Question

Water rises to a height h in a capillary tube of a certain diameter. This capillary tube is replaced by a similar tube of half the diameter. Now the water will rise to a height of
A. 4h
B. 3h
C. 2h
D. h

Answer 1

As a first step, you could recall the expression for surface tension for the liquid in a capillary tube. From that expression, you will understand the relation between the height at which the liquid rises in a capillary tube with the radius of the capillary tube. Now you could apply this relation to both the given cases and thus determine the answer.

Formula used:
Surface tension,
$\gamma =\dfrac{rh\rho g}{2\cos \theta }$

Complete step-by-step solution
In the question, we are given the height at which water rises in a capillary tube of a certain diameter. Now the capillary tube is replaced by a similar tube of half the diameter of the previous tube. We are asked to find the height at which water will rise.
In order to answer this let us recall the expression for a surface of the liquid in a capillary tube which is given by,
$\gamma =\dfrac{rh\rho g}{2\cos \theta }$
$\Rightarrow h=\dfrac{2\gamma \cos \theta }{r\rho g}$
Where, h is the height at which the liquid rises in the capillary,$\gamma$ is the surface tension of the liquid, r is the radius of the capillary tube, $\rho$ is the density of the liquid, g is the acceleration due to gravity and $\theta$ is the angle of contact with the surface.
From this relation we see that the height at which the liquid rises in the capillary is inversely proportional to the radius of the capillary tube, that is,
$h\propto \dfrac{1}{r}$
Let the diameter of the first capillary tube be${{d}_{1}}$ , its radius be ${{r}_{1}}$ and the height at which the liquid rises in the tube be${{h}_{1}}$, then,
${{h}_{1}}\propto \dfrac{1}{{{r}_{1}}}$
But,
${{r}_{1}}=\dfrac{{{d}_{1}}}{2}$
So,
${{h}_{1}}\propto \dfrac{2}{{{d}_{1}}}$ ……………………………………………. (1)
Now, let the diameter of the second capillary tube be${{d}_{2}}$ , its radius be ${{r}_{2}}$ and the height at which the liquid rises in the tube be${{h}_{2}}$, then,
${{h}_{2}}\propto \dfrac{1}{{{r}_{2}}}$
But,
${{r}_{2}}=\dfrac{{{d}_{2}}}{2}$
Also,
${{d}_{2}}=\dfrac{{{d}_{1}}}{2}$
Then,
${{h}_{2}}\propto \dfrac{4}{{{d}_{1}}}$………………………………………………………… (2)
Combining (1) and (2), we get,
$\dfrac{{{h}_{1}}}{{{h}_{2}}}=\dfrac{\dfrac{2}{{{d}_{1}}}}{\dfrac{4}{{{d}_{1}}}}$
$\Rightarrow \dfrac{{{h}_{1}}}{{{h}_{2}}}=\dfrac{1}{4}$
$\therefore {{h}_{2}}=4{{h}_{1}}$
But in the question ${{h}_{1}}$ is given as h, therefore, the height at which the water will rise for a similar tube of half the diameter is found to be 4h. Hence, option A is the right answer.

Note: The rise and fall of liquid in capillary tubes can be explained by the existence of pressure difference. The liquid surface is known to be flat when the pressure difference between the liquid side and the vapor side is zero. When the concave surface the pressure on the vapor side is more and for the convex liquid surface the pressure on the liquid side is more.