Question

A capillary of radius r = 0.2 mm is dipped vertically in a liquid of density, (p = 1gm/cc). A small piston (x) is inserted inside the capillary which maintains the constant pressure of air ($P_1$ = 2 × $10^4$ dynes/cm²) above the hemispherical meniscus. If ambient pressure is $P_o$ = $10^5$ dynes/cm² and surface tension of liquid is T = 72 dynes/cm, then find the height h of liquid in cm. (weight of curved part of liquid is ignored)

Answer 1

89 cm

Answer 2

The pressure at the level of the liquid surface in the reservoir is the ambient pressure $P_o$.

Inside the capillary tube, consider a point at the same horizontal level as the liquid surface in the reservoir. The pressure at this point can be found by considering the pressure at the top of the liquid column and the pressure due to the height of the liquid column.

The pressure of air above the hemispherical meniscus is $P_1$.

Due to the hemispherical meniscus, there is a pressure difference across the meniscus given by the Laplace formula, $\Delta P = \frac{2T}{r}$, where $T$ is the surface tension and $r$ is the radius of the capillary. Since the meniscus is concave upwards, the pressure inside the liquid just below the meniscus is less than the pressure above the meniscus.

Pressure just below the meniscus $= P_1 - \frac{2T}{r}$.

The pressure at a depth $h$ below the level just below the meniscus is given by $P_{depth} = P_{surface} + \rho g h$.

So, the pressure at the bottom of the liquid column of height $h$ inside the capillary, at the level of the reservoir surface, is $P_{bottom} = (P_1 - \frac{2T}{r}) + \rho g h$.

By Pascal's principle, the pressure at the same horizontal level in a continuous liquid at rest is the same. Therefore, the pressure at the bottom of the liquid column inside the capillary is equal to the ambient pressure $P_o$.

$P_o = P_1 - \frac{2T}{r} + \rho g h$.

We are given the following values: $r = 0.2$ mm $= 0.2 \times 10^{-1}$ cm $= 0.02$ cm. $\rho = 1$ gm/cc $= 1$ gm/cm³. $P_1 = 2 \times 10^4$ dynes/cm². $P_o = 10^5$ dynes/cm². $T = 72$ dynes/cm. The acceleration due to gravity in CGS units is $g = 980$ cm/s².

We need to find the height $h$ in cm. Rearranging the equation for $h$: $\rho g h = P_o - P_1 + \frac{2T}{r}$ $h = \frac{P_o - P_1 + \frac{2T}{r}}{\rho g}$

Calculate the terms: $P_o - P_1 = 10^5 - 2 \times 10^4 = 100000 - 20000 = 80000$ dynes/cm². $\frac{2T}{r} = \frac{2 \times 72 \text{ dynes/cm}}{0.02 \text{ cm}} = \frac{144}{0.02} \text{ dynes/cm}^2 = 7200 \text{ dynes/cm}^2$. Numerator $= 80000 + 7200 = 87200$ dynes/cm². Denominator $\rho g = 1 \text{ gm/cm}^3 \times 980 \text{ cm/s}^2 = 980 \text{ gm/(cm}^2 \text{s}^2)$.

Note that 1 dyne/cm² = 1 gm/(cm s²). So the units are consistent.

$h = \frac{87200 \text{ dynes/cm}^2}{980 \text{ gm/(cm}^2 \text{s}^2)} = \frac{87200 \text{ gm/(cm s²)}}{980 \text{ gm/(cm}^2 \text{s}^2)}$. $h = \frac{87200}{980} \text{ cm} = \frac{8720}{98} \text{ cm} = 89 \text{ cm}$.