Problem Solving Strategies - 2018-12-11

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10:35 AM

@JohnRennie good morning

imgur.com/a/ha5RGKA

Can you help me with Q1

Morning :-)

To me all of the options seem related to wavelength

Well, what is the $k/\omega$ if you expand it in terms of wavelength and frequency?

Isn't it velocity

Yes, and the velocity is a constant

10:38 AM

And velocity is releatee to wavelength

The velocity of an electromagnetic wave is $c$ regardless of its wavelength

But what about v= lamda x time

It's true that $v = f\lambda$, but that just means $f$ and $\lambda$ are inversely proportional. For all EM waves $\omega/k = c$ and $c$ is a universal constant.

All EM waves always travel at the speed of light

Okay. Got it

@JohnRennie imgur.com/a/0HkhdZq

Q3

Why isn't all the options correct

My textbook gives the answer to be intensity only

I suspect it's a poorly written question ...

10:45 AM

Okay

The speed of light in a vacuum is the same for all wavelengths and all frequencies, but if the light is travelling through a medium like glass this isn't true.

So it should be a. c and d?

If this was vaccum

Yes, but if only (c) is correct the question must expect you to consider light travelling in a medium as well as in vacuum.

Actually (c) is false for light travelling in a medium, though you need very high intensities to see the effect.

Okay. Got it

11:00 AM

@JohnRennie imgur.com/a/HxlnMG2

Q9

What if the wave is not " monochromatic "

What happens then

I don't think it matters. As I recall the energy is just $(E \times B)/\mu_0$

11:14 AM

@JohnRennie imgur.com/a/sCuhr19

Q2 can you help me with this

They say a small area, so I'd guess you're supposed to take the electric field as having a constant value over the whole area.

Do I have to integrate here

As x seems to be decreasing

Do you know how the displacement current is defined?

I know the formula only

And that is ... ?

11:19 AM

Also the electric field doesn't change with time

@harambe The electric field is $ke/r^2$. Yes?

Yes

And the charge is moving at constant velocity $v$, so $r = r_0 - vt$.

Okay

So the field at $A$ is $E = ke/(r_0 - vt)^2$ i.e. it is changing with time.

11:23 AM

@JohnRennie got it

:-)

Incidentally, the question says $x$ is small because for large $x$ you need to take into account that the electric field propagates at the speed of light. Taking $x$ small means the propagation time is effectively zero.

@JohnRennie the answer neglects "vt" term. Why so

What is the answer?

qAv/2π$x^3$

The flux linked is EA, so it's:

$$ \phi = \frac{kAQ}{r^2} = \frac{kAQ}{(r_0 - vt)^2} $$

OK so far?

11:32 AM

Yep

And the displacement current is:

$$ J = \epsilon \frac{d\phi}{dt} $$

Now we need to differentiate wrt time, but there's a sneaky trick we can use:

Yes

$$ \frac{d}{dt} = \frac{d}{dr} \frac{dr}{dt} = v \frac{d}{dr} $$

i.e. just the chain rule. OK so far?

Yes

So:

$$ J = kAQ\epsilon v \frac{d}{dr}(r^{-2}) $$

(oops, forgot $v$)

11:39 AM

@JohnRennie yes but this doesn't match with the answer

I got this answer too

$$ J = \frac{AQ\epsilon v}{4 \pi \epsilon} \frac{-2}{r^3} $$

$$ J = -\frac{AQv}{2\pi r^3} $$

And that matches the answer

But you assumed r=ro-vt.Right?

Really all I assumed was that $dr/dt = v$

10 mins ago, by John Rennie

$$ \frac{d}{dt} = \frac{d}{dr} \frac{dr}{dt} = v \frac{d}{dr} $$

I suppose it should really be $dr/dt = -v$

Yes

Okay got it

4 hours later…

Abhas Kumar Sinha

3:34 PM

user image

the question number 2.(ii) and 3

Need solutions

2 hours later…

5:29 PM

Hello!

user image

How to do this?

My teacher asked me to use LCAM and LCE

I don't understand how can we use LCAM since there will be torque due to the electrostatic force

1 hour later…

6:43 PM

@Dante there won't be

The force is radial and since you'll be conserving angular momentum about the center it's moment will be 0

These sort of questions are asked in gravitation

There's an example in dcp

6:55 PM

@sammygerbil please ping me when you are free

7:40 PM

@Abcd hello

@Dante There is no torque. The charge on the sphere is uniformly distributed, ie spherically symmetric. So this is a central force, exactly like gravitational attraction. You can get the answer using LCE, no need to use LCAM.

(Note : AB is a diameter. Diagram is misleading.)

(Sorry Avnish you already answered this.)

8:26 PM

@sammygerbil Hi, back.

@Abcd ok

uploading in 2 mins after a chem doubt.

user image

@sammygerbil

And then:

There's this set of problems that I am facing difficulty in:

@Abcd What is your difficulty here?

user image

user image

18, 20 in first image.

@sammygerbil not sure how to do it for $\Delta \phi_1$

$\Delta \phi_2 = -\pi/3 + 3\pi/2 = 7\pi/6$

8:44 PM

@Abcd Are you using $t=0$ for $\Delta \phi_2$?

@sammygerbil t will ultimately get cancelled by subtraction so doesnt matter.

@Abcd ok. What about $\Delta \phi_1$? What is the difficulty? Isn't $\Delta \phi_2$ harder?

@sammygerbil I feel $\phi_1$ is 0.

@sammygerbil $A\sin kx$ gives amplitude. Does it have anything to do with phase?

@Abcd It did for $\Delta \phi_2$. Why not for $\Delta \phi_1$?

@sammygerbil i dont know, $W_1$ is standing wave.

8:50 PM

@Abcd What difference does that make? Isn't it easier than a travelling wave?

@sammygerbil i find travelling waves easier.

@Abcd Wavelength is same for both. So phase difference is same. Phase difference depends only on distance between points.

@sammygerbil in standing wave phase can either be $\pi$ or $2\pi$. Nothing else.

Becuase all particles separated by two nodes arrive at their max positions together.

So phase = $2\pi$

@Abcd Hmm yes I see what you mean.

@sammygerbil whats the answer

8:59 PM

@Abcd Phase difference is $2\pi$ for $\Delta \phi_1$?

@sammygerbil why not $\pi$?

@sammygerbil Which hmm?

5 mins ago, by Abcd

So phase = $2\pi$

5 mins ago, by Abcd

Becuase all particles separated by two nodes arrive at their max positions together.

how do we know they are separated by 2 nodes?

So what do you think phase difference is for standing wave?

please tell the answer now...

i am getting confused by this discussion actually.

9:02 PM

8 mins ago, by Abcd

@sammygerbil in standing wave phase can either be $\pi$ or $2\pi$. Nothing else.

You have a choice of 2 answers! It cannot be difficult!

Why not $\pi$ man!?

Yes that gives D

@Abcd Exactly. Why not?

But there must be some logic

@sammygerbil Why not $2\pi$

@Abcd How do you work out whether it is $\pi$ or $2\pi=0$?

4 mins ago, by Abcd

5 mins ago, by Abcd

Becuase all particles separated by two nodes arrive at their max positions together.

leave that question actually.

@sammygerbil Mechanics now please.

9:05 PM

4 mins ago, by Abcd

how do we know they are separated by 2 nodes?

12 secs ago, by Abcd

@sammygerbil Mechanics now please.

25 mins ago, by Abcd

user image

25 minssss :((((((((

I cant give more than 5 minutes while practicing.

JEE Main is 23 days way. VERY worried and time is important.

@Abcd You are not practising! No internet help in the exam!

@sammygerbil by practicing i meant discussing doubts.

I have practiced and collected doubts to ask quickly.

Thats what I do always.

Practice -> Ask doubts -> sleep. (for physics)

@Abcd Then your goal should be to clear doubts, not to pass over them.

@sammygerbil oho

@sammygerbil I saw the solution and got the trick.

They graphed it at $t=0$

They graphed the cosine function.

$n\pi/2k$ are nodes.

Simple.

Mechanics now, please?

9:11 PM

@Abcd ok. What have you tried?

@sammygerbil 15 attempt:

Force equations:

$F = 2ma$

$F - f = ma$

But

I am also worried about the slanted support of top pulley.

@sammygerbil all the 3 problems are quite similar so its a single troubling concept that I want to understand.

my force equations dont bring out the significance of $\mu$

@Abcd How do you bring $\mu$ into the problem?

@sammygerbil $f_{max} = \mu mg$

@sammygerbil are my equations correct?

3 mins ago, by Abcd

I am also worried about the slanted support of top pulley.

@Abcd Not quite. $F=3ma$

@sammygerbil why

9:18 PM

@Abcd The blocks move together, the only external force is $F$.

@Abcd Gotit?

@sammygerbil why is it not coming from my equations?

12 mins ago, by Abcd

$F = 2ma$

lower block^

13 mins ago, by Abcd

$F - f = ma$

^ upper block.

sorry, lower block:

$F + f = 2ma$

@Abcd $F=2ma$ is wrong. $F=3ma$.

@sammygerbil i have corrected it

42 secs ago, by Abcd

$F + f = 2ma$

@sammygerbil please try to understand I have written separate equations for upper and lower block.

@Abcd No. Only force on lower block is $f$.

$f=2ma$

@sammygerbil will pulley's support not exert force on it?

9:28 PM

Forces on pulleys cancel out.

@sammygerbil i was referring to the long stick that supports the pulley.

String slips round pulleys, resultant force on lower block is zero.

Wont it exert force on block?

No.

y

9:30 PM

There is a torque on lower block but zero resultant force.

Reaction forces at each pulley are equal and opposite.

user image

y does this part not exert force on block^

At lower pulley string exerts force on lower block in NE direction, at upper pulley string exerts force is SW direction on lower block.

The two forces are equal and opposite but displaced (not in same line of action). Therefore zero resultant but non-zero torque.

@sammygerbil so the blue part exerts force but is cancelled?

The force on the block is determined by the tension in the string and the amount by which it changes direction. The directions of the supports for the pulleys are irrelevant.

ok?

Same principle applies in Q19.

Force down plane is $2mg\sin30=mg$. This accelerates both masses so $a=mg/3m=g/3$. Force on mass $m$ is $mg/3$ which equals tension in string.

doubts are 18 and 20 ...

9:45 PM

@Abcd What about 19?

@sammygerbil please answer the previous message, "so the blue part..."

@sammygerbil i was able to solve it... you can see a tick there.

What do you mean "the blue part"? The string exerts force on the pulley which transmits that force to the block. What the blue part does is irrelevant. It is rigidly connected to the lower block. It is part of the lower block.

@sammygerbil ok

If you understood 19 you should be able to understand what is happening at the pulleys in 15.

What next? 18?

@sammygerbil yes

9:50 PM

@Abcd What is your doubt here?

@sammygerbil 'tis hard

@Abcd Why?

What is hard about it?

@sammygerbil one second. A different doubt just came up

@sammygerbil 1:45

Doubt is: Why doesnt he bring the circular scale to 0 when he is putting the sphere in between?

@Abcd Why do you think he needs to do that?

@sammygerbil because 0 scale reading needs to start at 0, otherwise he'll have to subtract the 0 error?

10:03 PM

@Abcd Don't understand what you are asking. There is no "error".

i think i misunderstood the working of screw gauge

we kind of work in a backward direction.

We place an object and then move scale towards it...

I thought we would make room for the object between the jaws slowly and slowly

@sammygerbil ^

@Abcd To bring the jaws closer together you have to rotate the drum. The position of the moving jaw is determined by the rotation of the drum.

@sammygerbil not sure how to approach it man

@Abcd ?

:48010559 Q 18

10:09 PM

Sorry, back to Q18.

You have an object pulled in opposite directions by 2 forces. What do you think happens?

Start by calculating maximum friction forces.

@Abcd ?

@sammygerbil 6 N,10 N

Yes. Now what are the forces on the blocks at $t=0$? Do the blocks move?

@sammygerbil how to determine that if they'd move or not. I was facing problem with it because of the presence of the string in between

@Abcd Is the applied force big enough to overcome friction?

Yes the string does make it more difficult. Imagine the string is initially slack.

Then RH mass moves bcos 15N force is larger than 10N friction limit.

Resultant force on RH mass is 5N.

@sammygerbil i thought so but tension is there and you didnt consider it...

10:22 PM

Not yet!

Friction on LH block is initially zero until string becomes taut.

Then friction increases to 5N and stops RH block from accelerating.

(Yes it is a very messy problem. There are several possibilities to consider.)

At $t=2s$ force F'=4N. Then resultant force on blocks is 11N which is not enough to overcome friction.

@sammygerbil tension?

... I'm not sure we can get the values of friction by solving. It seems to me the amount and direction of friction on each block depends on whatever initial motion occurs, which depends on unknown length of string. But we could check the options to see if any are impossible.

ie use given values of friction to find tension in string attached to each block.

If these values are different option is not possible.

Aside question: For vernier calliper you get mean diameter as $0.5666666...$ cm. What would your writing on paper be : 0.56 or 0.57 cm ?

10:38 PM

@Abcd Which do you think?

@sammygerbil 0.57

ok

@sammygerbil whats the answer

@Abcd why?

@sammygerbil round off

10:40 PM

@Abcd Round up? Round down?

@sammygerbil up obviously

@Abcd To the nearest 0.01 cm?

@sammygerbil ???

(Which is sometimes up, sometimes down.)

I am just rounding off using mathematical rounding off rules

10:41 PM

@Abcd What does "rounding off using mathematical principle" mean?

What mathematical rule?

@sammygerbil academic.brooklyn.cuny.edu/geology/leveson/core/linksa/…

There is no mathematical or scientific rule for rounding off. It is entirely a choice, a convention, which depends on your purpose.

If you are satisfied that you have followed this convention, why do you need an answer from me?

@sammygerbil well options only have 0.56 and not 0.57 in the question.

I got the mean as 0.5666....

@Abcd What is the question?

user image

10:54 PM

@Abcd Maybe you are using the VS incorrectly?

@sammygerbil Please see attempt:

0.5 + 0.08 + 0.03 = 0.61

0.5 + 0.04 + 0.03 = 0.57

0.5 + 0.06 + 0.03 = 0.59

Mean = (0.61+0.59+0.57)/3

hell

calculation error.

@sammygerbil am sorrry. got it right this time. calculation error!!

ok. Back to Q18?

Its 4:28 am

I need some sleep :(

@Abcd ok. But I think we've nearly got it. Check if the unmatched forces on the blocks match.

24 mins ago, by sammy gerbil

ie use given values of friction to find tension in string attached to each block.

23 mins ago, by sammy gerbil

If these values are different option is not possible.

eg for (a) applied force on LH block is 4N to left and friction is 4N to right so tension in string must be 0. However for RH block applied force is 15N to right and friction is 5N to left so tension is 10N. Tension doesn't match, so option is wrong.

Repeat for other options.

11:18 PM

@harambe Question?

@sammygerbil no though I am solving questions. If I get stuck I will ping you

@harambe ok

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Dec '1811

Problem Solving Strategies

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