Problem Solving Strategies

5:17 AM

hi! can anyone help me with this

https://physics.stackexchange.com/questions/461108/derivation-of-mean-free-path?noredirect=1#comment1035219_461108

Zerix

@AnshumanSinha wait for John Rennie. I am sure he will be able to help you

John Rennie

5:28 AM

@Zerix hi, did you sort that enthalpy of vapourisation question. I reallised literally moments after you left what the mistake was.

Zerix

@JohnRennie good morning.

No I couldn't....

John Rennie

It's actually really simple.

The enthalphy of vapourisation is defined as the enthalphy change for one mole of water at 373K and 1 atm changing to one mole of steam at 373K and 1 atm. So the enthalpy for the process described in the question is just 2 x 40.66kJ. OK so far?

Anshuman Sinha

Shouldn't it be equal to change in enthalpy only, as it is a constant temp process

Zerix

Ok

John Rennie

Remember that H = U + PV and because this is a constant pressure process that means $\Delta H = \Delta U + P\Delta V$

Zerix

5:32 AM

Okay

@AnshumanSinha looks like it's constant pressure process

John Rennie

So $\Delta U = \Delta H - P\Delta V$

Zerix

Yes

Anshuman Sinha

Isn't vaporisation of water a constant temp. process ?

John Rennie

We are given $\Delta H = 81.32$kJ so you just need to calculate $P\Delta V$ and subtract it.

Zerix

@JohnRennie okay

John Rennie

5:35 AM

If you do that it gives the answer 75.12kJ

Zerix

Okay. Let me try then

@AnshumanSinha It's an open system I think. So. Pressure is constant and temperature can vary. At least that's how I interpret

John Rennie

Pressure and temperature are both constant.

Zerix

Both. Okay

John Rennie

But volume changes so enthalpy and internal energy are different

Anshuman Sinha

okay

how to calculate delta(V) ? using density relation ?

@JohnRennie may you help me with my problem physics.stackexchange.com/questions/461108/…

John Rennie

5:42 AM

@AnshumanSinha that's a standard derivation that you'll find all over the internet.

Anshuman Sinha

I only find the derivation assuming a cylinder which gives different result for mean free path.

something like l = 1/pi*d^2*n.

but this result seems to be derives from spherical considerations.

Zerix

Finally got the answer...... Whew

John Rennie

@AnshumanSinha the volume swept out by the particle is a cylinder of radius $d$. It's radius $d$ because the centre to centre distance between it and any particle it hits is $2r$ i.e. the diameter.

And the volume of this cylinder is $4\pi d^2\ell$

You just set this equal to the volume per particle $1/n$ giving you $\ell = 1/(4\pi d^2 n)$.

Anshuman Sinha

are you getting the same expression as I've posted in the question ?

@JohnRennie

John Rennie

Hmm, there's a factor of 3 difference ...

Anshuman Sinha

5:56 AM

yeah, maybe because you're not considering the equal probability consideration for all directions

John Rennie

Oops, there's a rogue factor of 4 in there

I must have been thinking of the area of a sphere.

Jacob P.J

Hello

John Rennie

@AnshumanSinha I guess the factor of 3/4 comes from removing some of the simplifications used in the standard analysis, but I don't know how that factor is derived.

Jacob P.J

I have a sort of homework qn on capacitors

For the circuit shown in the figure q1, q2, and q3 are the charges stored in the 3 microfarad, 4 microfarad, 5 microfarad capacitor respectively. Find the ratio of q1:q2:q3

Anshuman Sinha

okay.

I think it'll come once we consider the molecule to be moving in all possible directions, thus giving us a sphere instead of a cylinder.

YUSUF HASAN

6:04 AM

@JohnRennie Have you looked through my question?

John Rennie

Hi Yusuf, no I haven't. Give me a moment to find it ...

Jacob P.J

ibb.co/WnbFHDH check this link for the image, please somebody reply asap i have an exam tmrw 😢

Anyone 😟

John Rennie

@YUSUFHASAN it looks straightforward enough, if a bit tedious. How far have you got with it?

Jacob P.J

Hello

For the circuit shown in the figure q1, q2, and q3 are the charges stored in the 3 microfarad, 4 microfarad, 5 microfarad capacitor respectively. Find the ratio of q1:q2:q3
ibb.co/WnbFHDH check this link for the image, please somebody reply asap i have an exam tmrw 😢
Anyone 😟

YUSUF HASAN

@JohnRennie I don't understand the feasibility of this happening. If lambda is varying with time,then how is the power being delivered is constant?

Zerix

6:12 AM

@JacobP.J ibb.co/WnbFHDH

Link is dead in your comment here

Jacob P.J

Sorry

ibb.co/WnbFHDH

Zerix

No. Problem

Nobody recognizeable

@JohnRennie are you free now?

Zerix

You can ask after Yusuf @JacobP.J

Jacob P.J

Okay

John Rennie

6:15 AM

@YUSUFHASAN as the wavelength changes the energy per photon changes. If the power is being kept constant then the number of photons per second must be changing.

Step one is to work out the cutoff wavelength for photoemission i.e. take the 2eV work function and calculate the corresponding wavelength.

YUSUF HASAN

Yeah did that... Kind of stuck after that

John Rennie

You are told the light is on for 120 seconds, so you can work out how many of those 120 seconds the wavelength is below the cutoff wavelength.

YUSUF HASAN

Ohh... I see...Because the cycle is 2 minute on and 1 minute off?

John Rennie

Yes, I get the cutoff wavelength to be 620.4nm. Do we agree?

YUSUF HASAN

Yep... Got that as well..

John Rennie

6:19 AM

And the equation for the wavelength (in nm) is $\lambda = 300 + 4t$

And I get $t = 80$s as the time when the wavelength exceeds this. So photoelectrons are only produced for the first 80s of each 3 minute cycle.

Are we agreed so far?

YUSUF HASAN

Yeah.. I agree to this as well

John Rennie

So now all you have to do is come up with an equation for the number of photons per second as a function of time.

You know $\lambda = 300 + 4t$ and the photon energy is $E = hc/\lambda$

And the total power is 100W = 100J/sec

YUSUF HASAN

So... No. of photons= 100/(hc/lamda)?

Jacob P.J

Hey @JohnRennie

John Rennie

So the number of photons per second is $N = 100/E = 100\lambda/(hc)$

YUSUF HASAN

6:25 AM

Okay... Nd this should be integrated from 0 to 80,i believe? @JohnRennie

John Rennie

@JacobP.J hi Jacob. There's a bit of a queue at the moment but I'll be with you in a bit ...

Jacob P.J

@JohnRennie alright im waiting .

John Rennie

@YUSUFHASAN exactly, yes. That will give you the number of photons capable of creating a photoelectron in each 3 minute cycle. Then multiply that by the 1% efficiency.

YUSUF HASAN

@JohnRennie Okay thanks a lot!

3

John Rennie

@YUSUFHASAN it's a bit of a tedious calculation, but basically straightforward.

← 5 messages moved from The h Bar

YUSUF HASAN

6:28 AM

@JohnRennie Yeah.. The final answer looks quite dirty here, but still, I got the concept! :)

John Rennie

Jacob P.J

@JohnRennie im here

John Rennie

19 mins ago, by Jacob P.J

For the circuit shown in the figure q1, q2, and q3 are the charges stored in the 3 microfarad, 4 microfarad, 5 microfarad capacitor respectively. Find the ratio of q1:q2:q3
ibb.co/WnbFHDH check this link for the image, please somebody reply asap i have an exam tmrw 😢
Anyone 😟

Jacob P.J

I got q1 as 30 micro columb

I am confused about the upper half

John Rennie

@JacobP.J the 3uF capacitor is easy isn't it, because you just use Q=CV and you know C and V. Yes?

Jacob P.J

6:31 AM

Yep

Thats why i got it

John Rennie

To do the top half you need to work out the voltage across the 4uF capacitor

Jacob P.J

Is the voltage across 4 same as that across 3uf

I guess not

John Rennie

To do this you combine the 1uF and 5uF capacitors into a single equivalent capacitor.

Can you do this, or do you need a hint?

Jacob P.J

The combined voltage across 4uf and that of the parallel equals battery voltage 10 v

John Rennie

Correct

Jacob P.J

6:33 AM

I dont know how to proceed with the calculation

John Rennie

You have a 1uF and 5uF capacitor in parallel. And capacitors in parallel add. So the total capacitance of those two capacitors is 6uF. Yes?

Jacob P.J

Is the charge across each arm the same since potential is same ?

@JohnRennie yea

John Rennie

So in effect you have a 4uF and a 6uF capacitor in series, and you know that the voltage across the two of them is 10V.

Jacob P.J

Yep

So i can calcukate net charge

Calculate*

Is this net charge equal to q1

John Rennie

No, it's not as simple as that, but it's not very hard.

The key is that the charge on the 4uF and 6uF capacitors must be the same.

Can you see why it must be the same, or should I explain why?

Jacob P.J

6:38 AM

Why

John Rennie

Let me do a quick diagram ...

Jacob P.J

Bcz they are In series?

I guess in series current is the same, so charge accumulated on each plate is same

John Rennie

Look at the point I've marked with the blue arrow.

There is no way for the total charge at this point to change because it's between two capacitors.

Jacob P.J

Yep i see it, is my series reasoning correct?

John Rennie

Yes :-)

Jacob P.J

6:41 AM

Oh yeah it wouldnt change

So q2 =q3

John Rennie

Yes. So the voltage on the 4uF capacitor is $V_4 = Q/4uF$ and the voltage on the 6uF capacitor is $V_6 = Q/6uF$.

Jacob P.J

But its further complicated , they have asked for the capacitor inside one of the parallel arrangement

John Rennie

And we know that the two voltages must add up to 10V so $V_4 + V_6 = 10$V. Yes?

So now you can work out Q, and that gives you the answer for the 4uF capacitor. Now, just one more step to go.

OK so far?

Jacob P.J

Ok

Then solve the parallel 1uf and 5uf

John Rennie

Now you know Q you can work out the voltage on the 4uF capacitor, which should come to 6V.

And that means the volatge across the 1/5 pair is 4V. Yes?

Jacob P.J

6:47 AM

Ao that across the parallel must be 4

Yes

John Rennie

BOOM! :-)

Jacob P.J

Lemme work it out

On an unrelated topic, i have a question about displacement current

Nobody recognizeable

@JohnRennie are you free now?

John Rennie

@Nobodyrecognizeable hi, yes

Jacob P.J

Inorder to explain displacement current my textbook used dc current in a circuit with conducting wires and capacitor, the wire has no net EF inside and hence no displacement current

I wonder what happens if ac is allowed to pass through the wire

Nobody recognizeable

6:50 AM

Jacob P.J

Will there be displacement current created in addition to conduction current?

Zerix

@JohnRennie hi. Are you busy

John Rennie

@Zerix I have two questions queued up :-)

Zerix

Yes. I would like to ask after those ofc

John Rennie

@Nobodyrecognizeable you've snipped off the top of the question ...

Zerix

6:59 AM

@JacobP.J so is it your physics test tomorrow?

Jacob P.J

@Zerix Yep

Zerix

All the best!

Jacob P.J

Its saturday 12:29 pm here

@Zerix where are you from, im from India

Zerix

India

Nobody recognizeable

Jacob P.J

7:01 AM

@Zerix thanks

Aha 😆

Nobody recognizeable

Jacob P.J

@Nobodyrecognizeable its not working why dont u type it out

Nobody recognizeable

@JohnRennie there are three charges as you see.

John Rennie

@Nobodyrecognizeable OK, Three charges at the corners of a triangle in the xy plane?

Nobody recognizeable

@JohnRennie the origin is defined tp be the centroid.

@JohnRennie although that is a triangle. Triangle is in xy plane .

John Rennie

7:07 AM

Are we given the dimensions of the triangle?

Nobody recognizeable

@JohnRennie yep sides are of length- L

John Rennie

Equilateral triangle then. OK.

And you're asked to find the z component at a height L above the triangle.

Nobody recognizeable

@JohnRennie yep component of electric field.....

John Rennie

That doesn't look all that hard, though the calculation is a bit fiddly.

Nobody recognizeable

Basically what I was saying that the potential at (0,0,L) is 0.

Now if the electric field is the gradient of the potential should not it be zero at that point ?

John Rennie

7:13 AM

Ah, I see what you're saying. Because that point is the same distance from all three charges, so the total potential should add up to zero.

Nobody recognizeable

@JohnRennie basically I can do the Calculation and I can find the answer what the thing which confuses me is that a b is equal to $E=-\partial V/\partial x$ then why the electric field should not be zero?

@JohnRennie yep that's for sure

John Rennie

Hmm ...

Jacob P.J

@Nobodyrecognizeable nice question btw which website?

Nobody recognizeable

@JacobP.J creepymind.com :p

Jacob P.J

😑 really?

Nobody recognizeable

7:18 AM

@JacobP.J it's my personal domain so you can understand now

Jacob P.J

I see

Good question tho

John Rennie

@Nobodyrecognizeable what is the correct answer? C?

Nobody recognizeable

@JohnRennie +4. But what's the answer to the doubt?

John Rennie

+4 ??

Nobody recognizeable

@JohnRennie ordinarily JEE main questions have 4 marks on them if you have them correct

John Rennie

7:22 AM

OK, but I was wondering which of the answers is correct

Nobody recognizeable

@JohnRennie yep c is correct and as always you are correct.

John Rennie

But doesn't that mean when L=0, i.e. at the centroid, the vertical component of the field goes to infinity? That can't be correct?

Nobody recognizeable

Ok let me send the solution again.

I think if l is equal to zero then you don't have a triangle you have a point and you are asking what is the electric field at that point who is necessarily needs to be infinity.

@JohnRennie ^^^^

So I think that predicts the accurate results

@JohnRennie hey professor are you here?

John Rennie

Give me a moment

Ah, that's what confused me. $L$ is both the edge length of the triangle and the distance above the plane.

I'll have to leave this for now as I have to work for a while now. Back later.

Nobody recognizeable

7:39 AM

@JohnRennie if you find out an answer please just post it.

Zerix

9:29 AM

@JohnRennie are you free?

John Rennie

@Zerix hi, yes

The chat room has quietened down now :-)

Jacob P.J

Hello

John Rennie

@JacobP.J hi

Jacob P.J

@JohnRennie when u have time will you please go through my second question about ac and wire

Zerix

@JacobP.J I can wait. You should ask first. Your exam are also there

Jacob P.J

9:34 AM

@Zerix no you go on, you've been waiting since morning

Besides you came first 😀

John Rennie

@JacobP.J we tend to assume the wire is ideal and has zero resistance. In that case the electric field inside the wire is always zero whether we apply AC or DC.

Jacob P.J

Okay i see

John Rennie

Since the displacement current is $dE/dt$ (ignoring polarisation) that means the displacement current is always zero.

In a real wire the resistance isn't zero and there will be a non-zero field and hence a non-zero displacement current. But it will be small.

Jacob P.J

Will ac produce em waves as it passes through an ideal wire i guess no

John Rennie

Yes it will, after all that's precisely what a radio aerial does :-)

But under normal circumstances the radiation is negligible.

Jacob P.J

9:39 AM

Hm okay thanks

John Rennie

@Zerix your turn

Zerix

A rod of length 3L is suspended from both ends with 2 ideal strings. Label the ends of the rod A and B. At a distance L from A lies a mass m1 and at a distance L from B lies a mass m2. The string in B is cut. What is the tension in the other string exactly when the string is cut?

How to proceed here

There will be fixed axis rotation about A

John Rennie

This was a question posted earlier, wasn't it?

Zerix

@JohnRennie yes but we didn't get the solution

It's @Don'tbeax-tripledot question's

@JohnRennie how to write translation equation....

John Rennie

the rod starts accelerating downwards, and it also starts rotating about its centre of mass. You calculate the downwards acceleration just by taking the net vertical force and dividing by the mass. Note that the expression for this will include the unknown tension in the string $T$.

Now you take moments about the centre of mass to calculate the angular acceleration. Again the result will involve $T$.

Finally you require that the acceleration at the tied end of the rod be zero. This gives you an equation that you can solve for $T$.

Zerix

9:46 AM

@JohnRennie doubt

The rod is massless

Why would tension affect the com then?

John Rennie

@Zerix The rod mass is the total mass of all the stuff that's falling so it's m1 + m2

Zerix

Okay. That would make sense

John Rennie

The rod moment of inertia is the moment of inertia of the 2 masses.

And the centre of mass isn't in the middle of course.

Zerix

Yes

John Rennie

All this makes the calculation tedious, but it's still straightforward.

I find it hard to get excited about calculations that are basically straightforward but involve lots of tedious calculations ...

Zerix

9:54 AM

I was confused about the rod being massless and tension won't be there in translation equation

But Looks like I was wrong

Don't be a x-triple dot

Hmm, I am confused about the axis of rotation

I assumed it would be the tight end rather than the COM

Zerix

Yes... Something to do with rod being massless

@JohnRennie why is the reason for the rod not rotating about fixed end

John Rennie

Ultimately you're going to be equating the linear acceleration with the angular acceleration. The linear acceleration is the acceleration of the centre of mass, so we use the centre of mass as our reference point.

The instantaneous centre of rotation is indeed the tied end of the rod, but choosing that as the reference point isn't going to help you do the calculation.

Zerix

Okay

Don't be a x-triple dot

How should one write the equation of the angular acceleration?

I get $TL\dfrac{2m_2+m_1}{m_1+m_2}=(m_1+m_2)\alpha$

Am I correct?

John Rennie

10:07 AM

Calculate the torque and the moment of inertia, then use $\tau = I\alpha$

Don't be a x-triple dot

Ok

I get $I=L^2(m_1+4m_2)$

I haven't really worked a lot with rotation before so I'm quite unconfident

John Rennie

For point masses $I = m\ell^2$

Don't be a x-triple dot

Wait, should I concentrate the whole mass in the com and use $I=(m_1+m_2)x_{com}^2$?

John Rennie

Oops, I forgot to label the arrow pointing to the COM - it's the dashed one in the middle.

So you're going to get $I_1 = m_1L^2(m_2/(m_1 + m_2))^2$ and $I_2 = m_2L^2(m_1/(m_1 + m_2))^2$

Maybe everything is going to simplify, but it looks like tedious algebra to me ...

Don't be a x-triple dot

Oh, so you take $I$s about the com... I'm a rotation noob

Now the translation eq should be $(m_1+m_2)g-T=(m_1+m_2)a$

John Rennie

10:23 AM

Yes

Jasper

I think it should be "don't be an x-triple dot" instead, @Don'tbeax-tripledot.

Don't be a x-triple dot

40 mins ago, by John Rennie

Now you take moments about the centre of mass to calculate the angular acceleration. Again the result will involve $T$.

John Rennie

And the upward acceleration of A due to the rotation is $(L + x)\alpha$

Don't be a x-triple dot

@Jasper That sounds quite weird to me

@JohnRennie Hold on a second, let me digest this :)

Jasper

@Don'tbeax-tripledot That's because you pronounce x as ax, and we use "an" when there is a vowel sound. It's not the letter but the sound that matters.

John Rennie

10:25 AM

Focus gentlemen, this is a physics chat room not a use of English chat room.

Jasper

Sorry. =)

2

Don't be a x-triple dot

Now you told me that I should use $\tau=I\alpha$. $\tau$ about the com just $T(L+x)$ because the two masses obviously have no effect wrt. the COM, right?

John Rennie

The total torque about the COM (we'll take clockwise to be positive) is $\tau = +T(L+x) - m_1gx + m_2g(L-x)$

Don't be a x-triple dot

Hold on a second, isn't $m_2(L-x)-m_1x=0$?

$\iff m_2L=(m_1+m_2)x\iff m_2L=(m_1+m_2)Lm_2/(m_1+m_2)=Lm_2$

I feel like I'm missing something here...

John Rennie

@Don'tbeax-tripledot Ah, good point, yes of course. You're correct that they will sum to zero because that's how the centre of mass is calculated.

That will simplify the algebra a bit

Don't be a x-triple dot

10:34 AM

So now we basically have: $$T(L+x)=(m_1L^2(m_2/(m_1 + m_2))^2+
m_2L^2(m_1/(m_1 + m_2))^2)\alpha$$

John Rennie

Yes

Zerix

It looks so hard lol

John Rennie

@Zerix it's messy, but not hard.

Don't be a x-triple dot

I'll try to shorten that down a bit to simplify our existence

Which boils down to $$T(L+x)=\dfrac{m_1m_2L^2\alpha}{(m_1+m_2)^2}(m_1+m_2)=\dfrac{m_1m_2L^2\alpha}{m‌_1+m_2}$$ I think

To sum up: $$\begin{cases}T(L+x)=\dfrac{m_1m_2L^2\alpha}{m‌_1+m_2}\\ (m_1+m_2)g-T=(m_1+m_2)a\end{cases}$$

John Rennie

I often like to check calculations by making a simplifying assumption. Suppose $m_1 = m_2$ so the COM is in the middle. Put $m_1 = m_2$ in your equation and check it gives the right result.

@Don'tbeax-tripledot yes

Don't be a x-triple dot

10:41 AM

@JohnRennie I do this all the time as well :)

To proceed...

57 mins ago, by John Rennie

Finally you require that the acceleration at the tied end of the rod be zero. This gives you an equation that you can solve for $T$.

John Rennie

Yes, if the angular acceleration is $\alpha$ then the linear acceleration at A is just $\alpha(L+x)$.

For A to remain stationary this (upwards) linear acceleration must be equal and opposite to the downwards acceleration of the centre of mass.

Don't be a x-triple dot

So $\alpha(L+x)=g-T/(m_1+m_2)$

I am curious if we get the correct result

Zerix

@JohnRennie so calculate com. Then find moment of inertia about COM

John Rennie

@Zerix yes

Zerix

Use torque equation to find angular acceleration

Torque due to tension only?

Then use the translation equation to calculate acceleration

And finally equate translation equation to angular avveleration at B

@JohnRennie ^^

Is this all we have to do

John Rennie

10:48 AM

At A not B. A is the tied end.

Zerix

Okay. So A

Don't be a x-triple dot

Wait a second...

Does $L$ cancel?

John Rennie

@Zerix Remember that for a rotating object a distance $\ell$ from the pivot the linear acceleration is $a = \ell\alpha$

So you are equating $\ell\alpha$ to the downward acceleration of the COM

Zerix

Got it

Don't be a x-triple dot

But $L$ is not given

John Rennie

10:50 AM

@Zerix when you start doing the mechanics stuff in JEE you'll run into problems like this, though they won't be as messy.

Zerix

I have solved similar question but the rod wasn't massless. It was easy in that case

Don't be a x-triple dot

From eq. (1), $$\alpha=\dfrac{T(L+x)(m_1+m_2)}{m_1m_2L^2}$$ so $$\dfrac{T(L+x)^2(m_1+m_2)}{m_1m_2L^2}=g-\dfrac{T}{m_1+m_2}$$

John Rennie

@Don'tbeax-tripledot The result is obviously going to depend on L because the rotation depends on L while the downward acceleration doesn't ...

Don't be a x-triple dot

Well, the "result" given by them does not contain $L$ whatsoever

:|

Oh we also have to substitute $x$...

John Rennie

On the left you have a factor of $L^2$ in both the numerator and denominator so they will cancel.

Don't be a x-triple dot

10:56 AM

Yeah Yeah

Let me do the calculations

John Rennie

@Zerix the principle is the same. This is just messy because it's two point masses so all the quantities involved are functions of the masses.

Zerix

Yeah. The equation are used in the same way

But I calculated alpha from the tied end and substituted it in the translation equation

Don't be a x-triple dot

:C

It still doesn't give the correct result, I'll redo the calculations

John Rennie

@Don'tbeax-tripledot does it look as though it has the right form? i.e. that it could just be a calculation error?

Don't be a x-triple dot

Mhm, not even close

Perhaps I've made multiple calculation errors

Let me redo them

Here's what I get... If you have time to check that would be great

2 days ago, by Don't be a x-triple dot

If I recall correctly the solution is among the lines $m_1m_2g/(4m_1+m_2)$

(they probably had masses swapped as well but that doesn't matter much)

Thanks a lot for your help @JohnRennie! If you spot the mistake let me know, but the point is that I've understood the method

Zerix

11:23 AM

@Don'tbeax-tripledot that's a scary answer

@JohnRennie are you free for 2 minutes

John Rennie

I'm free if it's quick

Avnish Kabaj

What's the question

2 hours ago, by Zerix

A rod of length 3L is suspended from both ends with 2 ideal strings. Label the ends of the rod A and B. At a distance L from A lies a mass m1 and at a distance L from B lies a mass m2. The string in B is cut. What is the tension in the other string exactly when the string is cut?

This?

Zerix

Yes

John Rennie

@Don'tbeax-tripledot if $m_1 = m_2$ both your result and the answer give the same result of $mg/5$

Zerix

@JohnRennie imgur.com/a/h9UXSOL

Q31

I don't know what to follow here

Avnish Kabaj

11:27 AM

P(V)^x= constant