Ye displacement wala method best hai. Badi aasani se ques. Ho jata hai, thanks a lot.
@AshishKumar-xb7dr5 ай бұрын
1 saal se mereko wedge problem samjh nhi aa rha tha mgr aap 16 min. Me samjha diye thanks a lot ❤❤❤❤❤❤❤❤
@doubtdestroyer5 ай бұрын
Glad ...I could help you.. thanks for your feedback...it keeps me motivated.
@AshishKumar-xb7dr5 ай бұрын
@@doubtdestroyer iam in class 10 and preparing for Olympiad i started my preparation from class 9 sir please aap advanced ke related ach ache illustrations upload kijiye ki jisse hme apna concept building me madad ho .💕💕💕💕
@lakshayraj293914 күн бұрын
wonderful explanation sir... 11th me ye concept samajh nhi aaya thaa , but aaj jee mains 2025 se 11 din phle aache se samajh aa gya..❤❤❤ Thank you so much.
@doubtdestroyer14 күн бұрын
Glad ...you understand it 👍
@Mathematics-y4j2 ай бұрын
Wow amazing explanation guru ji❤❤❤
@doubtdestroyer2 ай бұрын
Thank you for your appreciation 🙏
@rabikashyaP173Ай бұрын
Thanks a lot bhaiya ❤
@sandeeppopalghat91435 ай бұрын
We want more type of this videos
@doubtdestroyer5 ай бұрын
Sure...you will get them...
@sandeeppopalghat91435 ай бұрын
Nice explanation sir
@doubtdestroyer5 ай бұрын
Thank you Sandeep ji... please share and make sure your friends also subscribe...it is motivation for me.. keep smiling and keep learning..
@Ashu-w8t5 ай бұрын
Sir plz take question no. 64 on cengage on nlm in which a cylinder is present between the wedges moving with different velocity
@doubtdestroyer5 ай бұрын
You will get it soon...
@doubtdestroyer5 ай бұрын
Hey...you can solve it by this very simple approach. First let's see the situation by sitting on any wedge . Simple one is sit on wedge which is moving with u velocity in backward direction. To this observer the other wedge will appear to move by 3u velocity. And cylinder will move on inclined plane. Now since cylinder and wedge are in contact. This means horizontal component of cylinder velocity is 3u. This means v cos30 = 3u V = 2√3 u. Now since this observer is also moving with u velocity in backward direction so from ground ( for ground observer) The net velocity of cylinder should be vector addition of these two velocities. If you add these two velocities u and 2√3 u by vector addition You will get net velocity= √7u. If you still not get it...join telegram channel and ask this question there I will post it with diagrams.
@Ashu-w8t5 ай бұрын
Sir is there is any other approach to solve it
@doubtdestroyer5 ай бұрын
@@Ashu-w8t it is simplest ...you can directly solve by taking ground as reference frame but that will be slightly difficult to imagine so to solve.