Question

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 $70\;{\rm{dynes/cm}}$, Take $g = 9.8\;{\rm{m/}}{{\rm{s}}^2}$.

Answer 1

Hint:Use the conversion $1\;{\rm{dyne}} = {10^{ - 5}}\;{\rm{N}}$ 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 $h = 7\;{\rm{cm}}$, the surface tension is $T = 70\;{\rm{dynes/cm}}$ and for water, the angle of contact is $\theta = 0$.

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

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

Now we substitute the value $7 \times {10^{ - 2}}\;{\rm{m}}$ as $h$, $70 \times {10^{ - 3}}\;{\rm{N/m}}$ as $T$, $9.8\;{\rm{m/}}{{\rm{s}}^2}$ as $g$, $1000\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}$ as $\rho $ and $0$ as $\theta $ in the above equation, we get,
$
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 $r = {\rm{0}}{\rm{.2}}\;{\rm{mm}}$.

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

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 $r$ is dipped into the liquid of density $\rho $ and the surface tension $T$, the rise and fall of the liquid is expressed as $h = \dfrac{{2T\cos \theta }}{{r\rho g}}$.