Question

In a capillary tube, water rises up to a height of 3 mm. What is the height up to which water will rise in a capillary tube of radius one-third the first one?
A) 1 mm
B) 3 mm
C) 6 mm
D) 9 mm

Answer 1

The liquid given in the question is water. The angle of contact for glass-water interface is 0°. The height up to which liquid rises in a capillary is given by h = $\dfrac{{4\sigma \cos \theta }}{{\rho dg}}$ where d is the diameter of the capillary tube and $\theta$ is the angle of contact.

Complete step by step answer:
It is given that water rises up to a height of 3 mm in the capillary tube. However, the radius of the capillary is not given. So we assume that the radius of the first tube is r mm.
The angle of contact ($\theta$) for glass-water interface is 0°.
The cosine of angle of contact is, cos $\theta$ = cos 0° = 1.
The diameter of the capillary tube is, d$_1$ = 2r.
Using the formula for height of capillary rise for the first tube,
$\Rightarrow$ h$_1$ = $\dfrac{{4\sigma \cos \theta }}{{\rho {d_1}g}}$= 3 mm … equation (1)
Substituting the value of d$_1$ and cos $\theta$ in equation (1), we get,
$\Rightarrow$ 3 = $\dfrac{{4\sigma \times 1}}{{\rho (2r)g}}$
Taking (2r) to the left side of the equation,
$\Rightarrow$ 3(2r) = $\dfrac{{4\sigma }}{{\rho g}}$ …equation (2)
Since the fluid is not changing, the terms on the right side of the equation remain constant. This equation will be used later to simplify our equation and find out the height of water rise in the second tube.
In the case of the second capillary tube, the radius of the tube becomes one-third of the first tube. Let us take the radius of the second capillary tube as r$_2$ mm.
$\Rightarrow$ r$_2$= $\dfrac{r}{3}$ mm
The diameter of the second capillary tube d$_2$becomes, d$_2$= 2r$_2$= 2$\left( {\dfrac{r}{3}} \right)$ mm
Writing the formula for height of capillary rise for the second tube, where the height up to which water rises in the second tube is h$_2$mm.
$\Rightarrow$ h$_2$= $\dfrac{{4\sigma \cos \theta }}{{\rho {d_2}g}}$ mm …equation (3)
Since there has been no change in the liquid, so the angle of contact also remains the same along with other constants. Substituting the value of cos $\theta$ in equation (3), we get,
$\Rightarrow$ h$_2$ = $\dfrac{{4\sigma \times 1}}{{\rho {d_2}g}}$
Separating the constant terms and d$_2$in the right side of the equation, we obtain,
$\Rightarrow$ h$_2$ = $\dfrac{1}{{{d_2}}} \times \left( {\dfrac{{4\sigma }}{{\rho g}}} \right)$ …equation (4)
Substituting equation (2) in equation (4) and putting the value of d$_2$ in terms of r, we obtain,
$\Rightarrow$ h$_2$ = $\dfrac{1}{{\left( {\dfrac{{2r}}{3}} \right)}} \times 3\left( {2r} \right)$
Upon simplifying,
$\Rightarrow$ h$_2$ = $\dfrac{3}{{({2}{r})}} \times 3({2}{r})$ = 9 mm
Hence, the height up to which water rises in the second tube is 9 mm.
Therefore, the correct option is D.

Note: The question can also be solved by substituting the values of surface tension, density and acceleration due to gravity to find the value of the radius of the first capillary tube and then using one-third of the obtained value of the radius to find the height of capillary rise in the second tube. However, it would become very calculation intensive and thus, the answer might be susceptible to calculation error.