Question

A capillary tube of radius $r$ is immersed in water and water rises in it to a height $h$. The mass of water in the capillary tube is $m$. If the radius of the tube is doubled, the mass of water that will rise in the capillary tube will now be-
A) $m$
B) $2m$
C) $\dfrac{3}{2}$
D) $4m$

Answer 1

When on end of the capillary tube of radius $r$ is immersed in water (liquid) of density $\rho$. Then water rises to some height $h$ in the tube and the shape of the liquid meniscus in the tube is concave upwards, making an angle of contact $\theta$ with the tube.

Complete step by step answer:
Step 1: The radius of the capillary tube is $r$ so the radius of curvature of the liquid meniscus will be $r$.
The height by which the liquid level rises in the tube can be given by the formula-
$h = \dfrac{{2S\cos \theta }}{{\rho gr}}$ ……... (1)
Where $S =$Surface tension of liquid, $\theta =$angle of contact, $\rho =$density of liquid, $r =$radius of curvature i.e. equal to the radius of tube, and $g =$acceleration due to gravity (used to calculate the weight of liquid)

Step 2: But we know that the mass of the liquid can be given by the formula-
Mass$=$Volume$\times$Density
$\Rightarrow m = Ah \times \rho$
Where $A = \pi \mathop r\nolimits^2$=area of the circle from which concave curvature is taken, $h =$height of the liquid in the tube, $\rho =$density of liquid
So, $m = \pi \mathop r\nolimits^2 h \times \rho$ …….. (2)

Step 3: Now from equation (1),
$\pi \mathop r\nolimits^2 h\rho = \dfrac{{2S\cos \theta }}{{\rho gr}} \times \pi \mathop r\nolimits^2 \rho$ …... (3)
From equation (2) and (3)
$\Rightarrow m = \dfrac{{2S\cos \theta }}{{\rho gr}} \times \pi \mathop r\nolimits^2 \rho$
$\Rightarrow m = \dfrac{{2S\cos \theta }}{g} \times \pi r$
$\Rightarrow m \propto r$ ………... (4)
From equation (4), it is clear that mass is directly proportional to the radius. So, if radius is doubled then the mass of liquid in the tube will be doubled i.e. $m = 2m$.

$\therefore$ Hence, the correct option is (B).

Note:
(i) If the angle of contact is obtuse then $h$ will be negative. It shows that the liquid meniscus will be convex upwards. In this situation the liquid level will be pressed in the tube.
(ii) Liquid level in the tube will rise when $\theta < 90^\circ$, liquid level will fall when $\theta < 90^\circ$, and liquid level will remain unchanged when $\theta = 90^\circ$.