@JohnRennie good morning

@Scáthach morning :-)

Are u free now? You are early today

@Scáthach yes I'm free

@JohnRennie When its my turn ping me.....Btw..Good morning

@JohnRennie imgur.com/a/1FrySYF

Couldn't solve

@Scáthach 26 or 27?

26

∆U=nCv∆T.

∆H=+ve

I get only these information

Give me a moment to have a think about it

Okay

If you start with $TP^{-1/3}$ you can substitute for $T$ using $PV=RT$

So you'll end up with $PV^{3/2} = constant$

Yes that gives gamma to be 3/2

@JohnRennie hi?

for some reason I'm struggling this morning.

Okay

The answer isn't (1) because (1) would be an isothermal process and for that $\gamma=1$.

@JohnRennie yes

But I can't remember which way the temperature change goes for $\gamma$ greater or less than 1.

Should I post solution @JohnRennie

OK, let's see the solution

imgur.com/a/iNN4nph

Q26

OK so they are using the equation for the specific heat in a polytropic process

@JohnRennie You done?...

@ayc in a moment

@JohnRennie but we want specific heat at constant volunlme, right

@Scáthach no, you want the specific heat for the process.

@JohnRennie okay. I thought we had to calculate Cv for internal energy

Got it

@ayc done

I guess you just have to remember the equation for the specific heat in a polytropic process $PV^z$ is:

$$ C=\frac{R}{\gamma-1}+\frac{R}{1-z} $$

@JohnRennie yes.

@JohnRennie imgur.com/txC2da8 ...Anything on that page would be great..But,lets begin with Q.66

You're told that the force the water exerts on the ship is proportional to the velocity of the ship relative to the water. So if we call the velocity of the ship $v$ then the force is $F = k(v_0 - v)$ for some constant $k$. Yes?

@JohnRennie Yes

And we're told that initially when $v=0$ the force is $T_0$, so $T_0 = kv_0$ i.e. $k = T_0/v_0$.

So our equation is:

$$ F = \frac{T_0}{v_0}(v_0 - v) $$

The acceleration of the ship is $a = dv/dt = F/m$ so the equation for the speed of the ship is:

@JohnRennie I get it..I just didnt think of T0=kV0

$$ \frac{dv}{dt} = \frac{T_0}{mv_0}(v_0 - v) $$

@JohnRennie Now Q.64

Start by calculating the tension in the wires. The vertical components of the two tensions must add up to 15g, so for each wire the vertical component is 7.5g so the tension is $7.5g/\cos30 = 15g/\sqrt{3}$

Then the horizontal force is the tension times sin30 which is $7.5g/\sqrt{3}$, and there are two wires to the total horizontal force is $F_h = 15g/\sqrt{3}$.

OK so far?

@JohnRennie Yes

Ah, I need to work for a few minutes ...

@JohnRennie Ping me when you are back

@ayc back

@JohnRennie Let's continue

@ayc rather than go through the problem in detail, let me outline the approach.

First work out the horizontal force due to the strings.

Then assume the block doesn't slide and calculate the acceleration of the platfrom+block $a$

Now calculate the force on the block due to this acceleration, i.e. F=ma, and see how this force compares to the frictional force. If F is less than the frictional force then the block doesn't slide and you already have the acceleration.

@ayc OK so far?

@JohnRennie I did exactly the same....ok so far!

If the force on the block is greater than the static frictional force (and I'm guessing that it is) then the block slides. In that case the acceleration is the dynamic frictional force divided by the mass of the block i.e. $a=\mu_d g$.

@JohnRennie I think you are not able to find one equation!, right?....

@ayc huh? You can certainly find an equation but first you have to figure out whethr the block slides or not.

@JohnRennie hi. Are you free

@Scáthach hi, yes I'm free

@JohnRennie imgur.com/a/ntpf3th

Q7

How to conclude the small block moves left

@Scáthach The way you do this is to start by assuming the small block doesn't slide, then calculate the acceleration of the two blocks due to the applied force. So take the 62N force, subtract off the friction with the floor to get the net force, then divide by the total mass of 9kg to get the acceleration $a$. OK so far?

Ok

@JohnRennie I'm not able to proceed.,,stuck in finding an equation...Could you solve?

@Scáthach Now, if both blocks accelerate with the acceleration $a$ that means the small block must accelerate at $a$ so the force on it must be $ma$ i.e. $5a$. Yes?

Yes

@Scáthach now compare that force $5a$ with the frictional force $\mu mg$. If $5a$ is less than the frictional force then the block won't slide. If $5a$ is greater than the frictional force then the block will slide.

@JohnRennie doubt

@Scáthach yes?

@JohnRennie first we calculated the acceleration for which small block stays at rest which is a. Then u said if both blocks move together, the avcelerwtion on small block is 5a

Wouldn't accelerstion change in second case

I said if the acceleration of the small block is $a$ then the force on the small block has to be $5a$.

So nothing to do with the first case.. Okay got it

@JohnRennie Hi ! Please ping when its my turn :-)

@JohnRennie okay let me give this a try then

@ayc @tatan you can ask

@tatan @ayc is next, but right now I need to work for a few minutes

ok. I will wait

@JohnRennie My turn after tatan XD

@ayc @tatan @Dante @Scáthach OK, I'm back. Who's around?

me

I am.

Damn! A second late XD

ayc first I guess

same question!

Sure

First lets calculate the horizontal force on the platform when the string AB is cut.

@ayc The vertical force in the wires is $2T\cos30$ and this has to equal $15g$, so the tension in the wires is $T = 15g/\sqrt{3}$. Yes?

And the horizontal force is $2T\sin30 = T = 15g/\sqrt{3}$

@JohnRennie wait

OK ...

@JohnRennie Once the string is cut the tension's in both the wires wont be same and moreover the rod has acceleration downwards and hence net force in the y direction wont be zero..right?

The tension in the wires will be the same. I agree the tension in the strings will change when AB is cut.

@JohnRennie Ok,lets just move on assuming it will be same

OK. If the horizontal force is $15g/\sqrt{3}$ then the acceleration of the two masses (assuming they move together) is $F/m = g/\sqrt{3}$. That means the force on the block is $12g/\sqrt{3}$.

The static coefficient of friction is 0.6 so the static frictional force is $0.6 12g = 7.2g$

And $12g/\sqrt{3} = 6.93g$

That means the frictional force is larger than the force due to the acceeration of the two block and platform, and therefore the block won't slide.

@JohnRennie vertilcal force wont be equal to 15g right?..because the rod has downward motion also?.......

Yes. And that means the tension in the strings is less than we calculated, which means the horizontal force on the platform is less than we calculated.

So that means the force on the block due to the acceleration of the platform and block is less than we calculated. So it doesn't affect our conclusion that the block won't slide.

@JohnRennie ooh..just like we did before,,I get it

so...how do we find actual accelration then?(i.e accelration of block+platform)

@ayc that should be easy. It would be easy if it was a point mass instead of two blocks ...

Hmm I get the horizontal acceleration to be $g \sqrt{3}/4$

@JohnRennie Btw,answer's option a=5

@JohnRennie hi. I will be available after 2 hrs from now. Are you done by then?

Oh, wait, the normal force will be reduced as well.

@ayc I'm afraid I give up on this one. Either it's really complicated or I'm missing the simple trick for doing it.

@Scáthach I should be around then.

@Dante do you want to ask now?

Yeah, just a minute

The stress is the (net) force per unit area. The net force is $2F - F = F$. So the stress is $F/A$.

@JohnRennie ohk ..its alright.

@JohnRennie Option 2?

Nope, I marked the same in the exam...

Is it mark all that are correct, or is only one answer correct?

One answer

It's D

Have you got the solution?

Yeah

Let's have a look

Solution is useless.

It says "The stress is compressive on the left end, tensile on the right."

That's all.

Let's think about this. Suppose we hang the bar vertically from one end, then attach a weight that has the same mass as the bar to the other end. So it looks like this (diagram incoming):

OK I think I get it. The net force is $F$. Suppose the bar was fixed at one end. Then we'd have a situation like this:

@Dante In that case the strain would be $F/AY$. Yes?

Yes

But the end isn't fixed, so when we apply a force F to the rod it moves:

Ok, strain will be non uniform now.

So the stresses in the rod are less than they are with a fixed end, and the strain is less than $F/AY$. We can calculate the stress and strain using an analogy:

If you hang the rod from one end then we have a force F at one end and zero force at the other, just like in the case of applying a horizontal force to one end.

And this is a the standard problem of calculating a spring extension when the spring is hung from one end. From memory the extension works out at half $F/k$ i.e. the average strain would be $F/2AY$.

And as you say the stress increases as we move upwards through the rod.

Right

@JohnRennie Understood :-)

Thanks!

@JohnRennie

That's a fairly routine circuits calculation isn't it? Either user phasors or a complex impedance as you prefer.

Is it? I never saw such a problem before...

The answer is (C)

It is!

@Dante there are two ways to approach this. A quick shortcut way or a rigorous way.

I'd like to listen to both if there isn't any problem.

@Scáthach Are you in the queue?

The quick shortcut way is to know that the impedance of a capacitor deceases with increasing frequency. That means when you have a resistor and capacitor in series the total impedance of the two decreases with increasing frequency, so the current increases with increasing frequency.

Free for 1 hr

@Scáthach I'm free as soon as Dante and I have done this problem.

@Dante Is that OK so far?

Yeah

In this problem you have two signals $E_1$ and $E_2$ with equal voltages but different frequencies, and those two signals produce currents $i_1$ and $i_2$.

Since the impedance decreases with increasing frequency the signal with the higher frequency will produce the larger current.

Oh....right.

We can do this rigorously, but the problem is that it sounds as if you haven't learned how to use phasors or complex impedance yet.

No, I haven't. The above reasoning is enough I guess!

Thanks!

You won't be able to tackle questions like this until you've learned how to do AC circuits. The rigorous answer to the problem would involve stuff you don't know.

I've heard about complex impedance.

It's probably not there in the syllabus.

Couldn't find it in any JEE book.

Is it @Scáthach ?

I'm sure I've had JEE students asking about phasors, so I assume phasors are in JEE.

Yeah phasors are there, gotta study that properly ;_;

Phasors and complex impedance are basically the same thing. Phasors are done using diagrams while complex impedance just uses algebra, but they are basically the same thing. Phasors are probably easuer to use.

@Scáthach do you want to ask now?

@Dante haven't heard about it in being syllabus but lot of people use it for solving

@JohnRennie I see, will study it soon then.

@JohnRennie probably in half an hour

@Scáthach Yeah cuz lot of people write FIITJEE exam and they ask it XD

@JohnRennie hi

@JohnRennie hi.

You need to write down the equation for the flux $\Phi$ through the loop. This will mean integrating the magnetic field due to wire, which should be straightforward.

Yes.

Then just use Faraday's law $E = -d\Phi/dt$

@Dante is your problem done?

Can i ask?

I'm on it, but yeah you can ask.

@JohnRennie what should be the field outside the spheres?

@JohnRennie are you here?

@Nobodyrecognizeable The field outside the spherical shell is just given by Gauss' theorem.

In the solution they say 0.

@JohnRennie ^^ I think due to the positive charges the charge induced in the in a Cell should be negative but the outer surface of the shell what charge should that have?

@JohnRennie Got it :-) had forgotten the fact that induced emf will be at peak when velocity will be highest .

@Nobodyrecognizeable I'll have to get back to you. I don't remember offhand how to do this so I'll have to research it, and right now I don't have time for that.

@Dante Cool :-)

@Nobodyrecognizeable I could help a bit, I had researched these problems few days ago XD

@Dante ok . You got the question.

Assume the charge is q inside the space due to the charge density. Now the charge on the inner surface will be -q from gauss law right?

@Dante yeah. What's with the outer surface?

Okay, now since shell is grounded it's total potential is zero. Let charge on outer surface be x.

$Kx/R_2 + Kq/R_2 + K(-q)/R_2 =0$ Right?

@Dante so the charge on outer surface is zero.

Yep

@Dante thanks. Whats the field inside R_1 and R_2?

Between r1 and r2? It should be zero.

Between center and r1, it's not zero

That can be found out using guass law again.

@Dante your jee preparation looks quite complete. Great. Never expected a 12th student to answer these questions.

No lol, I'm one of the dumbest kids here.

@Dante how'd you gain the electrostatic energy. No need to explain.

Integrate energy density due to field

@Dante whats the volume element?

A small shell will do right?

Yeah it'd be fine

@Dante fine . If we had two shell instead of one annular shell. What should be the charges on the shell?

*shells?

With outer one grounded?

@Dante yep..

Outer one would have 0 on both surfaces.

If the density has total charge q, inner one would have -q on inner surface and q on outer surface.

@Dante inner -q outer +q.

Oops, right!

Sorry

@Dante yeah.

@JohnRennie hi

@Scáthach hi

@Dante fine.. you have 7 days for the final battle . Best of luck.👍👍👍👍

Thanks dude.

@JohnRennie have a nice day professor good bye.

@Dante cya. Now here . In iits later.

Cya hehe.

@JohnRennie I have a very small (yes or no ) question...are you free?(question other than :are you free?)

@JohnRennie imgur.com/a/6zr1LxI

My answer is not matching..

@JohnRennie the angle at B from A is {2}/{√5 }

For maximum, I think it should be 90

@JohnRennie The question is this:Would the weight of a charged ball be greater than its intrinsic mass?.I think its should be more because the self energy adds to the mass?

@Scáthach you want the force to be applied normal to the red line. Yes?

Yes

So the angle $\theta$ has to be equal to the angle $\theta$ that I've marked in red.

$\tan\theta = 1/2$

I didn't understand what u did

We know the angle between $F$ and the red diagonal line must be 90° to get the largest moment. Yes?

Yes

And the angle between F and the horizontal is $\theta$

@JohnRennie got it

@Scáthach cool :-)

hello @JohnRennie

@ayc yes, a charged ball will have a larger mass than an charged ball. The difference will be the electrostatic energy divided by $c^2$.

@user8718165 hi

@JohnRennie and that electrostatic energy is self energy only right?

Yes

@JohnRennie i have one more quetsion:my book says that capacitance of each conductor increases in presence of the other i.e in asystem of two conductors each conductor can stpre more charge than when they are isolated...I have difficulty in understanding this..could you explain?

I'm not sure what there is to explain.

If you have a single isolated wire with a charge Q and a surface potential V then the capacitance of that wire is $C = Q/V$

You can calculate $V$ using Gauss' theorem.

Though I forget the result.

If you have two wires with charge densities Q and -Q and the potential difference between them is $V$ then the capacitance of the two wires is again $Q/V$.

($Q$ is the charge per unit length and $C$ is the capacitance per unit length)

@tatan is k the rate constant or the equilibrium constant?

I think it should be rate constant

@tatan $$ k = Z \exp \left(\frac{-E_a}{RT}\right) $$

The activation energy is smaller for the second reaction, but we don't know $Z$.

@tatan is the answer C.

@Abcd I don't have the answer

Sorry for the edits. But it should be C because:

C is the fastest step of the reaction

So it has maximum rate constant.

other step is smaller.

B--> C process is very fast.

You have to cross very LESS energy to reach C from B.

Why not D then?

While from A to B you have crosss VERY MUCH energy

@tatan yes my bad. It should be D.

It cant be C because the rates are no where comparable.

Ok

@JohnRennie Do you have any rigorous explanation why is it D over C?

Could the fact that second peak is lower than first contribute to it?

@tatan It is very bad to do questions you don't have answers for.

yes^^^

:-/

I would have guessed the answer was (C). We don't know the pre-exponential factor $Z$, but if $Z$ is roughly the same for both reactions it comes down to the activation energies. The activation energy for the second reaction is smaller than for the first reaction but not much smaller. So $k_2$ will be greater than $k_1$ but not much greater.

I see.

@JohnRennie Isn't activation energy of B is far less ? as B is a intermediate hence the Ea for the conversion is very low low as also seen from the distance between the B and the hill compared to distance between A and the 1st hill

@JohnRennie Ea1 is measured from bottommost point

that is from point A

@AdvilSell it looks to me as if $E_{a1}$ is about twice $E_{a2}$

@JohnRennie but Ea1 is activation energy for A to B shouldn't it be measured from A to B

@AdvilSell ah, OK, yes I guess it should. In that case $k_1 \ll k_2$

@JohnRennie Okay , Thanks :D

👍

Thanks all

@AdvilSell assuming that the pre-factors are roughly the same.

@JohnRennie oh , yeah ,okay :D

@JohnRennie can I ask a question , you are also related to IT , aren't you ?

@JohnRennie can we explain interaction between two magnetic monopoles using inverse square law....?

@AdvilSell yes

@JohnRennie My laptop is showing a hard drive crash message from time to time repeatedly , but nothing bad has happened till now , what is the real issue or what is the solution to stop this , these message don't come in morning but start coming at around 5:00 pm IST

@AdvilSell what is the message? What does the message say?

@ayc yes, if magnetic monopoles existed the force between them would obey an inverse square law.

@JohnRennie It says you have a vital problem in your hard drive and you may lose all your data and program , then it goes on to show 2 options either to ignore the message or to start a backup , starting a backup give instruction of a hard drive replace , clicking ignore , does what it says

Is that the exact wording of the message? If so it sounds like malware to me.