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A vertical U-tube (communicating tube) contains a liquid of density rho and surface tension T. The tube has inner radius r. When the entire system accelerates horizontally with acceleration a, the pressure at point A (at the liquid surface in the left arm, taking the meniscus correction into account) is given by which of the following expressions? (Take atmospheric pressure as P0, and assume the contact angle is 0 deg.)
- P0 + rho*g*h - 2T/r
- P0 - rho*a*L - 2T/r
- P0 + rho*g*h - rho*a*L - 2T/r
- P0 + rho*a*L + 2T/r
Correct answer: P0 - rho*a*L - 2T/r
Solution
When the U-tube accelerates horizontally with acceleration a, a pseudo-force acts in the opposite direction on the liquid. For the left arm (arm A), the pressure is reduced by the term rho*a*L (where L is the horizontal separation contributing to pressure difference). The capillary pressure for a wetting liquid with contact angle 0 is 2T/r, which is subtracted from atmospheric pressure (meniscus is concave, so pressure inside is less than outside). Therefore P_A = P0 - rho*a*L - 2T/r.
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