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(D) The air-water interface between two parallel plates forms a cylindrical meniscus. For a cylindrical surface,the excess pressure $\Delta P$ is give

Vedclass · Accepted Answer

$P_0 - \frac{2T}{d}$

Vedclass · Answer

(D) The air-water interface between two parallel plates forms a cylindrical meniscus. For a cylindrical surface,the excess pressure $\Delta P$ is given by $\Delta P = \frac{T}{R}$,where $R$ is the radius of curvature of the meniscus. Since the contact angle is zero,the meniscus is a semi-cylinder with diameter $d$. Thus,the radius of curvature $R = d/2$. The excess pressure is $\Delta P = P_{atm} - P_{inside} = \frac{T}{R}$. Substituting $R = d/2$,we get $\Delta P = \frac{T}{d/2} = \frac{2T}{d}$. Since the meniscus is concave towards the air,the pressure inside the water is less than the atmospheric pressure. Therefore,$P_{inside} = P_0 - \Delta P = P_0 - \frac{2T}{d}$.

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