Before watching Booster Checklists for JEE Adv, students are advised to watch Revision Checklists on all topics from links given below - All topics of Class 11 - th-cam.com/play/PLYVDsiuOZP5okVjaOFGuRkLT51-WgG95_.html All topics of Class 12 - th-cam.com/play/PLYVDsiuOZP5pp1ak489L0oaMulGzuQWQ3.html Link to all Booster Checklists for JEE(Adv) - th-cam.com/play/PLYVDsiuOZP5o7T58PG8SUiKECCzbY37J9.html Physics Galaxy Team wishes all JEE Adv aspirants BEST to perform their best in finals !
Sir ....if u can please provide the pdf of the pages which u r writing on...then, this will be very much benificial for last min. revision.....also at this time if we make notes it consumes more time..... please sir🙏
Each of the sub-topics taken up in this checklist is a clear indication that Ashish Sir is providing quality Advanced content in this last moment. He is a true inspiration for us!! 🙏🙏
This is the best thing for us, sir one request is that please put two lectures a day because most of us are 2021 aspirants and time is v less Thanks for the efforts sir
All concepts in this video are already covered by sir in concept videos. I completed all concept videos of electrostatic.and now I feel confident. Thank you very much sir for your hard work.
@@harshyadav3150 Pres. inside is excess by 4T/r from "outside pressure" initial outside press is Patm and radius is R finally outside press is Patm - Pe and radius is R'
@@thehealingguy1503 Can you please explain how to go about the problem a little more? I understand that Excess pressure wants to pull the sphere inside. So P2=Patm +4T/r -Pe. But why are we not considering Pe in the initial condition? Also why are we not considering atmospheric pressure itself? I'm very confused.
Would like to give some more names helpful for other Rkh sir Janardhan sir Mka sir(best chemistry teacher for organic and inorganic) And all Mohit Tyagi sirs channel teachers(Absolute gold) Asish sir(Our God) Akash Goyal sir
@@saikhadloya8911 I am unacademy plus student I studied mka sir in class 11 but in class 12 I am currently studying skm sir organic chemistry and vishal joshi sir inorganic chemistry and brijesh jindal sir physical chemistry Physics from nitin Sachin sir and maths anshul singhal sir Asm mentioned name I know janardhan sir he has best collection of problem if you want awsm problem for physics visit Indian school of physics you will be amazed to see level of questions
@@Shivansh0264 OH no ! Sorry wrong shivansh 😅, mei aur Shivansh pandey plus-mates hai..... aur Guruji ke students hai tho isiliye aisa bola..... sorry 🙏
28:19 charge distribution on concentric charged metal shells due to charge induction 39:36 application of electric pressure (illustration 16 in PG advanced book) 42:25 application no. 2
Sir itni *subtle-ta* se topic samjhate hai fir lagta topic mein toh kuch tha hi nahi🙏🙏🙏 Respect for The *LEGEND* Mai bache hue video puri *punctual-ta* se dekhunga
For the soap bubble problem, excess pressure due to surface tension is equal to the electric pressure at equilibrium. 4T/x = (sigma)^2/2€. Where x is the radius of bubble at equilibrium
@@imPriyansh77 we will not take it inward or outward. Equilibrium is attained when pressure inside the bubble is equal to the pressure outside it. So that it stops expanding
44:00 as the charge will exert pressure both inwards and outward at eqm Px= Patm+4T/R - Pe and Py= Patm+ Pe and hence as at eqm Px=Py =>2Pe=4T/R from here we can get the radius and also we can use gas law if we consider growth of bubble isothermal then P1V1= P2V2 here v1 is know and p1= patm +4T/r while p2 = patm+ 4T/r - Pe. show we can calculate R from there also
39:12 Sir this question can easily solved by COM concept we will break sphere in 2 hemispheres and then consider their charges at COM and use force bet 2 charges simply.
Thank you sir for all your help and lectures . These lectures have been instrumental in clarifying concepts and increasing the application power so as to attack jee advanced questions. Thank you for everything.
Day 4 Present Sir. Just completed this checklist also. And today I'll complete advanced illustrations of electrostatics also. Thankyou sir and PG TEAM and ofc POGI.👽❤️
Thank you sir Iam 2022 aspirant but I regularly watch your booster checklist because it gives me lots of confidence and determination towards my goal after seeing your consistency🙏🙏
This one lecture has covered almost all the concepts that are important for jee advanced . Best content at the last moment. A great work , Sir. Thank you for the checklists!!
Sir you're God to me 🙏🏻 Such difficult concepts of electrostats now seems crystal clear all credits goes to sir. Now I can solve Pyqs very easily of JEE ADV. Earlier I used to hate this chapter but now i can finally solve such difficult questions of jee adv. Thank you so much sir!!!
Sir it will be very much beneficial if you will upload all parts together in one day only.. like part-1+ part-2 at same day.. As very few days are remaining for advance
Thank you sir for your hard work I recently solved your adv illustration book in July And now it is like I am getting full revision of book Thank you sir
In 11th ohysics was my weakest bt you and e saral videos helped me a lot and last mock test i got 58 out 66 inphysics jee adv type and 35 in chem and 35 in maths so total was128 out of 198
Sir at 41:17 kya ham force ko equate nhi kar skte hai means force on half sphere (by pressure) is equal to force due to qo on (columns law) half, sphere , sir dono sai same answer aa rha . PLZ help kigiye sir
there was also one more hidden concept in this video that when the ink was wet it was behaving as a reflecting surface but as soon as it get dried it turned into an opaque surface
the excess pressure inside the bubble and the electric pressure are acting in the same direction(away from centre)..... So how can u balance them ... Please elaborate....
@@mananarora141 in simple words vo charge usko bahar push krre honge . Surface tension membrane jse usko piche ki trf khichegi as a membrane to force balance .
@@vanshkapoor8371 but if charge is absent then in that case that excess pressure prevents the bubble from collapsing. So here the role of excess pressure is reversed ?
Sirrr please this checklists fasstt ASAP.....optics , modern , electrodynamics atleast ...days are very less and frequency of videos are lesss than that also🙏🙏 These booster checklists are helping me very much that's why 🙏
@@sumeetdubey9616 what?? he will complete it but he already told that these are booster videos not regular class lectures like onion phy , concept videos , where he will cover every single chap separately , he will cover them in booster in broad sections , like mechanics , optics etc u have to watch revision checklist and concept videos before it yourself .
15:17 If no electric field lines due to the charge q inside the cavity is going out of the sphere then why is there any electric field intensity due to q at point P which is outside the sphere ?? Electric field intensity due to q at P should be zero with this logic.
Net electric field is zero,not individual field due to +q So net flux via metal cavity(outside of metal cavity ,i.e. in metal,and at further distances) will be zero.
Bruh that excess pressure is which will be balanced by electric pressure Initially bubble wasn't charged and it had that excess pressure due to surface tension( that excess pressure is excess to atmospheric pressure) now when the bubble is given some charge it experiences electric pressure too which is initially large so it tends to expand the bubble. You see excess pressure tends to decrease with increase in radius and at equilibrium point electric pressure will balance out the excess pressure.
16:32 sir if we have a dielectric constant K of outer region of sphere (keeping the cavity at vacuum ) how will it affect the answer to the first question ?? @Physics Galaxy
Before watching Booster Checklists for JEE Adv, students are advised to watch Revision Checklists on all topics from links given below -
All topics of Class 11 - th-cam.com/play/PLYVDsiuOZP5okVjaOFGuRkLT51-WgG95_.html
All topics of Class 12 - th-cam.com/play/PLYVDsiuOZP5pp1ak489L0oaMulGzuQWQ3.html
Link to all Booster Checklists for JEE(Adv) - th-cam.com/play/PLYVDsiuOZP5o7T58PG8SUiKECCzbY37J9.html
Physics Galaxy Team wishes all JEE Adv aspirants BEST to perform their best in finals !
Sir ap hindi medium students ko bhi padhte ho kya ? Sir plz reply ....plz sir help me plz
I hope and know that booster checklist will take me a millenia ahead in my Adv prepration 👍👍
Sir iss video mey jitne concepts aapne explain kiye sabmey doubt tha, ab sab clear hogya, thank youu veryy much
Sir ....if u can please provide the pdf of the pages which u r writing on...then, this will be very much benificial for last min. revision.....also at this time if we make notes it consumes more time..... please sir🙏
sir for home work ques at 20:20 i am thinking of resloving the forces along tension and mg
Jee aspirants 2024 also watching this😅😅
The quality bro!!!😅😅😅
Each of the sub-topics taken up in this checklist is a clear indication that Ashish Sir is providing quality Advanced content in this last moment. He is a true inspiration for us!! 🙏🙏
Where are you preparing Mathematics and Chemistry from?
@@AkashVerma-ec2zs yt
@@souviksarkar6706 channel??
@@AkashVerma-ec2zs for chemistry you can go with IITian Explains(MKA sir) and for maths Mohit Tyagi sir's channel .
Thanks
Kitni rank aayi thi??
same que as above
Best physics teacher in the world
Fr
I swear this channel is a blessing I've never revised electrostatics in a better way
This is the best thing for us, sir one request is that please put two lectures a day because most of us are 2021 aspirants and time is v less
Thanks for the efforts sir
Thank you sir for these gems. You are truely a great teacher and an even better person for providing them at free.
Now where r u bro.
@Memes shorts he was probably in class 12
🎯Jee advanced 2024🔥🔥🔥
The previous 3 lectures were very helpful....i bet this one will also be motivating😃😃
Another milestone by physics galaxy on youtube by giving best advance level content 👌
It not just makes concepts clear instead it makes e feel STRONGER.
Thank You SIR
All concepts in this video are already covered by sir in concept videos.
I completed all concept videos of electrostatic.and now I feel confident.
Thank you very much sir for your hard work.
44:07 done sir
It can be solved by using gas law,
P1V1=P2V2
P1=Patm+4T/R V1=4/3π R^3
Similarly P2=Patm+4T/r- Pe V2=4/3π r^3
Thanks a lot sir ❤️
Bro I think P2 should be Patm +4T/r +Pe coz Pe and 4T/r are in same direction
@@harshyadav3150 Pres. inside is excess by 4T/r from "outside pressure"
initial outside press is Patm and radius is R
finally outside press is Patm - Pe and radius is R'
@@harshyadav3150 the excess pressure is acting inwards and electric pressure is outwards.
@@harshyadav3150 Patm+Pe+4T/R=P2 because Excess pressure and Pe will try to expand bubble while P atm of outside will try to stop.
@@thehealingguy1503 Can you please explain how to go about the problem a little more? I understand that Excess pressure wants to pull the sphere inside. So P2=Patm +4T/r -Pe. But why are we not considering Pe in the initial condition? Also why are we not considering atmospheric pressure itself? I'm very confused.
IN TH-cam ONLY ARORA SIR AND MOHIT TYAGI SIR PROVINDING CONTENT OF JEE ADVANCE
AND ONLY SERIOUS STUDENTS KNOW THIS 😊👍
Would like to give some more names helpful for other
Rkh sir
Janardhan sir
Mka sir(best chemistry teacher for organic and inorganic)
And all Mohit Tyagi sirs channel teachers(Absolute gold)
Asish sir(Our God)
Akash Goyal sir
@@saikhadloya8911 I am unacademy plus student I studied mka sir in class 11 but in class 12 I am currently studying skm sir organic chemistry and vishal joshi sir inorganic chemistry and brijesh jindal sir physical chemistry
Physics from nitin Sachin sir and maths anshul singhal sir
Asm mentioned name I know janardhan sir he has best collection of problem if you want awsm problem for physics visit Indian school of physics you will be amazed to see level of questions
@@Shivansh0264 Oye Power Stone 💜 tu yaha kya kr rha hai 😁😂.........
@@shyamkarthikeya4769 aap jante ho mujhe?
@@Shivansh0264 OH no ! Sorry wrong shivansh 😅, mei aur Shivansh pandey plus-mates hai..... aur Guruji ke students hai tho isiliye aisa bola..... sorry 🙏
28:19 charge distribution on concentric charged metal shells due to charge induction
39:36 application of electric pressure (illustration 16 in PG advanced book)
42:25 application no. 2
MashaAllah, Great Explanation, Rich Content, No Comedy, Pure n Focussed, No Revolution Kind of Things , Amazing teacher
Sir itni *subtle-ta* se topic samjhate hai fir lagta topic mein toh kuch tha hi nahi🙏🙏🙏 Respect for The *LEGEND*
Mai bache hue video puri *punctual-ta* se dekhunga
From
unacademy and vedantu ...................
To
""PHYSICS GALAXY"" and ""MOHIT TYAGI"" SIR
We all grew up!!!!
Thnk u so much sir!!!!
And also consider mka sir iitian explains
44:07 due to repulsion between charges pressure inside will decrease by €^2/2e and then we can apply PV = constant
WILL IT BE CORRECT TO EQUATE Patm=Patm+4T/R+Pelectrostatic and calculate R ? (the external and internal pressure)
@@noamchomsky729 No
@@noamchomsky729 i think it could be Patm=Pe+4T/R
@Yash Giri Yes🥲
Thank you for making Advanced Physics so simple...
Booster checklist not only boosts our concepts but they also boosted our confidence.ty sir.
For the soap bubble problem, excess pressure due to surface tension is equal to the electric pressure at equilibrium. 4T/x = (sigma)^2/2€. Where x is the radius of bubble at equilibrium
why excess pressure acts inward ?
@@imPriyansh77 we will not take it inward or outward. Equilibrium is attained when pressure inside the bubble is equal to the pressure outside it. So that it stops expanding
44:00 as the charge will exert pressure both inwards and outward at eqm Px= Patm+4T/R - Pe and Py= Patm+ Pe and hence as at eqm Px=Py =>2Pe=4T/R from here we can get the radius and also we can use gas law if we consider growth of bubble isothermal then P1V1= P2V2 here v1 is know and p1= patm +4T/r while p2 = patm+ 4T/r - Pe. show we can calculate R from there also
Amazing lectures.Wish I had a chemistry and math GALAXIES
Mohit Tyagi
@@aksaxena1000 exactly
39:12 Sir this question can easily solved by COM concept we will break sphere in 2 hemispheres and then consider their charges at COM and use force bet 2 charges simply.
Great thinking, can be used
Yep , had seen it in GMP too
Sounds game changing 😏
Sir Ashish Arora... You are a legend. I appreciate and know what freat work you are doing as I myself am a physics teacher.
Thank you sir for all your help and lectures . These lectures have been instrumental in clarifying concepts and increasing the application power so as to attack jee advanced questions. Thank you for everything.
Day 4 Present Sir. Just completed this checklist also. And today I'll complete advanced illustrations of electrostatics also. Thankyou sir and PG TEAM and ofc POGI.👽❤️
pogi meaning? and why sir says that??🤔
Thank you sir Iam 2022 aspirant but I regularly watch your booster checklist because it gives me lots of confidence and determination towards my goal after seeing your consistency🙏🙏
This one lecture has covered almost all the concepts that are important for jee advanced . Best content at the last moment. A great work , Sir. Thank you for the checklists!!
Best channel to revise your concepts ❤️❤️❤️
Sir you're God to me 🙏🏻 Such difficult concepts of electrostats now seems crystal clear all credits goes to sir. Now I can solve Pyqs very easily of JEE ADV. Earlier I used to hate this chapter but now i can finally solve such difficult questions of jee adv. Thank you so much sir!!!
Now you can get promoted to A class
Sir please keep on this please continue to know because this will be helpful for JEE 2022 aspirant thank you so much sir..🙏🙏
Sir it will be very much beneficial if you will upload all parts together in one day only.. like part-1+ part-2 at same day..
As very few days are remaining for advance
@@dhruvkothari7611 btw my friend DHRUV is more intellectual than me
TH-cam is not mature enough to handle this level content ❤❤
Thank you sir for your hard work I recently solved your adv illustration book in July And now it is like I am getting full revision of book Thank you sir
THANK you so much sir!! Only you are the one providing 100% free highly efficient and quality content on TH-cam. ❤️🙏 Bless me for jee advanced 2021.🙏
Thx a ton sir, was confused a lot with this topic , but now everything is crystal clear
Superb Content As Always😁😁
Hats off sir🙏🙏
You are great
your greatness cannot be described in words🙏🙏
This is best use of time not even a single second wasted
♥️
Thank you so much sir ❤❤❤❤
After watching every vdo I got confidence thank you sir
sir you are doing wonderful job amazing level of content seen first time
Sir plz upload schedule for booster checklist so that we can plan accordingly...
Thank you so much sir
This is the best content for JEE Adv on youtube
In 11th ohysics was my weakest bt you and e saral videos helped me a lot and last mock test i got 58 out 66 inphysics jee adv type and 35 in chem and 35 in maths so total was128 out of 198
Same here bro
Thank you sir these sessions are knowledge as well as motivation boosters🙂
Hats off sir for you dedication and selflessness shown towards non paid students🙏🙏👏👏
i have learn many new concept by this thank you sir
Sir, please share the schedule of booster checklist so that we can prepare accordingly also try to increase the frequency of lectures (2 in a day)
Sir at 41:17 kya ham force ko equate nhi kar skte hai means force on half sphere (by pressure) is equal to force due to qo on (columns law) half, sphere , sir dono sai same answer aa rha . PLZ help kigiye sir
50-80 concepts in just 55 min . 🥵🥵
The only person...... ........
there was also one more hidden concept in this video that when the ink was wet it was behaving as a reflecting surface but as soon as it get dried it turned into an opaque surface
💥That's it. No one can beat this advanced level lecture...
💙 Extremely helpful at last moment
🧞♂️ Miss you PAUG
POGI**
miss kyu
what does it mean? and why sir says that🤔🤔
@@manmayb6959 All have a some nick name ;)
44:22
4T/R = Electric Pressure
R = 8T E° / R (sigma)^2
the excess pressure inside the bubble and the electric pressure are acting in the same direction(away from centre)..... So how can u balance them ... Please elaborate....
@@mananarora141 pressure has no direction . Pressure multiply by perpendicular area gives force.
@@mananarora141 in simple words vo charge usko bahar push krre honge . Surface tension membrane jse usko piche ki trf khichegi as a membrane to force balance .
@@vanshkapoor8371 thanks
@@vanshkapoor8371 but if charge is absent then in that case that excess pressure prevents the bubble from collapsing. So here the role of excess pressure is reversed ?
Back 2 back video with full dedication. Thanq sir ❤️
Thank you so much Sir.
I think without you what I would've done. You are a 'Farista' for many students. ❤
These videos are great sir... Pls continue this series 🙏
Sirrr please this checklists fasstt ASAP.....optics , modern , electrodynamics atleast ...days are very less and frequency of videos are lesss than that also🙏🙏 These booster checklists are helping me very much that's why 🙏
😭😭😭😭😭 I think we are cheated sir told to complete syllabus by 26 but I don’t think it will happen
@@sumeetdubey9616 what?? he will complete it but he already told that these are booster videos not regular class lectures like onion phy , concept videos , where he will cover every single chap separately , he will cover them in booster in broad sections , like mechanics , optics etc u have to watch revision checklist and concept videos before it yourself .
@@devangvasishth2778 okay but do you think all main topics will be covered by 26?
TILL DATE AMAZING videos sir, HATS OFF TO YOUR UNFLINCHING SERVICE TOWARDS US..!!!!!!!!!
Thanks sir booster checklist helping us very much!!🔥👍🏻
ok
LAST WAS AWESOME AND THIS IS (AWESOME)ⁿ
Sir will checklist come tomorrow?
Sir please upload fast as you can , I know you are busy too but if you can then please upload at least two videos per day.
Thank you sir i am watching 3 day before adv exam i wish i watched it earlier cleared all adv concepts ✨ thankyou from bottom of my heart
If I go to IIT ,then credit will go to Sir 💯
sir please increase the frequency of videos so that we can complete on time. Thank you
thank u so much sir bhagwan app ko padhane ke liye shakti de
15:17 If no electric field lines due to the charge q inside the cavity is going out of the sphere then why is there any electric field intensity due to q at point P which is outside the sphere ?? Electric field intensity due to q at P should be zero with this logic.
Net electric field is zero,not individual field due to +q
So net flux via metal cavity(outside of metal cavity ,i.e. in metal,and at further distances) will be zero.
Thankyouusomuch Sir!! It's really helpful for us!👏👏👏
At 44:16 radius is 2 e°/sigma^2 4 T
Bhai sigma ki terms me answer nhi de skte.....usme bhi radius involved h
thank you sir
coaching me thode doubts reh gaye the conductor se related
aapne sab dho daale :)
Thank u soo much sir, it really boosts our brain
Sir your revision checklist and ncert solutions are really very helpful sir, thank your sir
Thank you for all your efforts 🙏👏
sir please share the timetable of this checklist series..so that we can shape our prepration accordingly..please sir
Sir I am big fan of you ,more than any superstar 👌👌👌👌
One request - Sir plz Attach a pdf of the same content
Beside this Thank u very much For Quality Adv content 🙏
It is tooo good checklist
Sir please make these available on PG app too. Please sir. 🙏🙏
Yes sir we come to watch your channel on youTube only
And there are lot of distractions here
Yes sir please.
There is no pg app in google play store
sir, can you please upload the schedule of the booster checklist in community section
Sir you are real hero for us thank you for this amazing content 🙏🙏🙏
Thankyou sir
Sir kindly tell how to improve confidence level in these last days........
Pls make a video on it Sir
Thank you sir and pg team
Sir jaldi se complete kara doh phir advanced for denge ❤️❤️❤️❤️❤️❤️
26:00 for my revision
Hey are u targeting 2023 adv?
Thanks a lot sir and Pg team 🙏🙏
Thank u sir and pg team❤️
44:11 inside pressure is always greater ,how will the bubble attain equilibrium
Edit: Anybody
Bruh that excess pressure is which will be balanced by electric pressure
Initially bubble wasn't charged and it had that excess pressure due to surface tension( that excess pressure is excess to atmospheric pressure) now when the bubble is given some charge it experiences electric pressure too which is initially large so it tends to expand the bubble. You see excess pressure tends to decrease with increase in radius and at equilibrium point electric pressure will balance out the excess pressure.
@@cyborgsharma6824 both excess pressure and electric pressure are radially outward, they cannot cancel each other out.
@@ilickcatnip yah.... that's I am asking 😅🤟
@@chahal6754 The excess pressure and electric pressure will try to expand the bubbly until it the pressure inside becomes equal to outside pressure.
@@ilickcatnip Exactly! and 10 people who liked his comment got their concept wrong 🤦🏻♂️
Thnk u so much sir 💖💖💕💕💕💕💕💖💖
Sir please add this lecture app
It will be great help
Thank u
44:05 R=8£T/sigma^2
How plz tell?
Sir please share the schedule of this 2021 jee advance booster checklist
Helps alot 😊😊😊😊😊
16:32
sir if we have a dielectric constant K of outer region of sphere (keeping the cavity at vacuum ) how will it affect the answer to the first question ?? @Physics Galaxy
east and west ashish sr is best love you sir
Fun fact - every one watching it at 1.5X + 😂😂😂😂😂🤣🤣🤣
I am watching at 1X
@@anirudhac2004 me too
2x 😂