

5:50 AM

Ok

Dante

@JohnRennie Hi, good morning!

John Rennie

@Dante morning :-)

Scáthach

@JohnRennie okay but still it's not giving anything to me. We can proceed after dante

Dante

@Scáthach I'll wait.

I thought you were done.

Scáthach

@Dante okay. Will just take 2 minutes i guess

John Rennie

5:55 AM

The volume between $r$ and $a$ is:

$$ V = \tfrac{4}{3}\pi r^3 - \tfrac{4}{3}\pi a^3 $$

The charge is $\rho V$ so it is:

Scáthach

Yes

John Rennie

$$ Q = \frac{A}{r} \left ( \tfrac{4}{3}\pi r^3 - \tfrac{4}{3}\pi a^3 \right) + Q_0 $$

Where $Q_0$ is the charge at the centre.

Scáthach

Yes

John Rennie

And from Gauss' law $Q = E 4\pi r^2 \epsilon$ so:

$$ E = \frac{A}{r} \frac{1}{4\pi r^2 \epsilon} \left ( \tfrac{4}{3}\pi r^3 - \tfrac{4}{3}\pi a^3 \right) + \frac{Q_0}{4\pi r^2 \epsilon} $$

Scáthach

OK

@Dante you can ask

Dante

6:06 AM

@JohnRennie Is restriction motion of particles always associated with decrease in entropy?

John Rennie

@Dante I'm not sure what that means. Can you give an example?

ayc

@JohnRennie Should the mass of a blackhole be quite big for us to make an experimental observation of hawkings radiation.....i.e what are the conditions at which we can make an observation of H.R?

Dante

I wanted to ask, why does freezing of water have positive $\delta S$ at $-1^{\circ} C$

John Rennie

@Dante When water freezes heat is transferred from the water to the environment so the entropy of the water decreases while the entropy of the environment increases. The total entropy change is the sum of the water decrease and the environment increase. Yes?

Dante

yes

John Rennie

6:13 AM

@Dante If the freezing is a reversible process then we know the total entropy change is zero i.e. the decrease in the entropy of the water is equal to the increase in entropy of the environment. Yes?

Dante

Yes

John Rennie

@Dante So can you see where I'm going with this?

Dante

Yep, entropy of the surroundings increase more than magnitude of decrease in entropy of water while freezing, right?

at -1

Heat is flowing from hotter body to colder body.

which is supposed to be spontaneous.

Am I getting it right?

John Rennie

Yes. Freezing of water at -1°C is not a reversible process because the water is supercooled and therefore not in equilibrium.

And in any irreversible process the total entropy increases.

Dante

Ok, I get it thanks, Why is entropy called 'unavailable energy' of the system?

John Rennie

6:23 AM

The work done in an irreversible process is always less than the work done in a reversible process. The decrease in work is associated with an increase in entropy.

That's why an entropy increase is associated with a decrease in the work that can be done.

Dante

Oh....k.

Irreversible process happens on it's own right?

Why do we have to do work there? Any example?

John Rennie

For example in a car engine the pistons are driven by the expansion of the gas in the cylinders, and the work done in that expansion makes the car go. The more nearly the expansion is reversible the more work will be done and the better the fuel economy will be.

Dante

Oh, I see.

ayc

6:43 AM

@JohnRennie My question now......The one I already asked!

John Rennie

@ayc the Hawking temperature of all macroscopic black holes is so low we could never observe it. We'd need to make a tiny black hole, and that's far beyond our current technology.

You can Google the Hawking temperature. I can't remember the equation offhand, but it's quite simple.

ayc

@JohnRennie Is it correct to say that we've discovered it in some other world by using Many-worlds interpretaion of Q.mech?

YUSUF HASAN

@JohnRennie Hi

ayc

@JohnRennie Is my interpretation of the interpretation wrong?

YUSUF HASAN

Are you free rn? @JohnRennie

John Rennie

6:53 AM

@YUSUFHASAN hi Yusuf, yes I'm free.

YUSUF HASAN

John Rennie

@ayc it's hard to say anything concrete about quantum theory interpretations. To be hnest I get impatient with the subject.

YUSUF HASAN

@JohnRennie I know this is about proportionality of self-inductance..I just don't get the factors for it

John Rennie

@YUSUFHASAN I'll just be a few minutes. Back real soon.

YUSUF HASAN

Kay..I'll be waiting here

ayc

6:56 AM

ohk!

John Rennie

@YUSUFHASAN I'm not sure to be honest. Self inductance is proportional to the area isn't it?

YUSUF HASAN

@JohnRennie Yep..That's what...I have only derived it for 2 coaxial coils in my study on inductance..and there it is proportional to length, square of number of turns per unit length and area..Is that stuff valid here as well?

John Rennie

The self inductance is proportional to the flux through the coil so you'd need to work out how the flux depends on the dimensions of the coil.

But I've never calculated that for a triangular coil.

YUSUF HASAN

Okay...and would area come into consideration for this coil?

John Rennie

Googling found this ...

YUSUF HASAN

7:09 AM

This is the soln given..@JohnRennie ^^

@JohnRennie That pdf seems to have a very complex relation for this..But the solution seems too simple and it neglects area..What am I missing?

John Rennie

@YUSUFHASAN to be honest I have no idea. I'd have to Google for the equation to get the inductance of a triangular coil.

YUSUF HASAN

@JohnRennie Okay..So can we say that the solution provided above ^^ is wrong?

John Rennie

@YUSUFHASAN I don't know.

YUSUF HASAN

Okay...Cuz I also didn't understand much from the pdf you linked above.. :(

John Rennie

There's a simpler article here

But it's still more complicated than the answer implies.

YUSUF HASAN

7:14 AM

Okay..Well this was a JEE mains question..and it wasn't striked off as bonus or something...so I thought...But still, thanks! @JohnRennie

Scáthach

@JohnRennie hi

John Rennie

@Scáthach hi

Dante

@YUSUFHASAN I never understood mutual induction clearly ;_;

Will have to go through cengage once.

Scáthach

@JohnRennie hi

imgur.com/a/PHsRZaZ

I tried to go with virial theorem

That gave me: K. E= -4P E

So total energy is -3P. E

That would indicate towards (c) .. But it's not correct

John Rennie

@Scáthach The virial theorem tells us that if the potential is $V(r) \propto r^n$ then $2T = nV$. Yes?

Scáthach

7:24 AM

Oops. Yes

John Rennie

Can you take it from here?

Scáthach

@JohnRennie but this gives us To=-2V right

John Rennie

$n = -2$ so we get $2T = -2V$

Scáthach

Got it

Zero

John Rennie

Yes!

Or you could do it the hard way i.e. differentiate $V$ to get the force and use that to calculate the orbital velocity.

Scáthach

7:28 AM

Yes

John Rennie

I need to work now for about 45 minutes. Back later.

krauser126

7:55 AM

Hello I had a question

gyazo.com/3114dd7a1dbd6b9e9f68a9d7df8cf6a8

Suppose we had a mass at point P that was being accelerated towards the right due to the positive charge on ring 1 and a negative charge on ring 2. At the center of ring 2, how would you determine the speed of the mass?

I know it involves taking energy conservation (KEi + Ui = KEf + Uf) but I'm confused on what r distance to use for the potential energy formula. Is it the distance from the mass to the center of mass of each respective ring or the distance to the edge? If so, why?

John Rennie

8:35 AM

@krauser126 it's simpler than you think. Potential energies simply add, so take the potential energy due to the left ring at a distance $x$ and the potential energy due to the right ring at a distance $3R-x$ and just add them together to get the total PE.

Then the change between the PE of the start point and the end point gives you the change in the kinetic energy.

Dante

9:08 AM

@JohnRennie hi

John Rennie

@Dante hi

Dante

Is total entropy change for a reversible iso-thermal process always zero?

That's what my textbook says.

John Rennie

The total energy change for any reversible process is always zero.

Dante

I didn't get you..

Could you explain please?

John Rennie

9:24 AM

For a reversible process the entropy change for any component is give by $dS = dq/T$.

In a reversible process the temperatures of all parts of the total system are the same because by definition in a reversible process the whole system is in equilibrium.

And when heat flows between any two parts of the system, $a$ and $b$, $dq_a = -dq_b$ because heat is a conserved quantity.

That means $dS_a = -dS_b$, so the total entropy change is necessarily zero.

Dante

Okay, then why do we have a formula for entropy change of real gas in isothermal process? $\delta S=nR\ln(\dfrac{V_2}{V_1})$. Does this represent change in entropy of the gas only and not the total change?

John Rennie

> Does this represent change in entropy of the gas only and not the total change?

Exactly.

Dante

Ah, ok.

Dante

10:01 AM

@JohnRennie Hi

John Rennie

@Dante hi

Dante

My textbook says zener breakdown is reversible while avalanche breakdown is irreversible. What does it mean?

John Rennie

That's not strictly true because some zener diodes use avalanche breakdown.

Suppose you take a regular diode and apply an increasingly large reverse voltage to it.

At some point an avalanche breakdown starts, and in a regular diode that means the resistance of the diode falls to almost zero. The result is a very large current that dissipates so much power in the diode that it burns out.

So it's not the breakdown mechanism that causes the damage, it's the large current you get when the breakdown starts. If you used a power supply with a high internal resistance to limit the current then the diode wouldn't be damaged.

@Dante OK so far?

Dante

Yeah

John Rennie

Some zener diodes work by a quantum tunnelling method and some work by avalanche breakdown. It depends on the design of the diode.

Where zener diodes use an avalanche breakdown they are designed to beak down in a more controlled way so you don't get the massive currents and resulting heating and damage.

Dante

10:09 AM

Quantum tunneling? I heard that's something like teleportation of sub-atomic particles.

@JohnRennie Oh, I see. Understood, thanks

John Rennie

@Dante the electrons can tunnel through the depletion layer.

Dante

Oh,

John Rennie

Zener diode

A Zener diode is a type of diode that allows current to flow not only from its anode to its cathode, but also in the reverse direction, when the Zener voltage is reached. Zener diodes have a highly doped p-n junction. Normal diodes break down with a reverse voltage, but the voltage and sharpness of the knee are not as well defined as for a Zener diode. Normal diodes are not designed to operate in the breakdown region, whereas Zener diodes operate reliably in this region. The device was named after Clarence Melvin Zener, who discovered the Zener effect. Zener reverse breakdown is due to electron...

> The device was named after Clarence Melvin Zener, who discovered the Zener effect. Zener reverse breakdown is due to electron quantum tunnelling caused by a high-strength electric field. However, many diodes described as "Zener" diodes rely instead on avalanche breakdown. Both breakdown types are used in Zener diodes with the Zener effect predominating under 5.6 V and avalanche breakdown above.

Dante

10:44 AM

@JohnRennie Hi

John Rennie

@Dante hi

Dante

John Rennie

@Dante The loop is going to create a dipole normal to the page. Yes?

Dante

Yes.

John Rennie

And the field due to the wire is also going to be everwhere normal to the page.

Dante

10:49 AM

Yes

John Rennie

And the force on a dipole aligned with a field is ... ?

Dante

Oh, wait.

I have was using torque formula :facepalm:

Force should be mdB/dr right?

John Rennie

There are two possible forces on a dipole in a field. One is the torque, so you were right to look at the torque. So what is the torque on a dipole aligned with a field?

Dante

0

John Rennie

Correct.

the other force originates from having a dipole in a non-uniform field.

Specifically the field has to be non-uniform in the direction the dipole is pointing. If you imagine the dipole as a pair of equal and opposite charges then the force arises because the field is slightly greater at one charge then the other.

In this case the field from the wire is symmetric about the plane of the paper i.e. the field at the same distance above and below the plane of the paper is the same.

And the dipole points normal to the plane of the paper, so the forces on the two ends of the dipole will be equal.

Dante

10:57 AM

Yes.

How does it experience net force then?

John Rennie

So this type of force is also zero.

So in the question the force on the ring is zero.

Dante

No, it's not...

3rd option is the answer.

John Rennie

Ah, I oversimplified a bit. The ring isn't exactly a dipole.

Dante

Yeah, part closer to infinite wire will experience greater force, so there should be a net force..

But idk the approach properly.

John Rennie

The field from the wire is proportional to $1/d$ so $B = k/d$ for some constant $k$ that we aren't worried about.

So for the left side of the coil $dF = kI/(d - a) dL$ and for the right side of the coil $dF = kI/(d + a) dL$

But this leading towards option 4 not 3 ...

Dante

11:10 AM

@JohnRennie I just saw a solution online.

They used mdB/dr for force

where m is magnetic moment

John Rennie

$m$ is the dipole moment of the coil?

Dante

But idk how that formula is derived.

@JohnRennie Yeh, magnetic dipole moment.

John Rennie

OK, I think I see how that is derived.

I think with the currents shown the dipole is aligned with the field. That is the dipole and the field point in the same direction, and that's the high energy alignment. The low energy alignment is the field and dipole pointing in opposite directions. OK so far?

Dante

@AdvilSell I never saw this formula while studying, is this a standard one?

High energy alignment? you mean unstable equilibrium?

John Rennie

@Dante yes

Dante

11:17 AM

Umm, how? Looks like stable equilibrium to me.

John Rennie

@Dante Out of the two possible alignments parallel to the field one is stable and the other unstable. Yes?

Dante

yes

John Rennie

And the energy difference between the two alignments is $2Bp$, where $B$ is the field and $p$ is the dipole moment.

@Dante OK so far?

Dante

Is it different for magnetic dipole?

John Rennie

Oops, yes, I got the diagrams the wrong way round.

Suppose the dipole is in the high energy alignment so the energy is $E = 2Bp$

And suppose the field varies like this:

If you move the dipole left then you're moving it into a region of higher field so the energy increases. The increase in the energy is $dE = 2p dB$

If the energy increases when you move it left then you must have done some work to move it left i.e. you had to apply a force to move it left.

The work you did moving it a distance $dx$ is $F dx$ so $dE = F dx$ so $F dx = 2p dB$

And that gives $F = 2p\frac{dB}{dx}$

22 mins ago, by Dante

They used mdB/dr for force

That's where the equation for the force you saw in the solution comes from.

(I think)

Dante

11:34 AM

@JohnRennie I got it.

Thanks

John Rennie

I have to go now. I may be around later - not sure.

Dante

ok

PolarBear

1:55 PM

Here maximum electric field is √2E na, because in the initial formula E is E_rms.

Got that ^

Abcd

3:33 PM

@AdvilSell

Advil Sell

@Abcd yup ?

Abcd

@AdvilSell Q 10 of JEE M

Advil Sell

>A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The
support of the pendulum starts oscillating up and down with a small angular frequency of
1 rad/s and an amplitude of 10–2 m. The relative change in the angular frequency of the
pendulum is best given by:

@Abcd this one ?

Abcd

@AdvilSell yh

Advil Sell

@Abcd k , trying , a min.

WTF why this is so hard ...arghhh...

ahh...okay got it @Abcd

Abcd

3:45 PM

@AdvilSell tell

Advil Sell

see $\omega = \sqrt{g/l}$

now $\Delta \omega / \omega = 1/2 \Delta g / g$

Abcd

@AdvilSell yh

Advil Sell

now $\Delta g = \omega_1^2 \cdot A_1$

Abcd

@AdvilSell ?????

Advil Sell

@Abcd see the above support is oscillating and it's acceleration max. is the equation above and hence the change in acc. is the same

Abcd

3:49 PM

@AdvilSell oh :( ... got it.

@AdvilSell you did it urself or understood from soln?

Advil Sell

@Abcd did myself , took SOOOOOOO....much time

:(

@AdvilSell good

@Abcd Score?

@Dante Urs?

Ab count karuna.

3:54 PM

Count nd ping

Dante

-_-

Abcd

@AdvilSell V simple one: 18

Dante

Send the key

Abcd

@Dante Scroll 2 the bottom of the page

$40-f = ma$

Dante

Where?

Abcd

3:55 PM

$(f+40)R = I\alpha $

yeah

Can;t find anything

$a = R \alpha $

Ah, understood

@AdvilSell these 3 dont give d right answer

try

Advil Sell

3:56 PM

@Abcd huh...the equation are correct tho. , maybe a calculation mistake

Abcd

@AdvilSell really they dont, just try 1ce

Advil Sell

@Abcd what did you take I ??

Abcd

$(80 - ma)R = I\alpha$

@AdvilSell $\dfrac{1}{2}MR^2$

or wait is it $MR^2$ :( ?

Advil Sell

@Abcd HOLLOW cylinder

latter one

Abcd

@AdvilSell Where did u practice Comm Sys from????

Too many modulation questions

Advil Sell

4:03 PM

@Abcd DCP , I have just done DCP and nothing else in Comm. Sys.

Did you get the answer now ?

Abcd

@AdvilSell are you able to do Q 22 in 2-3 minutes. Please try

Advil Sell

@Abcd Not now , But it's a fairly straightforward question just formula based

Abcd

@AdvilSell not now??

Advil Sell

@Abcd I I will revise the formulae ( which I will do just the day before ) it's will be easy

Abcd

@AdvilSell Is that same ques there in DCP

Ill attempt DCP right away if it is

Advil Sell

4:08 PM

@Abcd Nope

yeah but the formula is

Abcd

@AdvilSell i also kn the formula

Advil Sell

@Abcd oh....okay

krauser126

5:23 PM

@JohnRennie Right, I get that part, but I'm not sure if the distance I should be taking should be from the point to the edge of the ring (In which case you'd have to take the hypotenuse of aright triangle) or if it's just the distance to the center of the ring.

Abcd

11:03 PM

@AdvilSell AITS FT 8 --- 1

Advil Sell

Can you send QP ?

Abcd

@AdvilSell See pprs room

Advil Sell

Ya got it , downloading

@Abcd No Idea , now will see in morning again

Abcd

@AdvilSell K

Abcd

11:20 PM

@AdvilSell Q 4

See soln if you dont get it, and after seeing the soln tell me if you have read it ever?

Advil Sell

@Abcd Not exactly , But I knew that stokes law is only valid for a infinite body of fluid , hence I would have marked correct ( I didn't knew that force will be increased I knew it will change )

Abcd

@AdvilSell Where is this given man? I never read it.

Advil Sell

@Abcd My physics teacher , told about the exceptions also stokes law is only valid for spherical objects

Abcd

@AdvilSell thanks.

Advil Sell

@Abcd cool , Good Night :D

Abcd

11:26 PM

@AdvilSell GN

Transcript for

Problem Solving Strategies