A capillary tube of radius \( 0.25 \, \text{mm} \) is dipped in ethanol (\( S = 0.0227 \, \text{N/m}
\), \( \rho = 806 \, \text{kg/m}^3 \), \( \cos \theta = 1 \)). What is the capillary rise? (Take \( g =
9.8 \, \text{m/s}^2 \))
\( h = \frac{2 S \cos \theta}{\rho g a} \).
\( S = 0.0227 \, \text{N/m} \), \( \rho = 806 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( a =
0.25 \times 10^{-3} \, \text{m} \).
\( h = \frac{2 \times 0.0227 \times 1}{806 \times 9.8 \times 0.25 \times 10^{-3}} = 0.02298 \, \text{m} =
2.3 \, \text{cm} \).