IMPORTANT: Students should note that at 32:00 in case of U-Tube by mistake in figure sir has taken left side level rise by considering rotation axis on right side but the concept and analysis remain same. Download Black Board Sets for JEE Advanced 2022 from PG Telegram. Join at t.me/physicsgalaxyworld Read eBook on PYQ Models on Physics Galaxy Mobile app. Download link : bit.ly/PhysicsGalaxyApp Boost your JEE Advanced preparation with #1 Bestseller book of 700+ advance illustrations. Check at - amzn.to/3qyIK8H. Watch all Booster checklists - playlist link for JEE Advanced Revision : th-cam.com/play/PLYVDsiuOZP5o7T58PG8SUiKECCzbY37J9.html
50:08 , controversy comes from application of Bernoulli theorem, if we remember that what Bernoulli theorem says, it's validity is only for similar streamline. In rotating fluid the streamline at A and B is different because effect of centrifugal force drifts the liquid more in backward direction, as highest pressure is in the middle and it decrease gradually while going to B. But rotational effect compensate and pressure overall inc. So the first relation is valid and the Bernoulli one is not.
18:55 Assumptions: 1) Cylinder does not fully submerge in the container. 2) In case 3 increase in ht of two liquids is taken equal. #) Also NOTE is applied to both cases 2 & 3. {😉pheww.. saved a lot of calculation here.} @) In case 3 m'g is same as mg but with rho2 $) Btw open to any corrections required. {Ptss.. good luck calculating😅} Case 1: - large surface area if Work Done to be found for, He to 2He it comes to be 0.5MgHe. Case 2: - surface area of container if Work done to be found for, He to 2He it comes to be 1.5MgHe. NOTE [for cases 2 and 3]: - here I took y as He and dy as y and dx as x to get Fb=(rho)gS(He+y+{S/S1-S}) and 2He is height from surface to contact surface between rho1 and rho2.Also He is width of rho2 liquid Case3: - surface area of container if Work done to be found for, He to 3He it comes to be W2 + W3 = =2.5mgHe + mgHe(S/S1-S) + 0.25mgHe(1/S1-S) + 0.5m'gHe + 0.25m'gHe(1/S1-S) {# Finally done with the calculations.... long due to variables}
you cannot directly take (rho)gh as both fluids have different densities... and the integration part will also be bifurcated as the densities of the liquids are not same throughout the length
We can seperately write the equation sir derived for both the liquids assuming pressure at junction as Pj and just add the two equations ... It will come in terms of Pa and Pb and we know it's values
50:04 Bernoulli's equation is an equation derived from conservation of energy whuch is taken in ground frame only since it is considered as conserved frame But in the rotation part it was considered in in the frame of liquid. Now we know that direction of centripetal force is opposite to centrifugal Also in Bernoulli's equation work done per unit volume by centripetal force was considered which was towards point A. Therefore w.r.t. ground, force was more on A and w.r.t liquid force was more on B Rest of the thing can be calculated by you.
In 50:15 Bernoullis equation cant be applied as it is ROTATIONAL flow of water. P + 1/2pv^2 + pgh = E, in our course if we switch streams lines under the condition that flow is newtonian than E remain same for 2 different stream lines but if rotational flow is concern so E remain same for a particular stream line.. Btw advance form of bernoulli is adapted to find out all type of cases(but alas that isnt in our course
50:42 bernoulies condition applicable for points in same streamline but points A,B are on different streamline so that gives an false result ....and thats the reason of controversy....thank you sir for this awesome content...
47:34 In this case the point B is moving along the rotating fluid (as frame of reference is rotating vessel) so velocity of B is 0 w.r.t. fluid so Pb is greater than Pa 50:38 In this case the frame of reference is ground and point B is moving with respect to rotating liquid hence it has some velocity so Pb is less than Pa (Difference is just of frame of reference) (Do point out if wrong)
So this message to thank physics galaxy team and ashish sir as seeing the concept and revision video just helped me to overcome the fear of fluids and now i can solve jee advanced question and feeling much confident
38:30 after this I again tried the inverted tube question asked in jee adv paper 2 stem question and I hit that right in first attempt without any problem thank you sir...
50:30 , since there is no rotational motion of fluid and fluid is not flowing the fluid statics is applicable here whereas when we apply bernoullis theorem to this case it gives us false results as the fluid is not flowing since the container and the liquid both are rotating with the same angular velocity. Hope it's correct❤
Sir the controversy with Pressure difference using rotating fluid and by Bernoulli's theorem because we are missing out 'dha' factor in it where d is density, h is height on the direction of net acceleration and a is acceleration. As here including g also we have xw² as centripetal force so on accounting that we can say that both points A and B are not on the same line perpendicular to the jet acceleration of the fluid. The one analysed with rotation of vessel is correct one.
50:28 When Sir says gap-the main thing to notice is that in Bernoulli Method, the point B is actually at a depth of (h+y) below the liquid and point A is only h below. Assuming them to be along the same Horizontal, means, we nglected the pressure that the Extra Volume of Water created on B(to visualize, think the area enclosed by the Parabola and the X-Axis case). This is the reason that we get a different SIGN as per me(if I am wrong, then do point out)
I think other explainations are much better You argument is kind of similar that you can't apply Bernoulli But you can't deduce from your argument that sign will be opposite We can only conclude we can't use Bernoulli in rotating liquids
50:00 Sir ji bernoullis eqn can be used on 2 pts lying on the same streamline… whereas here A&B are on differnt streamlines As said by sir: “fundamental point”
If we consider centripetal force in frame other than liquid , pressure difference will still be same but this time centripetal force will act in opposite direction .
50:13 i think bernoullis eqn should not be applied in this case ....since it is applied along a single stream , but here the stream line a A different from that of B
50:22 I think it can be because those 2 points are on different streamlines. And we can apply Bernoulli equations on two points only if they are on same streamlines .
we use derived bernoulli's equation in rotating frame due to radial accn towards center which is P - 1/2 (rho) (wx)^2 + (rho)gy = constant where P - 1/2 (rho) (wx)^2 is the reduced pressure at distance x from axis
At 50:08 I think Bernoulli's method is incorrectly used because the velocity omega * x of liquid is perpendicular to the plane and not along the streamline.
Sir I am requesting ki aap PG app par jo revision booster workshop hai vo over hone wali hai 12days ke approx mein,toh aap uss series ki day ko extend kar dijiye na please it was a very helpful series for Me and genuinely I wanted it to be extend till next year's JEE Mains.
50:20. Sir I think so that it is due to frame of reference as in pressure analysis we take frame of vessel which give (shows) centrifugal force on fluid but in bernoulli's analysis we take ground as reference which give centripetal force And centripetal and centrifugal have opposite direction
Centrifugal is considered only in frame of tube considered for pressure variation and in ground frame we use Bernoulli's equation but centripetal is not a separate force, net sum of inward real forces is considered as centripetal force
@@physicsgalaxyworld sir if a question comes like where the pressure should be greater at Pa or Pb? Suppose options will be like A)Data is insufficient B)depends on calculating from the frame of reference C) Pa>Pb D) Pb> Pa. Sir what can we say about it sir please reply as soon as possible as jee adv is very near.
5:40 sir yha pe side wall 1 ke liye avg pressure me atmospheric pressure ka contribution to liya hi nhi?????????? and wall 2 me to Patm and P1 dono ka contribution liya h please tell sir
@@lokeshjangir3089 mujhe samjao bhai please... 2nd wall mai surface ka pressure add kiya(Po+hp1g) ... But 1st wall mai surface pressure(Po) add nahi kiya... Aisa kyun
54:36 sir uppar wali liquid ko replace krke neeche wali liquid hi bhar denge such that height uska change hojaega ( such that it will create same pressure difference as initial )
50:00 i think we cant use bernoulis theorem here.. as it is for stream line flows.. and here we are talking about diffrent streamlines.. i mean the point A doesnt belong to the streamline flow of the point B.. please correct me if im wrong
I think that Bernoulli's equation is applicable for two points in a streamline laminar flow. In the given situation, points A & B are not in the same streamline (imagine rotating streamlines), so ..?
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37:15 if the cross section area of container and tube is given, do we need to consider the rise of water level due to weight of gas in the tube and than calculate the pressure.
50:00 sir here total energy of particle A and particle B is not conserved! As the velocity pressure is increased at point B..the hydrostatic pressure should have been decreased..but here that is not the case
35:41 Sir cavities liquid displace nhi krengi fir aapne uski wajah se force kyu consider Kiya hai??? Thankyou Sir for this extraordinary high quality content....
at 5:21 we should not add P atm in calculating avg force on side wall of 2nd liquid as on wall external Patm is also so it gets cancelled out similarly what had done in 1st fluid . sir please tell me if i am wrong
I had same confusion, my best guess is that total force includes P0 but experienced force does not take P0 because there is external atmospheric force acting on it
Bernaulis equations holds good for that situations where we can apply equation of continuity. And in the equation of continuity velocity is parallel to that area through which liquid passes and reches other point where also area vector and velcity vector are are parallel and also number of streamlines is also same for both the areas but here as liquid is moved from A to B and velocity of fluid is perpendicular to the movement so i think bernaulis result is not correct.
50:35 Sir I think Bernoulli equation is not applicable in horizontal direction as if we think of an imaginary cylindrical liquid layer of width dx then all such layers are different and Bernoulli equation is applicable between two points of same liquid layer ...but it will be applicable in vertical direction
Sir at 50:20 is the contradiction created because we have not considered the sign of wx when we applied the Bernoulli eqn Or is there any other reason,pls clarify PG Team.
I think it can be because those 2 points are on different streamlines. And we can apply Bernoulli equations on two points only if they are on same streamlines . Can anyone please verify me ?
sir are these the same illustrations that are in your Advanced illustrations book , because i do not have the book so if i watch these and solve the black board problems will it be ok !!
Is the controversy because of the frame sir? In the previous case we worked out the equations in the frame of the container and so accounted for the centrifugal force in the outward direction While applying Bernaulli's we took the velocities of points A and B in the ground frame itself Maybe that is reason for the controversy
50:33,,bernoulli's eqn cannot be used as this is not the case of fluid dynamics ,,it is the case of fluid statics only, as container and liquid are moving together ,or we can say that liq is at rest wrt container like the case of accelerated containers
Sir is explaining the concept of pressure by liquid... Patm you can consider if it is asked in the question due to liquid and if it is net force on wall then don't consider
Sir me advance dekar iit gandhinagar jana chahta hu or mere board me 67% hi aaye hai, me 2022 aspirant hu kya mere liye vo criteria remove ho gaya hai na Plz bataye......🥺🥺
IMPORTANT: Students should note that at 32:00 in case of U-Tube by mistake in figure sir has taken left side level rise by considering rotation axis on right side but the concept and analysis remain same.
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50:08 , controversy comes from application of Bernoulli theorem, if we remember that what Bernoulli theorem says, it's validity is only for similar streamline. In rotating fluid the streamline at A and B is different because effect of centrifugal force drifts the liquid more in backward direction, as highest pressure is in the middle and it decrease gradually while going to B. But rotational effect compensate and pressure overall inc. So the first relation is valid and the Bernoulli one is not.
Kinda stuff could be seen in adv 22 Question.
@@Aesthetic_Skyline I think you are talking about that question in which Bernoulli had to be applied in gas which was getting compressed.
@@aspirant6924yep i think it was incorrect type statement
@@aspirant6924 wasnt it an SAT question too
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18:55 Assumptions: 1) Cylinder does not fully submerge in the container.
2) In case 3 increase in ht of two liquids is taken equal.
#) Also NOTE is applied to both cases 2 & 3. {😉pheww.. saved a lot of calculation here.}
@) In case 3 m'g is same as mg but with rho2
$) Btw open to any corrections required. {Ptss.. good luck calculating😅}
Case 1: - large surface area if Work Done to be found for, He to 2He it comes to be 0.5MgHe.
Case 2: - surface area of container if Work done to be found for, He to 2He it comes to be 1.5MgHe.
NOTE [for cases 2 and 3]: - here I took y as He and dy as y and dx as x to get Fb=(rho)gS(He+y+{S/S1-S}) and 2He is height from surface to contact surface between rho1 and rho2.Also He is width of rho2 liquid
Case3: - surface area of container if Work done to be found for, He to 3He it comes to be W2 + W3 =
=2.5mgHe + mgHe(S/S1-S) + 0.25mgHe(1/S1-S) + 0.5m'gHe + 0.25m'gHe(1/S1-S)
{# Finally done with the calculations.... long due to variables}
33:15 we will integrate (rho)w^2 x dx separately for 2 liquids , putting limits accordingly and then equate with (rho)gh
Yes,I did the same thing
All the best for adv
you cannot directly take (rho)gh as both fluids have different densities...
and the integration part will also be bifurcated as the densities of the liquids are not same throughout the length
Use pressure equation
Ek quadratic ayegi x me sidha solve ho jaega
We can seperately write the equation sir derived for both the liquids assuming pressure at junction as Pj and just add the two equations ... It will come in terms of Pa and Pb and we know it's values
50:04 Bernoulli's equation is an equation derived from conservation of energy whuch is taken in ground frame only since it is considered as conserved frame
But in the rotation part it was considered in in the frame of liquid.
Now we know that direction of centripetal force is opposite to centrifugal
Also in Bernoulli's equation work done per unit volume by centripetal force was considered which was towards point A.
Therefore w.r.t. ground, force was more on A and w.r.t liquid force was more on B
Rest of the thing can be calculated by you.
In rotation it was centrifugal force
Yup
I literally gone through books but missed conceptual point written in my notes only ,👍
In 50:15 Bernoullis equation cant be applied as it is ROTATIONAL flow of water. P + 1/2pv^2 + pgh = E, in our course if we switch streams lines under the condition that flow is newtonian than E remain same for 2 different stream lines but if rotational flow is concern so E remain same for a particular stream line..
Btw advance form of bernoulli is adapted to find out all type of cases(but alas that isnt in our course
yaa that's why in course we were told... "irrotational flow"
50:42 bernoulies condition applicable for points in same streamline but points A,B are on different streamline so that gives an false result ....and thats the reason of controversy....thank you sir for this awesome content...
The best teacher for jee advanced physics
Rajkumar sir pw author of irodov solution book is best
No@@Hello_fools
@@Hello_foolshell naw
47:34
In this case the point B is moving along the rotating fluid (as frame of reference is rotating vessel) so velocity of B is 0 w.r.t. fluid so Pb is greater than Pa
50:38
In this case the frame of reference is ground and point B is moving with respect to rotating liquid hence it has some velocity so Pb is less than Pa
(Difference is just of frame of reference)
(Do point out if wrong)
So this message to thank physics galaxy team and ashish sir as seeing the concept and revision video just helped me to overcome the fear of fluids and now i can solve jee advanced question and feeling much confident
sir i think the first case was right because this situation is analogous to a cyclone and there is a low pressure zone inside the cyclone, so PA
38:30 after this I again tried the inverted tube question asked in jee adv paper 2 stem question and I hit that right in first attempt without any problem thank you sir...
50:30 , since there is no rotational motion of fluid and fluid is not flowing the fluid statics is applicable here whereas when we apply bernoullis theorem to this case it gives us false results as the fluid is not flowing since the container and the liquid both are rotating with the same angular velocity.
Hope it's correct❤
but if we use concept of fluid statis then Pa=Pb (which is wrong)
45:57 option A and Pb1 also equal to P°+ rho(w²x²) /2
Sir the controversy with Pressure difference using rotating fluid and by Bernoulli's theorem because we are missing out 'dha' factor in it where d is density, h is height on the direction of net acceleration and a is acceleration. As here including g also we have xw² as centripetal force so on accounting that we can say that both points A and B are not on the same line perpendicular to the jet acceleration of the fluid. The one analysed with rotation of vessel is correct one.
50:28 When Sir says gap-the main thing to notice is that in Bernoulli Method, the point B is actually at a depth of (h+y) below the liquid and point A is only h below. Assuming them to be along the same Horizontal, means, we nglected the pressure that the Extra Volume of Water created on B(to visualize, think the area enclosed by the Parabola and the X-Axis case). This is the reason that we get a different SIGN as per me(if I am wrong, then do point out)
I think other explainations are much better
You argument is kind of similar that you can't apply Bernoulli
But you can't deduce from your argument that sign will be opposite
We can only conclude we can't use Bernoulli in rotating liquids
Sir, bernoullis is applicable for steady flow, here particles at points A and B are in different velocities
Nice
33:12 Use pressure equation
Ek quadratic ayegi x me sidha solve ho jaega
50:00 Sir ji bernoullis eqn can be used on 2 pts lying on the same streamline… whereas here A&B are on differnt streamlines
As said by sir: “fundamental point”
Thats correct.... Actually its a rotational flow and we cant apply bernoulli directly however some advanced form of bernoulli adapted to find that out
@@5dots297 ohh where can I find that adv form ? Any link that you can share ?
@@kshitijvispute6792 did you got that adv form of bernaulli. If yes then plz tell
If we consider centripetal force in frame other than liquid , pressure difference will still be same but this time centripetal force will act in opposite direction .
50:13 i think bernoullis eqn should not be applied in this case ....since it is applied along a single stream , but here the stream line a A different from that of B
50:22 I think it can be because those 2 points are on different streamlines. And we can apply Bernoulli equations on two points only if they are on same streamlines .
Exactly I thought abt the same scenario. That's I also searched abt bernoulli's equation in rotating fluid but I didn't get any appropriate result
the real guru whom i trust the most for my academic carrer
we use derived bernoulli's equation in rotating frame due to radial accn towards center which is P - 1/2 (rho) (wx)^2 + (rho)gy = constant where P - 1/2 (rho) (wx)^2 is the reduced pressure at distance x from axis
At 50:08 I think Bernoulli's method is incorrectly used because the velocity omega * x of liquid is perpendicular to the plane and not along the streamline.
@50:30 pressure is inversely proportional to velocity, Va Pb
31:58 in right side wala pressure badna chahiye
44:54 a option
@50:35 sir maybe the controvery is because the flow inside the cylinder due to roation is turblulent and hence bernoulli will not be applicable
At 4:32 dont we need to add force due to atmospheric pressure in avg pressure on first wall?
I have the same doubt as well.
it should be added
Sir I am requesting ki aap PG app par jo revision booster workshop hai vo over hone wali hai 12days ke approx mein,toh aap uss series ki day ko extend kar dijiye na please it was a very helpful series for Me and genuinely I wanted it to be extend till next year's JEE Mains.
we cannot take height same , g effective at both point a and b is different . At b , g eff is resultant of gravity and centrifugal acc
Reuploaded???
Sir 50:35 ka reason bata dijiye sign controversy
Ashish Sir has already asked students to discuss and post their views in comment section
50:40 Sir yaha kyuki aap container ke reference mein Bernoullis theorem lga rhe hai toh velocity bhi relative honi chahiye
50:20. Sir I think so that it is due to frame of reference as in pressure analysis we take frame of vessel which give (shows) centrifugal force on fluid but in bernoulli's analysis we take ground as reference which give centripetal force
And centripetal and centrifugal have opposite direction
Well explained
Centrifugal is considered only in frame of tube considered for pressure variation and in ground frame we use Bernoulli's equation but centripetal is not a separate force, net sum of inward real forces is considered as centripetal force
@@physicsgalaxyworld sir if a question comes like where the pressure should be greater at Pa or Pb? Suppose options will be like A)Data is insufficient B)depends on calculating from the frame of reference C) Pa>Pb D) Pb> Pa. Sir what can we say about it sir please reply as soon as possible as jee adv is very near.
5:43 in expression of force on wall 1 why force due to P0 is not included
Outside pressure is also P0
Same doubt
At 55:00, why sir took the velocity of point just inside as zero? It should be something nonzero.
Imp for me
very op concept✨ 9:56 ✨
50:00 sir bernoulli nhi laga skte vo stream line flow me lagta hai ....whirling ya circling fluids me nhi lagta ? Bolo sir sahi hai kya
I can see sir these are all from prev year papers of jee advanced and mains. But it's great to study them all together relatively 💯
50:31 @PhysicsGalaxy sir please tell the gap. could not find that . better make a youtube short on this.
17:34 , should we not write dFB as S ( dy+ dx ) rho g as Sy(rho) g was already acting on it the additional buoyant force must be countered by f
yes
Sir 4:44 me (Pavg + Pnot)*Area hona chaiya na kyoki Pnot bhi to pure surface pe distribute hoga
Yes
Rotating liquid in a cylinder m centrifugal force h or Bernoulli m centripetal force include h centripetal or centrifugal are opposite direction 50:37
Don't know how sir knew my thoughts. I just really wanted lecture on fluid. Thank you very much sir
reason may be that beronouli theorem is not applicable in this case because flow is not streamline.
5:40
sir yha pe side wall 1 ke liye avg pressure me atmospheric pressure ka contribution to liya hi nhi??????????
and wall 2 me to Patm and P1 dono ka contribution liya h
please tell sir
P atm will not be needed here also
@@physicsgalaxyworld i got it it sir❤❤
@@lokeshjangir3089 mujhe samjao bhai please... 2nd wall mai surface ka pressure add kiya(Po+hp1g) ...
But 1st wall mai surface pressure(Po) add nahi kiya... Aisa kyun
@@rahuldravid8414 second wall me bhi nhi karna Patm ko add .
@@lokeshjangir3089 vo nai... Sirface ka pressure kyun add kiya... Po+pgh1 vala
54:36 sir uppar wali liquid ko replace krke neeche wali liquid hi bhar denge such that height uska change hojaega ( such that it will create same pressure difference as initial )
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Rajkumar sir best
31:58 pe u tube me left side ka level kyu badega ? Mujhe lagta hai right side ka badna chahiye due to centrifugal force
Yes that’s correct. Sir has confirmed that right level will rise but the process will be same
@@physicsgalaxyworld thanks for letting me know
50:00 i think we cant use bernoulis theorem here.. as it is for stream line flows.. and here we are talking about diffrent streamlines.. i mean the point A doesnt belong to the streamline flow of the point B.. please correct me if im wrong
Right bro even I am thinking the same
I think that Bernoulli's equation is applicable for two points in a streamline laminar flow. In the given situation, points A & B are not in the same streamline (imagine rotating streamlines), so ..?
Mujhe bhi yahi lagata hai
35:33
Here due to cavities total volume of liq displaced should decrease so buoyant force should decrease?
Bernolies equation Always valid in steady, frictionless streamline flow only, so that's why result is getting wrong
Sir pls ek jee advanced level mock test ki pdf
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50: 00 sir i think , psudo force ke work ki vajah se energy aayi he ,by work energy thorme, enery is negative, so RHS me aapne negative sign lena hoga
Sir 9:52 pe wo density ki inequality kaisi aayi? According to me it should be p3>P2>p1
@@manishshewale1040 arey to 3>2>1 hona chahiye na…sir ne 3>1>2 liya hai
You saw it uploaded yesterday night too????
Yes due to some mistake it got published at night and got removed... we are sorry for the inconvenience
@@physicsgalaxyworld no inconvenience sir how can we complain when you are giving so much to us thank you sir
37:15 if the cross section area of container and tube is given, do we need to consider the rise of water level due to weight of gas in the tube and than calculate the pressure.
50:00 sir here total energy of particle A and particle B is not conserved! As the velocity pressure is increased at point B..the hydrostatic pressure should have been decreased..but here that is not the case
Sir in upper side wall
Pavg = P0 + rho1gh1/2
35:41 Sir cavities liquid displace nhi krengi fir aapne uski wajah se force kyu consider Kiya hai???
Thankyou Sir for this extraordinary high quality content....
bhai hinduism k naam pr dhongi andhvishwas se lut rhe h.
. uspe kch kro vivek saad
Sir in case of large compression expansion will we consider constant pressure?42:04
at 5:21 we should not add P atm in calculating avg force on side wall of 2nd liquid as on wall external Patm is also so it gets cancelled out similarly what had done in 1st fluid . sir please tell me if i am wrong
I had same confusion, my best guess is that total force includes P0 but experienced force does not take P0 because there is external atmospheric force acting on it
Bernaulis equations holds good for that situations where we can apply equation of continuity. And in the equation of continuity velocity is parallel to that area through which liquid passes and reches other point where also area vector and velcity vector are are parallel and also number of streamlines is also same for both the areas but here as liquid is moved from A to B and velocity of fluid is perpendicular to the movement so i think bernaulis result is not correct.
Most waiting this chapter generally capillaries aand viscosity
Thanks a lot sir for continuing the Booster Series. Please add this video to the Playlist of Advanced Boosters
50:35 Sir I think Bernoulli equation is not applicable in horizontal direction as if we think of an imaginary cylindrical liquid layer of width dx then all such layers are different and Bernoulli equation is applicable between two points of same liquid layer ...but it will be applicable in vertical direction
yes ,perfect
Hope so they don't make another controversial question from fluid in jee advanced 2024. As usually they did.
Sir at 50:20 is the contradiction created because we have not considered the sign of wx when we applied the Bernoulli eqn
Or is there any other reason,pls clarify PG Team.
I think it can be because those 2 points are on different streamlines. And we can apply Bernoulli equations on two points only if they are on same streamlines . Can anyone please verify me ?
50:18 Because bernoullis theorem is not applicable for rotating fluids so P(B) - P(A) is correct!!
sir are these the same illustrations that are in your Advanced illustrations book , because i do not have the book so if i watch these and solve the black board problems will it be ok !!
All adv illustration of that book are already on this channel ...see playlist :)
These are different illustrations used in this video
Is the controversy because of the frame sir?
In the previous case we worked out the equations in the frame of the container and so accounted for the centrifugal force in the outward direction
While applying Bernaulli's we took the velocities of points A and B in the ground frame itself
Maybe that is reason for the controversy
Only few know it was uploaded last night ....😁😁😁😁
Yes due to some mistake it was uploaded and got removed... we are sorry for inconvenience caused
Sir please increase the frequency of boosters please bring booster on rbd sir
Coming soon
Sir workdone on cylinder wale case me agar hme work done by liquid on cylinder likhna hoga to hum same formula use kr skte hai????
50:33,,bernoulli's eqn cannot be used as this is not the case of fluid dynamics ,,it is the case of fluid statics only, as container and liquid are moving together ,or we can say that liq is at rest wrt container like the case of accelerated containers
Thank you sir each and every seconds during this moment utilised fruitfully 😊
sir how can we access PYQ models e book in pc.
At 37:38 how sir wrote pfinal in eqn 1??
same doubt
Rotating liquid in cylinder : The results are different because of different frame of reference.
Think more carefully
4:27 pe who Patm ka bhi term ayega na ?
yes I think so
ha wall 2 me to Patm ka bhi liya h
Sir is explaining the concept of pressure by liquid... Patm you can consider if it is asked in the question due to liquid and if it is net force on wall then don't consider
sir i m following your 700 jee advance illustration book is that only enough
At 41:50 sir why are considering only change in pressure ????
Also effect of surface tension if given
50.42 the motion of fluid is turbulent bernouli not applicable
In force and torque calculation when to take p° and when not...
Sir me advance dekar iit gandhinagar jana chahta hu or mere board me 67% hi aaye hai, me 2022 aspirant hu kya mere liye vo criteria remove ho gaya hai na Plz bataye......🥺🥺
@25:44 tan ∆= a/g or g/a ????
Amazing session sir !
Sir mechanics me kinematics ka nahi aya PLZZ upload soon
Coming soon
@@physicsgalaxyworld thanks sir
But before advance PLZZ 🙏🏻
Sir surface tension par bhi ek booster checklist le lijiye 🙏
Just needed it❤✅
Wow sir this checklist is awesome 🔥and the content is extraordinary, Such a great teacher of physics on earth!
At 28:36 shouldn't we consider g effective (√g^2+a^2) for vertical tube part in calculation of P°
No it should not be because only vertical pressure variation is considered... watch concept videos to clear this fundamental...
@@physicsgalaxyworld understood 👍 thank you
Thank you for the session sir!
PG Team can u pls tell that are the notes of these sessions available anywhere?Pls do reply..🙏
Thank u sir. Please provide booster checklist for mechanics as well.
Coming soon
Sir pls teach capillary action