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Does capillary rise violate hydrostatic paradox?

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Does capillary rise violate hydrostatic paradox?

Question on the hydrostatic paradoxIt's about capillary rise of waterForces causing capillary riseIf a hole is drilled at the bottom of a vessel, why is the pressure of the liquid leaving the vessel equal to atmospheric pressure?Is hydrostatic pressure independent of temperature?Pressure on horizontal levels same?About hydrostatic pressure affecting measured weight on a scaleIs Pascal's law incorrect?Hydrostatic pressure in a gasDerivation of height of a liquid in a capillary tube

3

$begingroup$

If $p$ is a pressure and $p_A = p_textatm + hdg,,$ $p_B = p_textatm$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?

fluid-statics capillary-action

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edited 30 mins ago

Qmechanic♦

106k121961223

asked 5 hours ago

Lelouche LamperougeLelouche Lamperouge

604

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So P(A) need not be equal to P(B)???
  $endgroup$
  – Lelouche Lamperouge
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You must have meant hydrostatic principle rather than hydrostatic paradox. ;) A paradox cannot be "violated"--rather, it appears to violate principles. And true principles resolve the paradox rather than violating it. :P
  $endgroup$
  – Dvij Mankad
  5 hours ago
  
  

  
  
  
  

add a comment | 

3

$begingroup$

If $p$ is a pressure and $p_A = p_textatm + hdg,,$ $p_B = p_textatm$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?

fluid-statics capillary-action

share|cite|improve this question

edited 30 mins ago

Qmechanic♦

106k121961223

asked 5 hours ago

Lelouche LamperougeLelouche Lamperouge

604

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So P(A) need not be equal to P(B)???
  $endgroup$
  – Lelouche Lamperouge
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You must have meant hydrostatic principle rather than hydrostatic paradox. ;) A paradox cannot be "violated"--rather, it appears to violate principles. And true principles resolve the paradox rather than violating it. :P
  $endgroup$
  – Dvij Mankad
  5 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

If $p$ is a pressure and $p_A = p_textatm + hdg,,$ $p_B = p_textatm$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?

fluid-statics capillary-action

share|cite|improve this question

edited 30 mins ago

Qmechanic♦

106k121961223

asked 5 hours ago

Lelouche LamperougeLelouche Lamperouge

604

$endgroup$

share|cite|improve this question

edited 30 mins ago

Qmechanic♦

106k121961223

asked 5 hours ago

Lelouche LamperougeLelouche Lamperouge

604

share|cite|improve this question

edited 30 mins ago

Qmechanic♦

106k121961223

asked 5 hours ago

Lelouche LamperougeLelouche Lamperouge

604

edited 30 mins ago

Qmechanic♦

106k121961223

edited 30 mins ago

Qmechanic♦

106k121961223

asked 5 hours ago

Lelouche LamperougeLelouche Lamperouge

604

asked 5 hours ago

Lelouche LamperougeLelouche Lamperouge

604

2 Answers
2

active

oldest

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5

$begingroup$

The pressures at A and B are indeed equal. However, the pressure in the fluid immediately below the curved meniscus is equal to $p_atm-hdg$ as a result of surface tension. So the pressure at A is $$p_A=p_atm-hdg+hdg=p_atm=p_B$$That is, there is a discontinuous change in pressure across the meniscus as a result of the surface tension in combination with the curvature. The pressure on the upper side of the interface is $p_atm$ and the pressure on the lower side of the interface is $p_atm-hdg$.

share|cite|improve this answer

answered 5 hours ago

Chet MillerChet Miller

15.8k2825

$endgroup$

add a comment | 

2

$begingroup$

$p_A$ is equal to $p_B$ here. The disparity is arising due to the fact that pressure just outside the meniscus is greater than the pressure inside. This is due to the curvature of the meniscus and surface tension.
This difference is compensated by $hdg$ to make $p_A=p_B$.

share|cite|improve this answer

edited 53 mins ago

Sebastiano

310119

answered 5 hours ago

himanshuhimanshu

353

$endgroup$

add a comment | 

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5

$begingroup$

The pressures at A and B are indeed equal. However, the pressure in the fluid immediately below the curved meniscus is equal to $p_atm-hdg$ as a result of surface tension. So the pressure at A is $$p_A=p_atm-hdg+hdg=p_atm=p_B$$That is, there is a discontinuous change in pressure across the meniscus as a result of the surface tension in combination with the curvature. The pressure on the upper side of the interface is $p_atm$ and the pressure on the lower side of the interface is $p_atm-hdg$.

share|cite|improve this answer

answered 5 hours ago

Chet MillerChet Miller

15.8k2825

$endgroup$

add a comment | 

5

$begingroup$

The pressures at A and B are indeed equal. However, the pressure in the fluid immediately below the curved meniscus is equal to $p_atm-hdg$ as a result of surface tension. So the pressure at A is $$p_A=p_atm-hdg+hdg=p_atm=p_B$$That is, there is a discontinuous change in pressure across the meniscus as a result of the surface tension in combination with the curvature. The pressure on the upper side of the interface is $p_atm$ and the pressure on the lower side of the interface is $p_atm-hdg$.

share|cite|improve this answer

answered 5 hours ago

Chet MillerChet Miller

15.8k2825

$endgroup$

add a comment | 

$begingroup$

The pressures at A and B are indeed equal. However, the pressure in the fluid immediately below the curved meniscus is equal to $p_atm-hdg$ as a result of surface tension. So the pressure at A is $$p_A=p_atm-hdg+hdg=p_atm=p_B$$That is, there is a discontinuous change in pressure across the meniscus as a result of the surface tension in combination with the curvature. The pressure on the upper side of the interface is $p_atm$ and the pressure on the lower side of the interface is $p_atm-hdg$.

share|cite|improve this answer

answered 5 hours ago

Chet MillerChet Miller

15.8k2825

$endgroup$

share|cite|improve this answer

answered 5 hours ago

Chet MillerChet Miller

15.8k2825

answered 5 hours ago

Chet MillerChet Miller

15.8k2825

answered 5 hours ago

Chet MillerChet Miller

15.8k2825

2

$begingroup$

$p_A$ is equal to $p_B$ here. The disparity is arising due to the fact that pressure just outside the meniscus is greater than the pressure inside. This is due to the curvature of the meniscus and surface tension.
This difference is compensated by $hdg$ to make $p_A=p_B$.

share|cite|improve this answer

edited 53 mins ago

Sebastiano

310119

answered 5 hours ago

himanshuhimanshu

353

$endgroup$

add a comment | 

2

$begingroup$

$p_A$ is equal to $p_B$ here. The disparity is arising due to the fact that pressure just outside the meniscus is greater than the pressure inside. This is due to the curvature of the meniscus and surface tension.
This difference is compensated by $hdg$ to make $p_A=p_B$.

share|cite|improve this answer

edited 53 mins ago

Sebastiano

310119

answered 5 hours ago

himanshuhimanshu

353

$endgroup$

add a comment | 

$begingroup$

$p_A$ is equal to $p_B$ here. The disparity is arising due to the fact that pressure just outside the meniscus is greater than the pressure inside. This is due to the curvature of the meniscus and surface tension.
This difference is compensated by $hdg$ to make $p_A=p_B$.

share|cite|improve this answer

edited 53 mins ago

Sebastiano

310119

answered 5 hours ago

himanshuhimanshu

353

$endgroup$

share|cite|improve this answer

edited 53 mins ago

Sebastiano

310119

answered 5 hours ago

himanshuhimanshu

353

edited 53 mins ago

Sebastiano

310119

edited 53 mins ago

Sebastiano

310119

answered 5 hours ago

himanshuhimanshu

353

answered 5 hours ago

himanshuhimanshu

353