Question

An incompressible liquid is filled in a container. There is air above the liquid and a massless movable piston with a hole is also placed on the air. A thin capillary tube of inner radius 0.1 mm is dipped vertically into the liquid through the hole on the piston as shown in the figure. The air in the container is isothermally compressed from the original volume V₀ to \(\frac{100}{101}\)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 T_water =0.075N/m, P₀=10⁵N/m² , g=10m/sec²,  \(\rho l\)=10³ kg/m³, \(\theta=0\))

Answer 1

Correct Answer: 25

 

Pressure-Volume Relation and Height Calculation

Given that \( T \) is constant: \[ PV = \text{constant} \]

From this, we get the relation between pressure and volume: \[ P \propto \frac{1}{V} \]

If the volume changes by a factor of \( \frac{100}{101} \), the pressure changes by a factor of \( \frac{101}{100} \). Thus, the pressure becomes: \[ P_f = 1.01 \times P_0 \]

The equation for the pressure difference is: \[ P_0 - \frac{2T}{R} = 1.01 P_0 - \rho g h \]

Rearranging this to solve for \( \rho g h \): \[ \rho g h = 0.01 P_0 + \frac{2T}{R} \]

Now, substituting the given values: \[ (10^3)(10)h = (0.01)(10^5) + \frac{2 \times 0.075}{0.1 \times 10^{-3}} \]

Solving for \( h \): \[ h = \frac{2500}{10^4} \, \text{m} = 25 \, \text{cm} \]