An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner radius 0.1 mm is dipped vertically into the liquid through the airtight piston hole, as shown in the figure. The air in the container is isothermally compressed from its original volume 𝑉0 to 100 101 𝑉0 with the movable piston. Considering air as an ideal gas, the height (ℎ) of the liquid column in the capillary above the liquid level in cm is_______. [Given: Surface tension of the liquid is 0.075 N m−1 , atmospheric pressure is 105 N m−2 , acceleration due to gravity (𝑔) is 10 m s −2 , density of the liquid is 103 kg m−3 and contact angle of capillary surface with the liquid is zero]

 To determine the height 

$h$ of the liquid column in the capillary tube, we need to consider the balance of forces due to surface tension and the pressure difference caused by the compression of the air in the container.

Step 1: Calculate the Pressure Difference

The air in the container is compressed isothermally from volume $V_0$ to $\frac{100}{101}{V}_0$. For an ideal gas undergoing isothermal compression, the pressure $P$ and volume $V$ are related by:

$$P_0{V}_0=PV$$

where $P_0$ is the initial pressure (atmospheric pressure), and $P$ is the final pressure after compression.

Given:

$$V=\frac{100}{101}{V}_0$$

Thus:

$$P=P_0\frac{V_0}{V}=P_0\frac{V_0}{\frac{100}{101}{V}_0}=P_0\frac{101}{100}$$

The pressure difference $\mathrm{\Delta }P$ is:

$$\mathrm{\Delta }P=P-P_0=P_0(\frac{101}{100}-1)=P_0\left(\frac{1}{100}\right)=\frac{P_0}{100}$$

Given $P_0=10^5{\text{N/m}}^2$:

$$\mathrm{\Delta }P=\frac{10^5}{100}=10^3{\text{N/m}}^2$$

Step 2: Relate Pressure Difference to Capillary Rise

The height $h$ of the liquid column in the capillary tube is given by the Jurin's law:

$$h=\frac{2\gamma \cos \theta }{\rho gr}$$

where:

• $\gamma$ is the surface tension of the liquid ($0.075\text{N/m}$),

• $\theta$ is the contact angle ($0$ degrees, so $\cos \theta =1$),

• $\rho$ is the density of the liquid ($10^3{\text{kg/m}}^3$),

• $g$ is the acceleration due to gravity ($10{\text{m/s}}^2$),

• $r$ is the radius of the capillary tube ($0.1\text{mm}=0.1\times 10^{-3}\text{m}$).

Substitute the values:

$$h=\frac{2\times 0.075\times 1}{10^3\times 10\times 0.1\times 10^{-3}}$$

Simplify:

$$h=\frac{0.15}{1}=0.15\text{m}=15\text{cm}$$

Step 3: Adjust for Pressure Difference

The pressure difference $\mathrm{\Delta }P$ due to the compression of the air affects the height of the liquid column. The additional height $\mathrm{\Delta }h$ due to $\mathrm{\Delta }P$ is:

$$\mathrm{\Delta }h=\frac{\mathrm{\Delta }P}{\rho g}=\frac{10^3}{10^3\times 10}=0.1\text{m}=10\text{cm}$$

Step 4: Total Height

The total height $h$ of the liquid column in the capillary tube is the sum of the capillary rise and the additional height due to the pressure difference:

$$h=15\text{cm}+10\text{cm}=25\text{cm}$$

However, this approach seems to double-count the effect of pressure. A more accurate method is to consider that the pressure difference directly affects the capillary rise. The correct height $h$ is given by:

$$h=\frac{2\gamma \cos \theta }{\rho gr}+\frac{\mathrm{\Delta }P}{\rho g}$$

Substitute the values:

$$h=\frac{2\times 0.075\times 1}{10^3\times 10\times 0.1\times 10^{-3}}+\frac{10^3}{10^3\times 10}=15\text{cm}+10\text{cm}=25\text{cm}$$

Thus, the height of the liquid column in the capillary tube is:

$$\boxed{{\displaystyle 25}}$$