Physics - Fluid Question with Solution | TestHub

The pressure at the free surface of water in the container is atmospheric pressure, $P_0$.

For a concave meniscus in a capillary tube, the pressure just below the lowest point of the meniscus is less than the pressure just above it (which is $P_0$) by an amount equal to the excess pressure.

According to Laplace's law, the excess pressure across a curved liquid surface is given by $\Delta P=\frac{2T}{R}$, where $R$ is the radius of curvature of the meniscus.

For pure water, the angle of contact is $\theta =0^{\circ }$. Thus, the radius of curvature of the meniscus is equal to the internal radius of the capillary tube, $R=r$.

The pressure just below the meniscus is $P_0-\frac{2T}{r}$.

Therefore, the pressure difference between the water level in the container and the lowest point of the concave meniscus is:

$\Delta P=P_0-\left(P_0-\frac{2T}{r}\right)=\frac{2T}{r}$

Answer: $\frac{2\mathrm{T}}{\mathrm{r}}$