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Hii Guyss, I studied physics from Ashish sir through youtube and solved adv illustrarion book and followed blocked strategy.. So , i'm Deepesh from IITD here. If you have got more than 97 %ile in jee main, then turant advanced ki preparation me lag jao. Meri mains rank 25k thi, and advanced me 4 digit rank. Doing mechanical engineering from IITD. ALL THE BEST FOR JEE ADVANCED ❤❤❤
Q 14 Genuine Doubt Why is the point of contact between finger and the ring taken as axis of rotation? It is given in the question that the finger rotates about the center of the smaller circle and the larger ring stays in contact with the finger. So isn't the center of the smaller ring the axis of rotation? (As it is in rest, while finger is not in rest since it is continuously rotating about the center of the smaller ring) Please smn explain
yes you are right iaor is defined as the axis whose all point remain at instantaneous rest in this case the centre of circle is at instantaneous rest therefore the answer would be (K)=1/2(MR^2 + M(R-r)^2)w^2 therefore there was no option satisfying this answer therefore the question was particularly dropped in the first part of paragraph.
@physicsgalaxyworld Sir, but in the official answer key of jee advanced 2017 paper 2 no answer was found correct for Q17 Physics which you were solving at 8:42 and the nearest possible option was found to be option 2 and not 1
please correct your video and short about frame dependancy of force force is always independant of reference frames clearly written in kleppner kolenkow also and its the fundamental
@@jeevang750 Physics aur maths me yhi difference H Cause of force is attraction F=gm1m2/r^2 and f=kq1q2/r^2 in sb me mass charge and distance between particles independant of frame hota h that is why force is always frame independant
@@DebayuChakraborti-eq8cq or u can say that force dependency on frame is based on the concept that ur talking about. Whatever frame u see, the distance between two charges don't change, sooo.
A thin uniform disc of mass 'M' and radius 'R' is rigidly fixed to a vertical wall at point 'O' with a massless hinge. The disc is free to rotate in the vertical plane about the hinge. A small particle of mass 'm' (much smaller than 'M') sticks to the rim of the disc at point 'P' which is diametrically opposite to the hinge (point O). The disc is released from rest in a horizontal position (point P is at the lowest point). Find the angular velocity ω of the disc just before the particle loses contact with the disc (point P reaches the highest point). Ye ai ne create kiya hai problem, valid hai kya ? Agar solution mila toh pls koi reply karo
Yeh wala ek illustration se uthaya gaya hai... So basically at the maximum amplitude, i.e when disc is in horizontal position. Therefore writing the SHM eqn, α = π/2 sin(wt + Φ) As, α = π/2 , at t =0. I.e Φ = π/2 Now when the point of contact loses, ie N =0.. I.e 2mRw² = mgcosα Cosα = 1 -2sin²(α/2)... Then we can find the omega from abv FBD eqn. And to find theta at which this happens, we can diffn the SHM eqn, and then, W = -(π/2)w sin(wt) Insert value of W, i.e √g cosα/2R As we know sin can be negative only in the 3rd or 4th quadrant, i.e using principal value of trigonometry formulae, u can solve the abv sin exp, and find 't', after which put this ' t', in the amplitude eqn to find ur theta... Hope it helps Edit: to find angular freqn, using the TBD eqn 3mgR β = (11MR²/2)α α is angular accn
I am in 12th now. PC/IOC ke liye Alk sir ke lectures (paid tg se) achhe rhenge ya phir akk sir/vj sir ke lectures (apni kaksha vale)??? please suggest seniors. Baaki subjects sort ho gye bas PC/IOC Bach rhe h.
I do not understand this definition. $dQ/dt$ represents the rate of CHANGE of charge flow at an instant even though current is defined as only the charge flow per unit time. Please reply
jee advance qweschion were never wrong they can be done by practical knowledge even if they are theoriticallyy incorrect example that chimney qweschion of jee advance
@@IITDMECHANICAL15 yes bro u are right, that ques was conceptually correct and options were not matching but sir said that one of the options is correct despite JEE ADV gave tht as BONUS
@@RohitRaj-n7g tum jaise logo ko sharam ani chahiye/. ek insaan ne itna knowledge hone ke bad bhi students ke liye itna sab kra and tum ese kah rhe ho.
@@shrey---------- mains 1 me yr chem bilcul nhi ati thi . negative me number the .. although other subject thik hai. like mains 1 me maths me 98.25 percentile the. now mains 2 ka result 25 april ko hai. isme jyada expectation hai as me chem krke gaya tha ioc and thodi physical
Can anyone pls explain why Normal is in horizontal direction and friction is upwards? Should it be opposite? Like normal in vertical and friction in horizontal?
mana ki maine lucknow city choose kiya hai exam centre ke liye ham kaise jane ki lucknow ki koun sa thau(place) me centre hai koi ko thaha hai kya plz reply how to find
@@IITDMECHANICAL15 should have used "came here with the speed that is obtained when magnetic and electric forces are balanced or equal" My mind went on the Lorentz equation