Problem Solving Strategies

Scáthach

04:58

@JohnRennie good morning

John Rennie

05:11

@Scáthach morning :-)

Scáthach

@JohnRennie hi

imgur.com/a/n1TeVQo

Q27 help

John Rennie

@Scáthach the sound is reflecting off the cardboard and travelling back towards the source, and this creates interference. So the changes in intensity are due to constructive and destructive interference as the cardboard is moved.

Scáthach

@JohnRennie standing waves

John Rennie

I'm not sure I'd describe it as a standing wave because it only has one boundary (at the cardboard). It's just down to the path difference between the incident (red) and reflected (blue) ray.

Scáthach

@JohnRennie how is interference established? Path difference between incident and reflevted rays?

John Rennie

05:28

@Scáthach Yes. If the phases of the incident and reflected waves are the same at the point D then they reinforce and you get constructive interference. If the phases are in antiphase (a phase difference of $\pi$) at D they cancel each other out and you get destructive interference.

Scáthach

Okay

John Rennie

When you move the cardboard 20cm you change the path length of the reflected ray by 40cm.

That means 40cm is equivalent to a phase difference of $\pi$ i.e. half a wavelength.

Scáthach

Yes

@JohnRennie got it

Scáthach

05:40

@JohnRennie imgur.com/a/NSHLFiI

Q30

I can't follow what is happening with the detector here

John Rennie

Let me draw a diagram ...

With the detector at the initial position the path lengths from S1 and S2 are the same so the sound waves are in phase. The question asks how far the detector has to move along the line for the path difference to become half a wavelength.

Scáthach

@JohnRennie yes. The diagram makes it so clear. I get it now

tatan

@JohnRennie Hi!

John Rennie

05:55

@tatan morning :-)

Abcd

← 4 messages moved from JEE/High School Maths Problems

tatan

I need help with this one

John Rennie

@tatan The voltage drop across the diode is 50V. Yes?

tatan

yes

John Rennie

So the voltage drop across the 5k resistor is 70V. This immediately tells you the total current flowing.

tatan

06:01

Yes

John Rennie

The voltage drop across the 10K resistor is 50V, because it's in parallel with the diode.

tatan

Yess

John Rennie

So that tells you the current through the 10K resistor ...

tatan

Ok... then we subtract it from the total

thanks

Scáthach

@JohnRennie hi

amanuel2

06:09

I actually got it thanks!

But i got a quick conceptual question, that my teacher went over I do not understand

tatan

@amanuel2 For Q2. isn't it 2mvcos(45) ?

amanuel2

@tatan Yeah, i still dont understand tho

tatan

Why?

amanuel2

Why was my method wrong, and -2mvcos(45) right

it rebounds so it has a x and y direction right?

tatan

Wait

Initially $p=mv {cos(45) i+sin(45) j}$ and finally $p=mv{- cos(45) i+sin(45) j}$. So, $P_initial-p_final=2mv cos(45) i $ where i,j are unit vectors in positive x and y directions

@amanuel2

amanuel2

06:17

Yeah hold on

@tatan Seems like i didnt

get the inital velocities right

How does the ball have a y velocity when it just hit the wall staright without rebounding to an angle?

tatan

Yes that was just what I was also thinking

amanuel2

Cause i 100% understood what i did with the x velocity and was so dumb of me, but not sure about the y

tatan

A ball thrown straight at a wall cannot rebound at an angle of 45

amanuel2

^^ yeah that too lmao

tatan

If you throw it at an angle of 45 then only it can rebound at an angle. I mean in that case the normal to the wall at the point of impact bisects the angle between initial and final velocity vector

amanuel2

06:26

Wait so in that case it must have been thrown at a 45 degree angle

Ahh i see why there is an inital y velocity now!

tatan

Yes... that was what I assumed

amanuel2

I completely painted the wrong image in my head for the problem "
A ball thrown straight at a wall cannot rebound at an angle of 45
"

John Rennie

@Scáthach hi. Sorry for the delay. I've got a dead server I'm trying (and failing!) to fix.

amanuel2

@tatan got one more question before i go off if ur still available

tatan

Post it... someone around here will solve

amanuel2

06:29

Here you mind explaining the (L/2)-D ?

tatan

@amanuel2 Do you realise why -"A ball thrown straight at a wall cannot rebound at an angle of 45"

amanuel2

@tatan Yeah its actually impossible lmao that I think of it.

tatan

Its not mentioned what L/2 and d mean but assuming L is length of the board and d is the distance moved its pretty clear that if you move a distance of L/2 (relative to the board) to the center the board also moves backward by a distance of d (relative to the ground)... so, relative to the ground your net displacement is L/2-d . I mean $x_{y,g}=x_{y,b}+x_{b,g}$ where x is displacement and y means you, b means board and g means ground.

(Note that this is a vector equation.. so be careful about signs)

amanuel2

Oh i forgot to show the first part

Abcd

@JohnRennie Morn

John Rennie

06:41

@Abcd morning :-)

amanuel2

@tatan oh! this concerns lab frames and center mass frames

our prof said he will omit that from the exam

Abcd

amanuel2

Ill see if thats a question we should worry about. thanks for the help tho!

Abcd

@JohnRennie $I_c = 100 I_b$

@JohnRennie 1) $I_CR_L + 6 = 10$

@JohnRennie 2) $I_B R_B = I_C R_L$

@JohnRennie Which equation out of 1 and 2 is wrong and why??

John Rennie

Equation 2 is wrong. It's wrong because $V_{BE} \ne V_{CE}$

Abcd

06:46

@JohnRennie We will take it as 0 ?

John Rennie

Equation (1) is the way to do the problem. (1) gives you $I_C$ and you know $I_B = I_C/100$

The question says neglect $V_{BE}$ so $I_B = 10/R_B$

Abcd

@JohnRennie What if first polariser is kept horizontal in polarisation experiment?

John Rennie

@Abcd you'll need to explain what you mean by the polarisation experiment.

Abcd

@JohnRennie Light is sent through Polariser +Analyser system.

John Rennie

OK. So as drawn no light gets through the analyser.

Abcd

06:56

@JohnRennie No even polariser is kept horizontal. (its vertical in above diagram)

Then will light pass through polariser?

John Rennie

Yes. If the polariser is horizontal it allows only light from the source polarised in the horizontal plane (I forget what fraction of the total that is). Then all the light allowed through the polariser passes through the analyser.

Abcd

@JohnRennie You didnt get me. Let me mark on the diagram and show.

John Rennie

Like that?

Abcd

@JohnRennie yes

i feel it shouldnt pass through polariser in that case

John Rennie

@Abcd why?

The source consists of all polarisations. The polariser allows through only the subset of the light from the source that is polarised in the horizontal plane.

Abcd

07:02

@JohnRennie the red arrow is in ZX direction

And in original it is in XY direction

John Rennie

The dark red arrow just shows the direction of travel of the light ray. The direction of polarisation is always normal to the direction of travel.

If you think of the light wave as a lateral wave then the displacement is always normal to the direction of travel (because that's what a lateral wave is). The direction of polarisation is just the direction of the displacement.

The unpolarised light is a superposition of many individual waves with their directions of polarisations in random directions. The polariser allows through just the waves with their polarisations in the direction that matches the polariser.

(it's a little more complicated than that, but that's the basic idea)

Fawad

07:53

Question no.7 how to solve?

Our am frequency has lower band frequency 2250khz and upper band frequency 2750khz

John Rennie

@Fawad It tells you the modulation frequency, 250kHz, is 10% of the carrier so the carrier is 2500kHz. That means the station occupies the range from 2500-250kHz to 2500+250kHz. OK so far?

Fawad

Now I think we can allot 2000khz and 2900khz frequency because they are out of our frequency band

@JohnRennie yes

John Rennie

@Fawad suppose you allocated 2900kHz to the new station, then the new station's range would be from 2900-250kHz to 2900+250kHz.

Fawad

@JohnRennie isn’t it 2900-290 to 2900+290?

John Rennie

In real life you would allocate the same bandwidth (250kHz) to stations regardless of their base frequency. Anyway, for the purposes of this question is doesn't matter.

Whether you allocate 250kHz or 290kHz if you place the new station at 2900kHz its range is going to overlap the existing station.

Fawad

08:01

@JohnRennie Yesh. So we can allot 2000khz. Thanks

Scáthach

08:26

@JohnRennie hi

John Rennie

@Scáthach hi

Scáthach

@JohnRennie imgur.com/a/Vzlvpqz

Q32

How will you proceed

John Rennie

There's a really simple way to understand this using a diagram. The diagram itself is very simple, but explaining how and why the diagram works will take a little while.

I can try and explain if you would like, or I can just show you the diagram.

Scáthach

Diagram

John Rennie

We can represent a wave as a vector of fixed length rotating anticlockwise. So S1, S2 and S3 are shown as three vectors. We are told the phase difference between them is 120° and that means on our diagram they are spaced 120° apart.

The total wave is given by adding up the three vectors, and obviously when you add the three vectors they cancel each other out and you get zero.

Scáthach

08:36

Yes

John Rennie

So the intensity at the point P is going to be zero.

The question is whether you feel happy you know why this diagram works.

Scáthach

@JohnRennie Don't know how this wotks

user8718165

@Scáthach are these JEE problems?

Scáthach

@user8718165 yes jee practice problems more

John Rennie

@Scáthach I can explain but it will take a few minutes. Do you know about using phasors in electrical circuits? Basically it is the same idea as that.

Scáthach

08:42

@JohnRennie yes

user8718165

@Scáthach :-) Well, may be I didn't come across these as of yet:-)

Scáthach

@user8718165 it's given in hcv. If you want to try, you will find these there

John Rennie

Suppose you have a vector of length $A$ rotating anticlockwise at an angular frequency $\omega$. Then at a time $t$ that vector makes an angle $\omega t$ with the horizontal.

So the vertical component (the $y$ component) of the vector is $A\sin\omega t$, and that is the equation for a wave. OK so far?

Scáthach

Yes

user8718165

@Scáthach :-) will begin hcv(both I and II) in april

2

John Rennie

08:49

So we can describe our wave as the $y$ component of the rotating vector.

Scáthach

Okay

John Rennie

Now suppose we have a second wave at a phase $\phi$ ahead of our original wave. In the question the angle $\phi$ was 120°.

Scáthach

Okay

John Rennie

For the second wave we draw a vector an angle $\phi$ ahead of our first vector, so the angle with the horizontal axis is $\omega t + \phi$. Yes?

Fawad

$\sqrt{{{I}_{s1}}^2+{{I}_{s2}}^2+{{I}_{s3}}^2 +2{I}_{s1}\cdot{I}_{s2}\cos{120}+2{I}_{s2}\cdot{I}_{s2}\cos{120}+2{I}_{s3}\cdot{I}_{s1}\cos{240}}$

John Rennie

08:51

Then the $y$ component is $A_2 \sin(\omega t + \phi)$, and again that is exactly the equation for a wave an angle $\phi$ ahead of our first wave.

Scáthach

@JohnRennie yes

John Rennie

If we add our two waves we get $A_1\sin(\omega t) + A_2\sin(\omega t + \phi)$, and this is the same as the $y$ component of the vector we get from adding our two vectors.

Scáthach

Yes

John Rennie

So we can add our two waves by first adding the vectors, then taking the $y$ component of the result.

Likewise if we have three waves we can calculate the sum by adding the three vectors representing those waves and then taking the $y$ component of the result.

Scáthach

@JohnRennie If we add our two waves we get $A_1\sin(\omega t) + A_2\sin(\omega t + \phi)$, and this is the same as the $y$ component of the vector we get from adding our two vectors.

John Rennie

08:56

Sometimes this is useful and sometimes not - it depends on the question. In this case it is very useful because the sum of our three vectors is zero, and the $y$ component of a zero length vector is zero.

@Scáthach yes ... ?

Scáthach

If adding vectors give ycomponent then why should I take y component after adding them

John Rennie

@Scáthach In this question adding the y components gives $A\sin(\omega t) + A\sin(\omega t + 120°) + A\sin(\omega t + 240°)$. Is it obvious to you what the result of this sum is going to be? (because it isn't obvious to me :-)

Scáthach

No

John Rennie

But if you add the three vectors first, it is obvious what the sum of the three vectors is. Yes?

Scáthach

Yes

John Rennie

09:00

You don't need to be a maths genius to see that the sum of the three vectors is the zero vector.

@Scáthach And what is the $y$ component of the zero vector?

Scáthach

0

John Rennie

And that immediately tells us $A\sin(\omega t) + A\sin(\omega t + 120°) + A\sin(\omega t + 240°) = 0$

Scáthach

Got it

@Fawad thanks to you too

John Rennie

@Scáthach you could of course just expand out the sum using the sin(a+b) equation and you'd get the same answer - eventually.

Scáthach

09:18

@JohnRennie yeah. I think that's the best shot

John Rennie

@Scáthach the diagrams are a quick way to solve some problems, but you do need to feel confident about using them.

Scáthach

@JohnRennie what is the significance of S1S2=S2S3

John Rennie

When you start doing circuits you'll use diagrams like this, where those diagrams are called phasors.

@Scáthach actually the spacing doesn't matter. The question tells you that at $P$ the phases differ by 120° and that's all you need to know.

Scáthach

Hmm. I thought it had some meaning. Got it

John Rennie

The phases at $P$ will depend on the relative phases of the three sources and the distances between them, but in this case we don't care as we are told what the phases at $P$ are.

Scáthach

09:24

Okay. Got it

Abcd

10:05

@JohnRennie

John Rennie

@Abcd hi

Abcd

in JEE/High School Chemistry Problems, 10 mins ago, by PolarBear

Which of the following combination of statements is true regarding the interpretation of the atomic orbitals ?
(a) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the
orbital of lower angular momentum.
(b) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the
azimuthal quantum number.
(c) According to wave mechanics, the ground state angular momentum is equal to $\frac{h}{2\pi}$
(d) The plot of $\psi$ Vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value.

@JohnRennie please answer, multi correct

John Rennie

(a) is kind of true. Only the s orbitals have non-zero values at the nucleus so the higher angular momentum p, d, etc do stay away from the nucleus more than the s orbital. But is it correct to say d orbitals have lower density near the nucleus than p orbitals. I'm not sure.

I'm not sure what (b) means, though I'd guess (b) and (d) are basically the same question.

For (c), it depends on the atom. For a hydrogen atom the ground state has zero angular momentum not $\hbar$.

Abcd

@JohnRennie answer given is a and d.

@JohnRennie not able to understand a and d properly

John Rennie

Is this for a hydrogen atom?

Abcd

10:14

@JohnRennie not mentioned

John Rennie

@Abcd in that case it can't be answered as the atomic orbitals differ from atom to atom. There is no analytic expression for the orbitals except for hydrogenic atoms.

Dante

@JohnRennie What d) probably means is that, highest electron density in higher orbitals like d and f is found farther from the center than others.

John Rennie

@Dante yes, that would be my interpretation.

Abcd

10:28

@JohnRennie I feel $x_m$ measurement is erroneous

@JohnRennie MSR = 12.

And there's some gap between 12 and 0 of VS

John Rennie

The reading is 12.7mm ...

Abcd

@JohnRennie I want to know why $x_m = MSR + VSR\times MSD$

John Rennie

I assume MSR means the reading on the main scale i.e. 12.

Abcd

yup

John Rennie

And the VSR is the reading on the vernier scale i.e. $x_v$ that is 7.

Abcd

10:33

@JohnRennie yes

John Rennie

I don't understand $x_m = MSR + VSR\times MSD$

It should be $x_m = MSR + VSR/10 \times 9MSD$

Abcd

@JohnRennie He has also given 10VSD = 9MSD

John Rennie

i.e. $x_m = MSR + VSR \times 0.9 MSD $

Abcd

@JohnRennie I am not getting anything

concepts-of-physics.com/pdf/vernier-calipers.pdf

Please see 2nd page of pdf

John Rennie

I don't understand what you're asking. The reading is obviously 12.7. What's with the arbitrary equations?

Abcd

10:38

@JohnRennie its the derivation of D = MSR + VSR * LC

@AdvilSell Are you able to understand why $x_m = MSR + VSR * MSD$

Advil Sell

@Abcd nope Is it even correct ? , I don't feel it's correct !

Abcd

@AdvilSell can you see the pdf please

Advil Sell

@Abcd I saw it still no idea , will tell you if I think i get it

Abcd

@JohnRennie Can you give me some insight/ intuition for the formula $D = MSR + VSR\times LC$

John Rennie

@Abcd Start with the MSR, which is 12 in this case.

The reading D is the MSR plus a bit, and we need to work out what that bit is.

Abcd

10:53

Ok

John Rennie

The VSR is the reading on the vernier scale where the two scales coincide, which is 7 in this case.

Abcd

yes

John Rennie

The point on the main scale where the two scales coincide is the sum of the two integers MSR + VSR. 19 in this case.

Abcd

@JohnRennie dint get

19 is just the point of coincidence how is it MSR +VSR

John Rennie

It seems kind of obvious from just looking at the diagram. I wonder if I can come up with a more mathematical argument for it ...

Suppose you start with MSR = 12 and VSR=0 (I'm using 12 as an example)

Abcd

11:00

@JohnRennie 0 of VS coincides with 12 of MS?

John Rennie

@Abcd The MSR is the largest integer less than the point on the main scale where $VSR=0$.

In this case that point is $D=12.7$ so the highest integer less than that is $MSR=12$

Abcd

@JohnRennie you said VSR = 0. That means?

John Rennie

@Abcd It means the position on the main scale immediately opposite zero on the vernier scale.

Abcd

@JohnRennie VSR = 7 means 7th div of VS is coinciding with some point on MS.

@JohnRennie still didnt understand formula D = MSR + VSR*LC

John Rennie

@Abcd Let's take this specific example and see if that helps.

In this example MSR = 12. Yes?

Abcd

11:09

yes

John Rennie

Use this algorithm:

Start with a variable $v=0$ - don't worry what $v$ is for the moment.

1. Does the position $v$ on the vernier scale match up with a marking on the main scale. If yes stop now.

2. If no, and it's no in this case add 1 to MSR and add 1 to $v$. Then go back to (1).

Abcd

why add 1 to MSR

John Rennie

In this case the readings coincide after seven steps i.e. when $v=7$ and we've added 7 to the initial MSR to get 19.

Now, the value on the main scale we've reached, 19, is what your diagram labels as $x_m$.

And the distance D being measured is $x_m - x_v$, so if we know what $x_v$ is we can find $D$.

Now, that number $v$ is equal to the number of divisions on the vernier scale that we moved to get to the point where the scales match.

Abcd

ya

John Rennie

So the distance $x_v$ is just $v$ (7 in this case) times the spacing between divisions on the vernier scale. And that spacing is $0.9 MSD$ (0.9 mm in this case).

On the main scale we started at $MSR$ and moved on division $v$ times, so $x_m = MSR + v MSD$.

And $x_r = v VSD = v (0.9 MSD)$

So $D = x_m - x_r = MSR + v MSD - v (0.9 MSD) = MSR + v (0.1 MSD) $

And $LC = 0.1 MSD$ so $D = MSR + v LC$

5 mins ago, by John Rennie

Now, that number $v$ is equal to the number of divisions on the vernier scale that we moved to get to the point where the scales match.

i.e. $v = VSR$

Hence $D = MSR + VSR \times LC$

Abcd

11:24

@JohnRennie very cool, understood. Can you please tell why LC = MSD - VSD ??

John Rennie

3 mins ago, by John Rennie

So $D = x_m - x_r = MSR + v MSD - v (0.9 MSD) = MSR + v (0.1 MSD) $

$LC$ is just a symbol being defined as $MSD - VSD = MSD - 0.9MSD = 0.1MSD$

Abcd

@JohnRennie But definition of LC is minimum distance that can be measured accurately

John Rennie

In the algorithm I described above we used the variable $v$ that is an integer. Since $v$ is an integer the smallest difference we can measure is between $v$ and $v+1$.

If you increase $v$ by one you increase $D$ by $MSD - VSD$

That's why $MSD-VSD$ is the smallest distance we can distinguish.

@Dante so your gauge exam question was screwed?

Dante

Yeah XD

Abcd

@JohnRennie 0 error for above caliper means "This much it will always show extra, so subtract it from the value you get while using the caliper" . Am I right?

John Rennie

11:34

The reading on that vernier is 0.3 mm. Yes?

Abcd

yes

John Rennie

So the vernier will always read 0.3 mm more than the actual measurement.

Abcd

ok

John Rennie

So you're correct that you need to subtract the zero reading, 0.3mm, from all the readings.

I'm not sure exactly how the term zero error is defined but the obvious interpretation is that it's +0.3mm.

Abcd

@JohnRennie I know that MSR is like adjacent integer but in this case how do we decide (instead of guessing)

How can we surely claim that its -1 and not -2 because there are no markings??

John Rennie

11:41

The zero on the vernier scale lies between two divisions on the main scale. How far left of the zero on the main scale can it be?

Abcd

didnt get you

John Rennie

Imagine the main scale continues to the left of zero, so the first division to the left of zero is -1, then -2 then etc etc

We know the zero on the vernier scale is to the left of the zero on the main scale because we can see that.

So the possibilities are:

a) the zero on the vernier scale could lie between 0 and -0.1 on the main scale.

b) the zero on the vernier scale could lie between -0.1 and -0.2 on the main scale.

c) and so on

Ok so far?

Abcd

1 sec

@JohnRennie does option C include between -1.0 and -1.1

John Rennie

That first sentence should be: Imagine the main scale continues to the left of zero, so the first division to the left of zero is -0.1, then -0.2 then etc etc

Dante

@AdvilSell @Abcd

Are X-rays in main syllabus also?

Abcd

11:46

@Dante ??

Dante

I mean I know it's there.

But have they ever asked questions from it?

Abcd

@Dante its easy$^\infty$

John Rennie

@Abcd opttion (c) would be between -0.2 and -0.3, option (d) would be between -0.3 and -0.4. Between -1.0 and -1.1 would be option, erm, don't know.

Abcd

@Dante check archive, IIRC yes

Dante

@Abcd Ok, lemme just study it then.

Abcd

11:47

@Dante just read from HCV its the easiest thing on earth

Dante

OKie

Abcd

@JohnRennie Ok, then?

John Rennie

@Abcd Looking at the diagram it's immediately obvious that 1 on the vernier scale lies to the right of zero on the main scale. Yes?

Abcd

@JohnRennie yes

John Rennie

@Abcd and to get to zero on the vernier scale we move left by a distance VSD i.e. by a distance of 0.9MSD.

Abcd

11:51

@JohnRennie ??

John Rennie

So the position of the vernier zero on the main scale cannot be less than zero - 0.9MSD.

@Abcd if you're at the 1 marker on the vernier scale what physical distance do you have to move left to get to the 0 marker on the vernier scale?

It would be more helpful to write VSD = 0.9 MSD

Abcd

@JohnRennie Yes

John Rennie

OK so if you're at the 1 marker on the vernier scale what physical distance do you have to move left to get to the 0 marker on the vernier scale?

Abcd

-0.9MSD

John Rennie

Yes. And we know the 1 on the vernier scale is to the right of the zero on the main scale.

Abcd

11:57

so??

John Rennie

So moving from 1 to 0 on the vernier scale cannot move you any farther left of 0 on the main scale than -0.9MSD i.e. the position of the vernier zero on the main scale has to be greater than or equal to -0.9 MSD on the main scale.

Abcd

you mean :

Let x be the distance at right of 0 on MS.

$x < 0.9 MSD$

John Rennie

Abcd

@JohnRennie So we deduce that blue spot is less than 0.9 MSD < 1 MSD => -1 MSD is to the right of it right??

John Rennie

@Abcd yes

@Abcd we deduce that the left red line is to the right of 0 - 1 MSD on the main scale.

Dante

12:09

@JohnRennie imgur.com/a/zZ6p8BJ What's the reason intensity of $K_{\beta}$ rays is less? I had thought higher shells will have more electrons, so they'll have higher probability to jump to lower shell. But, they'll have to travel long way down. Does the second factor dominate which results in this? ^

John Rennie

There isn't a simple explanation. The transition probability depends on how much the electric field of a photon mixes the two orbitals involved. You end up having to calculate an integral that is vaguely like:

$$ \int \psi_{final}^* x \psi_{initial} $$

It turns out that this is greater for the 2p to 1s orbitals than for the 3p to 1s.

I suppose if you want a arm waving argument you could say that the 3p and 1s overlap less than the 2p and 1s because the 3p is on average farther away from the nucleus than the 2s, and the 1s orbital is concentrated round the nucleus.

I need to go.

Dante

Oh, ok.

ayc

13:43

@JohnRennie Why is this definition of time inadequate?......"We perceive increase in entropy as time".......I mean:That's what time is right?

Abcd

14:03

@PolarBear @AdvilSell @Avka

Avka

I always solve using

Y + V.c*n = diff between measures

Abcd

what is Y

what is n ?

someone explain

@AdvilSell try

Advil Sell

@Abcd C..??

PolarBear

Didn't calc but isn't it easy? What's the prob

Abcd

14:23

@AdvilSell wrong

Advil Sell

@Abcd Final B is the answer ??????

Abcd

@PolarBear c2 is unconventional

@AdvilSell yes

Avka

@Abcd c2 was asked in advance 2016

Abcd

@AdvilSell explain

Avka

You can check it's solution out

Abcd

14:25

@Avka this full question is adv 2016's

@AdvilSell back in 10 mins.... ill read your solution if u post it...

Avka

Oh lol

Advil Sell

@Abcd for negative LC vernier Callipers (C2) always take MSR as MSR + MSD and do your stuff

John Rennie

14:38

@ayc that's really a question for the h bar room rather than here.

Abcd

36 mins ago, by Abcd

@JohnRennie C2

John Rennie

@Abcd for C2 the readings match at 36mm on the main scale. Yes?

@Abcd hello?

Abcd

@JohnRennie back 1 sec

@JohnRennie yes, then?

John Rennie

@Abcd and the match on the vernier scale is at 7. So the distance from zero to the match on the vernier scale is 7 times the vernier spacing i.e. 7.7 mm.

Abcd

@JohnRennie K?

John Rennie

14:52

Abcd

@JohnRennie Ok, then?

John Rennie

So the reading on the main scale is 36 - 7.7 = 28.3 mm

Abcd

13 mins ago, by John Rennie

@Abcd for C2 the readings match at 36mm on the main scale. Yes?

@JohnRennie But for MSR we take the one before 0 of VS right??

which is 28 in this case

John Rennie

The reading on the main scale is where the left red line is. So it's greater than 28 and less then 29, but we can't read exactly what it is.

But we know that 36 on the main scale matches up with 7 on the vernier scale.

So if we start at 36 on the main scale and move left by 7 divisions on the vernier scale we will have reached the reading.

Abcd

@JohnRennie In 12.7 question also the 0 didnt coincide

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