Question

A glass capillary tube of internal radius \[r = 0.25\,{\text{mm}}\] is immersed in the water. The top end of the tube projects by \[2\,{\text{cm}}\] above the surface of the water. At what angle does the liquid meet the tube? Surface tension of water= \[{\text{0}}{\text{.07}}\,{\text{N}} \cdot {\text{m}}\] and its contact angle is 0.
A. \[{\sin ^{ - 1}}\left( {0.5} \right)\]
B. \[{\cos ^{ - 1}}\left( {0.5} \right)\]
C. \[{\cos ^{ - 1}}5/14\]
D. \[{\sin ^{ - 1}}5/14\]

Answer 1

Hint:Use Jurin’s law for the height of the liquid column to determine the angle at which the liquid meets the tube. This expression gives the relation between the angle of contact, surface tension of the liquid, density of the liquid, acceleration due to gravity and the radius of the capillary tube.

Formula used:
The equation for Jurin’s law for the height of the liquid column is
\[h = \dfrac{{2T\cos \theta }}{{\rho gr}}\]
Here, \[h\] is the height of the liquid column, \[T\] is the surface tension, \[\theta \] is the angle of contact, \[\rho \] is the density of the liquid, \[g\] is the acceleration due to gravity and \[r\] is the internal radius of the capillary tube.

Complete step by step answer:
Rewrite the equation for the Jurin’s law.
\[h = \dfrac{{2T\cos \theta }}{{\rho gr}}\]
Rearrange the above equation for angle of contact of the water with the glass of the capillary tube.
\[\theta = {\cos ^{ - 1}}\left( {\dfrac{{h\rho gr}}{{2T}}} \right)\]
The density of the water is \[1000\,{\text{kg/}}{{\text{m}}^3}\].
\[\rho = 1000\,{\text{kg/}}{{\text{m}}^3}\]
The rough value of acceleration due to gravity is \[10\,{\text{m/}}{{\text{s}}^{\text{2}}}\] .
\[g = 10\,{\text{m/}}{{\text{s}}^{\text{2}}}\]
Substitute \[2\,{\text{cm}}\] for \[h\], \[1000\,{\text{kg/}}{{\text{m}}^3}\] for \[\rho \], \[10\,{\text{m/}}{{\text{s}}^{\text{2}}}\] for \[g\], \[0.25\,{\text{mm}}\] for \[r\] and \[0.07\,{\text{N}} \cdot {\text{m}}\] for \[T\] in the above equation.
\[\theta = {\cos ^{ - 1}}\left( {\dfrac{{\left( {2\,{\text{cm}}} \right)\left( {1000\,{\text{kg/}}{{\text{m}}^3}} \right)\left( {10\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right)\left( {0.25\,{\text{mm}}} \right)}}{{2\left( {0.07\,{\text{N}} \cdot {\text{m}}} \right)}}} \right)\]
\[ \Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{{\left( {2 \times {{10}^{ - 2}}\,{\text{m}}} \right)\left( {1000\,{\text{kg/}}{{\text{m}}^3}} \right)\left( {10\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right)\left( {0.25 \times {{10}^{ - 3}}\,{\text{m}}} \right)}}{{2\left( {0.07\,{\text{N}} \cdot {\text{m}}} \right)}}} \right)\]
\[ \Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{{0.5}}{{0.14}}} \right)\]
\[ \Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{5}{{14}}} \right)\]
Therefore, the liquid meets the tube at an angle of \[{\cos ^{ - 1}}\left( {\dfrac{5}{{14}}} \right)\].
Hence, the correct option is C.

Note: The actual values of the density of the liquid and the acceleration due to gravity are \[997\,{\text{kg/}}{{\text{m}}^3}\] and \[9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}\] respectively. For the sake of calculation, these values are rounded to \[1000\,{\text{kg/}}{{\text{m}}^3}\] and \[10\,{\text{m/}}{{\text{s}}^{\text{2}}}\] respectively.