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Question: A capillary tube of a uniform bore is dipped vertically in water which rises by 7cm in the tube. Fin...

A capillary tube of a uniform bore is dipped vertically in water which rises by 7cm in the tube. Find the radius of the capillary tube if surface tension is 70  dynes/cm70\;{\rm{dynes/cm}}, Take g=9.8  m/s2g = 9.8\;{\rm{m/}}{{\rm{s}}^2}.

Explanation

Solution

Use the conversion 1  dyne=10−5  N1\;{\rm{dyne}} = {10^{ - 5}}\;{\rm{N}} and use the formula for the rise in tube is directly proportional to surface tension and inversely proportional to density of the liquid and radius of the capillary tube.

Complete step by step solution:
We know from the question that the rise of water in the tube is h=7  cmh = 7\;{\rm{cm}}, the surface tension is T=70  dynes/cmT = 70\;{\rm{dynes/cm}} and for water, the angle of contact is θ=0\theta = 0.

We have to convert the units into MKS units to make our calculation easy and error free.
h=7  cm=7×10−2  mh = 7\;{\rm{cm}} = 7 \times {10^{ - 2}}\;{\rm{m}} and T=70  dynes/cm=70×10−3  N/mT = 70\;{\rm{dynes/cm}} = 70 \times {10^{ - 3}}\;{\rm{N/m}}.

We know the formula of rise of liquid in the capillary is,
h=2Tcos⁡θrρgh = \dfrac{{2T\cos \theta }}{{r\rho g}}
Here, rr is the rise in the capillary, ρ\rho is the density of the water, which we know that it is equal to 1000  kg/m31000\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}.

Now we substitute the value 7×10−2  m7 \times {10^{ - 2}}\;{\rm{m}} as hh, 70×10−3  N/m70 \times {10^{ - 3}}\;{\rm{N/m}} as TT, 9.8  m/s29.8\;{\rm{m/}}{{\rm{s}}^2} as gg, 1000  kg/m31000\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}} as ρ\rho and 00 as θ\theta in the above equation, we get,
7×10−2  m=2×70×10−3  N/m×cos⁡0r×1000  kg/m3×9.8  m/s2 r=2×70×10−31000×9.8×7 =0.2×10−3  m 7 \times {10^{ - 2}}\;{\rm{m}} = \dfrac{{2 \times 70 \times {{10}^{ - 3}}\;{\rm{N/m}} \times \cos 0}}{{r \times 1000\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}} \times 9.8\;{\rm{m/}}{{\rm{s}}^2}}}\\\ r = \dfrac{{2 \times 70 \times {{10}^{ - 3}}}}{{1000 \times 9.8 \times 7}}\\\ = 0.2 \times {10^{ - 3}}\;{\rm{m}}

After simplifying the above value, we get r=0.2  mmr = {\rm{0}}{\rm{.2}}\;{\rm{mm}}.

Hence, the radius of the capillary tube is r=0.2  mmr = {\rm{0}}{\rm{.2}}\;{\rm{mm}}.

Additional information: The phenomenon of fall or rise of the liquid in the capillary tube is termed as capillary action. For example: The tip of the nib of a pen is split to provide capillary action.

Note: When a capillary tube having radius rr is dipped into the liquid of density ρ\rho and the surface tension TT, the rise and fall of the liquid is expressed as h=2Tcos⁡θrρgh = \dfrac{{2T\cos \theta }}{{r\rho g}}.