Question

When a glass capillary tube is dipped in water, a 1.0 cm rise in the water level is observed at 18 °C. The internal radius of the capillary is ______ cm.
[Given: Surface tension of water at 18 °C = 73.2 dyne cm⁻¹ ; difference in the densities of water and air at 18 °C = 0.996 g cm⁻³ ; gravitational acceleration constant, g = 980 cm s⁻² .
Assume that water completely wets the glass capillary and the interface between the water and the air phase inside the capillary is a hemisphere.] (round off to two decimal places)

Answer 1

Correct Answer: 0.148 - 0.151

Capillary Rise Calculation 

To determine the internal radius of the capillary, we use the formula for capillary rise:

\[ h = \frac{2T \cos \theta}{r \rho g} \]

Given that water completely wets the glass, \( \theta = 0^\circ \), so \( \cos \theta = 1 \). Thus, the formula simplifies to:

\[ h = \frac{2T}{r \rho g} \]

Rearranging for \( r \):

\[ r = \frac{2T}{h \rho g} \]

Substituting the given values:

\( T = 73.2 \, \text{dyn/cm} = 73.2 \times 10^{-5} \, \text{N/m} \), 
\( h = 1.0 \, \text{cm} = 0.01 \, \text{m} \), 
\( \rho = 0.996 \, \text{g/cm}^3 = 996 \, \text{kg/m}^3 \), 
\( g = 980 \, \text{cm/s}^2 = 9.8 \, \text{m/s}^2 \).

Substitute these values into the formula:

\[ r = \frac{2 \times 73.2 \times 10^{-5}}{0.01 \times 996 \times 9.8} \]

Calculating \( r \):

\[ r \approx 0.001496 \, \text{m} = 0.1496 \, \text{cm} \]

Final Answer:

Rounding off to two decimal places, \( r \approx 0.15 \, \text{cm} \). The calculated internal radius of the capillary is \( 0.15 \, \text{cm} \), which is within the expected range of \( 0.148 \, \text{cm} \) to \( 0.151 \, \text{cm} \).