A glass capillary tube of internal radius $r=0.25\, mm$ is immersed in water. The top end of the tube projected by $2\, cm$ above the surface of the water. At what angle does the liquid meet the tube? Surface tension of water $=0.7\, N / m$.

Question

When two soap bubbles of radii $ a $ and $ b $ ($ b>a $) coalesce, the radius of curvature of the common surface is:

Answer 1

To solve this problem, we need to find the angle at which the water meets the capillary tube, i.e., the contact angle $ \theta $. We can use the formula related to capillary action:

h = \dfrac{2T \cos \theta}{\rho g r}

Where:

$ h $ is the height of the liquid column (in meters)
$ T $ is the surface tension of water $(0.7 \, \text{N/m})$
$ \rho $ is the density of water $(1000 \, \text{kg/m}^3)$
$ g $ is the acceleration due to gravity $(9.8 \, \text{m/s}^2)$
$ r $ is the radius of the capillary tube $(0.25 \, \text{mm} = 0.25 \times 10^{-3} \, \text{m})$

The tube projects $ 2 \, \text{cm} = 0.02 \, \text{m} $ above the water, which means the height $ h $ that the water can rise is given by the projection height:

h = 0.02 \, \text{m}

Substituting these values into our equation:

0.02 = \dfrac{2 \times 0.7 \times \cos \theta}{1000 \times 9.8 \times 0.25 \times 10^{-3}}

Rearranging gives:

\cos \theta = \dfrac{0.02 \times 1000 \times 9.8 \times 0.25 \times 10^{-3}}{2 \times 0.7}

Calculating this:

\cos \theta = \dfrac{0.02 \times 980 \times 0.25 \times 10^{-3}}{1.4} = \dfrac{0.0049}{1.4} \approx 0.0035

Using the cosine inverse function:

\theta = \cos^{-1}(0.0035)

Upon calculation, $ \theta \approx 70^\circ $.

Therefore, the angle at which the liquid meets the tube is $ \theta = 70^\circ $. The correct answer is:

$\theta = 70^{\circ}$ is the correct answer.

Answer 2