A capillary tube with a radius of 0.50 mm is vertically dipped into...

The pressure just below the meniscus in the capillary tube is given by $P_0 - \dfrac{2T}{r}$, where $P_0$ is the atmospheric pressure, $T$ is the surface tension, and $r$ is the radius of the tube.

The pressure $P$ at a depth $h$ below the meniscus inside the tube is:

$P = P_0 - \dfrac{2T}{r} + \rho g h$

The difference between this pressure and the atmospheric pressure is:

$\Delta P = P - P_0 = \rho g h - \dfrac{2T}{r}$

Given values are:

$\rho = 1000 \text{ kg/m}^3$

$g = 9.8 \text{ m/s}^2$

$h = 5.0 \text{ cm} = 0.05 \text{ m}$

$T = 0.075 \text{ N/m}$

$r = 0.50 \text{ mm} = 0.5 \times 10^{-3} \text{ m}$

Calculating the hydrostatic pressure term:

$\rho g h = 1000 \times 9.8 \times 0.05 = 490 \text{ N/m}^2$

Calculating the capillary pressure term:

$\dfrac{2T}{r} = \dfrac{2 \times 0.075}{0.5 \times 10^{-3}} = 300 \text{ N/m}^2$

Substituting these values into the pressure difference equation:

$\Delta P = 490 - 300 = 190 \text{ N/m}^2$