85% Students don’t know Basics of Friction | Eye opener JEE PHYSICS

[@physicsgalaxyworld]

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[@aryankataria6483]

I am sitting in Jossa 2023 and got NIT Allahabad also but still feels soo good of seeing our old good concepts ... FOR ONE WHO ARE IN PREPARATORY PHASE trust me it is one of the first whose gonna shape you and your career in a whole new direction... Good luck buddy's

[@anushkaverma3706]

Hey...I also got mnnit allahabad cse😊😅..Btw what's ur branch..

[@guider8504]

​@@anushkaverma3706how much percentile did you got?

[@aryankataria6483]

Mechanical

[@aamaspirantpartyJEE]

​@@anushkaverma3706what's your percentile and rank ...

[@nitrrstudent]

I got NIT RAIPUR the best for core branches after NIT JSR

[@gojousatoru4721]

9:43 sir the path traced by this object would most probably be the path 1 curve as the friction is acting opposite 2and 3 rd would certainly be not possible the path 4th is curved in outward direction which is possible only if the object was revolving in clockwise direction

[@ABCDef-w2j]

I think it should move along 2 paths as velocity of object = vel. w.r.t disk +vel of disk There resultant will be along 2 path and since tangential velocity of disk is changing motion of object will also change in that direction

[@anuragtewary5314]

Here is my analysis of last part: it moves through path 2. (Complete analysis is done assuming block is not at the edge.) As the block start moving, it does so radially outward wrt disk. Now as the disk rotates in clockwise sense, at every moment that the block slides along the line, there is some angular displacement in the clockwise sense. This implies, it cannot travel through path 1 or 4 as Sir mentioned (as in both cases, there is no angular displacement). Now, it moves through either path 2 or 3. But it must be noticed, that the clockwise rotation is causing the curvature. While staying in contact with the disk, the object tries to have a straight line path, but there is another velocity (that of the disk) that moves it "up" (basically, rotates it clockwise). Hence it gets a clockwise curvature. (Outside the disk, its path is a straight line tangential to the point leaving the disk. Hence my conclusions: 1. If the object were at the edge, it would follow path 1, but tangentially outward. 2. If the rotation were anticlockwise, it would follow path 3. 3. In current situation it would follow path 2. If anyone has any question/dispute behind this analysis, kindly reply. It would be really helpful 😃

[@GAMINGG122]

Bro I have one doubt When disc rotate in anticlockwise direction of tangetial fric get reverse As a result block now can take 1 or 4 and due curvature it follow 4 ???

[@User-k2v6o]

Straight line wala hi kyu isme same lekin curve wala kyu nhi banega

[@shreerajnandre]

​@@User-k2v6o as soon as the object leaves the rotating disc, its path is a perfect straight line but before that it has some angular displacement and thus its path is a bit curved when it is in contact with the table but after that it is a straight line

[@adityakrsingh-ur9ll]

actually i thought i would go vertically outward but not tangentially...but i did not think of vector sum of both sliding.. my answer for the path is 2 as the block will be having some velocity at the point of leaving and path number 2 is showing exactly that path.... Thank you so much ashish sir..for evolving my mind to this level♥

[@sahilkamble1491]

Even aftet paying 30 to 60 thousand for my tuitions they cant even explain me in this depth and fundamental level of thinking. really we all students are so lucky to have a teacher like Ashish sir. Happy guru pournima 😊 to Ashish sir and all PG team.

[@ravenking2458]

Sir according to me, it will go to path 3 in ground frame. Path 1 and 4 isn't possible as the block has sliding tendency but the direction of its motion will still be decided by the friction of the disc. For 2 & 3, at the moment when we increase the angular velocity of the disc, the new velocity of the block will initially have a tangential component only and the component of friction providing centripetal acceleration doesn't change. So the block goes through path 3.

[@karanjoshi6415]

Lekin ground ke respect mein toh disc angular velocity and centripetal ki direction mein move krega toh friction backward act krega na?

[@ravenking2458]

@@karanjoshi6415Yes, disc toh apni direction me hi move krra hai, block bhi usi se direction me move krega friction ki vajah se but path circular nahi hoga block ka.

[@poojasoni5630]

Sir, according to me it will go to path 3rd, As it has initial tangential component and there is constant friction component in it's opposite direction.

[@dancingfairy5371]

Sir as per my understanding, block will move in path 3 becoz to leave the disk block will have to cover some distance radially for that it will take some time, it can't leave it instantly so in that period of time eventhough block is in contact with disk means a friction is still acting on it thus the line along which block will move wrt disk will shift from South West drxn to North West drxn that's why it will follow path 3.

[@Sst._house_22]

That's right

[@aayushshrivastava2072]

I think the path 2 will be correct As the block slides, r increases so does the force due to increase in omega and centrifugal force. Hence, the block will accelerate in the direction of resultant force of the above mentioned forces untill it leaves the disc.... And when it leaves the disc it will stop accelerating and will move with a constant velocity ( not considering air friction for simplicity ) I am still not sure why the path appears to be curved as when block leaves the disc it should not be accelerating but the path says otherwise as the vector for position of block keeps changing Help would be greatly appreciated

[@AbhishekKumar-wo1yz]

My answer is also same

[@lootandoffers4567]

no bro 4 is possible

[@royalX555]

3 should be answer

[@lootandoffers4567]

@@royalX555 no bro 2 is correct just annalyse it with taking concept of centripetal force and v=rw velocity at that instant

[@lootandoffers4567]

yes bro doubt is correct why it will be curved path bcauz after leaving disc there will be no force on the particle

[@adarshkumarmeher6800]

The result direction of sliding with respect to ground (at an instant) of the block will be in the direction resultant of ( force of friction and centrifugal force ) at that instant That is most probably it will follow path 2 .

[@_justanotherhuman]

I don't if you noticed but this concept would be very easy if you just applied pesudo force in rotating frame in the dirn opposite to tengential acceleration

[@ginger0403]

Path 3 will be correct for observation w.r.t. ground. 1 and 4 are obviously wrong because the disc is rotating clockwise and hence path cannot be in opposite sense. Path 2 implies that "friction" is increasing and not "angular velocity" opposite of which is path 3 which should be the correct answer.

[@lomastripathi2595]

Physics skills ...............♾

[@kunaldas9543]

Meanwhile abj sir

[@BhaktDHRUVofficial]

Right 👍

[@target__iit]

Infinity symbol -> ∞

[@shashishekharmishraiitkanp6928]

​@@kunaldas9543abj sir is good but cant be compared to ashish arora sir☕☕☕

[@kunaldas9543]

@@shashishekharmishraiitkanp6928 yes ofcourse god is can't compared with king (human ). Hint Here asish sir is king ......xd

[@Yash-kt3jv]

Yes sir 🙏I have solved even advanced level questions on circular motion and pyqs of advanced and nsep but did not think in this deeply manner. I request you to start explaining video format for some good conceptual tbt questions. This would be very helpful for all PHYSICS enthusiasts

[@Homo_Sapien_earth]

Yesss

[@saurabhgyanani]

yes sir ye HC verma sir ne bhi exp. kar ke reason pucha tha apne schoolarship test me this year only

[@Babubhaiya-G2]

Anwer may be path 3rd as we have to calculate w.r.t ground. We use the concept of relative .... Motion

[@wahidzaman942]

I think the answer is path 3. I compared the friction with gravity and the tangential velocity with velocity of an oblique projectile. The path 3 which is a parabola, should be trajectory w.r.t ground.

[@jeeaspirant-jq2kk]

Sir path 2 as Velocity of A w.r.t B = Velocity of A w.r.t ground - Velocity of B w.r.t ground same for acceleration , so along that dotted line is the velocity of cube and and velocity of disc w.r.t so we add them and there is no acceleration of cube w.r.t disc but the acceleration of disc is along "at" and hence path followed is second

[@biranchidash374]

1. Sir pehle se thoda sa soch liya tha 2. Ans 3 hoga lgta hai (relative motion) after detach f=0

[@karthikp9438]

Sir it will follow path 2, as V(bd) is lets say -xi -yj and V(dg) is the tangential velocity of disc which will be something +wj, V(bg) will be V(bd)+V(dg), Thus it will outwards and also logically follow path 2

[@ankitabasak7233]

Since the disc is moving in clockwise direction and by looking the centrifugal acceleration and tangential acceleration,according to me path 2 is correct. This is an amazing concept sir Understand this concept for the first time😊 and Happy Guru Purnima sir. Please bless me😊😊

[@GuyWhoCooks]

I think so too

[@ADJEE2024]

How can you consider centrifugal force if you want to see the trajectory of block relative to ground

[@GuyWhoCooks]

@@ADJEE2024 Every force must be considered in my opinion. It's just that the reference frame for its displacement is different.

[@sangkritimajumder3936]

It helped me a lot sir. (JEE-2024 Aspirant)

[@piyushjayswal7734]

As acclrn req in disc frame radial and tangential is proportional to r . So with disc it's slipping direct will remain constant .and if we now comes to ground frame we see path 2 if mass is heavy and disc is spinning fast.

[@Nobody-xp6ip]

Sir answer is 4. Please confirm.

[@jeeaspirant36024]

Is the path 1 right because in relative motion we give the component of velocity of the table to the block in the opposite direction thus after that we will find the component of blocks velocity and the velocity given opposite to the block which comes as 1 option

[@satyampandeyII]

1, bcz if the object overcame the friction, it will slide opposite to the direction of angular vel, also the path will be less curved bcz of radial friction

[@Avani2-l7g]

The answer should be 1 because acc to me if the path of block will be in right direction [and downward as concluded] and the disc is moving opp to it hence the direction of block wrt disc must be as stated and yes sir before you explained i though it may move in tangential direction. Thank you sir

[@Samphysicsguy]

By solving we get x=e^-t and Y=e^-2t considering origin at the moment were v=rw which implies y=x^2 path 2 is correct

[@SantoshGupta-xw2yg]

Thank you sir ... Happy Guru Purnima 🙏 with your help I got into IIT KANPUR

[@IISCBANGALORE-ISMYDREAM]

3 case would be correct at 9:28

[@Ryesun87]

According to me block will move in the direction of path 3, kyuki jo friction ka kaam hota hai na wo ye hota hai ki jo relative velocity hoti hai na dono surfaces ke bich me use minimise kar sake ya ese bol sakte hai ki friction ki tendency dono surface ko sath chalane ki hoti hai but uski ek limit hoti hai or us limit se jyada wo kaam nhi kar sakta. Ab dekha jae toh alpha ki wajhe se A centripetal and A tangential badh jayega or ek time wo hoga jab wo 'Fs' max se jyada ho jayega toh ab jo centrifugal acc. hoga wo us friction se jyada ho jayega jo ki cetre ki taraf lag rha hai or same hi tangential ke case me hoga . Ab jo centrifugal acc. Hai na wo block ko bhar ki side leke jayega or jo tangential ki taraf jo friction hoga wo bhi lagta rahega tangential acc. Ki side lekin ab tangential friction ab relative motion ko rok nhi payega ab jo centrifugal acc. or 'f' tangential hoga unke net force ki wajah se block 3 path pe jayega 👍

[@TanmaySaini-kn4wh]

I think path 2 is correct because due to centrifugal force the block will start moving outwards but still will be moving tangential with the disk am i correct

[@AdityaJee-us1pq]

My explination for the problem asked by sir(Though i have the feel of it in my head, but may not be able to completly explain it with the clarity sir may explain. I would try my best though):- The initial conditions say that the disc is moving with constant angular velocity. Now, when we accelerate it, the block will tend to oppose the acceleration, and its velocity(instantaneous) will be less than the disc, thats why we see it slide backward W.R.T. the disc. But, it will not start moving in the backward direction(W.R.T ground). So, option 1, 4 are wrong. As to compare between 2 and 3, we can think of it as that the box though opposing the acceleration, still accelerates. The velocity of the box must increase as it moves more and more towards the edge of the disc. So the path must curve upwards. So the correct path must be path 2. Also sir, your TBTs and tips really helps. I would owe you just too much throughout this preperation. If ever possible, I will surely meet you once. I live in Jaipur only. Thanks for all your support, and i seek you blessings on this day of Guru Poornima. 🙏🙏

[@shashvatpandey3028]

abhi coaching me friction start bhi nahi hua hai, lekin tab bhi sab samajh aa gaya... wow sir!

[@C__YashKumar]

I think 4th path should be followed by block w.r.t observer on ground

[@THE_BLACK_ORDER]

B is right Top view me disc clock wise hai ghoom rahi block Slide bhi clockwise hi hoga

[@dancingfairy5371]

First thing that I learnt is that, Friction do not oppose relative motion, it opposes sliding tendency, thus always in opp drxn of relative sliding

[@SomethingUseful-se9jf]

This literally was a part of my doubt last year . Most teachers whom I asked simply ignored it and said to memorize it . Thanks to Ashish Arora sir for so much conceptual clarity ✨👍.

[@wdivyankop]

@@shashvatpandey3028 yyess

[@quantumfreak6506]

Already knew all this....just came up to check if there might be anything i am missing out to learn more about friction....thank u very much sir❤

[@bhoomi5577]

I think the answer is path 2 as in frame of reference of ground it will have tendency to move upwards and to avoid this a friction will act in vertically downwards direction. Now in horizontal direction it will have tendency to move inwards due to centripetal acceleration, again to avoid this a friction will act in outwards direction. hence it should follow path 2. Please confirm whether my thinking is right or not sir.

[@aninditadutta1660]

Happy guru Purnima sir Seeking your blessings 🙏 You are one of the best mentor in India And I am very grateful that I found this channel and able to follow your advice . From botton of my heart I thank you for your effort for us .

[@mishthiagarwal1680]

I think it can be path 2 and 1. It can be 2 for simple blick kept and 1 if it is a sphere kept

[@facthub9964]

Maine socha friction to cake walk but Aaj sir ne aanke khol di ❤❤❤

[@JEE322]

I directly imagined the trajectory without thinking of lifting my pen

[@HemantKumarpsu]

Ashish Arora, got lessons from him in Bansal Kota ages back. Good to see fit & fine

[@hindrajmalirollno3783]

path 3 block should follow because mw^r now acting inward direction the componet friction opposite but equal magnitude shown in diagram so certainly block can follow path 3 even if considering corollie forces or not

[@letsevolve9441]

Happy Guru Purnima Sirji..!! Inspite of being so fundamental why we weren't able to think it..? Thank u for providing such thought provoking content..!!

[@ansumand971]

Since the friction force isn't sufficient to keep the circular motion of the object so it should be accelerated in the opposite direction of force of friction, which is option 4

[@shihasnarvan]

Sir as per my analysis 4th option must be correct because here we are analyzing 2 motions at a time,one is of block along line of friction(let's class that ab),we are assuming that pseudo force is just greater than friction,therefore along ab it ,must have some acceleration, now as it moves away from center of disc,the r of a point,on which it has now reached along ab will increase,since omega is same the velocity of that point must be greater than the previous point of the block,therefore it must move a greater distance along direction perpendicular to ab,hence in the subsequent motion the path would be similar to option 4

[@shihasnarvan]

My assumptions are:- Omega is constant after motion along ab starts

[@ayu5h_Kumar]

I originally marked the answer "False" and thought what you just explained. But now, I'm confused about one possibility- If the disc originally has Omega=0 and then it starts rotating, i.e. we increase its angular speed, at the beginning of the motion, omega =0, but omega is increasing, so alpha = something. Since Omega=0, there is no friction component acting on the block along the radial direction, but since Alpha= something, there will be a tangential acceleration which must be provided by the frictional force. So, in this case, if Alpha is large enough so that the friction equals (mu)mg, the block will start sliding in the direction opposite to the direction of friction. Concludingly, we can say that friction is acting only in the tangential direction, so the block should slide back tangentially. s.n.: All of this is the description of just the instant after the commencement of the rotation of disk. Please Enlighten.

[@Anurag_jaiswaL]

i think the moment disk gets omega greater than 0 friction will start acting in radial direction and it will effect the motion of block.

[@ayu5h_Kumar]

@@Anurag_jaiswaL the question doesn't mention about the whole of motion. it just asks whther the block may slide along the tangent, no matter how long the time may be. it DOES slide tangentially if this case maybe considered. u r aabsoluuuutely correct- friction will start acting radially. but what i'm tryna highlight is JUST the instant.

[@pratyushprakash6998]

Sir, my analysis says that the block moves in path 2 wrt ground because as sliding starts the circular path in which block moves starts to get larger and wrt ground block has some max. acceleration although the point just beneath the block will have more velocity and acceleration so wrt disc block will seem to go behind of that point but wrt ground the block has a acceleration in tangential direction so it will tend to move forward with increasing circular radius and thus it will move in path 2.

[@ganeshkartikeyart6496]

He opened the eyes for early rise , of mine who never rise early just for study 😅. Thank you sir for your astonishing guide. Happy guru Purnima 🙏

[@_Future_Billionaire_]

Sir I want to add on in last case that when all qualities are given then i.e say r=5m Then in which quadrant block will leave the disc

[@ArchishmanNaskar-m1g]

Sir, I am very thankfull for your this kind of content. Please continue eye opener series like this vedio and wake up series.

[@raghavkarnatak9742]

thank you sir great thing to learn 🙂🙂

[@arpityadav6611]

I think ki option no 2 should be correct because in ground frame as the body start moving then firstly it will try to move in tangential direction and friction will oppose it but as the body move forward bofy will opse thrle dics in opposite direction of motion so body experience a foce in direction of circular disc so it will follow path 2

[@riyagovindrao1351]

If it start sliding then in ground frame, path of body will be in 2nd direction..

[@navkhr352]

4 might be the correct ans if rotation is clockwise and wrt observer on ground.......tried the same experiment at home

[@sahilpatil3428]

No sir, at first glance, I didn't thought that way Hw que : answer should be option 4.

[@kunalkale9346]

Happy Guru Purnima sir 🙏🏼❤️😌

[@BhargavShukla-yp2bh]

Sir path 3 because v disk vector will have greater magnitude than than v wrt disk and v wrt ground is vector sum of v wrt disk and velocity of disk

[@bhishamsaini3317]

sir please start question pratice for 2024 focusing mains and advanced

[@Anp137]

Sir ye to simple hcv ke questions ka thoda modification tha jesa mereko laga Ham agar disc ke frame mein chale jae to humko block kisi angle pe fisalta hua lagega normally nahi to grnf frame mein sidha khisakta hua lagega 9:18 sir mereko lagta hai ki 3 path mein jaega feel krke bola sir shi he kya 😅?

[@nitwalaatul]

I think path will be 2 , as when block is in disc it has both centripetal force and tangential force and when it will leave its centripetal acc and tangential acc will be zero but it have some velocity tangential and centripetal by this because of centripetal velocity it take curve and because of tangential velocity it move forwards. This is my thoughts. If wrong please make me help in correction.

[@studybits85]

I think it should be moving in path 2 bcz..the component of friction in horizontal is increasing as it depends on omega which is increasing..and the tangential acceleration is constant throughout ..which implies it will have constant velocity in vertical direction but increasing in horizontal dirn...so it should be following path 2..

[@ShivamGupta-or8xt]

at last, we have vel. of coin rel. to disc in ↙ direction, now if apply {vc/g=Vc/d+vd} and we know vd is ⬆ and add vectorially like (↙⬆) then vc would something lie like ↖,i.e option 2 & 3 came into consideration but as net fric. is more shifted to right side, so resultant is more close to mw^2r side, so it should follow 2nd path. Also I am not able to visulaize the ques at first, thank u sir for this brain teaser @Physics Galaxy Sir pls check my solution & pin it, if it is correct..it will also help others to imagine, if it is incorrect, pls corrct me

[@HobinderSinghA]

Its gonna be 2, I found by imitating the situation. Take your right hand put it on your right cheek such your fingers are pointing parallel to your nose (just keep them in plane of line of symmetry of your face). Twist your torso right hand side while extending them fully forward simultaneously. The path traced by your hand is the path of the object[ approximately of course], that is 2nd ;)

[@bruh_its_me]

Sir happy guru purnima

[@khushijindal10thatanmay18]

Happy Guru Purnima to such a fantastic teacher My revision notes of the guided checklist for chapters like electrostatics, current electricity gravitation ,etc comes in one sheet both sides . Sir please tell me are this much short notes ok or they have to be more short ??

[@LEGENDARY-P.M]

both 2 and 3 could be correct but by more specific terms 2 is more appropriate,,,, bas man lo ki tum yek chtia ho block ke upar baith jao tab aur ground pe khade ho kar ke bhi ,,,, SOCHO !!!!!! OMEGA ka direction zarrur dekhna .

[@AyushKumar-gx9tn]

Sir according to me it may slide in southwest but in some parabolic path

[@niu.06]

9:33 I tried to determine this practically using a utensil and a small toothpick box. The result seems to be 2nd situation. Now I can try explaining, due to the movement along the circular path the, sliding tendency is always away from the circular path, as the block moves away from the path, the radius r increases, increasing the accelerations on the body and it slides faster leading to a curved in throw out. (I am not sure this is a complete explanation, someone help me out with what I am faulty at)

[@somamaity2761]

Yeah I tried too and also getting path 2 but I am not able to explain/understand with fair degree of clarity

[@raghavdolyar4984]

I think 2nd path is right because particle will have sime tangential velocity and also the centripetal will balance friction after some time.

[@karann000]

HW Question - 2 will be the Answer.. 10:17

[@praveenpandey3589]

video dekhne se pahle mai yahi soach raha tha ki tangentialy hi hoga . thank you sir for clearing this concept

[@AKM-xl9eb]

I think the answer should be no.2 as we are observing from non inertial frame block will have a pseudo force in the direction of 2nd force

[@surendragodara1530]

No sir I was not think `may` situation like this and I think path 3 is correct because tangential velocity is wr and radially acceleration ke karan radially velocity are component of path 3

[@surendragodara1530]

If I am wrong please correct me

[@codedaily365]

5:24 sir @t (tangential acceleration) ke opposite friction hoga na? Why in the same direction?

[@SaurabhGupta08_]

Path of sliding is along 4.

[@jayesh210]

गुरुपौर्णिमा के अवसर पर आपको प्रणाम 🙏🙏

[@mitulrastogi1710]

Thank you sir for such a nice eye opener concept

[@ritojyotidebnath2023]

What a mind-blowing concept sir.

[@YashPandey-JEE2028]

Sir because definition of fric. tells us that fric. can only act opposite to vel. here fric is non tangential to path so velocity wouldn't be also.

[@vivekkumartripathi9990]

Please bless me sir for jee main 2024 !! 🎉🎉 HAPPY GURU PURNIMA🎉

[@gopalgajrani93]

Thank you sir for clearing the. Concept ❤❤

[@learnology9649]

How can we show that centrifugal force is a pseudo force?

[@theiigotriangularround4880]

Fs =mv2r ,fs-mv2r=0

[@Mandar225]

pathv4 is the answer because in ground frame it will also move in the tangential direction

[@mohammadarbazkhilji3939]

according to me path 2 will be more appropriate as in the ground frame we will also consider the velocity-vector due to the angular velocity of the disk

[@HamzaAli-df6fu]

Path3 acc to me Cuz at thodi der ke liye lagega Fir hat jayega To friction dhire dhire pura center ki taraf aa jayega To thoda curve aayega path me

[@mohammadarbazkhilji3939]

@@HamzaAli-df6fu please explain ki friction ka vertical component thodi der baad kyon khtm ho jayega?

[@studywithme1223]

Path 2 is correct

[@abhisheksonparote3900]

Sir I have framed a que so where I will ask that with diagram

[@tanmaykatiyar2821]

very good and conceptual video but ye sliding tendency wala samjh nahi aaya? 🤔vo bahar ki taraf centripetal force lagta hai aur ander ki taraf centrifugal. tangential force bhi samajh gaya but vo 2 aur 3rd path karte hue jata hai ye kaise hua?

[@vishnunavpute9747]

Nice elaboration sirji❤

[@varun27484]

Ans of 9:40 path-2

[@GallingEssay]

The Friction can be tangentially if the initial velocity of block wrt ground is 0, right? Because at that instant tendency is tangential and later after some time centripetal component will also come?

[@ranjanofficial3125]

Nice concept ... Love You sir ❤

[@BTSLover4069]

Happy Guru Purnima Guru Ji ....love from Bihar 💖🙏🙏🙏

[@bhargavsai8014]

In my view the answer should be path 2

[@ligasoofilms8033]

FROM THE NON INERTIAL FRANE OF REFERENCE OF THE DISC THE DISC HAVING BOTH TANGENTIAL AS WELL AS CENTRIFUGAL ACCELERATION, THE BLOCK EXPERIENCES A SLIDING TENDENCY RADIALLY OUTWARD AS WELL AS IN TANGENT DIRECTION , THUS NET SLIDING TENDENCY ARISES AT THAY SOME ANGLE AND NET FRICTION ACTS JUST OPPOSITE TO IT.

[@SSRXUSR]

HAPPY GURE PURNIMA SIR YOU ARE GOD OF PHYSICS GOAT OF IITJEE ONE DAY I WILL MAKE YOU FEEL PROUD AND YOU WILL THIS STUDENT IS MY ARJUN