@amanuel2 All the best

Thanks!

Anyone know how the prof got L/2 - D here?

@amanuel2 L/2- d comes because your origin which I suspect is the left most point of the cart gets shifted backwards

So coordinates of the com of boy and cart changes.. See the two diagrams and compare them. You will get the rough idea

Damn can't edit. The dashed line depicts your origin. The cart moves back so coordinate of man and cart changes but origin remains same

Ignore my first two messages

Sorry @Scáthach didnt see this was workin on problems :/

@Scáthach Oh!!!

Thank you so much! Even the TA's couldn't explain this. And a person from the net explained this well in a sentence.. SMH, that TA is trash

@JohnRennie good morning

@Scáthach morning :-)

@JohnRennie will you be free in 1 hr

@Scáthach yes, I'll be around until about 17:00 Indian time

@JohnRennie Need help ^

@tatan I'm working for a few minutes. Back soom.

Ok sir... please ping :-)

@tatan hi

hi

@tatan I would start with the mass/rod stationary at the top, then calculate the change in potential energy when it has rotated all the way to the bottom. This change will then be equal to the rotational kinetic energy $\tfrac{1}{2}I\omega^2$ so you can calculate $\omega$.

Okay.. the change in potential and gain in rotational kinetic energy will be due to both the rod and the ball right?

Yes. The ball moves $2\ell$ downwards and the centre of mass of the rod moves $\ell$ downwards so the change in GPE will be $2Mg\ell + mg\ell$.

@JohnRennie short question.why is core o the transformer laminated

yes

@Scáthach eddy currents

okay .how does it work though

@tatan for a rod pivoted at the end $I=\tfrac{1}{3}m\ell^2$ and for the mass $I=M\ell^2$. The total moment of inertia is just the sum of these two.

Don't we have to take into account the translational kinetic energy?

yes..

@tatan no, the motion is purely rotational.

How do we know its 'purely rotational' ? I mean what is 'purely rotational' ?

@Scáthach in a transformer you want a core with a very high $\mu$ because that gives large magnetic fields. A solid iron core is good, but the changing magnetic fields will generate eddy currents in the core and those eddy currents dissipate energy as heat. OK so far?

Wait, what do they mean by "full cycle"? Does that mean that the rod becomes vertical again but this time above the suspension point? (And so that v is minimum, the velocity In that state is null?)

ok

@tatan average linear velocity is zero. Which it obviously is since the pivot is fixed so the rod just goes round and round.

ok

@Scáthach the idea of laminating the core is that the plates have an insulator between them so eddy currents cannot flow from plate to plate.

You still get some eddy currents within the individual plates, but the eddy currents, and therefore the energy loss, are much reduced.

okay.got it

@Mr.Xcoder the way I interpret it is that the rod has just enough KE to reach the top. So the angular velocity at the top is zero.

I get (A) as the answer ;-)

@JohnRennie hi.

@JohnRennie Yeah ok at first I thought it would only reach the horizontal state (at 90 degrees from the initial state) and got a wrong answer

@tatan yes, I get the same.

ok.. then i have done it correctly

@Nobodyrecognizeable hi

@JohnRennie hi

If yall finished i got a question

@Scáthach hi

@amanuel2 just 2 minutes

@JohnRennie imgur.com/a/0UASUvo

Np ill ask after you

@Scáthach which question?

In Q5, do I have to calculate the whole current by finding the resistance and then dividing by voltage because it's coming awfully long

I doubt it will match after this

@Scáthach it shouldn't be that long. The resistance of the galvanometer and 5 ohm resistor is 4.75 ohms. So the total resistance is 20K + 4.75 which is approximately 20K ohms. Yes?

yes

So the current is 0.1mA.

so you ignored 4.75.hmm kay i canproceed now

@amanuel2 i am done

@Scáthach The currents in the answer are only given to 2SF so I think it's safe to ignore the 4.75 ohms :-)

@JohnRennie but voltage drop across galvanometer will be zero then

To work out the current we ignored the 4.75 ohms because that was a good approximation.

To work out the voltage across the galvanometer/shunt it is not a good approximation to ignore the 4.75 ohms. The voltage will be 0.1mA times 4.75 ohms = 0.475 mV.

So the current through the galvanometer will be 0.475mV/95 ohms = 0.005 mA

Alright

@JohnRennie but voltage drop across 20K ohm will be 2V.How is this possible

Oh yall didnt finish? my bad

@Scáthach the current is actually slightly less than 0.1mA so the voltage drop across the 20K resistor will be slightly less than 2V.

Hmm got it finally

@amanuel2 if you consider the woman and the beam then there are no external forces acting on them because there is no friction with the ground. Yes?

(no exernal horizontal forces - obviously there is gravity that acts vertically)

Yes

@amanuel2 if no external force acts on a system then the centre of mass of that system cannot move.

@JohnRennie Can you explain Newton's Law of Cooling please? I searched it on the internet but I am not able to properly understand it. In several places I have seen approzimations being made in problems related to this law but I can't make out where to apply approximation and where not to...

@JohnRennie Ah!

But how come this be ?

And it says its frictionless

@amanuel2 can you clarify what you are asking?

@tatan Newton's law of cooling isn't a fundamental law. In fact it's only an approximation. Cooling is a hugely complicated business because it depends on the convection currents in the air flowing around the body. When we do the experiment we find that Newton's law of cooling approximately describes the rate of heat loss.

I have a doubt in some specific problems

Yes... wait a sec.. I am posting

@JohnRennie Was confused on how it moved with no friction on the second one :/

In the second question, the method used is to find the "average" temperature at first... T=(97+87)/2 . Then using the fact that dQ/dt=KdT we solve it

@tatan if $\Delta T$ is the difference between the object temperature and the environment temperature then Newton's law of cooling tells us that:

$$ \Delta T(t) = \Delta T_0 e^{-t/A} $$

For some constant $A$ that we have to determine experimentally. Yes?

Can you please hold on for a minute?

@tatan OK, I'll answer @amanuel2's question

sure

@amanuel2 do you mean part (b)?

Yeah its the end of #9 which is g)

@amanuel2 now you've confused me. Are you asking about part (b) or part (g) - or both?

@JohnRennie hi

@Scáthach just dealing with the question from @amanuel2. Won't be long ...

@JohnRennie oh no i was asking abuot part b

sure

@amanuel2 Look at this diagram I drew for your previous question:

Yes.

@JohnRennie Yeah i see

(I didn't have a GIF of a woman so I drew a confused JEE student :-)

Yeah np lol, that makes a lot of sense

@Scáthach do you want to ask now?

I am back

After tatan I guess

@JohnRennie OH! I thought it asked for how much the dude moved!

Thats why i was so confused. if that was the question it would be L/2 right?

yes

@amanuel2 yes, L/2 is how far the person walked along the base, but because the base moved as well they only moved 3L/8 relative to the ground.

@tatan which of the two questions do you want to look at?

Are the methods of solving both the same?

I suppose not

@tatan Yes, you use the info you're given to calculate the constant $A$, then you assume $A$ doesn't change when you change the environment temperature.

Ok then solve the first one

@JohnRennie Ahh wait it does ask for the woman actual lol. And yeah I understand what u mean now, thanks

Yes

Initial difference is $\Delta T_0 = 40°$.

How do you bring back an old message like this?

yes

this opens the highlighted message in a new tab

@tatan click on permalink, copy and paste the permalink address in the new tab then just paste that address.

Ok ;-)

Yes :-)

Lets get back to the question

@tatan OK. The question has chosen the temperature change to make life easy because it cools to 1/4 of it's previous value i.e. 2 half lives.

ok

$$ 1/4 = e^{-t/A} $$

yes

$$A = t/\ln(4)$$

@JohnRennie we got this from $$10=40e^{-t/A}$$ right?

Yes

ok

now we substitute this value to the second equation

Yes

@JohnRennie are you guys done

@Scáthach I think so

@JohnRennie from the second equation we get 30=60e^-{t'/A} or t'=A ln2 . From the first equation we got t=A ln(4)=2Aln(2)=2t' or t'=t/2 right? Please answer afer you have helped @Scáthach

Lol. I deleted fast xD

I need to work for a few minutes ...

@Scáthach Why did you delete ;-|

didn't wanna interfere

Don't delete please...

srry.wont do next time

;-)

@Scáthach back!

@JohnRennie hi

@Scáthach hi

@JohnRennie imgur.com/a/erydBS6

Q18

Shortest wavelength of x ray? I don't understand this

@Scáthach In an X-ray tube you get the sharp lines, Ka and Kb, and also a broad background. Yes?

Yes

The broad background comes from bremsstrahlung scattering i.e. some fraction of the electron's kinetic energy is converted to an X-ray photon.

okay

The highest energy X-ray photon possible is when all the electron's KE gets converted to a photon.

The electron KE is eV so the highest energy photon has $eV = hf = hc/\lambda$

okay

So that highest energy photon has the shortest possible wavelength. That's what the question means.

@Scáthach is that OK? Shall I get back to @tatan's question?

yes

@tatan your turn

:-)

yes..

In the second equation we start with $\Delta T_0 = 60$ and end with $\Delta T = 30$.

Actually this is easy because it's one half life and the previous cooling took two half lives.

yes

@tatan So the time is half the previous time.

I got my answer as D in this problem

ok... That was a nice way to look at the situation

thanks

But the solution given is A

@tatan so your answer is correct!

:-))

@tatan are we done?

yes

for now

I may need help later

@ShubhraneelPal that looks straightforward enough

It's calculating the flux change then using Faraday's law $E = -d\Phi/dt$

The current in the outer coil is $I = E/R$ and the charge is the integral of the current so:

$$ Q = \frac{1}{R}\int E dt = \frac{1}{R}\int \frac{d\Phi}{dt} dt = \Phi/R $$

Alright ima go off, hopefully not too many center mass questions ;} on the xam

@amanuel2 you just need to get the hang of questions like that.

And the best way to get the hang of the questions is just do lots of them.

Im kinda in the hang of it but im so much more comfartable on the other topics

And when they put questions where the inital velocity is the inital velocity of center mass confuses meh :(

Like this : i.imgur.com/ZMp6zVH.png

Jeez its 4AM!

@JohnRennie hi

@Scáthach hi

@JohnRennie imgur.com/a/RVufv2F

Q2

My equation are

15= L+R

2=30/√{R^2+(wL)^2}

w=200π

Are these equation correct?

Can we start from the top.

The DC source gives you the value of the resistance because inductors have zero resistance for DC. Yes?

Yes. I didn't know this though

@Scáthach The voltage across an inductor is given by $V = -L dI/dt$. Yes?

Yes

And for DC $dI/dt = 0$

So for DC the voltage drop across an inductor is always zero.

OK

So we know that R = 15 ohms

Yes

You are quite correct that for a resistor and inductor in series the impedance is given by:

$$ Z^2 = R^2 + (\omega L)^2 $$

So this will give the value of L. Got it

And since for the AC 30V we get a current of 1.2A the impedance is $Z = 30/1.2$

@JohnRennie hi

@Scáthach hi. Sorry, I was working. I'm free now.

@JohnRennie imgur.com/a/zk9SX2k

In Q50, the frequency is not given for fundamental. How tonproceed without it

@Scáthach You are told that the wave in the column is the fundamental, and you know the relationship between the wavelength and the column length for the fundamental. Yes?

Yes

@Scáthach And you're told the speed of sound so you can calculate the frequency of the wave in the air column.

@JohnRennie hmm got it.

@Scáthach cool :-)

@JohnRennie imgur.com/a/aoWMNsU

First question

I forgot what to do when optical path changes..

The equation for the double slits experiment is $n\lambda = d\sin\theta$, where $d\sin\theta$ is the difference in the path length. Yes?

Yes

What you're doing is introducing an extra path length difference of $d(n-1)$ where $d$ is the thickness of the paper and $n$ is the refractive index of the paper.

Okay

So the equation becomes $n\lambda = d\sin\theta + t(n-1)$

(I've changed to using $t$ for the paper thickness as I already used $d$)

Okwy

The central fringe has $n=0$. So to find the angle of the central fringe with the paper just substitute $n=0$ to get $0 = d\sin\theta + t(n-1)$

Okay. What's next. How to calculate shift

You've worked out that the central fringe has been moved to the angle given by $d\sin\theta = t(n-1)$. The question is how many fringes is that away from the centre. Yes?

Yes

And to find that just substitute in your original equation $n\lambda = d\sin\theta$. We've worked out that $d\sin\theta = t(n-1)$ so we simply get $n\lambda = t(n-1)$.

Rats, I've used $n$ twice.

Make the RI $\mu$ so it's $n\lambda = t(\mu-1)$

Okay

Just solve for $n$, and that's how many fringes the central maximum is away from the centre.

@JohnRennie got it

@JohnRennie imgur.com/a/G1T6jQr

Help in second part

The phase difference between A snd Bhai is 120

How to calculate for A and C

I'm working I'm afraid.

@JohnRennie Hi :-)

Looks Like your are working for now , please Ping me once you are free :D

I won't be back for a few hours I'm afraid.

when you calculate the gravitational potential energy of an object, then can you just apply the formula of PE = Mgh with the height taken as the height of the center of mass?

which would return $$ a_r = a\sqrt3 $$

@MartianCactus Assuming( If the assumption is practical), that the change is gravitation potential is insignificant over the length of the body. Like, For a body like Pencil, when calculating its GPE with Earth, it can be taken as Mgh

no

i am talking about something more like a weird 3d shape that is not regular in any way at all

maybe a 3d amoeba

Could you tell me the exact problem?

can we take its center of mass's height and apply it into the formula to get its total GPE?

If $ h << R $

Theres a rod inclined at 30 degree from a surface and you have to find its potential energy. Its length is L.

Yup, you can do it.

what about in the case of an irregular shape like an amoeba instead/

what does r stand for here?

$h$ = Height of the object

$R$ = Radius of the earth

oh yeah ofcourse

how about if theres a giant earth sized amoeba and the earth?

you apply the formula for gravitational force between their center of masses right?

Nope

hmm..why is that?

i guess integration comes into play then huh

yup

$$ dV = \frac{G dm}{r} $$

so this stuff only works for symmetric objects

V is field here right?

Nope. Gravitational potential

$dm$ = Infinitesimal mass element, $r$ = distance from consideration to be mass element, $G$ = Gravitational constant

oh ya sorry

so when does exactly the center of mass approximation come into play? @AjayMishra

when h << R

@Scáthach For phase difference between $A$ and $C$,The path difference $$ x = \sqrt{(2d)^2+D^2} - D $$, $$= D\sqrt{1 + {\frac{2d}{D}}^2} - D $$ Then for small values of $d$ , $ {D\sqrt{1+ {\frac{2d}{D}}^2}} $ is approxiamately equals to $D(1 + \frac{(2d)^2}{2(D)^2}) $, which would make $$ x = \frac{2d^2}{D}$$ and from the part 1, $ \frac{d^2}{D} = \frac{2\lambda}{3} $, whic would make $x = \frac{4\lambda}{3} $ or $ \delta = \frac{2\pi}{3} $

@AjayMishra what about if 2 completely circular planets are separated by a distance? Cant you apply the formula for gravitational force by taking the distance between them as the distance between there center of masses? In that case, h is not smaller than r

No, you have to use integration.

hm..our teacher told otherwise, Maybe integrations also gives the same results by coincidence?

@MartianCactus "Only few"

How to do no.21??

Use definition of entropy and integration

@tatan $$ dS = \frac{dQ}{T} = \frac{C_pdT}{T}$$ Integrate them between integrals for both blocks.

Like for block $1$, Limits of integration would be an $T_1$ to $\frac{T_1 + T_2}{2} $

and then add the changes

@JohnRennie Are you there

@Abcd hi

@Abcd The two R2 resistors are in parallel, so they have a combined resistance of 750 ohms. Yes?

@JohnRennie yeah

So if we take the zener diode out the voltage at the point indicated is 12*750/(500+750) = 7.2V.

And that is less than the zener breakdown voltage of 10V.

@JohnRennie So??

Below its breakdown voltage a zener diode is just a reverse biased diode.

And no current flows through a reverse biased diode.

@JohnRennie Alright.

@Abcd Are you done ?

@AdvilSell Done with what?

I am done with life

@Abcd Prof. ?

@Abcd so do I

@AdvilSell yeah no more questions to JR atm

@Abcd JR

@AdvilSell hi

@JohnRennie Hello , I ahve a question wait :D

@JohnRennie

how to approach this one ?

@AdvilSell when you calculate the potential at the tip of a cone you start with the potential due to a ring and you integrate to get the cone. Yes?

@JohnRennie Yeah

so what should I do to remove some part of that ring ?

Now, for a ring at a point on the axis you are the same distance from every part of the ring. So if $r$ is the distance to the ring then the potential is just $kQ/r$, where $Q$ is the charge on the ring. I can draw a diagram if that isn't clear.

@JohnRennie yeah , Got that

But with the part cut out each one of those rings is missing one sixth of its length and therefore one sixth of its charge. So the potential for the remaining part of the ring reduces to $V = \frac{5}{6}\frac{kQ}{r}$.

@JohnRennie yeah

When you set up your integral for the cone it will look just like it did before but with a factor of 5/6. That factor can be taken out of the integral, so after you've finished tearing your hair out doing the integral you'll get the same result as before multiplied by 5/6.

@JohnRennie Okay then let me find the original integral first

@AdvilSell Can't you just Google potential at tip of cone

@JohnRennie I would Like to try it first myself

OK :-)

@JohnRennie till I calculate can you look at the given solution , I feel that it have some insight that I can't understand

@AdvilSell I think they are just unrolling the cut cone and flattening it. You get a sector of a circle with radius equal to the cone edge length $r$. The perimeter of the sector (the curved bit) is the 5/6 of 2\pi R you're left with after the wedge is cut out.

I guess you can integrate that, but I'm not sure that's any easier that just multiplying the potential for the whole cone by 5/6.

Ah, I get it.

@JohnRennie But wouldn't the Potential At point P be different for a cone and a Disc ??

Imagine that sector as part of a whole disk i.e. the centre of the disk is at the apex of the wedge.

@JohnRennie Okay

If you know the potential at the centre of a charged disk just multiply it by $\theta/2\pi$

Or equivalently multiply by $(5/3 \pi R)/(2\pi r)$

Though that assumes you know the equation for the potential at the centre of a charged disk ...

@JohnRennie I know that , But wouldn't the potential change for a disc and a Cone diffrent ??

No. Potential due to a charge element dQ just depends on distance from the charge element, and potentials always just sum.

So unrolling the cone is fine because it doesn't change the distance from the tip to any of the charge elements.

@JohnRennie Oh..Yes , Okay I got it now , I was silly

@AdvilSell to be fair it's quite a sneaky trick :-)

I have to go now. Back tomorrow.

@JohnRennie Okay thanks Cya !

@AdvilSell

@Abcd yo

@AdvilSell nvm :D

@Abcd Okay Cool

@AdvilSell What is MI of equilateral triangle?

@AdvilSell Do you rembr the formula??

@Abcd nope

← 5 messages moved from JEE/High School Maths Problems

@AdvilSell try 11th

Is it 4th option?

@Dante yes

ok

@Dante But I mean how will we do it if we dont know the formula?

@Dante you did it after seeing the formula right?

@Abcd I didn't even solve it. The way you can guess it is,

Out of 4 parts, 1 part is removed.

And that one part is very close to axis. (less contribution to moment of inertia)

So remaining moment should be > 3/4th of original one

@Dante Nyc method.

@AdvilSell