Question

What will be the maximum capillary rise for a tube having an internal diameter of $0.1$ mm, when it is held vertical with the bottom end dipped in pure water taken in a trough? Consider the temperature of water to be $20^\circ$C, the value of surface tension ($T_s$) is $72.8 \times 10^{-6}$ kN/m, and $\gamma_w = 9.79$ kN/m3.

• A. 0.2974 m
• B. 0.3867 m
• C. 0.5671 m
• D. 0.6782 m

Hint:

For capillary rise problems, always verify units and ensure the diameter is converted to meters for accurate results.

Answer 1

The Correct Option is A

The capillary rise ($h$) is calculated using the formula:
\[h = \frac{4 T_s}{\gamma_w \cdot d}\]
where:
$T_s = 72.8 \times 10^{-6} \, \text{kN/m}$ (surface tension),
$\gamma_w = 9.79 \, \text{kN/m}^3$ (unit weight of water),
$d = 0.1 \, \text{mm} = 0.0001 \, \text{m}$ (diameter of the tube).
Substitute the values:
\[h = \frac{4 \cdot 72.8 \times 10^{-6}}{9.79 \cdot 0.0001} = \frac{291.2 \times 10^{-6}}{0.000979} \approx 0.2974 \, \text{m}\]