Bro, pehle topic ko atche se padhke Jan ke yaha aoo.....ye derivation channel hai na ki tumko sab samjhane wala.....derivations are important for school so we come here to get a proper way . don't blame any teacher please. If you don't understand then go there are so many channels
Ascent formula Consider a capillary tube of radius r dipped in a liquid of surface tension S and density ρ. Suppose the liquid wets the sides of the tube. Then its meniscus will be concave. The shape of the meniscus of water will be nearly spherical if the capillary tube is of sufficiently narrow bore. As the pressure is greater on the concave side of a liquid surface, so excess of pressure at a point A just above the meniscus compared to point B just below the meniscus 2S p R Where R is the radius of curvature of meniscus. If θ is the angle of contact then from right angled triangle shown in figure, we have r cosθ R r or R cosθ 2Scosθ p r Due this excess pressure p, the liquid rises in the capillary tube t height h when the hydrostatic pressure exerted by the liquid column becomes equal to the excess pressure p. Therefore, at equilibrium, we have p hρg 2Scosθ or hρg r 2Scosθ or p rρg
Thanks is for this video now i can prove this formula!
Thanks sir all clear ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤
Very useful video thanks ❤❤
Thanks sir❤
Thnk u sir❤
Sir Isko exam m ese hi kark e aayenge
Description mein notes hain
Thank you sir ❤
Concept was good,
Cheeee 🤮🤮 ,
Derivation kaha hai
Mera ko nhi Aya samaj
To glti Teri h
@@Yuvankanand ok
Meh bhai ki side hu
Repeatedly dekho.. And write a couple of times.
2S/R kahan se aya?
It is excess pressure
@@VishalKumar-qt9em Okay.. But how is it equal to 2S/R?
This is the previous formula in this ch ..first you have toh understand that to do this@@rawskeleton876
Smjh mein nhi aaya fir se 🫤🥹
mere ko nhi aya 😑
Kya nhi aaya brother ?
To jaker phad
Kya poti nhi aa rhi kya
Ek do baar dhayan se fir se padho aa jayega bro😊
Bkwas bol full ra ha.but .betuki baate..👍🏻
Bhai kya teacher h ghnta smj nhi aya
Bro, pehle topic ko atche se padhke Jan ke yaha aoo.....ye derivation channel hai na ki tumko sab samjhane wala.....derivations are important for school so we come here to get a proper way . don't blame any teacher please. If you don't understand then go there are so many channels
Ascent formula
Consider a capillary tube of radius r dipped in a liquid of surface tension S and density ρ. Suppose
the liquid wets the sides of the tube. Then its meniscus will be concave. The shape of the
meniscus of water will be nearly spherical if the capillary tube is of sufficiently narrow bore.
As the pressure is greater on the
concave side of a liquid surface, so
excess of pressure at a point A just
above the meniscus compared to point
B just below the meniscus
2S
p
R
Where R is the radius of curvature of
meniscus. If θ is the angle of contact
then from right angled triangle shown
in figure, we have
r
cosθ
R
r
or R
cosθ
2Scosθ
p
r
Due this excess pressure p, the liquid rises in the capillary tube t height h when the hydrostatic
pressure exerted by the liquid column becomes equal to the excess pressure p. Therefore, at
equilibrium, we have
p hρg
2Scosθ or hρg
r
2Scosθ
or p
rρg