00:00 - Introduction 05:37 - Motion and rest 11:04 - Important terms 35:12 - Types of motion 45:57 - Equations of motion 1:46:28 - Motion under gravity 2:11:30 - Graph 2:15:53 - Slope 2:20:30 - Area under the curve 2:23:29 - Equations of motion (Graphical method) 2:36:07 - Graph analysis 3:06:30 - Derivation of equations of motion 4:03:51 -All the best all of you!
Shailendra sir, Kaas apke is tarah ke lecture 11th starting mein hote.... I missed it because on that time every youtube channels was teaching only for jee/neet content. None of one were found like you. 😊😊😊😊 Thanku so much sir. Love from Delhi NCR (Faridabad). ❤❤❤
@@urbestie_army yaa it enough but u need to focus on every line that sir write on the slides because he not repeat same thing again and again. Happy journey 😄😄
1:27:00 answer is 60km/h Sol- T1=30km /30kmh^-1 = 1hour T2=30/x Vavg=t.d.t/t.t.t 40=60km/total time taken Total time taken=40/60 =1.5hr Now, Let's fine the value of x T1+T2=total time taken 1hr+30/x=1.5 30/x=1.5-1 30/x=0.5 0.5x=30 x=60 answer
1:27:16 QUESTION: on a 60 km track a train travels the first 30km with a uniform speed of 30 km/h. How fast must the train travels next 30 km so as to average 40 km per hour for the entire trip? ANSWER: 1. Find the time taken for the first 30 km: * Time = Distance / Speed * Time = 30 km / 30 km/h = 1 hour 2. Find the total time for the entire 60 km journey: * Total time = Total distance / Average speed * Total time = 60 km / 40 km/h = 1.5 hours 3. Find the time remaining for the last 30 km: * Remaining time = Total time - Time taken for first 30 km * Remaining time = 1.5 hours - 1 hour = 0.5 hours 4. Find the speed needed for the last 30 km: * Speed = Distance / Time * Speed = 30 km / 0.5 hours = 60 km/h Therefore, the train must travel the next 30 km at a speed of 60 km/h to achieve an average speed of 40 km/h for the entire 60 km journey.
33:45 sir , acceleration negative hota ha to uska hamesa matlab retardation nhi hota ha , uska matlab hota ha opposite direction, as told by our KOP 😊❤
A__________O___________B 1st-take out time interval of A-O Then time interval of A-B Next, Subtract time interval of A-B from A-O Then, take out speed of O-B By putting distance(30km) and the time u just got from 3rd step. And it's done ✔
@@shruti893You didn't understand ? T1= 30/30 ho jayega aur T2=30/x 40(average Speed) = 60( Total Distance ) / 30/30 + 30/ x....T1 me same speed par same distance cover kiya hai toh 1 hojayega 😊
Sir there is correction To solve this problem, we'll break it down into two parts: the time it takes for the packet to reach the ground and the velocity at which it reaches the ground. 1. Time to reach the ground: We can use the equation of motion: s = ut + (1/2)gt^2 where: s = displacement (height) = 39.2 m u = initial velocity = 0 (since the packet is dropped) g = acceleration due to gravity = 9.8 m/s^2 t = time (which we want to find) Rearranging the equation to solve for t, we get: t^2 = 2s/g t^2 = 2(39.2)/9.8 t^2 = 8 t = √8 ≈ 2.83 seconds So, it takes approximately 2.83 seconds for the packet to reach the ground. 1. Velocity at which it reaches the ground: We can use the equation: v = u + gt where: v = final velocity (which we want to find) u = initial velocity = 0 g = acceleration due to gravity = 9.8 m/s^2 t = time ≈ 2.83 seconds (found earlier) v = 0 + 9.8(2.83) v ≈ 27.7 m/s So, the packet reaches the ground with a velocity of approximately 27.7 m/s. But you haven't clearly told why height was negative 😔😔
bro, when he says u=9.8 then why you take u=0 and if body moves in downward direction then h should be taken negative and sir is right and your answer is wrong. hope you understand👍
Sir I have purchased Arjuna Jee 2025 but jee preparation needs patience and I have it but school has no patience that's why for school preparation 😀 I'm here. Lectures are very helpful for us ❤❤❤❤ Thank you very much sir 😊😊😊
Before studying from him I was afraid so much of phy but he just made my day with his amazing lecture I am finding much difficulties in numerical only as it need more and more practise...
@@Nikhilthakur7466 haan Inka dekh lo aur bas ek topic sir ne Miss kra hai relative velocity aap ise kahi aur se padh lo aur inse pdhne ke baad mind map se pdh lena Lakshman sir ke
Sir jo aap homwork me question dete ho unke ans bta diya kro usse confusion nahi hogi hme 🤔 Only ans sir, solve ham khud ker lenge ✍️✍️ Who are agree with me 🙁
1:52:36 but in this case two types of acceleration due to gravity are there first is when the ball is thrown up that will be negative g, when ball reaches v=0 then it comes down there positive acceleration of g is applied then why did we only took negative acc for the whole case?
Sir Displacement is a vector quality and distance is scalar soo how we can say that Distance is greater than or equal to displacement I think this is wrong statement may be ye ho sakta hai Distance ≥ | Displacement | iske magnitude ke equal ya greater ho sakta hai
sir 1:52:50 wale question mein vertically upwards agar phenk rahe hai ball ko toh pehle ball upar jaegi then 0 velocity attain karne ke baad neeche ayegi toh uss hisaab se tower ki height ke saath voh ek extra distance bhi 2 baar cross kar rhi hai upar jaate waqt and neeche aate waqt
hello sir...sir apne gravity k first part me bola ki object neeche drop ho rhi hogi to positive g hoga...to fir baloon wale question me ut -1/2gt2 kyn???
Thanks a lot sir , you teach so well, i have a very weak base in physics and i don't understand physics at all bit today i understood it very well , i was revising this vhapter because tomorrow is my exam , Just praying i score well 🙏🏻🌷✨
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Very nice lecture ❤❤❤❤❤❤❤❤❤❤❤
Proper explanation and hardwork done by sir
2:36:02 Mamla ek dum chakachak hai😅
Sir please provide session pdf
Thankyou so much sir ❤
We want new One shot series for 2024-25 Batches 😢
Manzil series ❤
UDAY SERIES
Kon kon 2024. 2025 batch wala hai Like karo❤❤❤❤❤
Hi
Hey, jo ye sir padha rhe isme deleted syllabus nai haina ?
Nhi hai deleted topics@@I_ll_win
Me
2024-2025 batch attention pls😅
😢@@MaahiAadi-bc4pn
😮
🤔🤔🤔
uss question ka answer is 30km/h
❤❤❤❤❤
My search for the best physics teacher for boards ended here
Lots of love to Shailendra sir❤
Yes❤❤
@@Manshi98hii
Bilkul sahi kaha ❤❤
Same
Bro ya boards ka liya thk hai na
00:00 - Introduction
05:37 - Motion and rest
11:04 - Important terms
35:12 - Types of motion
45:57 - Equations of motion
1:46:28 - Motion under gravity
2:11:30 - Graph
2:15:53 - Slope
2:20:30 - Area under the curve
2:23:29 - Equations of motion (Graphical method)
2:36:07 - Graph analysis
3:06:30 - Derivation of equations of motion
4:03:51 -All the best all of you!
Thankuh bhai
Where is dif.and int. ???
Copy paste 💩💩💩
🙏🙏🙏🙏🙏
Thanks bro
Shailendra sir, Kaas apke is tarah ke lecture 11th starting mein hote.... I missed it because on that time every youtube channels was teaching only for jee/neet content. None of one were found like you. 😊😊😊😊 Thanku so much sir. Love from Delhi NCR (Faridabad). ❤❤❤
Bhai ncert ka level ka cover hojayega na?
@@I_eat_Childrens_hahahhaa Bhai hojaega😊
Hey guys i m not preparing for jee Or neet so for only grade 11 is it enough to score a pretty good marks?
@@urbestie_army yaa it enough but u need to focus on every line that sir write on the slides because he not repeat same thing again and again. Happy journey 😄😄
@@manish_coding ok do i need any reference book to score good or ncert + TH-cam teachers are enough to practice question
⏳Timestamps:
00:00 - Introduction
05:37 - Motion and rest
11:04 - Important terms
35:12 - Types of motion
45:57 - Equations of motion
1:46:28 - Motion under gravity
2:11:30 - Graph
2:15:53 - Slope
2:20:30 - Area under the curve
2:23:29 - Equations of motion (Graphical method)
2:36:07 - Graph analysis
3:06:30 - Derivation of equations of motion
4:03:51 - Thank You Bacchon!😊
Copy paste 💩
00:00 - Introduction
05:37 - Motion and rest
11:04 - Important terms
35:12 - Types of motion
45:57 - Equations of motion
1:46:28 - Motion under gravity
2:11:30 - Graph
2:15:53 - Slope
2:20:30 - Area under the curve
2:23:29 - Equations of motion (Graphical method)
2:36:07 - Graph analysis
3:06:30 - Derivation of equations of motion
4:03:51 - Thank You
sir's handwriting is just amazing
agree?👇
yes i also notice it
Like 👍 students watching this lecture in the academic session 2024-25
1:27:00 answer is 60km/h
Sol- T1=30km /30kmh^-1 = 1hour
T2=30/x
Vavg=t.d.t/t.t.t
40=60km/total time taken
Total time taken=40/60 =1.5hr
Now,
Let's fine the value of x
T1+T2=total time taken 1hr+30/x=1.5
30/x=1.5-1
30/x=0.5
0.5x=30
x=60 answer
Wrong the answer is 30km per hr
Yes write ans
Answer is right
My answer exactly same
Yes@@NOELSE
Anybody from 2024-2025 batch >>>>>>>>>>>>>>>>>
Here
Me
2:11:32 ----- 17.6 m÷s and -19.6
Same ans19.6
Same
Shouldn't it be -17.6 m/s as acc it - ?
Yes
@@NEWZ_EDITZ19.6
1:27 :01
Ans : t1= 30/30=1
t2=30/v2
Vavg=60(total distance)/1+(30/v2)=1200/20=60kmh^-1
1:27:16
Ans . The next 30 km is travelled with the speed of 60 km/h .
The teacher is here to change the full TH-cam history🤫🤫💥
भाइयों कभी हार मत मानना क्योंकि आपकी मां आपके जीत का आसाय लगा बैठी है ❤❤❤❤😢❤😢❤😢😢😢😢
❤❤❤❤❤❤❤❤❤❤❤ aap kinse school me ho
Kaha se ho sister , tumhari baatein ❤
🥹🥹🥹🥹🥹🥹🥹🥹🥹❤❤❤❤❤❤❤
Papa bhi
Thanks a lot sis❤
2:00:42 Maja aa gya sir ji bindas lecture hai 💜🙏🏻👌🏻
⏳Timestamps:
00:00 - Introduction
05:37 - Motion and rest
11:04 - Important terms
35:12 - Types of motion
45:57 - Equations of motion
1:46:28 - Motion under gravity
2:11:30 - Graph
2:15:53 - Slope
2:20:30 - Area under the curve
2:23:29 - Equations of motion (Graphical method)
2:36:07 - Graph analysis
3:06:30 - Derivation of equations of motion
4:03:51 - Thank You likeeeeee
Copy paste 💩
1:27:16 QUESTION: on a 60 km track a train travels the first 30km with a uniform speed of 30 km/h. How fast must the train travels next 30 km so as to average 40 km per hour for the entire trip?
ANSWER:
1. Find the time taken for the first 30 km:
* Time = Distance / Speed
* Time = 30 km / 30 km/h = 1 hour
2. Find the total time for the entire 60 km journey:
* Total time = Total distance / Average speed
* Total time = 60 km / 40 km/h = 1.5 hours
3. Find the time remaining for the last 30 km:
* Remaining time = Total time - Time taken for first 30 km
* Remaining time = 1.5 hours - 1 hour = 0.5 hours
4. Find the speed needed for the last 30 km:
* Speed = Distance / Time
* Speed = 30 km / 0.5 hours = 60 km/h
Therefore, the train must travel the next 30 km at a speed of 60 km/h to achieve an average speed of 40 km/h for the entire 60 km journey.
2:10:00. (I)= -17.6
--------------(ii)= 19.6
33:45 sir , acceleration negative hota ha to uska hamesa matlab retardation nhi hota ha , uska matlab hota ha opposite direction, as told by our KOP 😊❤
Sir relative baat kar rhe hai bhai
1:27:14 ANSWER =60km/h.. sir equation of motion agyi smj thanks for amazing lectures ❤❤❤❤ *KOP*KING OF PHYSICS ❤❤
How can you explain??
A__________O___________B
1st-take out time interval of A-O
Then time interval of A-B
Next, Subtract time interval of A-B from A-O
Then, take out speed of O-B
By putting distance(30km) and the time u just got from 3rd step.
And it's done ✔
@@Uzma-vic ok thank you
@@shruti893You didn't understand ? T1= 30/30 ho jayega aur T2=30/x
40(average Speed) = 60( Total Distance ) / 30/30 + 30/ x....T1 me same speed par same distance cover kiya hai toh 1 hojayega 😊
@@Reaper-px6lw ouu Okie got it thenku😃👍🏻
1:26:54 Yes sir sare equation aur questions samajh aa gya hai ❤️
Teaching style wow
Explanation wow ❤
Hair cut and sir you looking a wow
Explanation wow wow wow 😅
Sir there is correction
To solve this problem, we'll break it down into two parts: the time it takes for the packet to reach the ground and the velocity at which it reaches the ground.
1. Time to reach the ground:
We can use the equation of motion:
s = ut + (1/2)gt^2
where:
s = displacement (height) = 39.2 m
u = initial velocity = 0 (since the packet is dropped)
g = acceleration due to gravity = 9.8 m/s^2
t = time (which we want to find)
Rearranging the equation to solve for t, we get:
t^2 = 2s/g
t^2 = 2(39.2)/9.8
t^2 = 8
t = √8 ≈ 2.83 seconds
So, it takes approximately 2.83 seconds for the packet to reach the ground.
1. Velocity at which it reaches the ground:
We can use the equation:
v = u + gt
where:
v = final velocity (which we want to find)
u = initial velocity = 0
g = acceleration due to gravity = 9.8 m/s^2
t = time ≈ 2.83 seconds (found earlier)
v = 0 + 9.8(2.83)
v ≈ 27.7 m/s
So, the packet reaches the ground with a velocity of approximately 27.7 m/s.
But you haven't clearly told why height was negative 😔😔
bro, when he says u=9.8 then why you take u=0 and if body moves in downward direction then h should be taken negative and sir is right and your answer is wrong. hope you understand👍
@@arpitsrivastav1118 Brother I said there is correction ig
u=0 because object is dropped/released
@@reetasingh9672 I think you should search this question on google
@@arpitsrivastav1118 I've checked on Meta AI😅 dono hi way glt ni h but thnks
Hi
Homework question 1:27:20 answer is 60km/h
Kaise bata do
Is me kya find karna hai kaise
Who is watching on November 😊
1:26:56 x = 60
t1=30km/30km per h= 1 hour
t2=30/x
Vavg=40km per hour
40=60/1+30/x
solve for x we get x=60
Wrong hai bhai
Answers is correct but process is wrong
शिक्षा से बड़ा कोई ज्ञान नहीं और गुरु से मिले आशीर्वाद इससे बड़ा कोई सम्मान नहीं ❤❤❤❤😢😊😊😊 ______ love you sir ♥️🥰
Ha
You are absolutely right 🙏
Hello ji navya
Yes
Itna gyaan kyo de rahe ho
Sir, 2:09:57, mujhe displacement 4+19.6=23.6 mila, par doosro ko 15.6 mila? Sir, shouldn't g be +9.8m/s
Who is watching in September 😅
Me
Me 😅😅
@@Shekhawatakshu all the best 👍
@@Anas_khan_69 thnx✨
1:27:09 hour ka question ka ans 60km/h..
💯 done sir thanks for the amazing session JAI HIND SIR 🎉🎉🎉🎉❤❤❤❤❤❤❤❤❤
Are you a NDA ASPIRANT...,?
2:11:22
Velocity= (+17.6ms-1)
S=(+19.6m)
00:00 - Introduction
05:37 - Motion and rest
11:04 - Important terms
35:12 - Types of motion
45:57 - Equations of motion
1:46:28 - Motion under gravity
2:11:30 - Graph
2:15:53 - Slope
2:20:30 - Area under the curve
2:23:29 - Equations of motion (Graphical method)
2:36:07 - Graph analysis
3:06:30 - Derivation of equations of motion
4:03:51 - Thank You Bacchon!
Who is seeing this video on November
Me
Me
Me
Me
Me in December broh 😢😢
Sir I have purchased Arjuna Jee 2025 but jee preparation needs patience and I have it but school has no patience that's why for school preparation 😀 I'm here.
Lectures are very helpful for us ❤❤❤❤
Thank you very much sir 😊😊😊
same here bro...
@@KhushiSahani-hk4rq Ek thali ke chate bate hi hain :)
2:00:39 sir isme shuru me equation of motion ki 2nd equation bhi to lag sakti thi ?? Any help??
Haa laga skte hai
Kis kis ke 10 th me 80%se upar aaye hai
❤
93.2
Systummmmmmmm
97.8❤
94.5%😊
U were a very much needed teacher❤
2:11:45 sir first answer is 20.4m/s
2nd answer is 22.4m below from helicopter 🚁
Send total solution
Fastly
I think it's wrong
X=60 km track
X=30kmh speed
Train travel next.
60/40= 1.5 t.D.t/ t.t.t
1.5=we right (2).
30/2
=60 m/s speed Hp
Who is watching in October
27oct..🙌
31 oct
Today ☠️ for first time
@@somu_sempai matlab 11th barbaad ho jayega
Matt krna yarr
1:38:40 imp ques
1:49:53 motion under gravity questions start
1:27:18 solution 60km per hour
Before studying from him I was afraid so much of phy but he just made my day with his amazing lecture I am finding much difficulties in numerical only as it need more and more practise...
Bhai inke one shot dekh kr scl exams mei acche marks score kr skte hai kya?
@@Nikhilthakur7466 haan Inka dekh lo aur bas ek topic sir ne Miss kra hai relative velocity aap ise kahi aur se padh lo aur inse pdhne ke baad mind map se pdh lena Lakshman sir ke
@@Nikhilthakur7466 mind map kevl 50 min max rhega vo bhi 2× me krke dekhoge to kaam ban jaayega
@@atul-vh3zv hey can you provide me with the link.
@@STUDYNEZUKO-js2wr what kind of link??
Sir AAP jitna bhi time le lo 7-12 ghante but pura concept padha do questions k sath koi bhi topic skip mat Karo 🥰🥰🥰 love you sir
Hlo
In the end we all become stories ✨❤️🩹💐
Sir bhut aacha lecture tha love from uttarakhand
SIR I AM IN CLASS 11TH FROM 2024-25 BATCH.. AND I WANT TO PREPARE FOR SCHOOL LEVEL EXAMS.. IS THIS VIDEO ENOUGH FOR DETAILED EXPLANATION OF NCERT?
2:11:08 queston ans -17. 6m/s ans hight 19.6m
2:11:24
Helicopter ki bhi distance add krni hai usmai
19.6 kaise aa skta hai
Sir jo aap homwork me question dete ho unke ans bta diya kro usse confusion nahi hogi hme 🤔
Only ans sir, solve ham khud ker lenge ✍️✍️
Who are agree with me 🙁
H=19.6m
V=-17.6m/s
Same😊
Thanx bro
Thank you so much sir the lecture is awesome ❤❤❤❤❤❤❤❤❤❤
1:52:36 but in this case two types of acceleration due to gravity are there first is when the ball is thrown up that will be negative g, when ball reaches v=0 then it comes down there positive acceleration of g is applied then why did we only took negative acc for the whole case?
Who is watching in October😅❤
SIR AAP KI HAIRSTYLE KING JONG KI TARAH HAI😘😘
Sir your lecture is awesome I have no doubt for this chapter but that DPP is provide has relative velocity questions so how can we solve that question
Arjuna neet class 11th 2.0 2025 attendance here 👇
Sir Displacement is a vector quality and distance is scalar soo how we can say that Distance is greater than or equal to displacement I think this is wrong statement may be ye ho sakta hai Distance ≥ | Displacement | iske magnitude ke equal ya greater ho sakta hai
Yes bro
You are right 👍
Bro distance can be greater or equal to displacement this is the correct statement
sir 1:52:50 wale question mein vertically upwards agar phenk rahe hai ball ko toh pehle ball upar jaegi then 0 velocity attain karne ke baad neeche ayegi toh uss hisaab se tower ki height ke saath voh ek extra distance bhi 2 baar cross kar rhi hai upar jaate waqt and neeche aate waqt
Amazing lecture 🙏☺️ thank u sir jiii
Shalindar sir ko kaun kaun dil se pasand karta h
are behen padho 😅
Hmm
@@devshukla9440 haa gjb 😊😊
Sir acha nhi bhut shi pada te hai
Thik padate h
1:34:00 and 2:10:00 best question to solve
Answers kitne aaye
1:47:04 value of g is always - 9.8 or -10 hota hai chahe body upar jaye ya phir niche aye !!!!
at 2:04 hr ballon question why are you taking acceleration in negative sign....you are wrong here
Yess
Is this lecture enough for class 11 boards???
1:27:20 ans = 60 km/h
Same
@@AdityaPratapsingh-h5k same
Neet 2025 fodna h bhaii ❤❤❤❤
Sir pls ek week m 3 lecture physics ka plz sir 12 jan se exam hai plz plz sir ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤
How was your exam? 😊
Sir aap time ka tension mt lo please lecture ko lamba krdo
2:04:00 sir g ki value positive lenge na agar object niche aa raha hai to
1:25:44 if we solve 2AB/AB ,, it will be AB not 2..... 💔💔💔
AB is numerator and denominator
What are u talking about
1:27:11 - homwwork answer- 60 km /hour
Who is watching in december 😢
2:11:10
i) -17.6 m/s
ii)-19.6 m
Jo log arjuna jee 2025,lene vale hai jyada se jyada like kare 🎉🎉🎉🎉🎉🎉
From arjuna neet cz of backlogs for half yearly 🤧🙃🍃
2024 Me kon dekh raha hai
me
Mee
This is sufficient video for us,
Thank you sir
We all proud of you
velocity ... 17.6 m per second and the packet is 19.6 m away from the helicopter.
Can solved it
Sir please ek shedule bataye ki kis tarah se padhe
Bhai ase koi ni btata pdna to suru kr apne ap samajhne me a jayega
2:46:52 I was literally watching the video at 2 am ... 🌚💀
Sir aapane yah chapter bahut acche tarike se padhaanaa hai❤❤
Sir your handwriting is very very good 👍
1:55:08 sir mene 2nd method bhi kar liya
2:07:56 t=2,-4 hoga na sir
Thanku so much sir for this amazing lecture ;;; love from kashmir ❤❤❤
2:10:03 -18m/s
Not 16 m but 20 m sorry😅
your teaching is marvelous
:D
Attendance for december
hello sir...sir apne gravity k first part me bola ki object neeche drop ho rhi hogi to positive g hoga...to fir baloon wale question me ut -1/2gt2 kyn???
Me BHI yah I soch Raha hu Abhishek sir me negative bataya hai😢😢
Handwriting chumeshwari ☺️💯
1:27:17
Answer is 60kmh-1
Thanks a lot sir , you teach so well, i have a very weak base in physics and i don't understand physics at all bit today i understood it very well , i was revising this vhapter because tomorrow is my exam ,
Just praying i score well 🙏🏻🌷✨
How much did you scored ?
90+ ?
Mam class 11th school exam walo ka liya full year batch bhi launch kar dena agala saal please hame bhi pass hona hai 🥺🥺
Very best teacher 😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊❤❤❤❤❤
Sir you are the best teacher in the world ❤❤❤❤❤❤❤
Sir 2:7:13 pe question mein 1/2 ke baad 9.8 Kahan gaya 😢😢😢😢😢😢😢😢😢😢😢
Puri equation ko 9.8 se decide Kiya he
Nhi samj ayaa mujhe
Agyaa samjj
9.8 ko common liya haii
thank you very much sir
this has really build my basic concepts
Are u preparing for boards from this video or rivision
hi