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 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 (\( g=10 \text{ ms}^{-2} \)):
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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.
Updated On: Jun 9, 2026
- \( 0.022 \text{ cc} \)
- \( 0.066 \text{ cc} \)
- \( 0.099 \text{ cc} \)
- \( 0.033 \text{ cc} \)
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The Correct Option is C
Solution and Explanation
Concept:
Water rises in a capillary due to surface tension. The height \( h \) is given by Jurin's Law, \( h = \frac{2T \cos \theta}{r \rho g} \). Assuming water wets the glass perfectly, \(\theta \approx 0^\circ\), so \(\cos \theta = 1\). The volume \( V = \pi r^2 h \).
Step 1: Combine the formulas for height and volume.
$$ V = \pi r^2 \left( \frac{2T}{r \rho g} \right) = \frac{2 \pi r T}{\rho g} $$
Step 2: Substitute numerical values.
\( r = 1.5 \text{ mm} = 1.5 \times 10^{-3} \text{ m} \) \( T = 7 \times 10^{-2} \text{ Nm}^{-1} \) \( \rho = 1000 \text{ kg/m}^3 \) \( g = 10 \text{ ms}^{-2} \) $$ V = \frac{2 \times \pi \times (1.5 \times 10^{-3} \text{ m}) \times (7 \times 10^{-2} \text{ Nm}^{-1})}{1000 \text{ kg/m}^3 \times 10 \text{ ms}^{-2}} $$
Step 3: Calculate and convert units.
$$ V = \frac{2 \times 3.14 \times 1.5 \times 7 \times 10^{-5}}{10^4} = 65.94 \times 10^{-9} \text{ m}^3 $$ To convert \(\text{m}^3\) to \(\text{cc} (\text{cm}^3)\), multiply by \( 10^6 \): $$ V \approx 0.099 \text{ cm}^3 $$ $$\boxed{0.099 \text{ cc}}$$
Step 1: Combine the formulas for height and volume.
$$ V = \pi r^2 \left( \frac{2T}{r \rho g} \right) = \frac{2 \pi r T}{\rho g} $$
Step 2: Substitute numerical values.
\( r = 1.5 \text{ mm} = 1.5 \times 10^{-3} \text{ m} \) \( T = 7 \times 10^{-2} \text{ Nm}^{-1} \) \( \rho = 1000 \text{ kg/m}^3 \) \( g = 10 \text{ ms}^{-2} \) $$ V = \frac{2 \times \pi \times (1.5 \times 10^{-3} \text{ m}) \times (7 \times 10^{-2} \text{ Nm}^{-1})}{1000 \text{ kg/m}^3 \times 10 \text{ ms}^{-2}} $$
Step 3: Calculate and convert units.
$$ V = \frac{2 \times 3.14 \times 1.5 \times 7 \times 10^{-5}}{10^4} = 65.94 \times 10^{-9} \text{ m}^3 $$ To convert \(\text{m}^3\) to \(\text{cc} (\text{cm}^3)\), multiply by \( 10^6 \): $$ V \approx 0.099 \text{ cm}^3 $$ $$\boxed{0.099 \text{ cc}}$$
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