Live Classes, Video Lectures, Test Series, Lecturewise notes, topicwise DPP, dynamic Exercise and much more on Physicswallah App. Download the App from Google Playstore ( bit.ly/2SHIPW6 ) Physicswallah Instagram Handle : instagram.com/physicswallah/ Physicswallah Facebook Page: facebook.com/physicswallah Physicswallah Twitter Account : twitter.com/PhysicswallahAP?s=20 Physicswallah App on Google Play Store : bit.ly/2SHIPW6 Physicswallah Website: physicswallahalakhpandey.com/ LAKSHYA Batch(2020-21) Join the Batch on Physicswallah App bit.ly/2SHIPW6 Registration Open!!!!
What will you get in the Lakshya Batch? 1) Complete Class 12th + JEE Mains/ NEET syllabus - Targeting 95% in Board Exams and Selection in JEE MAINS / NEET with a Strong Score under Direct Guidance of Alakh Pandey. 2)Live Classes and recorded Video Lectures (New, different from those on TH-cam) 3)PDF Notes of each class. 4)DPP: Daily Practice Problems with each class having 10 questions based on the class of JEE Mains/NEET level. 5)Syllabus Completion by end of January, 2021 with topicwise discussion of Last 10 Years Problems in Boards, JEE Mains/NEET within Lecture. 6)The Complete Course (Video Lectures, PDF Notes, any other Study Material) will be accessible to all the students untill JEE Mains & NEET 2021 (nearly May 2021) 7)In case you missed a live class, you can see its recording. 8)You can view the videos any number of times. 9)Each chapter will be discussed in detail with all concepts and numericals 10)Chapterwise Approach towards JEE Mains/ NEET & Board Exams. ****Test Series for XI & XII**** We provide you the best test series for Class XI,XII, JEE, NEET chapterwise, which will be scheduled for whole year. The test series follows very logical sequence of Basic to Advance questions.& Evaluation of Test and Solution to all the questions at the end of the test. Class 11 Physics Chapter 03 : KINEMATICS : Motion in a Straight Line 01: Introduction || Average Speed th-cam.com/video/XIJAZM5G5Fg/w-d-xo.html Class 11 Chapter 3 Kinematics: Differentiation || Calculus part 01 || Mathematical Tool th-cam.com/video/UiVYFFRs_BE/w-d-xo.html Class 11 Physics Chapter 03 : KINEMATICS :Motion in a Straight Line 02 || Instantaneous Velocity || || IIT/ NEET th-cam.com/video/9nBWFM7XTEw/w-d-xo.html Class 11 Chap 3: KINEMATICS || INTEGRATION || ||Calculus Part 02 || Mathematical Tools || th-cam.com/video/_dmHLxqbXYU/w-d-xo.html Class 11 Physics Chapter 03 : KINEMATICS : Motion in a Straight Line 03 || Answer Batao Salute Pao #Physicswallah || th-cam.com/video/-PKUJCIJSfc/w-d-xo.html Class 11 Physics Chapter 03 : KINEMATICS : Motion in a Straight Line 04 ||Derivation Of Equations Of Motion Using Integration|| th-cam.com/video/AxhkozLNDO4/w-d-xo.html Difference Between Diaplacement and Distance || IIT JEE MAINS AND ADVANCE CONCEPT || th-cam.com/video/4IkIGhuhuxo/w-d-xo.html 11 chap 03 : Kinematics 05 | Displacement time Graph -Velocity time Graph - Acceleration time Graph th-cam.com/video/4OgPa7MKjRA/w-d-xo.html NEET Best Questions 01 || Motion in a Sraight Line 1 || KINEMATICS NEET || Motion in One Dimension th-cam.com/video/oq-TjVH-sXc/w-d-xo.html 11 Chap 03 :Kinematics 06 || Motion Under Gravity || Motion in a Straight Line || Class 11 / JEE || th-cam.com/video/G0ooGZA_gyc/w-d-xo.html Relative Velocity || Kinematics|| Motion in a Straight Line 08 || Class 11 Chapter 4 || JEE MAINS th-cam.com/video/3sLGCWGm610/w-d-xo.html Best Method For Rain Man Problems | Relative Velocity | Motion in a Plane | Kinematics JEE NEET th-cam.com/video/tEltpY8vBPM/w-d-xo.html River Boat Problem || Relative Velocity in 2D || River Man Problem || Motion in a Plane || JEE NEET th-cam.com/video/2QFRRL50-ds/w-d-xo.html Projectile Motion 01 || Class 11 chap 4 || Motion in a Plane|| Motion in 2-D || th-cam.com/video/iUi1M7YkDe4/w-d-xo.html Projectile Motion 02 || Class 11 chap 4 || Motion in a Plane || Projectile from a Height || th-cam.com/video/dEqKhmTTXJk/w-d-xo.html Projectile Motion 03|| Equation Of Trajectory || Derivation Of Equation Of trajectory|| Range Form th-cam.com/video/js_G3G02AkU/w-d-xo.html Mock Test 01 | Kinematics Best Questions for IIT JEE NEET | IIT JEE Questions Kinematics th-cam.com/video/jYLGziSgI-Y/w-d-xo.html 11 Chap 4 || Circular Motion 01 || Angular Velocity and Angular Displacement || IIT JEE /NEET th-cam.com/video/JxttUGnNuCU/w-d-xo.html 11 Chap 4 | Circular Motion 02 | Centripetal and Tangential Acceleration | Angular Acceleration | th-cam.com/video/pOvYwchAvCA/w-d-xo.html Circular Motion 03| Centripetal and Centrifugal Force IIT JEE/ NEET | Conical Pendulum |Death Well | th-cam.com/video/Wt9_bvjCFf0/w-d-xo.html 11 chap 4 | Circular Motion 04 | Derivation of Centripetal Acceleration or Centripetal Force | th-cam.com/video/QYbyobdXlrg/w-d-xo.html 11 chap 4 | Circular Motion 05 | Banking Of Road IIT JEE NEET | Banking of Road with Friction | th-cam.com/video/C0uVC700EIA/w-d-xo.html 11 chap 4 || Circular Motion 06 || Motion in a Vertical Circle IIT JEE / NEET || Critical Velocity th-cam.com/video/e0Tty0V1q2U/w-d-xo.html 11 chap 4 || Circular Motion 07 || Motion in a Vertical Circle On a Bowl || IIT JEE MAINS / NEET th-cam.com/video/_WMTtzIZQR8/w-d-xo.html
Answer of last question: Given, a = x² We know that a = dv/dt a = dv/dx × dx/dt a = v dv/dx v dv = a dx v dv = x² dx Integrating the above equation; v²/2 = x³/3 + C ... (1) Given when x = 0, v = 0 Putting in Eq. (1), 0 = 0 + C C = 0 Putting in Eq. (1), v²/2 = x³/3 At x = 2 m, v²/2 = 8/3 v² = 16/3 v = ± 4/√3
Who want Alakh sir as own school teacher ?? 😌😌😌💓💓💓 Edit: it's amazing to see that, students are still watching the old lectures of this "dariwale inshaan 🌝'❤️.... while he has launched so many new batches....this is the power of marker nd white board 😎😎
Sir I joined a coaching recently (fees-3.5 lacs) and the teacher is taking online classes from last 1 month and he is teaching physics too complex to understand . I cried too much as I was not able to understand the concepts then I preferred your video to clear my concepts ...THANK YOU SIR ...you are the GOD of students ...😇
After reading the comments from many of our seniors.. I realised that I have chosen a right platform for my neet exam prep. Million tons of thanks to you sir !!
Till date no one can explain better than this ...lakhs of coaching are nothing in front of these short videos...anyone preparing for JEE NEET has to just watch this and practice and excell...best ever teacher in century..in future it is going to be huge huge wonder to students...
ALAKH SIR>>>>>>>>>> ITS BEEN 4 AND HALF YEARS SINCE THIS VIDEO AND HE HAS LAUNCHED HIS OWN APP AND OPENED INSTITUTES ALL OVER INDIA ....BUT STILL NO ONE CAN BEAT HIS METHOD OF TEACHING AND EXPLAINING SO WE GET THE FEEL OF THE SUBJECT THANK YOU SO MUCH SIR
Answer for the last question: We know that a=dv/dt (a=dv/dt ×ds/ds ) a=dv/ds × ds/dt a=(dv/ds)v {where ds/dt=v} a= vdv/ds Given a=x² We know that a=vdv/ds Integrating on the both the sides vdv/dx=x²dx v²/2=x³/3+c Given that v=0; x=0 On solving the equation (v²/2=x³/3+c) We get c=0 v²/2=x³/3 v²/2=2³/3 v²=2⁴/3 v=root of(2⁴/3)=-+4/root3 I know I am too late but ...still😃
@@shephalipatra4356 anyone who knew his channel before he was big would get the feeling. I was subscribed to him since 2017 and these type of videos gives strong nostalgia
You need to visit pradeep kshetrepal sir's channel 50years+ experienced teacher physics wallah is like a child on the front of the legendary pradeep kshetrapal
Given a=x^2 => (dv/dt)=x^2 now multiplying and dividing left side by dx => (dv/dt).(dx/dx)= x^2 Or can write (dv/dx).(dx/dt) = x^2 Since dx/dt=v v dv/dx = x^2 => v dv = x^2 dx Integrating both side (indefinite) => v^2/2 = x^3/3 +C By given condition at x=0, v=0 C=0 Now v^2/2 = x^3/3 At x=2m, => v^2/2 = 8/3 => v^2= 16/3 => v= +_4/√3 👍👍👍
Answer is 4/√3. Explanation - a = (dv/dx)(dx/dt) a = (v)dv/dx a dx = v dv Integrating both sides, we get, x^3/3 + c1 = v^2 /2 + c2 After applying the condition we will find that c1 = c2 So, x^3/3 = v^2/2 v^2 = 8 * 2 / 3 v^2 = 16/3 v = √16/√3 v = 4/√3 or v = 4√3/3. Thanks.
@Isha_ actually mera bhi pehle 8/3 hi aa raha tha ans but is ch ki playlist ke nxt video me sir ne iska solution karaya hai video ke end main....usme 4/√3 hi aa raha hai ans
Sir no words for you... Physics become a easiest subject for me ..on!y because of you... Take my respect sir.. Hats off to you.. What a lectures yrr... 😘😉😊😇🤗☺☺☺☺
Soo till 10th i had my fav teacher on TH-cam was bkp -bhai kii padhaai ...and in 11th it's physics wallah alakh pandey ... really great sirr..hats off to u 🙌🙌✨✨
The answer of solution would be 4underroot3/3 . a =dv/dt a =dv/ds.ds/dt ( in order to eliminate dt I have done this . a = v.dv/ds Now integrate this .(here a=x², ds =x ) Integration of x².dx = integration of v.dv It would be x³/3 = v²/2 +c Put the value when body initially at rest C=0 Now put x= 2 8/3 = v²/2 Underroot 16/3 = v²/2 Answer would be 4 underoot 3 would come . Thank you very much sir . I am feeling confident . 😊😊
sir i still watch these old videos of yours because i belive that knowledge is same and the dedication is still the same no matter taught on white board with that sasta caera on on smart board with studio. Hats off to you sir hope this comment reaches alakh sir and reply me
Bhaiya mjhe inse samjh ni ata tunning hi nii hoti .. As a junior apse puchna tha mai abhi hi 11th me aya hu konse teacher se pdhu kinematics 🙏 bta dijye
@@prajwalkulkarni4530 Bhaiya maine bhaiya maine kr liya physics galaxy se Aur mai shuru se hi mehnat kr rha full potential .. Results aaye n aaye uska nhi soch rha bs apna full potential de rha . Starting se.. Kia mera ye decision shi h .. (( Jee student - aakash me pdhta hu ))
@@taegukie yaa why not look its given that a= x^2 so if we will intergate this than v= x^3/ 3 now ita given x= 2 so putting value of x in velocity so 2^3/3 then v= 8/3 when x is 2
You're not supposed to disrespect the teacher who're actually spending their time and effort on teaching you. True, they might not be making as much efforts as some other teachers but that doesn't exclude the fact that they actually taught you alot of things which you wouldn't have known otherwise. Also, as for TH-cam teachers, They're also teaching students and have taught many students in past. So it isn't essentially right to label them as 'TH-cam teachers' while comparing them with yet another disturbing term 'Actual teachers'. It's best to not complain and get knowledge from wherever you're able to.
@@ginger0403 Yeah, students who study from TH-cam seem to overlook the fact that there are several teachers whom students are paying for in school and coaching and they're also putting in equal efforts. If students are not able to grasp what those teachers are teaching, they can discontinue their coaching and study solely from TH-cam. But disrespecting those teachers just because these students are not able to understand because of their efforts is indubitably deplorable!
@@ginger0403 Hardwork and good teaching are not so related, even if you work hard but your students dont understand anything then the hard work is in vain...
Last question answer is a=dv/dt dv/dt=x² dv/dx×dx/dt =x² dv/dx×V=x² Vdv=x²dx Integrating both sides V²/2=x³/3+C x=0(given) V=0(given) Applying in EQ.=>C=0 Therefore, V²/2=x³/3 If x=2m V²=(2³/3)×2 V=√(8×2/3) V=√(4×4/3) V=4/√3 m/s OR V= 4√3/3m/s This is the method😊
Preparing for neet 2024 Pr abhi tk 11th aur 12th physics clear nhi hua dekhyea aage kya hota hai mera! Abhi toh sir ke wjh se sb clear hota ja rha hai hope I'll crack neet this year or next ,🥺
Edit: it's amazing to see that, students are still watching the old lectures of this "dariwale inshaan 🌝'❤️.... while he has launched so many new batches....this is the power of marker nd white board 😎😎
i m 0 in phy at 9:47 i tried that myself ,and i did it without any help , for the first time i felt that much happiness !!! thankyou sir thankyou !! !! hope i crack neet !!
Sir u r the best teacher in the world your teaching style is very different for another teacher u r great sir in future we need this type of teacher for every students and in all educational system 👩🏫🤲🤲
The answer is a=x^2 V. (dV/dx) =x^2 Vdv=x^2dx Take Integration of both side V^/2=x^/3+c Put x=0, V=0 You get c=0 V=root over(x^3/3). 2 Put x=2 Ans:4/√3
@@aarusharya1544 where is time in the q? See, ({ is integration) a = x² v = {adx => v = x³/3 + C Now, it is given that, x = 0 when v = 0 => 0 = 0³/3 + C => C = 0 To find, v when x = 2, put 2 at x, v = 2³/3 m/s v = 8/3 m/s
a=x^2 vdv=x^2dx Integrating both sides with limits 0-v on left side and 0-2 on right side vdv becomes v^2/2 = x^2dx becomes x^3/3 Solving with limits it becomes v^2/2=8/3 v^2 = 16/3 (transposing 2 from left hand side to right hand side (8x2)/3) therefore, v=+- 4/root 3 m/s at x=2m
Given, a = x² We know that a = dv/dt a = dv/dx × dx/dt a = v dv/dx v dv = a dx v dv = x² dx Integrating the above equation; v²/2 = x³/3 + C ... (1) Given when x = 0, v = 0 Putting in Eq. (1), 0 = 0 + C C = 0 Putting in Eq. (1), v²/2 = x³/3 At x = 2 m, v²/2 = 8/3 v² = 16/3 v = ± 4/√3
Solution Step 1: Given that: Acceleration in the body; a=x^2 At x=0 , the velocity of the body(u) = 0 Step 2: Calculation of the velocity of the body: In differential form, acceleration (a ) is given as; a=dv/dt Thus, dv/dt=x^2...........(1) Now, the differential form of velocity is v=dx/dt Therefore, from equation (1), we have; dv/dx × dx/dt = x^2 dv/dx × v = x^2 vdv = x^2dx Now, integrating the above for position x=0 to x=2m and for velocity v=0 to v, we have; ∫vdv from v to 0 = ∫x^2dx from 2m to 0 [v^2/2] from v to 0 = [x^3/3] from 2 to 0 [v^2/2 − 0] = [2^3/3 − 0] v^2/2 = 8/3 v^2 = 16/3 v = 4/√3 Thus, The velocity of the body will be = 4m/√3s.
I have started watching your lectures last month....and really they were better than lectures of my institute ....now I don't have fear of physics anymore...
13:43 Sir , 👉Integration of 'a' with respect to 't' gives change in velocity i.e ( v - u ). because lower and upper limit of velocity are u and v respectively. 👉another prove - integration give area under curve. area under acceleration & time graph gives change in velocity 😘Who one agree with me , reply on this comment below 👇
after watching your video sir no one a student say that physics is hard or jee mains is hard you are a literally god of physics for those students who want to study but no able to complete their target because of their financial problem , thank you so so so much sir you are a great teacher and no on words are able to explain about your best teaching method and your greatness😍😍😄😄😇😇😇😇🤩🤩🤩
@@chandrasravanthi5123 a=x^2 Then we can write. a=dv/dt And we can also write a=dv/dt= dv/dx multiply by dx/dt(dx/dt=v) Then vdv/dx=x^2 then we can integrate this
Wrong pta hai kaise ??????? Mai bta ta hu Integration ke time humko x^2 ko totally integrate krna hoga . Na ki pehle hi x ki value put krna hai . Then aakhir me 8/3 +C
Physics Wallah sir the way u teach is mind blowing sir pls asse hi videos bnaye kyuku aap ke contents of kr book pdneki jrurat nhi hoti sirf apke notes pdta hu aur bs sir pls jee se related bhi bnao na mains aur advanced ke lia
sir, you're the best teacher ever!! never would have thought that physics can be this easy to understand. starting my class 11th journey with your videos is a blessing from god. thank you sir.
![แฉกลโกงงานวัด 2024 [ โกงมั้ยครับ ep.106 ] | DOM](http://i.ytimg.com/vi/taC5dpP3HXA/mqdefault.jpg)
Live Classes, Video Lectures, Test Series, Lecturewise notes, topicwise DPP, dynamic Exercise and much more on Physicswallah App.
Download the App from Google Playstore ( bit.ly/2SHIPW6 )
Physicswallah Instagram Handle : instagram.com/physicswallah/
Physicswallah Facebook Page: facebook.com/physicswallah
Physicswallah Twitter Account : twitter.com/PhysicswallahAP?s=20
Physicswallah App on Google Play Store : bit.ly/2SHIPW6
Physicswallah Website: physicswallahalakhpandey.com/
LAKSHYA Batch(2020-21)
Join the Batch on Physicswallah App bit.ly/2SHIPW6
Registration Open!!!!
What will you get in the Lakshya Batch?
1) Complete Class 12th + JEE Mains/ NEET syllabus - Targeting 95% in Board Exams and Selection in JEE MAINS / NEET with a Strong Score under Direct Guidance of Alakh Pandey.
2)Live Classes and recorded Video Lectures (New, different from those on TH-cam)
3)PDF Notes of each class.
4)DPP: Daily Practice Problems with each class having 10 questions based on the class of JEE Mains/NEET level.
5)Syllabus Completion by end of January, 2021 with topicwise discussion of Last 10 Years Problems in Boards, JEE Mains/NEET within Lecture.
6)The Complete Course (Video Lectures, PDF Notes, any other Study Material) will be accessible to all the students untill JEE Mains & NEET 2021 (nearly May 2021)
7)In case you missed a live class, you can see its recording.
8)You can view the videos any number of times.
9)Each chapter will be discussed in detail with all concepts and numericals
10)Chapterwise Approach towards JEE Mains/ NEET & Board Exams.
****Test Series for XI & XII****
We provide you the best test series for Class XI,XII, JEE, NEET chapterwise, which will be scheduled for whole year.
The test series follows very logical sequence of Basic to Advance questions.&
Evaluation of Test and Solution to all the questions at the end of the test.
Class 11 Physics Chapter 03 : KINEMATICS : Motion in a Straight Line 01: Introduction || Average Speed
th-cam.com/video/XIJAZM5G5Fg/w-d-xo.html
Class 11 Chapter 3 Kinematics: Differentiation || Calculus part 01 || Mathematical Tool
th-cam.com/video/UiVYFFRs_BE/w-d-xo.html
Class 11 Physics Chapter 03 : KINEMATICS :Motion in a Straight Line 02 || Instantaneous Velocity || || IIT/ NEET
th-cam.com/video/9nBWFM7XTEw/w-d-xo.html
Class 11 Chap 3: KINEMATICS || INTEGRATION || ||Calculus Part 02 || Mathematical Tools ||
th-cam.com/video/_dmHLxqbXYU/w-d-xo.html
Class 11 Physics Chapter 03 : KINEMATICS : Motion in a Straight Line 03 || Answer Batao Salute Pao #Physicswallah ||
th-cam.com/video/-PKUJCIJSfc/w-d-xo.html
Class 11 Physics Chapter 03 : KINEMATICS : Motion in a Straight Line 04 ||Derivation Of Equations Of Motion Using Integration||
th-cam.com/video/AxhkozLNDO4/w-d-xo.html
Difference Between Diaplacement and Distance || IIT JEE MAINS AND ADVANCE CONCEPT ||
th-cam.com/video/4IkIGhuhuxo/w-d-xo.html
11 chap 03 : Kinematics 05 | Displacement time Graph -Velocity time Graph - Acceleration time Graph
th-cam.com/video/4OgPa7MKjRA/w-d-xo.html
NEET Best Questions 01 || Motion in a Sraight Line 1 || KINEMATICS NEET || Motion in One Dimension
th-cam.com/video/oq-TjVH-sXc/w-d-xo.html
11 Chap 03 :Kinematics 06 || Motion Under Gravity || Motion in a Straight Line || Class 11 / JEE ||
th-cam.com/video/G0ooGZA_gyc/w-d-xo.html
Relative Velocity || Kinematics|| Motion in a Straight Line 08 || Class 11 Chapter 4 || JEE MAINS
th-cam.com/video/3sLGCWGm610/w-d-xo.html
Best Method For Rain Man Problems | Relative Velocity | Motion in a Plane | Kinematics JEE NEET
th-cam.com/video/tEltpY8vBPM/w-d-xo.html
River Boat Problem || Relative Velocity in 2D || River Man Problem || Motion in a Plane || JEE NEET
th-cam.com/video/2QFRRL50-ds/w-d-xo.html
Projectile Motion 01 || Class 11 chap 4 || Motion in a Plane|| Motion in 2-D ||
th-cam.com/video/iUi1M7YkDe4/w-d-xo.html
Projectile Motion 02 || Class 11 chap 4 || Motion in a Plane || Projectile from a Height ||
th-cam.com/video/dEqKhmTTXJk/w-d-xo.html
Projectile Motion 03|| Equation Of Trajectory || Derivation Of Equation Of trajectory|| Range Form
th-cam.com/video/js_G3G02AkU/w-d-xo.html
Mock Test 01 | Kinematics Best Questions for IIT JEE NEET | IIT JEE Questions Kinematics
th-cam.com/video/jYLGziSgI-Y/w-d-xo.html
11 Chap 4 || Circular Motion 01 || Angular Velocity and Angular Displacement || IIT JEE /NEET
th-cam.com/video/JxttUGnNuCU/w-d-xo.html
11 Chap 4 | Circular Motion 02 | Centripetal and Tangential Acceleration | Angular Acceleration |
th-cam.com/video/pOvYwchAvCA/w-d-xo.html
Circular Motion 03| Centripetal and Centrifugal Force IIT JEE/ NEET | Conical Pendulum |Death Well |
th-cam.com/video/Wt9_bvjCFf0/w-d-xo.html
11 chap 4 | Circular Motion 04 | Derivation of Centripetal Acceleration or Centripetal Force |
th-cam.com/video/QYbyobdXlrg/w-d-xo.html
11 chap 4 | Circular Motion 05 | Banking Of Road IIT JEE NEET | Banking of Road with Friction |
th-cam.com/video/C0uVC700EIA/w-d-xo.html
11 chap 4 || Circular Motion 06 || Motion in a Vertical Circle IIT JEE / NEET || Critical Velocity
th-cam.com/video/e0Tty0V1q2U/w-d-xo.html
11 chap 4 || Circular Motion 07 || Motion in a Vertical Circle On a Bowl || IIT JEE MAINS / NEET
th-cam.com/video/_WMTtzIZQR8/w-d-xo.html
u r my mentor sr
thank u so much sr.
v=4/root3 ...sir...is that answer right
Aap Sahi se donation bhi likhate h
Sir😂😂😂😂,ITNA halwa saval tha.maja nhi Aya..Mera ans Bina pen paper ke a gya
Answer of last question:
Given, a = x²
We know that a = dv/dt
a = dv/dx × dx/dt
a = v dv/dx
v dv = a dx
v dv = x² dx
Integrating the above equation;
v²/2 = x³/3 + C ... (1)
Given when x = 0, v = 0
Putting in Eq. (1),
0 = 0 + C
C = 0
Putting in Eq. (1),
v²/2 = x³/3
At x = 2 m,
v²/2 = 8/3
v² = 16/3
v = ± 4/√3
Tq
Thnx yrr...I do in the same way...👍👍😊😊
Nice bro👍👍👍 but I am not able to solve it 😩😩😩
Mene bhi krdiya yrr yr😝😝😝
Mahan
India actually need teachers like you sir !!!🙏🏻🙏🏻🙏🏻
Hey jangid bro
@@apiiiey o tu bhi jangid😃
Yes
Yes bro
what need he is teaching to india
Who want Alakh sir as own school teacher ?? 😌😌😌💓💓💓
Edit: it's amazing to see that, students are still watching the old lectures of this "dariwale inshaan 🌝'❤️.... while he has launched so many new batches....this is the power of marker nd white board 😎😎
Meee
i dont
Meee
Mee
S I wamt
17:00
C= -23/6
And Final ans v at 2s= 17/6 ms-¹
wont it be 11/6 ms?
Yes, I also got 17/6 answer.
Yes 17/6 m/s is a right answer
@@zo257 phele integration sikh
yep same
Sir I joined a coaching recently (fees-3.5 lacs) and the teacher is taking online classes from last 1 month and he is teaching physics too complex to understand . I cried too much as I was not able to understand the concepts then I preferred your video to clear my concepts ...THANK YOU SIR ...you are the GOD of students ...😇
Not a god of students but is GOD FOR STUDENTS🤗
Yar paise nhi waste kro bhut regret hota h bad me
3.5lakh fee😂😂😂😂
Ase konse Mahan guru ke pass ja Raha hai
Bhai great 👍👍👍👍😂😀😀😀
T.w left Kar bhai Accha rahe ga Maine bhi left Kiya paise wapas le
Jo wahan Diya tha(3.5 lakhs)uska 10% sir ke account me send karde
After reading the comments from many of our seniors.. I realised that I have chosen a right platform for my neet exam prep. Million tons of thanks to you sir !!
don't worry you will fail if you kept spending time in comments!
🙃
@@aryanantil4796 I see you are a hypocrite unless you really want to fail
@@harshadvikram7458 yes i am a hypocrite
@@aryanantil4796 tu bhi yhi kr rha h bhai...spending time in comments
@@aryanantil4796 Bro thinks just spending 3-5 hrs in a day doing nthing does anything ;-;
Till date no one can explain better than this ...lakhs of coaching are nothing in front of these short videos...anyone preparing for JEE NEET has to just watch this and practice and excell...best ever teacher in century..in future it is going to be huge huge wonder to students...
😑
It's true,
This man is single handedly saving my education
Exactly 💯
ur mom is double handedly giving me handjobs
Haa
From 50 thousand to 5
million,huge success!!
Hardwork Pays!
5 million 😁
@ANON THE DOGE yes
now near 7 million
Now a 1b$ company
@@devanshmadaan rupees not dollar 😐
Who are regularly watching physics wallah lectures nowadays ,btw THANK YOU SIR
Me
Mee
Me
Me
Hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm
ALAKH SIR>>>>>>>>>>
ITS BEEN 4 AND HALF YEARS SINCE THIS VIDEO AND HE HAS LAUNCHED HIS OWN APP AND OPENED INSTITUTES ALL OVER INDIA ....BUT STILL NO ONE CAN BEAT HIS METHOD OF TEACHING AND EXPLAINING SO WE GET THE FEEL OF THE SUBJECT
THANK YOU SO MUCH SIR
soo true!! 😃
Satish mishra sir se padha hai kabhi ?
My Class Teacher Also Good..
@@aryashree7 ha baut ghatiya padhate hai MR sir jyda accha padhate h
@@adarshsinghbaghel1640 toh mat aa padhne
I provide here solution 👇
a = x^2
dv/dt = x^2
(dv/dx).(dx/dt) = x^2
(dv/dx) . v= x^2
vdv = x^2 dx
Int ( vdv = x^2 dx)
v^2 /2 =x^3 /3 +c
at x=0 the particle at rest hence v=0
Hence c=0
v^2 /2 =x^3 /3
v^2 =x^3 ( 2/3)
At x= 2
v^2 = 8.2/3
v= √16/3
v =4/√3
😅
Answer for the last question:
We know that a=dv/dt
(a=dv/dt ×ds/ds )
a=dv/ds × ds/dt
a=(dv/ds)v {where ds/dt=v}
a= vdv/ds
Given a=x²
We know that a=vdv/ds
Integrating on the both the sides
vdv/dx=x²dx
v²/2=x³/3+c
Given that v=0; x=0
On solving the equation (v²/2=x³/3+c)
We get c=0
v²/2=x³/3
v²/2=2³/3
v²=2⁴/3
v=root of(2⁴/3)=-+4/root3
I know I am too late but ...still😃
Hey babes , what is the value of v ..( after put the value of t = 1 nd v = _2
Please answer 🙂🙂
V = 2.66 m/s 🙂🙂
@@Swastik_Debnath correct mera bhi same aya hai😂 8/3
Your answer is wrong
4️⃣➗🌱3️⃣
Actually he changed so much now... These videos give a better nostalgic feeling than his digital classes give...
Board and marker... ❤️
Yea bro
Whyy can u explain me plzzzz🙏🙏🙏
@@shephalipatra4356 bcoz that nostalgia
@@geeteshpaidi200 ohhh😭 it's true ,,,
@@shephalipatra4356 anyone who knew his channel before he was big would get the feeling. I was subscribed to him since 2017 and these type of videos gives strong nostalgia
The video is still trending in October 2019!!!!!
Who said
Even now also this video is trending....
In November also
February 2020
@@saswatisaha8155 today also means on 3 feb but I know it will soon become trending video in April or may of 2020 similarly 21,22,23......
Till now and will forever
from 50k to 10 million salute to you sir jii !!!!!!!!!!!!!!!!!!!!!!!!!!🧔🧔🧔🧔🧔🧔🧔🧔
17:41 The answer is 8/3 m/s. But sir we are not able to take your salute because you are the great legend
True answer is 4/√3
Yes
@@pritighosh851 I saw that it wrong 😔
@@bhartirajan9727 hm me vhi 😞
Bhai answer chahe galt aya ho ya aye lekin bhai try hi sab kuch dilati hai
Right answer
गुरुर्ब्रह्मा ग्रुरुर्विष्णुः गुरुर्देवो महेश्वरः । गुरुः साक्षात् परं ब्रह्म तस्मै श्री गुरवे नमः ॥
A big salute to you bro
@@ManuChauhan7 thanks bro it;s mean alought
In the era of trolling teachers by making fun of them this is so warm😃❤
0= answer
Please give the meaning
Sir i never know that physics teacher's are also like you...you are a masterpiece😇💞
i never knew ur mom was so hot bro wtf
Best
Physics
Teacher.
Who
Thinks
The
Same
👇
👇
👇
Like wala bhikari
Thanks
@@NoobGaming-hx9jq sahi khaa ab hum sab dislike karenge
You are right bro
Big dasssss______hai sala bikhari credit leta hai
Best teacher ever.. Thank you very much sir !.. Your level of teaching is simply awesome.. 🔥
Did your goal completed?
I really don't fear physics anymore..
You are the best sir
Yaaa
oi
Uff course
Ya I really don't fear physics because I am the student of Alakh sir
You need to visit pradeep kshetrepal sir's channel 50years+ experienced teacher physics wallah is like a child on the front of the legendary pradeep kshetrapal
Given a=x^2
=> (dv/dt)=x^2
now multiplying and dividing left side by dx
=> (dv/dt).(dx/dx)= x^2
Or can write (dv/dx).(dx/dt) = x^2
Since dx/dt=v
v dv/dx = x^2
=> v dv = x^2 dx
Integrating both side (indefinite)
=> v^2/2 = x^3/3 +C
By given condition at x=0, v=0
C=0
Now v^2/2 = x^3/3
At x=2m, => v^2/2 = 8/3
=> v^2= 16/3
=> v= +_4/√3
👍👍👍
😊 good 👍🏼 brother 😀
Thanks
1 marks minus hoga si unit nhi daali last main m/s
@@anchalshrivastav5167 board ka paper thodi hai..itna to chalta hi hai
Answer is 4/√3.
Explanation -
a = (dv/dx)(dx/dt)
a = (v)dv/dx
a dx = v dv
Integrating both sides, we get,
x^3/3 + c1 = v^2 /2 + c2
After applying the condition we will find that c1 = c2
So,
x^3/3 = v^2/2
v^2 = 8 * 2 / 3
v^2 = 16/3
v = √16/√3
v = 4/√3
or
v = 4√3/3.
Thanks.
❤️❤️❤️ right
Thanks
Can you explain wtapp plz
Reply me plz
@@jakkalashobha6526 yes
17:44 --- V =2.66 m/s
X displacement h velocity nhi h bhai
@@hackerdoctor5041 Tera bhi Galt hai vha a function of x de raha hai t nhi hai
मेरा भी यही आया hai
Mera bhi yahi a Raha hai
Po
Sir aap mere liye TEACHER Nahi ek GOD hain.
Thanks for the videos...
Nice
👍
Sab mere hi sahare rho
(17:54) Answer is 8/3...Thankyou sir awsome lecture 🤘🏼🤘🏼
Bro u have to put unit (m/s)
no its, 4/√ 3
@@AkankshaGupta-pr6ovwrong its 8/3 m/s
you are absolutely correct!!8/3m/s
Yes you are correct
17:50 answer is 8/3 m/s. Sir you are the best teacher in the world
How ??? Explain
no it should be 4/√3 m/s
@Isha_ actually mera bhi pehle 8/3 hi aa raha tha ans but is ch ki playlist ke nxt video me sir ne iska solution karaya hai video ke end main....usme 4/√3 hi aa raha hai ans
@Isha_ maine bhi same kiya h
pr sabka ans kuch or kyu h 💀😶
@Isha_ arey hn!! sir ne agli video k last m bta diya tha🕺
2:13, my father's reaction after My result.
😂🤣
🤣🤣🤣
😂🤣🤣
Aree bhai bhai bhai🤣🤣🤣🤣🤣🤣
Lol
India deserves teachers like you sir❤️🔥❤️
From 50k to a unicorn company amazing journey a baig big salute to yo respected sir thanks for teaching us so grea❤❤😊
Sir no words for you...
Physics become a easiest subject for me ..on!y because of you...
Take my respect sir..
Hats off to you..
What a lectures yrr...
😘😉😊😇🤗☺☺☺☺
Thanks for ur dedication sir
Ans)8/3 m/s
Right answer bro
Right 💯💯
Galat answer bro
Right 👍
@@Anuradhakumari97927 galat hai yr mera bhi same aaya tha lekin galat hai
anyone is learning now ?
Me😂
Me tooo@@RishikPatro-ou7qr
Me😂
Yesss!
Yesss😅
Homework answer
8/3 ms^-1
Thank you sir ❤❤❤
Correct, I did it in the same way..😊
Thats actually incorrect
@@curiousity003it is incorrect bro
@@aperson4543 yes, I watched lec 4 then I realised that my answer is incorrect.
Hu wants and like to join alakh sir offline classes imagine if sir open their offline classes and five million student come to attend class 😉🤔🤔
Haha 😄🤣
Opened
Maths wallah 🤣😂😂
Khul gyi bro
Hu or who😂🤣
Soo till 10th i had my fav teacher on TH-cam was bkp -bhai kii padhaai ...and in 11th it's physics wallah alakh pandey ... really great sirr..hats off to u 🙌🙌✨✨
Same here
Yehh same
Same to you 😀😀
Woah
Same thing is applicable wid meh 😃
Omg!! Same🤩🥰😍❤
ANSWER OF QUESTION:- V=3m/s or V=2.66m/s.
Mere bhi yahi aara h..
@@himanshupokhriyal6883ooh bhai...tum bhi meri raftaar se chal rahe ho😅😅😅
Han same aa ra. Shi h kya?
Mera bhi yahi aaya bhai
6 yrs and still counting✨…
How much more years this lecture will be as useful and as powerful as 6years ago and now ❤
Finally we get Jeetu bhaiya!!!
Jeetu bhaiya baad me aye h sir 4 saal se padha rhe h yt p
Yes bro
Sir ap mere liye god of physics and god of chemistry dono ho, thanks for fabulous teaching 🙏🙏🙏🙏🙏🙏
Same here
PHYSICS, CHEMISTRY,AND BIOLOGY ARE EASY WHEN ALLAKH PANDEY SIR...MRITUNJAY SIR AND RITU RATTEWAL MAM...IS TEACHER...respectively...
Hiii
Neha mam for maths ❤️❤️
@@fanofabd7613. Meeru ante unnollu bro
@@corzoncorzon7011 ??? Neha mam classes free eyy bro yt lo vuntai
@@fanofabd7613 meme bro nee dp choosi ala anna🤣
The answer of solution would be 4underroot3/3 .
a =dv/dt
a =dv/ds.ds/dt ( in order to eliminate dt I have done this .
a = v.dv/ds
Now integrate this .(here a=x², ds =x )
Integration of x².dx = integration of v.dv
It would be
x³/3 = v²/2 +c
Put the value when body initially at rest
C=0
Now put x= 2
8/3 = v²/2
Underroot 16/3 = v²/2
Answer would be 4 underoot 3 would come .
Thank you very much sir .
I am feeling confident .
😊😊
How tf xdx and vdv can be simultenously integrated?
@@KattarHinduShivendra YES both should be integrated {adt={dv
{=integration
Why ds=x??
17:06 8/3ms^-1
same
right
Yes bro
It will be 17/16 ms-¹ check ur answer
@@captainnk7633 isne last question ka answer btaya h
sir i still watch these old videos of yours because i belive that knowledge is same and the dedication is still the same no matter taught on white board with that sasta caera on on smart board with studio. Hats off to you sir hope this comment reaches alakh sir and reply me
Sir exactly 3:57 your face expression is osm i repeated almost 5 times 😂😂😂😂😂😂
Hi
Yeah same here😂😂😂😂
Hii
@@NITian_Sayantika yes 😊
Yes same here........
I just come to comment the same thing here and i get this
V=८adt
V=८x²dt
V=x³/3+c
According to question
X=0 and body is at rest
So, 0=0+c
C=0
v=x³/3+ c
v=2³/3+0
V=8/3+0
V=8/3
I solved it and I came to the same conclusion
@@dishakadam4054 but a is given x square , we cant integrate it using dt i guess
a = x^2
dv/dt = x^2
(dv/dx).(dx/dt) = x^2
(dv/dx) . v= x^2
vdv = x^2 dx
Int ( vdv = x^2 dx)
v^2 /2 =x^3 /3 +c
at x=0 the particle at rest hence v=0
Hence c=0
v^2 /2 =x^3 /3
v^2 =x^3 ( 2/3)
At x= 2
v^2 = 8.2/3
v= √16/3
v =4/√3
😅
@@nvm263 yes that's what i am thinking
Preparing for JEE 2023 and still going for his video's explanation because of his amazing teaching skills 🧡🧡👏
Same Brother Best of Luck
Same bro
Bhaiya mjhe inse samjh ni ata tunning hi nii hoti .. As a junior apse puchna tha mai abhi hi 11th me aya hu konse teacher se pdhu kinematics 🙏 bta dijye
@@prajwalkulkarni4530
Bhaiya maine bhaiya maine kr liya physics galaxy se
Aur mai shuru se hi mehnat kr rha full potential .. Results aaye n aaye uska nhi soch rha bs apna full potential de rha . Starting se.. Kia mera ye decision shi h .. (( Jee student - aakash me pdhta hu ))
Preparing for neet 2023 .. Doing the same... Inke alawa kisi s smjh hi nhi ata
I am following Sir since he had 50k subscribers and I was in 9th. Now it's 5.71 million And I am in 12th. Time flies so fast.
how was your preparation
9th m 11th ka kya padh rha tha bhai
And after 1 yr 9.13million
Me in class 11 with Alakh sir having 10.1M sub
@@IXADIBYANSHUJHA now 10.3 M 🎉
Sir you make whole physics an interesting subject
Bhai tum galatfaimi ke sikar ho gy ho
17:26 answer is 8/3 m/s
Same bro! Idk how come some ppl are saying 4 root3😕
@@pragattii_show can u explain pls?
@@taegukie yaa why not look its given that a= x^2 so if we will intergate this than v= x^3/ 3 now ita given x= 2 so putting value of x in velocity so 2^3/3 then v= 8/3 when x is 2
@@pragattii_s thank you so much
@@pragattii_s C kaha gaya
Why are TH-cam teachers always better than the actual ones ...
Why can't this kind of amazing ones actually be teaching us in person
You're not supposed to disrespect the teacher who're actually spending their time and effort on teaching you. True, they might not be making as much efforts as some other teachers but that doesn't exclude the fact that they actually taught you alot of things which you wouldn't have known otherwise. Also, as for TH-cam teachers, They're also teaching students and have taught many students in past. So it isn't essentially right to label them as 'TH-cam teachers' while comparing them with yet another disturbing term 'Actual teachers'. It's best to not complain and get knowledge from wherever you're able to.
@@ginger0403 Yeah, students who study from TH-cam seem to overlook the fact that there are several teachers whom students are paying for in school and coaching and they're also putting in equal efforts. If students are not able to grasp what those teachers are teaching, they can discontinue their coaching and study solely from TH-cam. But disrespecting those teachers just because these students are not able to understand because of their efforts is indubitably deplorable!
School teachers ko troll kar deti hu kool lagugi huehuehue
@@ginger0403 bangali.
@@ginger0403 Hardwork and good teaching are not so related, even if you work hard but your students dont understand anything then the hard work is in vain...
Last question answer is
a=dv/dt
dv/dt=x²
dv/dx×dx/dt =x²
dv/dx×V=x²
Vdv=x²dx
Integrating both sides
V²/2=x³/3+C
x=0(given)
V=0(given)
Applying in EQ.=>C=0
Therefore, V²/2=x³/3
If x=2m
V²=(2³/3)×2
V=√(8×2/3)
V=√(4×4/3)
V=4/√3 m/s
OR
V= 4√3/3m/s
This is the method😊
How can you solve this question 🙃
Mera to 8/3aa RHA h
Ha@@Divya123sharma
@@Divya123sharmamera bhi same aa rha hai
@@Divya123sharma mera bhi same aa rha hai
Preparing for neet 2024
Pr abhi tk 11th aur 12th physics clear nhi hua dekhyea aage kya hota hai mera! Abhi toh sir ke wjh se sb clear hota ja rha hai hope I'll crack neet this year or next ,🥺
Edit: it's amazing to see that, students are still watching the old lectures of this "dariwale inshaan 🌝'❤️.... while he has launched so many new batches....this is the power of marker nd white board 😎😎
i m 0 in phy at 9:47 i tried that myself ,and i did it without any help , for the first time i felt that much happiness !!! thankyou sir thankyou !! !! hope i crack neet !!
Bapu ji aap neet de rhe ??jethiya ko maalum h??aur tappu ke kya haal chaal??
@@vatsal512 😂😂😂
@@vatsal512 😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂
🤣🤣🤣🤣🤣🤣😂
At it why didn't we integrated dt into t as we did earlier
18 may 22
No unnecessary theory
Very nice understanding concepts
Class 11th 2024-25 attendance
😅
yeh video mene 10th ma dekhli thi T_T
Sir u r the best teacher in the world your teaching style is very different for another teacher u r great sir in future we need this type of teacher for every students and in all educational system 👩🏫🤲🤲
The answer is
a=x^2
V. (dV/dx) =x^2
Vdv=x^2dx
Take Integration of both side
V^/2=x^/3+c
Put x=0, V=0
You get c=0
V=root over(x^3/3). 2
Put x=2
Ans:4/√3
Yes this is correct
same here
Bhai ye V. (dV/dx) formula kaha se mila??
@@apurboghosal4875 a=dv/dt
a=dv/dt *dx/dx
a=dv/dx *dx/dt
as we know dx/dt = v
therefore , a=dv/dx * V
Bhai velocity root me nhi aaygi so v=2.7m/s
Answer = 8/3 m/s or 2.67m/s
x ko integrate kaise kiya wrt time
???
@@aarusharya1544 where is time in the q?
See, ({ is integration)
a = x²
v = {adx
=> v = x³/3 + C
Now, it is given that,
x = 0 when v = 0
=> 0 = 0³/3 + C
=> C = 0
To find, v when x = 2, put 2 at x,
v = 2³/3 m/s
v = 8/3 m/s
@@blaablaa-jx7hy are sorry...ye mai nhi mera bhai tha...abhi vo class 8th mei hai. To use pta nhi tha.
Aur mai nda crack kr chuka hu.
To sorry to my younger bro
Use mai smjha dunga.
Iska correct Ans +-4/underroot3 hoga
Abhi solve kiya maine.
17:33 8/3 m/s
maza a gaya
11/6
Ha ha didnt expect that introduction!
Best teacher in the world
Moreover, the way u teach is wonderful.. Easy to understand
U r so lucky; u got ❤️ from alakh sir 4 years ago!!!
@@Tanmayyy28 Shai bat hai bhai
a=x^2
vdv=x^2dx
Integrating both sides with limits 0-v on left side and 0-2 on right side
vdv becomes v^2/2 = x^2dx becomes x^3/3
Solving with limits
it becomes v^2/2=8/3
v^2 = 16/3 (transposing 2 from left hand side to right hand side (8x2)/3)
therefore, v=+- 4/root 3 m/s at x=2m
Is 8/3 m/s wrong?
Given, a = x²
We know that a = dv/dt
a = dv/dx × dx/dt
a = v dv/dx
v dv = a dx
v dv = x² dx
Integrating the above equation;
v²/2 = x³/3 + C ... (1)
Given when x = 0, v = 0
Putting in Eq. (1),
0 = 0 + C
C = 0
Putting in Eq. (1),
v²/2 = x³/3
At x = 2 m,
v²/2 = 8/3
v² = 16/3
v = ± 4/√3
How a= dv/dx*dx/dt.??😟
Thank you harshit.....
@@suhanisahoo5461 its Harshil
@@harshilpatel7171 sry sry it's by mistake 😅
@@suhanisahoo5461 Ok no problem 😂❤️
2:13 you had me😂😂😂😂😂
17:45 answer 4/root3
Hint: use a=vdv/dx and then integrate
Bro explain kardo jara plz
Me also
Bhai differentiate kyun kr rhe ho
17:21 answer is : 8/3=2.66 sir Bipul das
Acceleration ko velocity mei integrate Kiya uske baad c = 0
And x ki value 2 put kar dee
Yahi process Kiya na ??
@@GoodMan98rf haa
@@GoodMan98rf Limits are from 0-2 then why do we need c? It's a definite interaction
@@GoodMan98rf x meter me diya hai bhai .
Salute to the this teacher who does not hesitate to salute a student if he gets the right ans.. 🥲🔥🙏🙏🙏Physics wallah is an emotion.. 💕❤
Solution
Step 1: Given that:
Acceleration in the body;
a=x^2
At x=0 , the velocity of the body(u) = 0
Step 2: Calculation of the velocity of the body:
In differential form, acceleration (a ) is given as; a=dv/dt
Thus,
dv/dt=x^2...........(1)
Now, the differential form of velocity is
v=dx/dt
Therefore, from equation (1), we have;
dv/dx × dx/dt = x^2
dv/dx × v = x^2
vdv = x^2dx
Now, integrating the above for position x=0 to x=2m and for velocity v=0 to v, we have;
∫vdv from v to 0 = ∫x^2dx from 2m to 0
[v^2/2] from v to 0 = [x^3/3] from 2 to 0
[v^2/2 − 0] = [2^3/3 − 0]
v^2/2 = 8/3
v^2 = 16/3
v = 4/√3
Thus,
The velocity of the body will be = 4m/√3s.
🎉
finally someone answered
Is it necessary to integrate the dv in LHS ??
I have started watching your lectures last month....and really they were better than lectures of my institute ....now I don't have fear of physics anymore...
The answer of question at 15:00 is 17/6 {where C= (-)23/6}.
Concept crystal clear ^_^
C=-17/6 &fi al answer is 23/6m/s
@@mohitjangra8582 Nope c is -23/6 and final answer is 17/6
Correct
Dono galat
@khushi Agrawal but -23/6 kaise aa rha h can u explain
13:43
Sir ,
👉Integration of 'a' with respect to 't' gives change in velocity i.e ( v - u ). because lower and upper limit of velocity are u and v respectively.
👉another prove - integration give area under curve. area under acceleration & time graph gives change in velocity
😘Who one agree with me , reply on this comment below 👇
Mughe samgh he nhi aaya
@@OFF_GaMeR_MoHiT koi baat nahi...🤗
I totally agree
sir mein galti nikal rhe ho ????ya explain kar rhe ho
Smj nhi aaya but senior ne bola hai to agree
Yaha tak aagye aage bhi jaaoge ,
Physics wallah ke saath Dhoom machaoge .....
😁😁
3:35 to complete lecture
Yeh mere liye super teacher hai sir ko sat sat naman 🙏🙏🙏🙏🙏🙏🙏 agar meri physics acchi hai toh bas pura credit inkho jata hai ❤❤
3:56 to 3:59 expression dekho sir ke❤️❤️
Camera thoda turn hua bhi
2 saal baad with fear of coronavirus my answer is 4/√3 ...
Same here after completing board
@@XONTV286 Wish u live a safe life with ur family bro😎😎
@@devendershokeen3982 thanks bro u too
Kitna padhe bhai 12 ke baad??
@@KaushalYadav-cs3xw kya
after watching your video sir no one a student say that physics is hard or jee mains is hard you are a literally god of physics for those students who want to study but no able to complete their target because of their financial problem , thank you so so so much sir you are a great teacher and no on words are able to explain about your best teaching method and your greatness😍😍😄😄😇😇😇😇🤩🤩🤩
a=dv/dx ×dx/dt (dx/dt=v)
a=vdv/dx
vdv = adx
vdv=x²dx [puting value of 'a' from question (a=x²)]
*Now integrating both side*
v²/2=x³/3 (x=3)
v²=16/3
and hence[ v=4/√3m/sec ]
😅😅
NICE LECTURE SIR, i really enjoyed it
my ans of the last ques is 8/3 m per sec
same
yeah i also got this answer but is it the right answer someone confirm please
@@thissidesoniya no it is 4/root 3
@@joeldsilva3030 yeah I got it...thank you
@@thissidesoniya Yup
Your teaching method was awsmmm......finally I got a good platform of physics you made physics really interesting..... Thanx sir for your hard work.
Answer of last question is 4/√3 m/s. This is Very easy questionfor me sir because of you❤✨
ANS FOR LAST QUESTION IS 4✓3 M/S THANK YOU SIR!!!!
@@ksoh1oe276 pata hai 4√3 hai answer
how?
The last question is simple just use the formula
a= v.dv/dx
Then,
X^2=v.dv/dx
vl uu explain wth including steps ..?
@@chandrasravanthi5123 sir ne explain kiya hai ye question iske baad wala video dekho
@@chandrasravanthi5123 a=x^2
Then we can write. a=dv/dt
And we can also write a=dv/dt=
dv/dx multiply by dx/dt(dx/dt=v)
Then vdv/dx=x^2 then we can integrate this
4/√3 is the ans
a= vdv/dx
vdv/dx=x^2
vdv= x^2dx
N then integration...
Wrong ans completely wrong
Correct
Wrong hai
Last me ja kar thodi se mistake kr di
Wrong pta hai kaise
???????
Mai bta ta hu
Integration ke time humko
x^2 ko totally integrate krna hoga .
Na ki pehle hi x ki value put krna hai . Then aakhir me 8/3 +C
@@jvlogs6878 but in the next vdo sir declared 4/√3 as correct
Great teacher for all time
Sir your teaching skills are really great our hard concepts become easier thank you sir so much 👍🏻😊😊
Old is gold (because new is sold)😂😂
Sir your teaching style is A1
Even dull student like me can understand easily
Sir jee ki chapter wise questions ki alag series banadijiye
TECH INFO
Ji sirkar
Physics Wallah thanks sir
Physics Wallah
Physics Wallah sir the way u teach is mind blowing sir pls asse hi videos bnaye kyuku aap ke contents of kr book pdneki jrurat nhi hoti sirf apke notes pdta hu aur bs sir pls jee se related bhi bnao na mains aur advanced ke lia
8/3
sir, you're the best teacher ever!!
never would have thought that physics can be this easy to understand. starting my class 11th journey with your videos is a blessing from god. thank you sir.
Same me too
16:33:- C=-23/6 and V=2.83 😊
Wrong even after 6 years 😂😂
@@python9005 galat kya h
Sahi hai bro
17:52 Velocity is 8/3 m/s
riight
Right
As always sir explained it in 15 mins which I was struggling from 2 days 😂
Abey bhosadike agua toh answer likh do Gyan pelna hai akhiri ka answer nhi like paenge . Mujhe check karna hai
frfr
Thank you so much sir
Answer is 8/3 m
8/3m per second
Yes
Is it 8/3+c m
@@Vishal132tno answer is 4/√3 as v square= 8/3 so after taking under root ans is 4/√3
a=x²
Find v when x is 2m , given at x=0, body is at rest. 17:44
2.6m/s, am I right?
last question answer
integrate both sides and equate them
answer will be root 2^4/3 or 4/root 3
Correct. Same here.
:D
Ry8😀
This is the last step
Plz tell me what is the ans of Second last question
@@HarmandeepKaur- + - 4/√3