Solveeit Logo
Login

Question

Question: A glass capillary tube of internal radius \[r = 0.25\,{\text{mm}}\] is immersed in the water. The to...

A glass capillary tube of internal radius r=0.25 mmr = 0.25\,{\text{mm}} is immersed in the water. The top end of the tube projects by 2 cm2\,{\text{cm}} above the surface of the water. At what angle does the liquid meet the tube? Surface tension of water= 0.07 N⋅m{\text{0}}{\text{.07}}\,{\text{N}} \cdot {\text{m}} and its contact angle is 0.
A. sin⁡−1(0.5){\sin ^{ - 1}}\left( {0.5} \right)
B. cos⁡−1(0.5){\cos ^{ - 1}}\left( {0.5} \right)
C. cos⁡−15/14{\cos ^{ - 1}}5/14
D. sin⁡−15/14{\sin ^{ - 1}}5/14

Explanation

Solution

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=2Tcos⁡θρgrh = \dfrac{{2T\cos \theta }}{{\rho gr}}
Here, hh is the height of the liquid column, TT is the surface tension, θ\theta is the angle of contact, ρ\rho is the density of the liquid, gg is the acceleration due to gravity and rr is the internal radius of the capillary tube.

Complete step by step answer:
Rewrite the equation for the Jurin’s law.
h=2Tcos⁡θρgrh = \dfrac{{2T\cos \theta }}{{\rho gr}}
Rearrange the above equation for angle of contact of the water with the glass of the capillary tube.
θ=cos⁡−1(hρgr2T)\theta = {\cos ^{ - 1}}\left( {\dfrac{{h\rho gr}}{{2T}}} \right)
The density of the water is 1000 kg/m31000\,{\text{kg/}}{{\text{m}}^3}.
ρ=1000 kg/m3\rho = 1000\,{\text{kg/}}{{\text{m}}^3}
The rough value of acceleration due to gravity is 10 m/s210\,{\text{m/}}{{\text{s}}^{\text{2}}} .
g=10 m/s2g = 10\,{\text{m/}}{{\text{s}}^{\text{2}}}
Substitute 2 cm2\,{\text{cm}} for hh, 1000 kg/m31000\,{\text{kg/}}{{\text{m}}^3} for ρ\rho , 10 m/s210\,{\text{m/}}{{\text{s}}^{\text{2}}} for gg, 0.25 mm0.25\,{\text{mm}} for rr and 0.07 N⋅m0.07\,{\text{N}} \cdot {\text{m}} for TT in the above equation.
θ=cos⁡−1((2 cm)(1000 kg/m3)(10 m/s2)(0.25 mm)2(0.07 N⋅m))\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)
⇒θ=cos⁡−1((2×10−2 m)(1000 kg/m3)(10 m/s2)(0.25×10−3 m)2(0.07 N⋅m))\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)
⇒θ=cos⁡−1(0.50.14)\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{{0.5}}{{0.14}}} \right)
⇒θ=cos⁡−1(514)\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{5}{{14}}} \right)
Therefore, the liquid meets the tube at an angle of cos⁡−1(514){\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 kg/m3997\,{\text{kg/}}{{\text{m}}^3} and 9.8 m/s29.8\,{\text{m/}}{{\text{s}}^{\text{2}}} respectively. For the sake of calculation, these values are rounded to 1000 kg/m31000\,{\text{kg/}}{{\text{m}}^3} and 10 m/s210\,{\text{m/}}{{\text{s}}^{\text{2}}} respectively.