🔸To join our Problem-Solving course for JEE Advance {Set of 60} which ensures strengthening, revision, and concept coverage altogether, go here - courses.jeesimplified.com/courses
Bhaiyya agr main explain karu to dekho Maine mana ki kisi theta angle pe Jane pr vo contact chod rha hai which implies Tension ke 2 components honge Tsintheta aur T costheta aur kyunki Y direction me mg lg rha hai to vo nahi dekhenge hum Aur jb Tsintheta just greater than Mg hoga tb vo contact chod dega aur yaha pe centripetal force jo Mu²/l hoga vo Tension ke barabar hoga to fir Tsintheta =Mg ke barabr krde to hume milega u²=√gl/sintheta aur kyunki u minimum chahiye to sintheta ko maximum hona padega to fir theta=90 degree hi hoga aur jo X direction me Tcostheta lg rha tha vo zero hoga => Minimum value of u=√gl
😂 PW , Pw tests are one of the shittiest test I have seen i have given their tests believe me it's just mains level. The pattern the format the type of questions nah bro it ain't sufficient. I have taken allen test series... It's just amazing earlier I used to think I am very smart cause I got good marks in PW tests then I have allen tests and realised I stand nowhere... Further grinding led to even conquering the allen tests but difference was there believe me... I can guarantee that question can never come in PW tests
Got the ans root 3gl ! but i was sceptical as the length wasn't given. My method was that i considered an angle x where the centripetal force is equal to tension second eq made was cons of energy, similar to in que. Then centripetal force eq had cosx in it so found that angle to be around pi\6. Then got the. Ans
When B will start lifting, the angle between string and ground changes and Tension also changes as B decelerates and hence, Horizontal force on A = Tcosθ also changes so, it'll be a double variable force.
bhaiya ek doubt h ki agr ball b pe hum sirf vertical velocity hi manenge mtlb ki ball A drag krke ball B ke position m ajaye , mene yeh soch k solve kiya toh ENERGY CONSERVE , FORCE BALANCE yeh toh same agye h prr MOMENTUM IN X-DIRECTION nhi a parha h smjh . aap yeh bta do mera assumption vertical velocity ka galat h ?
u is not the velocity at any time of B, it is the initial velocity given to B. So, as B moves, its direction changes due to the tangential force applied by string... Hope this helps..
exact same method.. and i did it be myself(not the kind of person who doesnt know anything but rocks the comment section saying 5 min mein ho gaya).. yeh waala genuinely 5-10 mins mein ho gaya and mostly because of intuition.. by the way you should have given the length of the string in the problem statement atleast should have mentioned it.. because there is always a possibility that the answer could be taken out without the length of the string and then to verify our answer we have to look at your answer and if our answer is wrong and we see your answer it erases the whole point of the question.. so please mention all the parameters in the question beforehand
1) saying "This is a wrong solution" with such confidence, Sir Isaac Newton, you can be wrong too sometimes. 2) the question asked for "MINIMUM" initial velocity, if we would have taken any general angle jaha par A lift horha, then it would mean A has more energy than it requires to lift the ball (cuz it has a velocity in vertical direction also), but at the top it can have it's vertical velocity 0, (when the object will lift) ensuring minimum energy was given to ball A. 3)a part of the observation process, he didn't explain this as it's kinda obvious ig
@@jeesimplified-subject and even if we consider the tension force.. the string remains taut all along so no distance covered against force.. so work wont be done either way
Since we have velocity in tangential direction, centripetal force(tension) and no forces like friction, why not take it in circular motion? Also when the circular motion becomes invalid, we know it must have happened due to movement of a. So here we have the favourable situation.
Friction hota to energy conservation wali equation mein friction ka work done ko consider krna hota , also momentum in x direction will not be conseerve
It shud remain taut always cuz the COM_x (locus of COM will be a vertical line) cannot be touched, so the balls will never come towards each other, else the ball A wud have already lost contact. (he cud have mentioned these details in the video)
and by this u'll get that velocities of balls shudn't have a component along the line joining AB hence u may take a line passing thru A (perpendicular to the plane) as the instantaneous axis of rotation.
🔸To join our Problem-Solving course for JEE Advance {Set of 60} which ensures strengthening, revision,
and concept coverage altogether, go here - courses.jeesimplified.com/courses
Just do T=mv²/L=mg and use energy conservation. Solve in 15 seconds with this method
Nice... Did the same...
Explain your physics behind the first equation !
Same
Is this wrong? @@jeesimplified-subject
Bhaiyya agr main explain karu to dekho Maine mana ki kisi theta angle pe Jane pr vo contact chod rha hai which implies
Tension ke 2 components honge Tsintheta aur T costheta aur kyunki Y direction me mg lg rha hai to vo nahi dekhenge hum
Aur jb Tsintheta just greater than Mg hoga tb vo contact chod dega aur yaha pe centripetal force jo Mu²/l hoga vo Tension ke barabar hoga to fir Tsintheta =Mg ke barabr krde to hume milega
u²=√gl/sintheta aur kyunki u minimum chahiye to sintheta ko maximum hona padega to fir theta=90 degree hi hoga aur jo X direction me Tcostheta lg rha tha vo zero hoga
=> Minimum value of u=√gl
Same problem but with given Data came in our Test[PW]
11th
😂 PW ,
Pw tests are one of the shittiest test I have seen i have given their tests believe me it's just mains level. The pattern the format the type of questions nah bro it ain't sufficient. I have taken allen test series... It's just amazing earlier I used to think I am very smart cause I got good marks in PW tests then I have allen tests and realised I stand nowhere... Further grinding led to even conquering the allen tests but difference was there believe me...
I can guarantee that question can never come in PW tests
@@AyushKumarmaurya-r4b Yes bro You are Right normal test are really low in level but I am talking about the Top batch's test
@@BeastboyAKJ that's good then... Keep it up by the way I am currently in allen star batch the tests their is also good...
Got the ans root 3gl ! but i was sceptical as the length wasn't given. My method was that i considered an angle x where the centripetal force is equal to tension second eq made was cons of energy, similar to in que. Then centripetal force eq had cosx in it so found that angle to be around pi\6. Then got the. Ans
IDK about being an educator but you are definitely going to be a champ entrepreneur
When B will start lifting, the angle between string and ground changes and Tension also changes as B decelerates and hence, Horizontal force on A = Tcosθ also changes so, it'll be a double variable force.
usne throughout the motion baat hi nahi kari sirf final moment pe tension ko observe kiya
kyuki initially obviously tension is 0
Massless rod can't apply perpendicular forces so calculate velocity using Wet . Find tension using circular and equate
bhaiya ek doubt h ki agr ball b pe hum sirf vertical velocity hi manenge mtlb ki ball A drag krke ball B ke position m ajaye , mene yeh soch k solve kiya toh ENERGY CONSERVE , FORCE BALANCE yeh toh same agye h prr MOMENTUM IN X-DIRECTION nhi a parha h smjh . aap yeh bta do mera assumption vertical velocity ka galat h ?
u is not the velocity at any time of B, it is the initial velocity given to B. So, as B moves, its direction changes due to the tangential force applied by string...
Hope this helps..
exact same method.. and i did it be myself(not the kind of person who doesnt know anything but rocks the comment section saying 5 min mein ho gaya).. yeh waala genuinely 5-10 mins mein ho gaya and mostly because of intuition.. by the way you should have given the length of the string in the problem statement atleast should have mentioned it.. because there is always a possibility that the answer could be taken out without the length of the string and then to verify our answer we have to look at your answer and if our answer is wrong and we see your answer it erases the whole point of the question.. so please mention all the parameters in the question beforehand
0:57 bhaiya ye irodov ka sawaal hai mene bana rkha hai
Pf ka h,irodov m alg question h usme time pucha h aur wo do masses nhi ek cylinder fixed tha aur ek mass ko perp velocity di thi
This is wrong solution; how can you say that at 90-degree tension's vertical force equal to mg rather than at general angle
1) saying "This is a wrong solution" with such confidence, Sir Isaac Newton, you can be wrong too sometimes.
2) the question asked for "MINIMUM" initial velocity, if we would have taken any general angle jaha par A lift horha, then it would mean A has more energy than it requires to lift the ball (cuz it has a velocity in vertical direction also), but at the top it can have it's vertical velocity 0, (when the object will lift) ensuring minimum energy was given to ball A.
3)a part of the observation process, he didn't explain this as it's kinda obvious ig
tq so much bro
How do you justify the tension here is a non conservative force . Tension is not always conservative.
Tension is an internal force here
@@jeesimplified-subject and even if we consider the tension force.. the string remains taut all along so no distance covered against force.. so work wont be done either way
i didn't get what he is trying to say can anyone explain why its necessary for us to know if tension is conservative or non
thanks for this great video
Why B is moving in circular path and A as its centre , since A is not fixed ?
Circular motion is with respect to each other not with respect to ground
Since we have velocity in tangential direction, centripetal force(tension) and no forces like friction, why not take it in circular motion? Also when the circular motion becomes invalid, we know it must have happened due to movement of a. So here we have the favourable situation.
Bhaiya pathfinder ke solution k lie koi series start krdo
Not a series , but for a structured videos of such level questions you can join Set of 60
My approach was similar but I stuck on one thing. Length of string is not given in question so how can we consider it in final answer?
Basically usko as parameter consider krle uss form mein answer nikal skte ho
I have solved this question when you have posted and I got it right
Good, keep grinding
Sidha energy conservation kyu nai laga skte,,, 1/2 m v^2 = 2m g l/2 (COM) ??????
Observe.adapt.excel.
Bhaiya agar ground pe friction hota to kya ise mai rolling ke sath compare kar pata?
Friction hota to energy conservation wali equation mein friction ka work done ko consider krna hota , also momentum in x direction will not be conseerve
Yes bhaiyaa❤
@jeesimplified wont mv^2/l=2mg then tension would be mg???? then we can use energy conservation
i think galti samajh gaya relative velocity lenge to 4mv^2/l=mg aayga usse fir root3gl aayga
aise asaan question mai hag diya maine
Bhaiya how would you find the net workdone by tension over here
Dono balls pr individually Work Energy Theorem apply kro, you will work done by Tension
No string length given and mentioned that the string will remain taut throughout the motion or not??
It shud remain taut always cuz the COM_x (locus of COM will be a vertical line) cannot be touched, so the balls will never come towards each other, else the ball A wud have already lost contact. (he cud have mentioned these details in the video)
and by this u'll get that velocities of balls shudn't have a component along the line joining AB hence u may take a line passing thru A (perpendicular to the plane) as the instantaneous axis of rotation.
bhaiya but why will block A have a velocity? we should consider it to be still.
Then momentum in x direction will not conserve
i used T= mu^2/r-2mg+3mgcostheta and put T=mg.but got root 6gl
Did you understand your mistake then ?
Because A is not fixed
lekin bhaiya agr question he smjh na ayya ...mtlb uski language aur data to kya kareee?????
plss reply
Dm me on telegram
Answer in 2 root(gl)?
Nope, you made a mistake here
Bhaiya set of 60 mei live sessions one on one hote hai?
Not officially, but you can always connect with me and Pratham.
Bro helpful video thanks❤️❤️
Thanks bro
How is U' = 0.
Couldnt understand
There is no non conservative ya koi external force jiska koi work hai, so mechanical energy will conserve
1st