Class 11 Physics | Elasticity | #19 Elongation of Suspended Rod Under Self Weight | For JEE & NEET

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Physics Galaxy

Physics Galaxy

Күн бұрын

Пікірлер: 62
@RukdSG
@RukdSG 12 жыл бұрын
Thank you for this video. It was difficult for me to agree on the formula without derivation on my book. Now it is clear
@shwetasinha2506
@shwetasinha2506 3 жыл бұрын
Are you an IITian now?
@Professor_Ray
@Professor_Ray 2 жыл бұрын
HEY WHERE ARE YOU NOW
@_kshitij_j_
@_kshitij_j_ 2 жыл бұрын
WE NEED ANSWERS
@harshsingh22105
@harshsingh22105 Жыл бұрын
@@shwetasinha2506 wbu?
@harshsingh22105
@harshsingh22105 Жыл бұрын
@@Professor_Ray wbu?
@devanshgupta3473
@devanshgupta3473 8 жыл бұрын
Sir if rod is placed in lift with acceleration g downwards then their will be no elongation?
@physicsgalaxyworld
@physicsgalaxyworld 8 жыл бұрын
yes... you are right...
@ArvindSingh-dy4lh
@ArvindSingh-dy4lh 7 жыл бұрын
Devansh G how?
@kartikrana1641
@kartikrana1641 5 жыл бұрын
@@ArvindSingh-dy4lh Pseudo Acceleration On Rod Will be Upwards Net Acceleration Will be Zero So Effective Weight of Rod Will be Zero So No Tension Will be Produced And Body Will Not Stretch As No Force is on Body
@jugalshah9261
@jugalshah9261 4 жыл бұрын
Sir in the elemental dx part, there is also a stress due to the upward mass which is compressing the dx part, so why didn’t we take it into account
@harshsingh22105
@harshsingh22105 Жыл бұрын
X is to much small part it will not experience any stress
@kartikgkalita
@kartikgkalita 11 ай бұрын
​bro thats a really good point
@mohaksharma7658
@mohaksharma7658 8 жыл бұрын
Respected Sir, I am stuck on a question for quite a long time...If we are given with a rod with its dimensions and Young's modulus which is free to move on a smooth surface...Then how to find the elongation in rod if a constant force F is applied on it?
@physicsgalaxyworld
@physicsgalaxyworld 8 жыл бұрын
+Mohak Sharma Use acceleration of rod a=F/m then consider an element of width dx at a distance x from one end of rod and find the tension in this element using T=(mass of rod behind this element)x(acceleration) then stress =T/S and strain =dy/dx as dy we consider elongation in element of length dx then integrate dy for the whole length of rod to get the total elongation...
@mohaksharma7658
@mohaksharma7658 8 жыл бұрын
Physics Galaxy Thanks a lot Sir..I got the ans..FL/2AY.. Thank u once again.
@sanjughosh4826
@sanjughosh4826 8 жыл бұрын
sir, is it working on non uniform application??
@physicsgalaxyworld
@physicsgalaxyworld 8 жыл бұрын
what do you mean by non uniform applications...
@ujjwalchitransh6411
@ujjwalchitransh6411 6 жыл бұрын
Sir can we apply centre of mas concept to get ans?
@physicsgalaxyworld
@physicsgalaxyworld 6 жыл бұрын
As elongation varies with the distance from the suspension point... center of mass cannot be used in this case...
@SumaidSyed
@SumaidSyed 9 жыл бұрын
Sir,we know that fractional change in the transverse length is proportional to fractional change in longitudinal length,so area of cross section is also changing,so how can we use that equation?
@physicsgalaxyworld
@physicsgalaxyworld 9 жыл бұрын
+Syed Sumaid As the cross sectional area of the wire is very small, we neglect the change in this analysis... even if you include the effect will be very small in case of numerical analysis...
@SumaidSyed
@SumaidSyed 9 жыл бұрын
Thank you very much!
@MessiEnic_
@MessiEnic_ 2 жыл бұрын
Awaj sunke cry aagya
@smileeverytime1684
@smileeverytime1684 6 жыл бұрын
Sir how M/L*x*g is coming????? PL tell sir please
@mohamednouman227
@mohamednouman227 4 жыл бұрын
Why are we integrating if the force is constant?
@ownbrace9131
@ownbrace9131 4 ай бұрын
Force is mass dependent... It's not constant
@RohanKUMAR-tf6pp
@RohanKUMAR-tf6pp 2 жыл бұрын
Sir please tension bata dijeye mx/l kaise aaya
@namanmishra771
@namanmishra771 4 жыл бұрын
Thanks a lot sir 👍🎉🎉
@IITiann
@IITiann 2 жыл бұрын
same question appeared in jee mains 2020
@devanshgupta3473
@devanshgupta3473 8 жыл бұрын
Sir if force is acting inside the rod(not on the surface) then any effect is their on the dimension of rod ? And we consider that force for stress formula or not?
@physicsgalaxyworld
@physicsgalaxyworld 8 жыл бұрын
here we consider force is distributed uniformly over any cross section of the rod... it is used in stress analysis...
@adityaranjan6871
@adityaranjan6871 3 жыл бұрын
Jhakas sir ....
@harshmaurya5966
@harshmaurya5966 5 жыл бұрын
sir what will be the enery stored by self elongation (plz help )
@gautam9339513131
@gautam9339513131 11 жыл бұрын
Thx a lot
@suhanajamwal8217
@suhanajamwal8217 3 жыл бұрын
Thnx sir
@besidevlogs8098
@besidevlogs8098 Жыл бұрын
Watching in 2023 ......😊..
@adityasart2644
@adityasart2644 Жыл бұрын
Ol
@akashiseijuro4444
@akashiseijuro4444 5 жыл бұрын
or you can simply take the length L as L/2 from centre of gravity
@arbinndo4620_IIT
@arbinndo4620_IIT 8 жыл бұрын
Sir why is strain = dy/dx?
@physicsgalaxyworld
@physicsgalaxyworld 8 жыл бұрын
because in elemental length dx we are considering a deformation dy so strain is given as ratio of deformation to initial length...
@raphaelapologetics4499
@raphaelapologetics4499 3 жыл бұрын
Ok
@SumaidSyed
@SumaidSyed 9 жыл бұрын
How does the diameter of rod vary from top to bottom after elongation?
@physicsgalaxyworld
@physicsgalaxyworld 9 жыл бұрын
+Syed Sumaid This you can find by using the elongation in every element dx and by keeping the volume constant... here we neglect the slight change in the average separation between molecules... Even if you do not wish to neglect this... you can use bulk modulus of the material to calculate this... but this will be a long analysis...
@SumaidSyed
@SumaidSyed 9 жыл бұрын
Thanks you for explanation.
@manvisinghkrishnpremi1206
@manvisinghkrishnpremi1206 2 жыл бұрын
Right now..... u must be in foreign 🤣
@nirajkrn
@nirajkrn Жыл бұрын
​@@manvisinghkrishnpremi1206haha sahi m yaar
@kunalnarwal2753
@kunalnarwal2753 3 жыл бұрын
Itna purana concept 😂 8 years ago.....
@arbinndo4620_IIT
@arbinndo4620_IIT 8 жыл бұрын
thank you sir.
@refrainrestrainresist-3rs49
@refrainrestrainresist-3rs49 Жыл бұрын
Thank you so much sir!
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