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22:25 1) Answer is √3R Because it is an equilateral triangle of side 2R and when we take half of it as right angle triangle and apply Pythagoras theorem taking hypotenuse as 2R and base as R then we get the perpendicular i.e. r perpendicular as √3R. 31:02 2) Angle is 90° Because there are two types of accelerations i.e. tangential and centripetal and in uniform circular motion tangential acceleration is zero only centripetal acceleration acts and we know that centripetal acceleration is always directed towards centre and here it is perpendicular to the velocity. Thus, angle between velocity and net acceleration is 90°.

@sunitarani6100

8 ай бұрын

How this is √3r ? If you can tell pls

@AJ-ho6uy

8 ай бұрын

Thanks for explanation

@majorambadastiparatkar.

7 ай бұрын

Yes it's correct

@neetknowledgehub

7 ай бұрын

Use Pythagoras theorem in equilateral triangle of side 2R

@s.s.sshankarshansahoo6977

3 ай бұрын

Yes you are right

23:09 d=√3 R MOI = 9MR^2/2

Thanku so much sir ...these notes are even helping the droppers .... ❤️

23:18 √3r because it is forming an equilateral triangle

Sir apki tabiyat jaldi thik ho..yehi dua h..taki aagey ke mind map bhi aap hi karao..❤❤❤❤❤

Thanks you so much sir , for providing this lecture ,

Thank you sir for this wonderful series ❤❤❤❤❤❤🎉🎉🎉

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Abhishek sir and Aswarya wala example was 😂😂

Amazing Lecture sir❤

1. Perpendicular distance √3 R aayegi aur I ki value : 9mr²/2 aayegi 2. Velocity and acceleration mein angle 90° hogi uniform circular motion ki

@Neet-20-25

7 ай бұрын

Yes bro same mera bhi😊😊😊

D = √3R Ans: 9MR^2/2

31:06 in case of uniform circular motion there is not at, so anet = ac, which means the angle bw linear velocity and anet will be 90°

@krehamathbeerehamath733

5 ай бұрын

🎉 12:27

22:07 ..... Ans.= 4ML^2/3

In uniform circular motion angle between velocity and centripetal acc is equal to 90°

Sir perpendicular distance √3R ayegi aur Moment of inertia 9/2MR^2

22:27 perpendicular distance - root3

22:19 d= root 3 R

Mera bhi naam Abhishek h aapse bs aaj mila hu test ki revision kr reha tha to pr bonding aapke sath aachai bani🎉 thanks ❤❤❤

Amazing Content

@22:29 Sir upr wale ring ka perpendicular distance √3R hoga ..apne Arjuna me padhaya tha 😊❤

Kon kon 700+ plus Lana chahta hai 👍

@harryshetty2006

3 ай бұрын

Tmk

@no-one406

3 ай бұрын

​@@harryshetty2006bhai gali kyo bak raha

@Rahuljat687

2 ай бұрын

TMC gan..u

@surjankarvekar5639

2 ай бұрын

😇😇

@rakshitchanpuriya6034

Ай бұрын

Bhai tu toh baccha lag rha hai??

9MR²/2 aayega sir? Perpendicular distance R√3 aayega shayad. 10th class wale concept yaad dila diye😂

Thankyou sir for this awesome strategy revision ❤🎉

22:14 R perpendicular Is √3R

Hme kb moment of inertia com k lena hai kb perpendicular axis theorem k MOI lena hai multiple mass me totally confused

Really amazing lecture 🔥🔥🔥🔥🔥

Iss question me d=√3 R Hoga aur moment of inertia=9MR²/2

Thanks you sir❤❤❤❤

Thank you sir🤟❤

23:18 r = √3 r

Perpendicular distance R√3

That's like a mindmap💜Thankyou sir

@pranavgarje

8 ай бұрын

Mind map hi hai

@msiLegend

8 ай бұрын

😂@@pranavgarje

Rolling motion is deleted.

22:25 Sir the perpendicular distance of body 1 from axis of rotation will be √3 R which we can put in the place of D in the formula ( Icm + Md^2) and if we consider these body as ring the Inet = 9mr^2/2 31:12 in ucm the angle b/w anet and v is 90°

@swarajnarang4364

8 ай бұрын

Mera v same yhi ans hai

@BismitaBinay-bc8fd

6 ай бұрын

Can you please explain how it's √3

@_k.dash2796

5 ай бұрын

@@BismitaBinay-bc8fd Pythagoras theorem

31:03 90°

√3R 22:26 ❤❤

33 mins mein Rotational Motion revision........waaah thank you sir

23:12 d=R root3 I = (9MR^2)/2

Aaj Arjuna neet me start hua h❤

@adarshattitudesrivastva3681

9 ай бұрын

Apna score btana ok😊😊😊

@anshmishra2035

9 ай бұрын

😂

Thank you guru g ❤️🙏 for osm lecture ❤️🙏 u r the best guru g ❤️🙏❤️🙏

Arjuna wale ❤❤❤

Start 2:22

I=4/3ML²

Thank you so much sir 🙏❤️

Thankyou sir for your efforts ❤

22:13 ka answer 9mr²/2

Thankyou so muchh sirr maine ye chapter aapse hi bhut achhe se padha tha class me but tension ho rahi thi ki ab revision kaise karun but ise dekhne ke baad sir pura concept aur formulas apne aap yaad aate chale gayee really thankyou so muchh sirr From arjuna 2025 ....❤

Much better than 5 hrs lecture🍁

@MagicWorldCreation99

3 ай бұрын

No bro, 5hrs lecture me wo question jayada karwate hai

23.14........R root 3

R√3 hoga D ka value

Kisko ya method acha lga ❤❤❤ REVISION KARNA KA

R perpendicular=√3R M.O.I= 9MR²/2

Wonderful lecture ❤❤❤

D=R√3

Tnx u sir for this series ❤❤❤

Angular acceleration & angular velocity lies at 90°

ANS= ROOT 3 R AND 9MRSQUARE/2

R√5 =13MR²/2

D=√3. R

thank u sir

22.24 , 5MR 2

very helpful

d = √3r

Angle is 90°

Rroot3 is the perpendicular distance at 23:14

@provishuu9929

8 ай бұрын

How ??

d = 2R hogi because total length me both ring radiation add hongi

At22:√3r

Sir pendicular distance √3R hai

Thankyou sir ❣️❣️ bahut dino se preshan the Moment of inertia ko lekar ❤

√ 3R perpendicular distance hoga

D= R root3

Sir Kasam se pura samjh aaya thank you so much😄😄

Thankyou sir so much ❤❤❤❤😊😊😊😊😊😊😊

Sir d = root3r

23:58 how the Icm of rod = ML^2/12

23:06 √3R

r= Rroot 3

23:15 d= √3R

d =√3 R

Hw ans Rrt3

Root 3 R

√3R

11/2 MR^2

Hello Sir ji❤❤❤❤❤

What is difference between mind map and full lecture..? Because both are one shot...

@dilshad2393

7 ай бұрын

Mind map is just every thing in very concise manner matlb chapter pehle pdha hona chahiye fir mindmap dekho okk

@Anjali11179

6 ай бұрын

@@dilshad2393 thank you

29:08 -- How M(rotational) = I/R^2 ??? I for disc is mr^2/2

@user-bg8hv3tu5u

4 ай бұрын

right

Sir par mee toh yeha padhe aya thaa 😅 4:33

√3R 22:28

Complete chapter fullfill with improvement topic ✅😊❤

22:25 d=√3R ; I = 9MR²/2

@Vinay_kumar720

6 ай бұрын

🎉 kaise explain this

R×1.73

√3

Done❤

Nice lec

18:12 disc ka to mr square upon 2 hoti hai na

Iske notes kha milenge ??

thank you sir

3rd rivision done sir ji ❤️❤️❤️❤️❤️