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If $p$ is a pressure and $p_A = p_{\text{atm}} + hdg,\,$ $p_B = p_{\text{atm}}$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?

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  • $\begingroup$ So P(A) need not be equal to P(B)??? $\endgroup$ – Lelouche Lamperouge Mar 19 at 16:41
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    $\begingroup$ 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 $\endgroup$ – Feynmans Out for Grumpy Cat Mar 19 at 17:08
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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 $p_{atm}-hdg$.

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$p_A$ 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 $p_A=p_B$.

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