Question

If $M$ is the mass of water that rises in the capillary tube of radius $r$, then mass of water which will rise in a capillary tube of radius $2r$ is:
$A.4M$
$B.M$
$C.2M$
$D.\dfrac{M}{2}$

Answer 1

We know that capillary rise of a liquid is associated with the phenomenon of surface tension. The tendency of a liquid surface to shrink into the minimum possible surface area is called surface tension. Adhesive force of the walls is responsible for the capillary rise. The height of the capillary varies inversely with the radius of the tube. This approach is required to solve this particular problem.
Formula used:
We are going to use the following two relations to solve this problem.
$\rho =\dfrac{m}{V}$and $h\propto \dfrac{1}{r}$

Complete answer:
Capillary rise is defined as the rise of liquid above zero pressure due to cohesive forces of walls of the tube. Before dealing with these types of problems we must note down the basic formula used to find the rise or fall of the capillary, which is given as follows:-
$h=\dfrac{2T\cos \Theta }{\rho gr}$……………… $(i)$
Where symbols have their meanings listed below,
$\rho =$Density of water
$r=$Radius of capillary tube
$\Theta =$Angle of contact
$T=$Surface tension
$g=$Acceleration due to gravity
$h=$Height of capillary
From equation$(i)$, we get the fact that,$h\propto \dfrac{1}{r}$, this relation is also known as Jurin’s law.
Using $h\propto \dfrac{1}{r}$ we get that when the radius becomes double the height becomes half.
We have $m=\rho V$…………. $(ii)$ as $\rho =\dfrac{m}{V}$
Where $V$denotes volume of the liquid.
Using $(ii)$for the initial condition, we get
$M=\rho V$
$M=\rho \times \pi {{r}^{2}}h$…………….. $(iii)$($V=\pi {{r}^{2}}h$As capillary is considered as cylinder)
Now, using $(ii)$ for the second condition when radius becomes $2r$ and height becomes $\dfrac{h}{2}$ and represented as $r'$ and $h'$ respectively, we get
$M'=\rho \times \pi {{(r')}^{2}}h'$($M'$is the mass of the liquid when radius becomes $2r$)
$\Rightarrow M'=\rho \times \pi .{{(2r)}^{2}}.\dfrac{h}{2}$
$\Rightarrow M'=\rho \times \pi .4{{r}^{2}}.\dfrac{h}{2}$
$\Rightarrow M'=\rho \times \pi 2{{r}^{2}}h$……………. $(iv)$
Dividing (iv) by (iii), we get
$\Rightarrow \dfrac{M'}{M}=\dfrac{\rho \times \pi 2{{r}^{2}}h}{\rho \times \pi {{r}^{2}}h}$…………………… $(v)$
Cancelling the like terms in equation$(v)$, we get
$\Rightarrow \dfrac{M'}{M}=2$
$\Rightarrow M'=2M$

So, the correct answer is “Option C”.

Note:
It is very important to apply the relation $h\propto \dfrac{1}{r}$ to get the idea about the change in the height of capillary and radius of the tube. Relation of density should be applied carefully. The concept of angle of contact should be applied carefully. The value of angle of contact lies between ${{0}^{o}}$ to ${{180}^{o}}$. Surface tension is also defined as the work required to increase the unit area of the liquid film.