Question

A capillary tube of inner radius 1.5 mm is dipped vertically in water. If the surface tension of water is \( 7 \times 10^{-2} \text{ Nm}^{-1} \), then the volume of the water that rises in the capillary tube is (\( g=10 \text{ ms}^{-2} \)):

• A. \( 0.022 \text{ cc} \)
• B. \( 0.066 \text{ cc} \)
• C. \( 0.099 \text{ cc} \)
• D. \( 0.033 \text{ cc} \)

Hint:

The volume of liquid in the capillary is independent of the density in terms of total mass, but the height is strictly dependent on the density and surface tension.

Answer 1

The Correct Option is C

Step 1: Set up the capillary rise.
Water climbs to a height $h$ given by Jurin's law $h = \dfrac{2T\cos\theta}{r\rho g}$. Water wets glass well, so $\theta \approx 0^\circ$ and $\cos\theta = 1$.
Step 2: Write the volume of risen water.
The water column is a cylinder of radius $r$ and height $h$, so its volume is $V = \pi r^2 h$.
Step 3: Combine the two formulas.
Substitute $h$ into the volume so the unknown height cancels nicely: \[ V = \pi r^2 \cdot \frac{2T}{r\rho g} = \frac{2\pi r T}{\rho g}. \] Notice one power of $r$ survives.
Step 4: List the numbers in SI units.
$r = 1.5\times10^{-3}\,\text{m}$, $T = 7\times10^{-2}\,\text{N m}^{-1}$, $\rho = 1000\,\text{kg m}^{-3}$, $g = 10\,\text{m s}^{-2}$, and use $\pi \approx 3.14$.
Step 5: Plug in and compute.
\[ V = \frac{2\times 3.14 \times (1.5\times10^{-3}) \times (7\times10^{-2})}{1000 \times 10} = \frac{65.94\times10^{-5}}{10^{4}} \approx 6.59\times10^{-8}\,\text{m}^3. \]
Step 6: Convert to cubic centimetres.
Since $1\,\text{m}^3 = 10^{6}\,\text{cc}$, multiply by $10^{6}$: $V \approx 6.59\times10^{-8}\times10^{6} \approx 0.099\,\text{cc}$.
\[ \boxed{0.099\ \text{cc}} \]