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# A liquid rises to a height of $5\,cm$ in a glass capillary of radius 0.02 cm. What will be the height of the same liquid in a glass capillary of radius 0.04 cm?

Last updated date: 11th Aug 2024
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Hint: In this question, we need to determine the height of the liquid in the glass capillary of radius 0.04 cm. For this, we will use the concept of the rise of liquid in a capillary tube, and the relation of height of liquid column supported with the radius of the tube is used by substituting all the given values.

For a radius of glass capillary$= 0.02\,cm$
Height up to which liquid rises$= 5\,cm$
We know that height of the liquid column supported in a capillary tube is given by $h = \dfrac{{2\sigma \,\cos \theta }}{{r\rho g}}$
Where,
$h$$=$Height up to which the liquid rises in the capillary tube
$\sigma =$The surface tension of the liquid in capillary tubes.
$\theta =$The angle of contact between the liquid and capillary tube walls.
$r =$The radius of the capillary tube in which liquid is dipped
$\rho =$Density of liquid
$g$$=$Acceleration due to gravity, which is constant
Now, for the first case,
$\Rightarrow h = 5{\text{ }}cm \\ \Rightarrow r = 0.02{\text{ }}cm \\$
So,
$\Rightarrow h = \dfrac{{2\sigma \cos \theta }}{{r\rho g}} \\ \Rightarrow 5 = \dfrac{{2\sigma \cos \theta }}{{0.02\rho g}} \\ \Rightarrow 5 \times 0.02 = \dfrac{{2\sigma \cos \theta }}{{\rho g}} \\ \Rightarrow\dfrac{{2\sigma \cos \theta }}{{\rho g}} = 0.10.....(i) \\$
For the second case,
$\Rightarrow h = ?{\text{ and }}r = 0.04cm$
As the liquid is the same in both cases,$\sigma ,\theta ,\rho ,g$it will be the same for both
From (i), we have
$\Rightarrow\dfrac{{2\sigma \cos \theta }}{{\rho g}} = 0.10$
And r$= 0.04cm$
So, $h = \dfrac{{2\sigma \cos \theta }}{{r\rho g}}$
$\Rightarrow h = \dfrac{1}{\mu } \times \dfrac{{2\sigma \cos \theta }}{{\rho g}} \\ \Rightarrow h = \dfrac{1}{{0.04}} \times 0.10 \\ \Rightarrow h = \dfrac{{10}}{4} \\ \Rightarrow h = 2.5\,cm. \\$
Therefore, the liquid will rise to the height of $2.5\,cm$in glass capillary of radius $0.04\,cm.$

Note:Before applying this concept, one should remember that $\sigma ,\theta {\text{ }}and{\text{ }}\rho$it depends on the liquid, not on the radius of the capillary tube. From ascent formula $h = \dfrac{{2\sigma \cos \theta }}{{\mu \rho g}}$, it is clear that the height h to which a liquid rises the capillary tube is:
(i) Inversely proportional to the radius of the tube.
(ii) Inversely proportional to the density of the liquid
(iii) Directly proportional to the surface tension of the liquid.
Hence a liquid rises more in a narrower tube than in a wider tube.