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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
$begingroup$
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
edited 7 hours ago
Qmechanic♦
106k121961223
asked 12 hours ago
Lelouche LamperougeLelouche Lamperouge
754
$endgroup$
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
edited 7 hours ago
Qmechanic♦
106k121961223
asked 12 hours ago
Lelouche LamperougeLelouche Lamperouge
754
$endgroup$
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
12 hours ago
1
$begingroup$
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
11 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
edited 7 hours ago
Qmechanic♦
106k121961223
asked 12 hours ago
Lelouche LamperougeLelouche Lamperouge
754
$endgroup$
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
fluid-statics capillary-action
edited 7 hours ago
Qmechanic♦
106k121961223
asked 12 hours ago
Lelouche LamperougeLelouche Lamperouge
754
edited 7 hours ago
Qmechanic♦
106k121961223
asked 12 hours ago
Lelouche LamperougeLelouche Lamperouge
754
edited 7 hours ago
Qmechanic♦
106k121961223
edited 7 hours ago
Qmechanic♦
106k121961223
edited 7 hours ago
Qmechanic♦
106k121961223
106k121961223
asked 12 hours ago
Lelouche LamperougeLelouche Lamperouge
754
asked 12 hours ago
Lelouche LamperougeLelouche Lamperouge
754
asked 12 hours ago
Lelouche LamperougeLelouche Lamperouge
754
754
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
12 hours ago
1
$begingroup$
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
11 hours ago
add a comment |
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
12 hours ago
1
$begingroup$
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
11 hours ago
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
12 hours ago
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
12 hours ago
1
1
$begingroup$
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
11 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
11 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $p_atm-hdg$.
answered 11 hours ago
Chet MillerChet Miller
15.8k2825
$endgroup$
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 $p_A=p_B$.
edited 7 hours ago
Sebastiano
318119
answered 11 hours ago
himanshuhimanshu
603
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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oldest
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$begingroup$
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 $p_atm-hdg$.
answered 11 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 $p_atm-hdg$.
answered 11 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 $p_atm-hdg$.
answered 11 hours ago
Chet MillerChet Miller
15.8k2825
$endgroup$
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 $p_atm-hdg$.
answered 11 hours ago
Chet MillerChet Miller
15.8k2825
answered 11 hours ago
Chet MillerChet Miller
15.8k2825
answered 11 hours ago
Chet MillerChet Miller
15.8k2825
answered 11 hours ago
Chet MillerChet Miller
15.8k2825
15.8k2825
add a comment |
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 $p_A=p_B$.
edited 7 hours ago
Sebastiano
318119
answered 11 hours ago
himanshuhimanshu
603
$endgroup$
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 $p_A=p_B$.
edited 7 hours ago
Sebastiano
318119
answered 11 hours ago
himanshuhimanshu
603
$endgroup$
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 $p_A=p_B$.
edited 7 hours ago
Sebastiano
318119
answered 11 hours ago
himanshuhimanshu
603
$endgroup$
$p_A$ 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 $p_A=p_B$.
edited 7 hours ago
Sebastiano
318119
answered 11 hours ago
himanshuhimanshu
603
edited 7 hours ago
Sebastiano
318119
edited 7 hours ago
Sebastiano
318119
edited 7 hours ago
Sebastiano
318119
318119
answered 11 hours ago
himanshuhimanshu
603
answered 11 hours ago
himanshuhimanshu
603
answered 11 hours ago
himanshuhimanshu
603
603
add a comment |
add a comment |
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$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
12 hours ago
1
$begingroup$
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
11 hours ago