Question

A capillary tube of radius $R$is immersed in water and water rises in it to height $H$. Mass of water in the capillary tube is $M$. If the radius of the tube is doubled, mass of water that will rise in capillary tube water will be
A. $2M$
B. $M$
C. $\dfrac{M}{2}$
D. $4M$

Answer 1

We know that when one end of the capillary tube is immersed in any liquid, the liquid will rise to some height in the tube. The shape of the liquid meniscus in the tube is concave upwards which makes an angle of contact with the tube. We will use the formula for this rise in height of liquid and its relation with mass as we have to determine the relation between mass of water and radius of the tube to get the answer.

Formula used:
$h = \dfrac{{2S\cos \theta }}{{\rho gr}}$, where, $h$is the height of rise of liquid level in the tube, $S$is surface tension of liquid, $\theta$is angle of contact, $\rho$is density of the liquid, $g$is the gravitational acceleration and $r$is the radius of curvature which is same as the radius of the tube
$2M$, where, $m$is mass of the liquid, $\rho$is density of the liquid, $A$is area of the cross section of the tube and $h$is the height of rise of liquid level in the tube

Complete step by step answer:
We know that the height of the water rise in the tube is given by $h = \dfrac{{2S\cos \theta }}{{\rho gr}}$
We are given that a capillary tube of radius $R$is immersed in water and water rises in it to height $H$, therefore
$H = \dfrac{{2S\cos \theta }}{{\rho gR}}$
The mass of liquid in the tube is given by $m = \rho Ah$.
Here, Mass of water in the capillary tube is $M$, therefore
$M = \rho AH$
Now, we will put the value $H = \dfrac{{2S\cos \theta }}{{\rho gR}}$ and cross section area $A = \pi {R^2}$ in this equation of mass
$\Rightarrow M = \rho \left( {\pi {R^2}} \right)\left( {\dfrac{{2S\cos \theta }}{{\rho gR}}} \right) \\\ \Rightarrow M = \pi R\left( {\dfrac{{2S\cos \theta }}{g}} \right) \\\$
Here, the terms surface tension of water, angle of contact and gravitational acceleration remains unchanged when the radius changes.
$\Rightarrow M \propto R$
This means that the mass of water in the tube is directly proportional to the radius of the tube. Therefore, if the radius of the tube is doubled, the mass of water that will rise in capillary tube water will also be doubled and become $2M$.

So, the correct answer is “Option A”.

Note:
We have seen in the formula of height of liquid level rise that it is dependent on surface tension of liquid, angle of contact, density of the liquid, the gravitational acceleration and the radius of curvature which is the same as the radius of the tube. There is one important thing to consider about the relation of angle of contact and height of liquid level rise. Liquid level in the tube will rise when the angle of contact is acute and fall when the angle is obtuse.