JEE ADVANCED 2020 SOLUTIONS | INTENSITY OF X RAYS VERSUS WAVELENGTH | CURRENT ALPHA AND OMEGA SERIES

[@PHYSICSSIRJEE]

Please smash the like button for a target of 750 likes for the next video . Let's make it there in two days 🙂👍🏼. An awesome "rotation" video on the way ❤️

[@puspalpaul185]

Thank uu very much sir ..... Did liked it at once

[@Prodbybah]

sir pleaase make a video on how to extensively study for inpho road map so that i can end up in top 35 for ocsc camp?

[@omjamdar7094]

Sir do you look after your mails? I want to talk some important things so please reply

[@sahilpatil3428]

16:44 AC

[@kavi5990]

Sir @16:43 I think answer is D A-similar to cutoff wavelength idea so correct is what I thought but I learnt from a stack exchange question that says electrons must always have some kinetic energy acc to QM, please confirm the validity of the statement B, C-As there must be some fraction converted into photon if at all any interaction(aceeleration takes place) so not correct D-im not getting any reason for why this cannot happen

[@as.if_0077]

This channel is pure gold🏆

[@sanchitmukherjee2410]

No not gold, platinum++

[@puspalpaul185]

@@sanchitmukherjee2410 diamond ..... cannot be valued with anything

[@eshh183]

14:43 6.25 14:56 C 15:19 5.85 keV 16:26 ABC

[@eshh183]

Please confirm the key sir whenever free. Thanks for the video!

[@Amankumar-by9tw]

I can't able to solve 14.43 question plzz send solution.

[@ameya7233]

14:36 6.25Ω 15:00 (B) 16:11 5385.60eV (not sure If we have to take effective Z or not to find binding energy of n=4 e) 16:27 (A)(B)(C)(D)

[@vedarthbhawsar4286]

Alpha and omega are love❤❤.... Thank you sir jee 🤩🤩... I am struggling with these problem... After your video, I understand how to tackle it

[@sasi8391]

You are one of the best teacher really I have never seen....

[@Gauravkumar-om8sj]

14:48 i think B should be the ans , as i think that option c can be ruled out as xray is not a nucleus phenomena or any sort of nuclear decay

[@PHYSICSSIRJEE]

Well done ✅👍🏼

[@utkarshpal6109]

14:32 6.25ohm. first i calculated lambda when resistance was not shunted ,, then added0.155angstrom to it now potential difference corresponding to this wavelength should be equal to potential difference across that shunted resistance, then applied ohms law and got the ans please sir tell the answer

[@akashkaul1923]

Wow jee adv preparation need u sir.....👏👏

[@D_P279]

16:23 Option no B would be most preferable according to me. Pls confirm sir.

[@triggeredskeletonbankey8864]

Thanks sir mene request kiya tha modern pe.... thanks lot🔥🔥

[@Prodbybah]

sir which part of morin and griffiths is for inpho and ipho please sir tell me there is no guide . please sir after some coaching material what for inpho and ipho

[@BruceWayne-gz6eg]

Ye to bhot sahi channel h, shahrukh ko bhi bhejta hu.

[@PHYSICSSIRJEE]

JEE bhai 🙌

[@masoomladka8017]

@@PHYSICSSIRJEE 💯😂

[@sushantsrivastava7013]

Sir Please Make a video on Tensile Force ........ Its used too much in pathfinder

[@PHYSICSSIRJEE]

Noted 👍🏼🙂

[@sushantsrivastava7013]

@@PHYSICSSIRJEE Thank You Sir

[@Hap1921]

Sir at 11:25 When the current is halved , why can't the the graph which is lowered at all points still be completely above I mean the red graph can still be completely about the green one in the right diagram , how can we be sure of the intersection?

[@Hap1921]

Sir pls reply to this

[@kavi5990]

​​@@Hap1921 it's obvious that characterisic x rays must be less in intensity so we can expect wavelengths near to it to lie below , for your question of how we can be sure coz we experimented and found out it is that way and that is the only way u can be 100percent sure

[@chethanreddyamanaganti9295]

sir in 12:41 actually 3 changes are done including distance being decreased t0 d/2

[@soniamittal5756]

16:43 Sir option no. BCD

[@abhiramvanama3074]

Hello sir I want to study from other than youtube tell me I want to join there

[@Fighter_Believer_Achiever]

@2 REVISION Thank you very much sir🥳🥳🥳🥳

[@vilasjadhao730]

That was most needed 🙏🙏🙏

[@KrishnaKumar-if9mg]

THANK YOU SIR , GOT MUCH NEEDED CEARITY IN THIS CONCEPT.

[@jgchaudhary1035]

😎😎😍good explanation using graphs

[@kisansalunke6637]

16:43 option no. B&D

[@PHYSICSSIRJEE]

My answer is different

[@mohitshrivastava5412]

Thank you Sir

[@vinayaksharma9251]

14:32.....6.25 ohm 14:56 ....B 16:23...BD?... Sir is it correct?

[@vilgax9955]

again top class...

[@Hap1921]

Sir I more doubt everytime I have gone through this topic I hear that the electrons get slowed down its sort of a brake that is applied to the electrons and thus this energy lost is emitted out as a spectrum of continuous X rays but sir as the electrons tend to bend around the nucleus of target metal their ke is being converted into the system electric potential energy them which energy is it that is being lost here and radiated out ?

[@saivinayakakkaldevi2746]

15:06 C 16:31 ABD

[@PHYSICSSIRJEE]

My answer to 16:31 is different

[@saivinayakakkaldevi2746]

@@PHYSICSSIRJEE ok sir let me try again and edit the comment

[@hellohello4360]

Dear sir, Can you suggest some resources to practice surface tension at jee advanced level

[@Gauravkumar-om8sj]

14:32 sir ans coming was 6.25

[@gauravpatel2495]

superb explanation

[@Rohankumar-mr8rz]

16:43 is the answer b??

[@PHYSICSSIRJEE]

My answer is different

[@Rohankumar-mr8rz]

Then there is some fun in it please make video on that too as soon as possible

[@AnkushKumar-jf3wu]

14:34 Sir I am stuck in this question, not getting the right answer

[@mridulrathi7776]

Sir the answers according to me are: 14:46 - 6.25 ohm 14:56 - B 15:19 - 5385.6 ev(not so sure) 16:23 - I think AB(not sure about C or D)

[@PHYSICSSIRJEE]

16:23 , my answer is different , rest are correct 👍🏼

[@mridulrathi7776]

@@PHYSICSSIRJEE ok sir 😅

[@puspalpaul185]

Mridul orz ..... u r NRT sir's son naa ?

[@mridulrathi7776]

@@puspalpaul185 oh i see you know him...

[@Amankumar-by9tw]

@@PHYSICSSIRJEE sir plzz explain 14.43 question

[@Fighter_Believer_Achiever]

Thank you very much sir🥳🥳🥳🥳

[@amiyancandol4499]

To be fair the must have given A or AC atleast :(, also sir if we consider (hypothetically) that each electron gives a photon i.e efficiency=100% then intensity would decrease right ? As then the intensity would depend on number of electrons per second which is essentially prop to the current.

[@anantagarwal5380]

Yes

[@masoomladka8017]

Ok

[@masoomladka7198]

kok

[@shivasudhan165]

Finally ❤️

[@saivinayakakkaldevi2746]

thanks alot sir

[@vedbrahmbhattiit-j4497]

Amazing😃🔥

[@as.if_0077]

Ok

[@sivapriya2632]

16:26 I feel none of the options are right sir.. Majority is heat but we cannot say no X-rays as accelarating charges produce EM radiation

[@anantagarwal5380]

Sir can we go by the formula for drift velocity?(please correct me conceptually if I missed something) Assuming temperature to be same,and width of the electrodes to be constant, I can write I=enw(d)Vd Now Vd= e.tau.(V/md) Now putting the changes (I/2)=e²tau.w.(d/2).4V/d.m.n' From here We get n'=n/8(n is number of atoms per unit volume) So can we say as n is decreasing so the intensity is decreasing? And for cutoff wavelength normal way of calculating.

[@lakshitasharma4709]

thank u sir

[@vinaykarthik7765]

Please do such more topics as early as possible.

[@supriya1729]

Hello sir..how are you...😀 Sir can you please make a video on inherently inelastic process and the mysterious heat loss and final wave motion....the instantaneous impulse .. and all that elaborately?☺️

[@_JAISWAL_SANATAN_US_POLO]

Aap kis Coaching se hain sir

[@PHYSICSSIRJEE]

www.physicssirjee.in/About.html

[@ablcanybodycanlearncubing869]

Sir what should I do in pathfinder I am jee 2022 aspirant

[@PHYSICSSIRJEE]

You can start immediately , first attack objective type 🙂👍🏼

[@ablcanybodycanlearncubing869]

Ok sir thx

[@devashishthakkar6723]

Gajab

[@amanSingh-wg6kq]

Was waitung fr this 😀

[@satya-physicslover3025]

Nice

[@aadijain8090]

in my opinion should be a,b 16:54 option b 15:08

[@PHYSICSSIRJEE]

My answer to 16:54 is different

[@aadijain8090]

@@PHYSICSSIRJEE sir can you explain this one then

[@PHYSICSSIRJEE]

@@aadijain8090 I will , in resolved series

[@himanshukumargupta9993]

I think intensity only depends on filament current not on voltage across anode and cathode

[@harshgupta_india]

Jee 2020 paper was a total mess ... surface tension ques then optics ques and now this one😒

[@shashanks3492]

sir how to tune thinking for jee advanced , like in that surface tension question also, if more thought is given, we get correct ans byt wrong key, same as in this question, also, you told, even after applying for key change, iit didnt rectify, though the method was accurate,,.. how far should we think in the exam sir,?

[@PHYSICSSIRJEE]

Forget about previous exam 2020 , that was a sham . Focus on reading question carefully and apply your concepts , all JEE papers are not bad , eg : 2019,18 etc... You will do well

[@devashishthakkar6723]

Sir plz take up problems which u think are needed like this ...thanx a lot

[@aryansudan2239]

revised

[@AnkushKumar-jf3wu]

Sir why you are recording your meetings in Zoom, I mean that you can use better application like OBS studio

[@PHYSICSSIRJEE]

Time constraint . No editor . I do everything on my own after 10 hr offline job . I will consider your advice once JEE advanced is over

[@AnkushKumar-jf3wu]

@@PHYSICSSIRJEE ok Sir I can understand your problem, As I too study in a offline coaching(FIITJEE), and I see that how teachers are committed to hard work.

[@gtnation6641]

You make me feel physics… thanks a lot sir…

[@shashanks3492]

sir,15:07 B? for 14:35 , i m not able to visualize circuit, sir, can you please help

[@PHYSICSSIRJEE]

Correct 🙌

[@prateekbheda9802]

🙂💥

[@vedant6598]

Sir please make more video of advanced pyq with postmortem as u get time as it's just 40 days remaining

[@PHYSICSSIRJEE]

Suggestions for questions?

[@vedant6598]

@@PHYSICSSIRJEE Sir question no.5 in 2020 P2 paper

[@PHYSICSSIRJEE]

@@vedant6598 let me check

[@vedant6598]

@@PHYSICSSIRJEEthanks so much sir for asking ❤️

[@vedant6598]

@@PHYSICSSIRJEE sir due to teach for chaina students in kk batch actually I am student of chaina from ashok sir batch 😊

[@aadijain8090]

my answer came out to be 6.25 ohms am i correct??? 14:40

[@paneerpyaazparatha]

I'm getting stuck at calculating the effective V in the parallel combination Can you help !?

[@aadijain8090]

@@paneerpyaazparatha use the parallel equivalent voltage formula (E1/R1 + E2/R2)/(1/R1 + 1/R2)

[@jatishgupta3497]

Sir please join unacademy i want to study from you

[@yatharthgupta6468]

Why do they ask these type of questions 😑 we are not professors who will know these graphs and facts...and neither they are in ncert or any jee book😑😑

[@BruceWayne-gz6eg]

In my opinion,JEE advanced these days is also checking students on whether they are aware of their areas of ignorance, which is making the exam notoriously tough.

[@aaryansahu2003]

Thanks a lot sir..I fear what mindset of a student(with significant conceptual clarity) would be after these ambiguities in jee 2020 paper. Ans 15:08 (B)

[@utkarshpal6109]

to be honest same here,, sometimes i feel like to make paper difficult they these paper setters look up for these ambiguous problems,,, there is no problem with tough papers, but at least they can eliminate ambiguity being the prestigious institutes.

[@siddharthgupta2867]

1.8 k views but less than 500 likes.... Guys plz don't forget to like

[@oswald7994]

100th like ❤️

[@rishabhbhardwajiitb178]

Inte sare key glt dene se to big loss hota hoga bhut ko..

[@sreenivasarao7041]

Who is that one disliker

[@PHYSICSSIRJEE]

My fan 🙂 , I know him.

[@sreenivasarao7041]

@@PHYSICSSIRJEE sir are you joking truely thisvideo, any of your videos shouldn't have dislikers because the effort you put to explain clearly deserves a lot