Problem Solving Strategies

Madhuchhanda Mandal

05:24

@blue Can you please format the question?

You have edit privileges I guess

@blue I thought on the explanation given by you, but could not agree on it

Anonymous

06:06

1

A: Why Energy is not conserved?

This is really an inelastic collision type problem with kinetic energy not conserved. Energy is conserved but the difficulty is that having an ideal system of masses and string makes it appear that energy is not conserved. To this end you have to allow either the masses or the string or both to b...

Anonymous

@MadhuchhandaMandal See Farcher's answer

Anonymous

His explanation is basically the same as mine.

Anonymous

Read it carefully.

Madhuchhanda Mandal

06:40

@blue According to what Farcher said, the given question is an impossible situation. I also believe that. But both of your "making the question valid attempt" is what I still don't agree. If the string is inextensible, it cannot store any PE.

Madhuchhanda Mandal

06:51

@blue See my last comment to Farcher's answer. That's the reason for posting the question and I am still unable to understand that.

Anonymous

@MadhuchhandaMandal The question is wrong anyway...

Anonymous

If it is inextensible then it is unsolvable

Anonymous

And no string no matter how rigid it is cannot be "inextensible".

Madhuchhanda Mandal

If it is close to inextensible, is the question valid?

Anonymous

@MadhuchhandaMandal Yes

Madhuchhanda Mandal

06:54

How? That's exactly what I am not able to understand

?

@blue are you there?

Anonymous

@MadhuchhandaMandal Farcher explained it! Energy is stored at PE of string in such a case

Anonymous

But the extension is negligible

Madhuchhanda Mandal

But had the extension been negligible, (1/2)*stressstrainvol would have been negligible since strain tends to

0

Anonymous

@MadhuchhandaMandal It has a high value of $k$....or young's modulus if you call it so...

Madhuchhanda Mandal

Above expression is also valid. Isn't it?

So above expression clearly shows PE tends to 0

Anonymous

07:09

@MadhuchhandaMandal Stress and Strain are related by young's modulus...

Madhuchhanda Mandal

What about above exp.?? Isn't it valid??

Anonymous

@MadhuchhandaMandal It is. Here strain is small.

Anonymous

Not stress

Madhuchhanda Mandal

Stress is tension itself. = some finite value

(Related to m only)

Anonymous

Yeah, tension is very high at that instant...

Anonymous

07:12

I told you so

Madhuchhanda Mandal

T=Stress=Mg=Finite

Anonymous

@MadhuchhandaMandal There's where you are wrong...

Madhuchhanda Mandal

Why?

Anonymous

There is an instaneous change in velocity in a very small time

Anonymous

You need to include $ma$ in there...

Madhuchhanda Mandal

07:15

(I know that during taughting of string tension is very high, but that has nothing to do with stress in the above exp.)

Anonymous

@MadhuchhandaMandal Your equation doesn't hold during that instant.....

Anonymous

$T \neq mg$

Anonymous

You need to take into account acceleration and jerk etc etc

Madhuchhanda Mandal

Yes JERK. That's what I was thinking of

The String will immediately begin to flow with some velocity and would again eventually stop once the event of collision is over. And this energy would appear as Heat. That is the only possible explanation I can think.

Anonymous

@MadhuchhandaMandal Also as potential energy

Madhuchhanda Mandal

07:19

But PE WON'T GET STORED IN THE STRING. That's what I think

Anonymous

@MadhuchhandaMandal It will....

Madhuchhanda Mandal

Because you can see in the PE equation, Its very very close to zero

Anonymous

The string isn't fully inextensible

Anonymous

@MadhuchhandaMandal Nope.

Anonymous

$k$ is very high

Anonymous

07:21

For nearly inextensible string

Madhuchhanda Mandal

See the equation na in which all quantities are known

I mean stress * strain equation

Anonymous

@MadhuchhandaMandal Which?

Anonymous

@MadhuchhandaMandal Stress is not known.

Madhuchhanda Mandal

Its Mg

Anonymous

No^

Madhuchhanda Mandal

07:22

You can calculate from constraints

Anonymous

You didn't take into account accn and jerk and the higher derivatives

Anonymous

It will be like T=Mg+Ma+...

Madhuchhanda Mandal

That will be during the event of collision

But not after collision is over

Once collision is over, what's Stress ??

Stress=T=Mg

Anonymous

After collision is over the string is already elongated...

Anonymous

Its permanently deformed

Madhuchhanda Mandal

07:25

You mean it has lost elasticity?

Anonymous

The natural length changes

Madhuchhanda Mandal

Then its pe is 0

What ever may be stress

(1/2) k x^2

X=0

Anonymous

Oh god :/

Anonymous

The energy has already been stored

Anonymous

Due to deformation

Anonymous

07:27

After that why are you applying that equation

Madhuchhanda Mandal

A string at natural length having energy??

What ever this energy is, it cannot be termed ELASTIC POTENTIAL ENERGY

Anonymous

Try it at home. Take a metal wire and elongate it. The natural length changes. Where did the work done go?

Anonymous

It went into the wire

Madhuchhanda Mandal

HEAT

Anonymous

@MadhuchhandaMandal I just said Potential energy

Anonymous

07:29

@MadhuchhandaMandal Not fully.

Madhuchhanda Mandal

Ooooooooooooo... You mean PE of molecules??

Interspace

Of molecules

Anonymous

Sort of

Madhuchhanda Mandal

Increases??

Anonymous

yes

Madhuchhanda Mandal

BUT WE WERE ARGUING WHETHER ELASTIC POTENTIAL ENERGY IS POSSIBLE OR NOT :|

Anonymous

07:31

@MadhuchhandaMandal I never mentioned the word "elastic".

Anonymous

I just said potential energy.

Anonymous

That's a kind of potential energy only.

Madhuchhanda Mandal

I was talking about EPE ONLY. That's why I mentioned stess*strain relationship

Anonymous

I see

Anonymous

Anyhow I hope you got it

Madhuchhanda Mandal

07:33

Nope

My question still remains un answered

EPE=0 isn't it??

I mean tends to

Anonymous

Yeah, you can say that

Madhuchhanda Mandal

That's what I was talking of

Although, I am not sure whether permanent deformation will take place or not, because Upper Yield Point Stress is not known

Heat is for sure

Anonymous

@MadhuchhandaMandal What is the first law of thermodynamics?

Madhuchhanda Mandal

dq=du+dw

Anonymous

@MadhuchhandaMandal Yes, if du=0 and dw=0 then what is dq? (Since you assumed no deformation and no extension)

Madhuchhanda Mandal

07:46

I have not assumed du=dw=0 !!!! I have only assumed du=0

dq=dw

Anonymous

@MadhuchhandaMandal What is dw here?

Anonymous

And why is it not 0?

Madhuchhanda Mandal

dw is the work done by infinite seemingly stress during collision

Anonymous

@MadhuchhandaMandal And what is the fomula for work?

Anonymous

integral of F.dx?

Madhuchhanda Mandal

07:47

F.dx

Yeah

Anonymous

@MadhuchhandaMandal You said dx=0...

Anonymous

Since there is no extension

Madhuchhanda Mandal

When????????

I said dx tends to 0

Anonymous

@MadhuchhandaMandal Then dq tends to 0

Anonymous

No heat

Madhuchhanda Mandal

07:49

Why??? F tends to infinite

Anonymous

@MadhuchhandaMandal Then dx is finite which implies there is extension.

Madhuchhanda Mandal

dx is infinitely small. Not 0

We have assumed that

Anonymous

Then there is deformation.

Madhuchhanda Mandal

When mentioning "close to perfectly inextensible" -> dx tends to 0

Anonymous

Since dx is infinitesimally small.

Madhuchhanda Mandal

07:51

Yes ofcourse

Anonymous

So there cannot be heat loss without deformation.

Madhuchhanda Mandal

Who said there will be?

Anonymous

15 mins ago, by Madhuchhanda Mandal

Although, I am not sure whether permanent deformation will take place or not, because Upper Yield Point Stress is not known

Madhuchhanda Mandal

"PERMANENT" Please carefully notice the word

Anonymous

Hmm, its highly impractical that there won't be any permanent deformation at such high tension.

Madhuchhanda Mandal

07:55

UYP Not known, however impractical it may seem, there can be no permanent def.

Anonymous

@MadhuchhandaMandal "there can be no permanent def." ?

Anonymous

What?

Madhuchhanda Mandal

deformation

def. Implies

Anonymous

There can be permanent deformation

Anonymous

Now we are back to square one

Madhuchhanda Mandal

07:56

Cannot be also

Where?

Anonymous

Even heat loss may or may not take place

Madhuchhanda Mandal

Nope it MUST take place.

Back in 30 minutes. Lunch time.

Swapnil Das

Hello, anyone there?

Anonymous

yo

Swapnil Das

I needed some help in chem.

Anonymous

08:01

i have about 2-3 minutes

Swapnil Das

If I ask soluble in NaoH, that means it is an acid?

Anonymous

Not necessarily

Anonymous

Even NaCl is soluble in NaOH

Swapnil Das

Cr(OH)3 soluble?

Anonymous

Probably yes

Swapnil Das

08:02

It is A or B?

acidic or basic?

Anonymous

It is basic

Swapnil Das

Al(OH)3?

Anonymous

amphoteric

Swapnil Das

Pb(OH)2?

Fe(OH)3?

Anonymous

@SwapnilDas ampho

Swapnil Das

08:04

SnO2?

Anonymous

@SwapnilDas basic

Anonymous

@SwapnilDas amphoteric

Swapnil Das

Ampho is soluble?

Anonymous

@SwapnilDas Amphoteric means both acid and base

Swapnil Das

Yes Ik, then soluble?

Anonymous

08:06

See acidity basicity is not related to solubility'

Anonymous

They are different properties

Swapnil Das

Ic.

Madhuchhanda Mandal

08:32

@blue ?

Are u there?

Anonymous

08:45

yes

Aryabhatta

09:31

Hello @blue

Yashas

hi @Aryabhatta

Aryabhatta

@Yashas, hello.

What is tue fundamental theorem of Algebra?

Yashas

Every number can be represented as a product of prime numbers.

Oops, wrong.

That is the fundamental theorem of arithmetic.

Every polynomial of degree n has n roots is the fundamental theorem of algebra.

Aryabhatta

09:49

@yahas, so why is this en.m.wikipedia.org/wiki/Fundamental_theorem_of_algebra different?

Yashas

"The theorem is also stated as follows: every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots. The equivalence of the two statements can be proven through the use of successive polynomial division."

Note that all real numbers are complex numbers.

Aryabhatta

What is the derivation of this theorem.

Yashas

I don't know

Aryabhatta

10:47

What is momentum? How is linear momentum different than angular momentum?

Hello anyone here.@yashas, @blue, @sammygerbil

Yashas

too broad for an answer

Aryabhatta

What do you mean?

Yashas

the answer will be too big

Aryabhatta

Ok..Q. which of the following opeartions will not change a vector?

1. Rotation about the tail.

2. Rotation perpendicular to its plane

3. Rotation on its own plane

4. None of the above

@yashas

Aryabhatta

12:19

@yashas! !

Yashas

4

Aryabhatta

Q. A ball is thrown horizontally from the top of the tower. What happens to the horizontal component of its velocity?

1. Increases

2. Decreases

3. Remains same

4. First decrease then increase

Yashas

I don't do other's assignment.

Aryabhatta

Whats the purpose of this chat room? (JUST WANTED TO KNOW)
If you are not interested, no issues

Yashas

Discussion room

not solutions to assignments

you have to show your attempt

then we discuss

and then you get the correct answer

Aryabhatta

12:26

@yashas, nothing will happen. I guess

Yashas

reason

Aryabhatta

@yashas, i think.its projectile

Yashas

good, continue

what next?

Aryabhatta

@yashas, i think gravity accelerates the object downwards but doesn't affect the horizontal component of..

Yashas

100% correct

you got your answer

Aryabhatta

12:39

The modern mendeleev periodic table consists of ...groups.

1. 7

2. 8

3. 9

4. 10

@yashas

Yashas

I won't answer this one.

Google it

Easy one.

Aryabhatta

What is modern mendeleev table?

Yashas

I think that includes the nobel gas group?

Aryabhatta

IIs it 9 then

Anonymous

13:13

@MadhuchhandaMandal When you need you can just leave your message by pinging me. I am not always around but will have a look when I return.

Aryabhatta

14:54

Hello @blue

Anonymous

@Aryabhatta Hello. But please don't ask me to solve/answer any board level questions.

Aryabhatta

@blue, means??

Anonymous

Don't ask me any school board level questions.

Aryabhatta

So which level do you want?

Anonymous

JEE Advanced Level.

Anonymous

14:57

Or JEE Mains.

Aryabhatta

@blue, i am not from india

Anonymous

@Aryabhatta Fine. But I saw the questions you asked on hbar.

Anonymous

They were standard school level questions

Anonymous

And this chat room is not for that purpose

Aryabhatta

Really. Of which level?

Anonymous

14:59

If you have any conceptual questions you can ask them here

Anonymous

3 hours ago, by Aryabhatta

Q. A ball is thrown horizontally from the top of the tower. What happens to the horizontal component of its velocity?

1. Increases

2. Decreases

3. Remains same

4. First decrease then increase

Aryabhatta

@blue, so whats the fundamental theorem of algebra?

Anonymous

@Aryabhatta That's not a conceptual question. It falls into the easily googleable category.

Anonymous

We don't answer such questions here.

Aryabhatta

Whats hysteresis? And hysteresis phenomenon?

Anonymous

15:03

@Aryabhatta That's not a conceptual question. It falls into the easily googleable category.

Anonymous

We don't answer such questions here.

Aryabhatta

Then everything falls under easily.googleable.catrgory.

Anonymous

@Aryabhatta No.

Sakazuki Akainu

i am no one to make a statement indeed, but why ask concepts like hysteresis here when you have books?

Aryabhatta

@blue, so what's that?

Anonymous

15:06

@SakazukiAkainu Exactly!

Sakazuki Akainu

i'm sure any good book on class 11th/12th (i don't remember now) will provide ample knowledge on that

and then you do have good websites too

Aryabhatta

@SakazukiAkainu, @blue, what concepts which are not found in google are discussed here?

Sakazuki Akainu

i have no idea :P i just came up here cuz i found this area pretty interesting

but i'm sure this is a "doubt clearing" sort of room, not a "tutor" room

:)

Anonymous

Look at the transcript. We've had some very interesting discussions

Sakazuki Akainu

you're all preparing for JEE or BITSAT too? just curious

Anonymous

15:09

@SakazukiAkainu both :)

Sakazuki Akainu

nice :D good luck!

although you'll crack bitsat with ease if you do the preparation for JEE MAINS alright

Anonymous

Hopefully so. Thanks!

Aryabhatta

Al(OH)3--->Amphoteric

Is it a concept?

Anonymous

It is a fact

Aryabhatta

Ok. I find your concept and mine concept about "concept" quite different!

Koolman

15:52

@sammygerbil okay thanks , I got it

What the use of initial data we can directly use the data of lyman and balmer series

Anonymous

Z is unknown

Anonymous

No wait

Anonymous

It is not needed

Anonymous

The first part is useless

Koolman

Wtf , I wasted about 8 mins on this

Anonymous

16:11

its a bad question

Koolman

16:49

Get ready for the result !!

Jaideep Khare

18:54

Every one best of luck for their JEE Main'17 Result. Do share your marks and ranks here.

Transcript for

Problem Solving Strategies