Really amazing and wonderful problem. And want to thakyou janardhan sir from physics sir jee and nitin sir from insp for such a great and wonderful series of problem which are more useful for our student community ....thanx a lot..🙏🏻
Anybody wandering or finding it hard to understand manipulation of (dmgsina) could , directly write it as m = u × (l /v) And force = m × g × sina If you look carefully expression for m is (mass per unit time × total time taken by belt to transverse l distance ) which will give you total mass m on belt, since after that total mass on conveyer belt remains constant.
@@PratyushKumar-d9f yes, now , you have pointed clearly... It's on left top of the screen... That's not on the falling sand... that's on the sand that's already on the belt moving steadily... So equilibrium on such sand means static Friction should balance mg sin theta
I will do that 🙂👍. It's just that the response to modern physics videos is lukewarm. As most students have not finished that topic. Anyways , I will bring something in coming weeks
Sir, thanks for the wonderful question and ur hardwork!,👌🙏😁I have a doubt, at 12:38 question, in which part of the system is the energy getting lost as heat?
@@PHYSICSSIRJEE sir I dont know how is it possible for you to solve so many doubts of so many students on TH-cam INSTANTLY!Extremely Fortunate to have been taught by you Sir😃🙏🙏
In 2nd question if we use the method you used in last part of 1st question then when stedy state is reached d(Wext )+d(Wdiss)=0 and dW(cons)/dt = uglsin(theta) and dK/dt is (1/2)uv² which yields a different ans. Where did i go wrong?
sir if considering just touching condition then dm mass sand has no velocity but the belt have velocity v upwards so velocity of dm mass with respect to belt is -v ....down wards ....so force on dm is in down wards ....hence rxn force on the belt is in upwards direction ......
Velocity of sand is downwards relative to belt. To understand, if you sit on belt it looks like sand is going downward. Hence relative velocity is downward. Also mass is being added. Hence thrust force will be in the direction of relative velocity(downward)!
Sir why is heat dissipated in this process.. friction here does no work as there is no relative motion b/w sand and belt...so where is this heat comeing from..??🙏
Sir greatly explained.....Sir I have an alternate solution... can you please clarify that is this method conceptually correct or wrong... Sir the frictional force on sand provides thrust force to the sand and as minimum force we have to find that indicates that in steady state we have to pull and hence sand particle has accn=0 hence f=integral dmgsin∅=ugh/v=uv(as it's equal to thrust force)...thus v=√gh and now on conveyor belt force will be f+uv=>2f=2uv.... F=2u√gh.... (Ans)
Sir nicely explained.... Sir I have an alternate solution can you please clarify is it conceptually correct or not..... Sir here on the sand as sudden change in momentum is due to friction thus friction provides thrust force..... As we need force minimum that means steady state... So on sand net friction force must be equal to mgsin∅(m total mass accumulated) and also equal to uv... So hcosec∅/v that is time for one dm mass to reach top. Point that means mass accumulated=uhcosec∅/v So net friction=(uhcosec∅/v) gsin∅=ugh/v and as it equals uv... Thus it yields v=√gh and as force on conveyor belt must be F=f+uv=>2f=2ugh/v=>2u√gh... Sir can it be an alternate solution?
In first question's last part why can't we sat that frictional force is the only heat dissipative force/thrust force and hence power dissipation is 'ugh'
thanks sir ...sir one doubt ....how in exit situation in the first question there will not exist any reaction force .....as in the entry situation both belt and sand have v velocity upwards ......so intially there is v relative is also 0 .......sir please tell ......
@@PHYSICSSIRJEE sir but initially the dm mass sand has an velocity wrt to belt in down ward direction as belt is moving upwards .....then dm/dt(v) acting downwards on sand hence reaction force on belt will be in upwards direction.....sir please help ..........thanks sir.....
sir rotation mein confidence nahi aa raha hai ,almost this sunday i read for 4th time i even understand it and able to do ques but HAVE NO CONFIDENCE ,i usually overthink or underthink an situation (an extra torque in fbd or missing it) any suggestions...
Prepare a separate plan to finish these in steps in 30 days (2 hours) daily without effecting your current schedule 1)Go through the theory and solved examples of "Anurag Mishra" book. 2) Solve previous year JEE Questions (including subjective of 1990's) in ROTATION 3) Solve Complete irodov of ROTATION 4) Finish PATHFINDER objective 5) OPTIONAL : Finish PATHFINDER build up your understanding section
@@PHYSICSSIRJEE Sir is electricity and magnetism by Anurag Mishra (Balaji publications) a good choice if I want proper theory for concepts and difficult problems?
Sir thank you so much sir...your videos helped me a lot...and I got AIR 140(OBC) in Jee advanced. Cannu suggest me some college sir pls...for cse....is roorke good sir or should I take some other branch for good iit🙏🙏
Of course I like the effect you put sir and iam a regular follower of you vedios sir.please give some tips to solve pathfinder physics sir,it's somewhat tough to me,actually more😅
![[PATHFINDER SOLUTIONS] JEE ADVANCED | SPACESHIP ROCKET | EFFICIENCY CHECK-24 , 25 SCHOOL PHYSICS](http://i.ytimg.com/vi/qSVP9AfmBBY/mqdefault.jpg)
![[PATHFINDER SOLUTIONS] JEE ADVANCED | SPACESHIP ROCKET | EFFICIENCY CHECK-24 , 25 SCHOOL PHYSICS](/img/tr.png)
![[Jee Adv] Falling Sand On A Moving Conveyor Belt | Pathfinder | MCQ 22,23,24 | Work And Energy](http://i.ytimg.com/vi/uFcwf6LXtZg/mqdefault.jpg)
![[JEE ADVANCED] WORK DONE BY INTERNAL FORCES | MAN LADDER BALLOON | SCHOOL PHYSICS SIR JEE QUICKIES](/img/n.gif)
The ~Hey you guys~ vibe is really awesome
Scout😂??
@@tanveersharma2793 ha
I thought , the "real" one visited my channel. You got me there 😁. Your choice of DP is 🔥😂
😂yes I thought that too i was about to say wow but then I checked his previous comments 😂
tanveer sharma LoL haha
Was able to solve all the questions by myself 😊😊.......and sir explanation is OP 🔥🔥
This video has convinced me to give up preparing for advanced. Mains it is.
Thanks for reducing the competition 😂😅
#100 For My Revision (Learnt something great today 🙂👍)
How was ur adv
How much did u scored?,?
Bas video dekhe Jaa rha hai kya hua advance lg toh nhi rha halat dekh kar
Thank you sirji😊♥️
very very well presented sir 🙂👍
Really amazing and wonderful problem. And want to thakyou janardhan sir from physics sir jee and nitin sir from insp for such a great and wonderful series of problem which are more useful for our student community ....thanx a lot..🙏🏻
Great explanation sir
Thankyou
i am rajat Chaudhary
Beta doubt ho to mujhse kyu nahi puchte
Thank you sir...
Anybody wandering or finding it hard to understand manipulation of (dmgsina) could , directly write it as m = u × (l /v)
And force = m × g × sina
If you look carefully expression for m is (mass per unit time × total time taken by belt to transverse l distance ) which will give you total mass m on belt, since after that total mass on conveyer belt remains constant.
Dil.jit liya sir aapne
Thanks sir !
sir why did we not include the reaction force on the sand in its fbd at 5:15 but we cinsidered it in the belt's fbd
I didn't include it because I was not drawing FBD or writing separate equation for sand. That force you mentioned will be there on sand
@@PHYSICSSIRJEE but sir you wrote static friction = int dmgsintheta on sand
@@PratyushKumar-d9f yes, now , you have pointed clearly... It's on left top of the screen... That's not on the falling sand... that's on the sand that's already on the belt moving steadily... So equilibrium on such sand means static Friction should balance mg sin theta
so the thrust force only acts for some time and therefore friction balances the mg component
sir something on AC circuits,modern physics-that radiation pressure sir,,(watched your previous video on that)
I will do that 🙂👍. It's just that the response to modern physics videos is lukewarm. As most students have not finished that topic. Anyways , I will bring something in coming weeks
@@PHYSICSSIRJEE aye aye captain..
thank you
@@ishaanshekhawat4178 trxd Dr DVD’s,
@@jayasimhapadakandla3806 i didn't get u..
Sir, thanks for the wonderful question and ur hardwork!,👌🙏😁I have a doubt, at 12:38 question, in which part of the system is the energy getting lost as heat?
Sand+belt rupture against each other and kinetic friction does negative work on them together , which generates heat
@@PHYSICSSIRJEE sir I dont know how is it possible for you to solve so many doubts of so many students on TH-cam INSTANTLY!Extremely Fortunate to have been taught by you Sir😃🙏🙏
loved it like always sir
In 2nd question if we use the method you used in last part of 1st question then when stedy state is reached d(Wext )+d(Wdiss)=0 and dW(cons)/dt = uglsin(theta) and dK/dt is (1/2)uv² which yields a different ans. Where did i go wrong?
d(Wext) =0 in second question. As explained in the question, there is no motor here unlike first question. So your first equation is wrong
I think heat must be there...
sir if considering just touching condition then dm mass sand has no velocity but the belt have velocity v upwards so velocity of dm mass with respect to belt is -v ....down wards ....so force on dm is in down wards ....hence rxn force on the belt is in upwards direction ......
Velocity of sand is downwards relative to belt. To understand, if you sit on belt it looks like sand is going downward. Hence relative velocity is downward. Also mass is being added. Hence thrust force will be in the direction of relative velocity(downward)!
sir , in the FBD of sand , why don't we consider v (dm/dt) on the sand element which is just falling on the belt 3:40
Mass of sand particle is fixed
@@PHYSICSSIRJEE thank you sir, but due to change in velocity can we get any force on that element.
Easy one sir. sir can you please confirm the answer of rotation challenge question 4.
It may be easy for you 👍🙂. Btw, Rotation Answer given is correct.
Bhai fat ke hath me aa gayi aur tu keh raha hai ki easy tha? Tum log saalo kis mitti ke bane ho?
Sir why is heat dissipated in this process.. friction here does no work as there is no relative motion b/w sand and belt...so where is this heat comeing from..??🙏
At the start when dm of sand falls , in dt sec , it slips, KINETIC friction acts
Okay sir
@@PHYSICSSIRJEE sir same in second question also heat will be produced..
@@jitendrapandey1085 yes , it does. Never denied that in the Solution
Sir greatly explained.....Sir I have an alternate solution... can you please clarify that is this method conceptually correct or wrong... Sir the frictional force on sand provides thrust force to the sand and as minimum force we have to find that indicates that in steady state we have to pull and hence sand particle has accn=0 hence f=integral dmgsin∅=ugh/v=uv(as it's equal to thrust force)...thus v=√gh and now on conveyor belt force will be f+uv=>2f=2uv.... F=2u√gh.... (Ans)
Friction is a such a troublesome guy...But amazing video!
Welcome
how to judge direction of the thrust force? understood the rest , very nice one sir
opposite to relative velocity or you can directly use intuition
ponnapu got it , thanks
anytime..
Sir nicely explained.... Sir I have an alternate solution can you please clarify is it conceptually correct or not..... Sir here on the sand as sudden change in momentum is due to friction thus friction provides thrust force..... As we need force minimum that means steady state... So on sand net friction force must be equal to mgsin∅(m total mass accumulated) and also equal to uv... So hcosec∅/v that is time for one dm mass to reach top. Point that means mass accumulated=uhcosec∅/v
So net friction=(uhcosec∅/v) gsin∅=ugh/v and as it equals uv... Thus it yields v=√gh and as force on conveyor belt must be
F=f+uv=>2f=2ugh/v=>2u√gh...
Sir can it be an alternate solution?
😊😁🔥
In first question's last part why can't we sat that frictional force is the only heat dissipative force/thrust force and hence power dissipation is 'ugh'
Kinetic Friction acts only during dt sec and not for entire h height as explained in first part of first problem
Hello sir revision 😀
All the best 👍🏼😀. Keep going
thanks sir ...sir one doubt ....how in exit situation in the first question there will not exist any reaction force .....as in the entry situation both belt and sand have v velocity upwards ......so intially there is v relative is also 0 .......sir please tell ......
In Entry case at the bottom, sand initially just before falling in , is at rest. So Vrel is not zero
@@PHYSICSSIRJEE sir but initially the dm mass sand has an velocity wrt to belt in down ward direction as belt is moving upwards .....then dm/dt(v) acting downwards on sand hence reaction force on belt will be in upwards direction.....sir please help ..........thanks sir.....
@@swagatasubuddhi1907 first Question or second? Please mention the time stamp
@@PHYSICSSIRJEE sir first one ....
@@swagatasubuddhi1907 on sand upwards , so on belt FBD downwards
sir rotation mein confidence nahi aa raha hai ,almost this sunday i read for 4th time i even understand it and able to do ques but HAVE NO CONFIDENCE ,i usually overthink or underthink an situation (an extra torque in fbd or missing it) any suggestions...
Prepare a separate plan to finish these in steps in 30 days (2 hours) daily without effecting your current schedule
1)Go through the theory and solved examples of "Anurag Mishra" book.
2) Solve previous year JEE Questions (including subjective of 1990's) in ROTATION
3) Solve Complete irodov of ROTATION
4) Finish PATHFINDER objective
5) OPTIONAL : Finish PATHFINDER build up your understanding section
@@PHYSICSSIRJEE i will be keenly following it sir jee ,your priceless suggestions are like gem for me...
@@PHYSICSSIRJEE Sir is electricity and magnetism by Anurag Mishra
(Balaji publications) a good choice if I want proper theory for concepts and difficult problems?
@@sambhavmishra5423 yes, decent 👍, but finish irodov on that chapter afterwards
@@PHYSICSSIRJEE Ok sir thanks, I will.
Sir thank you so much sir...your videos helped me a lot...and I got AIR 140(OBC) in Jee advanced.
Cannu suggest me some college sir pls...for cse....is roorke good sir or should I take some other branch for good iit🙏🙏
Roorkee is good for cse. Are you getting anything in top 4 IITs?
@@PHYSICSSIRJEE should see sir last year IIT Kanpur 133 us last...if luck favours may get kanpur
😴
Sir I am thinking ece a bit tougher side because it includes all semiconductors part ....so..
Or else for electrical can I get good placements and package sir
Sir can we say whenever an impulsive force acts, there is heat loss involved?
Impulsive dissipative force
@@PHYSICSSIRJEE sir meaning?
Are you a chaiNA batch teacher sir?
Yes🙂👍
Sir I heard that sir chaitanya pressures the students a lot and also it can be found in TH-cam is that true sir?
@@thayanithirk1784 no it's not true as far as I know 🙂. But my knowledge could be limited 🙏.Btw , did you like the effort I put in the above video?
Of course I like the effect you put sir and iam a regular follower of you vedios sir.please give some tips to solve pathfinder physics sir,it's somewhat tough to me,actually more😅
@@thayanithirk1784 if you are looking at JEE advanced , please solve objective section first 👍
Pls show me a path sir🙏🙏