Problem Solving Strategies

4:18 AM

@JohnRennie It may be traumatic but whatever it is, things we learn in that process is super beneficial and fun too. Maybe getting a perfect result shouldn't be the goal and the learning should be the primary goal!

John Rennie

8:35 AM

@Snapdragon-X the education that preparing for the JEE gives you is better than students get in the UK, and probably better than anywhere in the world. But as long as getting to a good college means the difference between a good life and working as a taxi driver it's going to be traumatic.

@AkshatSharma isn't that just the requirement that charge is conserved?

@Wolgwang hi :-) Sorry for the late reply but I tend not to be around in the afternoons at the weekend. Can you explain what you mean by resolving a vector into infinite components? Have you got a link to where you read this?

Wolgwang

9:10 AM

@JohnRennie No need to be sorry :-)

I was just reading this :-/

John Rennie

@Wolgwang We normally resolve a vector into orthogonal components.

So for example for a 3D vector we can write is as v = x i + y j + z k

where the unit vectors i, j and k are all at right angles to each other.

And in 3D we can only have 3 unit vectors all at right angles, so we only have three components.

Wolgwang

@JohnRennie So if someone says to resolve a vector in $n$ components (not necessarily orthogonal), then one would have to find $n$ vectors whose resultant is equal to the given vector?

John Rennie

Yes. It just means you are writing the vector as a sum of n other vectors.

v = a + b + ... + however many vectors you want.

Wolgwang

@JohnRennie Is there an algorithm to do so?

John Rennie

@Wolgwang suppose the vector is v = (x, y, z)

Wolgwang

9:24 AM

Yes

John Rennie

Then you define n new vectors all equal to a = b = ... = (x/n, y/n, z/n)

Wolgwang

Ohk Thanks:-)

John Rennie

Then when you add all n vectors together the x component of the sum will be n*x/n = x. And the same for y and z.

Ashish Ahuja

@JohnRennie hi

John Rennie

@AshishAhuja hi :-)

Ashish Ahuja

9:31 AM

I have a few questions to ask, I'll paste a picture, give me a second...

John Rennie

Actually I need to go and attend to some Monday morning chores, so I need 15 minutes. Upload the question and I'll have a look as soon as I'm back.

Ashish Ahuja

@JohnRennie sure I will do that. Please ping me whenever you come back.

I think the screenshot is a bit cut out from the right, I'll take another picture.

These are the questions I'm not able to solve. Just looking for a way to start them out.

John Rennie

@AshishAhuja I'm back!

Ashish Ahuja

I'm here.

John Rennie

OK. When the sphere is on its own the rate of energy loss is the usual Stefan-Boltzman expression W = σT^4. Yes?

Ashish Ahuja

9:46 AM

Yup

And you need to include surface area as well?

John Rennie

Ah, we need the area don't we. The total power is W = 4πr² σT^4

Ashish Ahuja

Yeah

John Rennie

Now when we surround it by the spherical shell the sphere still radiates energy at the same rate, but it also absorbs the energy emitted from the inside surface of the shell. So the net rate of energy loss is now 4πr² σT^4 minus the energy radiated by the shell.

OK so far?

Ashish Ahuja

Yes

John Rennie

And the energy radiated inwards by the shell is W' = 4πR² σTs^4

Ashish Ahuja

9:50 AM

Uhh how? Ts is surrounding temperature right, not shell temperature?

John Rennie

I think it means the shell is at the same temperature as the surroundings.

So Ts is the shell temperature.

Ashish Ahuja

Hmm.

Ok, let me just solve it then and see if the answer comes...

John Rennie

It says the space between the sphere and shell is evacuated, so the space around the sphere can't have a temperature since it's a vacuum.

Ashish Ahuja

But the shell could still have a different temp. than the surrounding around the shell right?

John Rennie

If that was the case the question would have to tell you that, or at least give you the information you need to calculate the shell temperature. But it doesn't.

Ashish Ahuja

9:56 AM

I'm not getting the final answer, maybe I'm just not grasping what you've explained correctly. What would be the next step to solving this? Answer should simply be (W - W')/W right?

John Rennie

Yes

Ashish Ahuja

Ignoring the constants, I'm getting...

$$\frac{r^2T^4 - R^2T_{s}^4}{r^2T^4}$$

*cancelling the constants

Given answer is a = 1 b = 1, so it is not dependent on the temperatures at all

John Rennie

I get the same answer as you ...

I guess I must have misinterpreted the question ...

I don't get it. Sorry :-(

Ashish Ahuja

No issues. Do you know what can be done in the 2nd one?

John Rennie

I don't see how you can answer the second question unless you know the contact angle.

Ashish Ahuja

10:01 AM

contact angle = zero

John Rennie

Ah, OK.

It's like capillary rise. Where the water touches the plate there is an upwards force of 0.073N per unit length of the water.

Ashish Ahuja

That's what I thought too but I wasn't able to come up with a concrete equation

@JohnRennie what do we equate it to? In capillary rise we easily equate it to weight, I don't see how that can be done here.

John Rennie

To be honest I can't remember how the rise is calculated for a single plate, but we could attempt it using the pressure change.

Ashish Ahuja

Sure, I'm fine with any approach. This was the first time I'd seen rise on a single plate as well.

John Rennie

The water surface near the plate is a section of a cylindrical surface, and this curvature causes a pressure difference given by the Laplace equation. Yes?

Ashish Ahuja

10:09 AM

Yup, but I don't really understand how the Laplace equation works. Do we need it for this problem?

(i.e the young-laplace equation)

John Rennie

Ashish Ahuja

radius of curvature = height?

John Rennie

That's not a great drawing. It's supposed to show that the water surface near the plate has a circular cross section.

@AshishAhuja yes.

Ashish Ahuja

ok

John Rennie

It's a cylinder, i.e. only curved in one direction, so the curvature is just h.

So the pressure difference is ΔP = γ/h

And the pressure difference caused by the rise is ΔP = ρgh. So we equate the two pressure differences.

Ashish Ahuja

10:14 AM

let me solve and check...

John Rennie

OK ... :-)

Ashish Ahuja

So $$h^2 = \frac{\gamma}{\rho \times g}$$

John Rennie

Yes

Ashish Ahuja

It doesn't match with the given answer. I can send a picture of the solution if you'd like, it doesn't make any sense to me, give me a second...

John Rennie

OK ...

Ashish Ahuja

10:16 AM

I have no idea where the 2 comes from, there is no explanation.

John Rennie

$$ \rho g \frac{h}{2} \ell h = \gamma\ell $$

The right hand side is the upwards force i.e. if the length of the plate is $\ell$ then the total upwards force due to the surface tension is $F = \gamma\ell$.

Ashish Ahuja

Yes

oh and that's weight

a crude approximation?

John Rennie

I think what it's doing is assuming that the shape of the water near the plate can be approximated as a triangle of base $h$ and height $h$, then the area of the triangle is $A = h^2/2$. Yes?

Ashish Ahuja

Yeah I got it. But isn't that a very crude approximation? Or would be that accurate?

John Rennie

That's a pretty crude approximation, yes.

Ashish Ahuja

10:21 AM

So is the answer you got above from the laplace equation the correct one?

John Rennie

I wonder if that nice Mr. Google can find us a solution ...

Ashish Ahuja

I'll try searching for single plate rise ...

John Rennie

Wilhelmy plate

A Wilhelmy plate is a thin plate that is used to measure equilibrium surface or interfacial tension at an air–liquid or liquid–liquid interface. In this method, the plate is oriented perpendicular to the interface, and the force exerted on it is measured. Based on the work of Ludwig Wilhelmy, this method finds wide use in the preparation and monitoring of Langmuir films. == Detailed description == The Wilhelmy plate consists of a thin plate usually on the order of a few square centimeters in area. The plate is often made from filter paper, glass or platinum which may be roughened to ensure complete...

There

Ashish Ahuja

May I ask what terms you searched for? I often struggle with finding stuff if I don't know the technical terms for what I'm searching.

John Rennie

I used this search

And it led me to this web page

Oh, wait. That doesn't give the rise.

Sorry, red herring.

Ashish Ahuja

10:28 AM

physics.stackexchange.com/a/536287/104157 this seems to be related?

5

A: How far can water rise above the edge of a glass?

I can't answer your question because it depends on the shape of the rim, however I can answer a related question that should be easily adaptable to your problem. If you have a puddle of water on a flat surface the thickness of the water film, $h$, is given by: $$ h = \sqrt{ \frac{2\gamma_{al}(1...

I found an answer of yours which refers to the same final result :D

John Rennie

@AshishAhuja Hmm, that also gives the pressure difference as $\Delta P = \tfrac12 \rho g h$

Ashish Ahuja

@JohnRennie I'll try reading it once more, didn't understand it completely in the first read..

Ohh I think I got it.

Let me just find a book to verify what I remember is correct...

Yup

The pressure on an inclined surface can be shown to be pressure at the centroid times times surface area

Ohh but then we use the triangle approximation again, nevermind.

John Rennie

I give up. I can't see where that factor of two comes from.

Ashish Ahuja

this answer gets the factor of 2 from the young-laplace equation itself interestingly.

@JohnRennie no issues, I'll just use the triangle approximation.

Thanks.

Ashish Ahuja

12:12 PM

nvm I did realize what I was missing just as soon as I posted this, they are both in free fall..

John Rennie

:-)

EVO

3:00 PM

I can't understand this problem

Two identical uniform solid spherical balls A and B of mass m each are placed on a the fixed wedge as shown in figure. Ball B is kept at rest and it is released just before two balls collides. Ball A rolls down without slipping on inclined plane and collide elastically with ball B. The kinetic energy of ball A just after the collision with ball B is:

Ashish Ahuja

@EVO what would basically happen is that sphere A will roll down and collide with sphere B at a moment when B is at rest just about to start rolling.

EVO

As per the given Qn both the blocks are released. Since A is in pure rolling it will be a rough plane. So when A collides with B,shouldnt Rotational energy be also transferred(won't friction torque it?)

**the solution says that only translational energy is tranferred

Ashish Ahuja

that is true but I don't think you can solve for the angular impulse without the friction coefficient, so generally you would ignore that. And they haven't mentioned anything about B and pure rolling.

Only translational energy would be transferred since we have only a linear impulse, no angular impulse.

Also if you consider friction between the two spheres, I don't think energy can be conserved at all since the spheres will slip wrt each other, so that makes the problem unsolvable. Not 100% sure about this though.

EVO

So does that mean after some time B would initiate pure rolling?

Ashish Ahuja

Uhh the problem doesn't mention anything about the friction coefficient of B so we can't say.

I was talking about during the collision about the angular impulse, if it existed.

EVO

3:17 PM

I'm just tryin to understand the situation after the collision. I think the a part of ball B's energy (after collision) would redisturbute into rotational energy.Is that correct?

Ashish Ahuja

Maybe after some time it will if sphere B is not smooth so it will roll on the plane, but instantly after the collision, I don't think so

EVO

Oh..thanks a lot

Ashish Ahuja

I might be wrong about the friction stuff leading to loss of energy, I guess we'll have to wait for JR for that.

EVO

Mmm

Snapdragon-X

3:58 PM

@JohnRennie It's absolutely not that someone diligently learning and appears for JEE mains will have to go through a judgement day where they'll know their fate. The top 20% contains all people who learned well....and it's not a standard generalization, i have observed it well enough. Those in the top 20% end up in okayish to great colleges. Eventually, in college if they perform well, they get into Slab A placement, and if not then a placement scale right above Below poverty line..... (1/2)

So, the remaining 80% are assumed to be those who messed up real bad OR those who had to give the exam since their family forced them or a casual sign up. And the Taxi driver generalization is a fallacy

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Problem Solving Strategies