Question

When a capillary is immersed in a liquid, the liquid of mass M rises in the capillary tube. If a capillary tube of double the radius is taken then the mass of the same liquid rising in the tube is?
(A) $2m$
(B) $\dfrac{m}{2}$
(C) $4m$
(D) None of these

Answer 1

In order to answer this question we need to understand that, When on end of the capillary tube of radius $r$is immersed in 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 answer:
Let 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,
$h = \dfrac{{2S\cos \theta }}{{\rho gr}}{\text{ }} - - - (1)$
$S = surface{\text{ }}tension{\text{ }}of{\text{ }}liquid$
$\theta = angle{\text{ }}of{\text{ }}contact$
$\rho = density{\text{ }}of{\text{ }}liquid$
$r = radius{\text{ }}of{\text{ }}curvature$
$g = acceleration{\text{ }}due{\text{ }}to{\text{ }}gravity$
$(used{\text{ }}to{\text{ }}calculate{\text{ }}the{\text{ }}weight{\text{ }}of{\text{ }}liquid)$
But we know that the mass of the liquid can be given by the formula,
$Mass = volume \times density$
$\Rightarrow m = Ah \times \rho$
$A = \pi {r^2}$ = area of the circle from which concave curvature is taken.
$h = {\text{height of the liquid in the tube}}$
$\rho {\text{ = density of liquid}}$
$m = \pi {r^2}h \times \rho {\text{ }} - - - (2)$
Now from equation (1),
$\pi {r^2}h \times \rho = \dfrac{{2S\cos \theta }}{{\rho gr}} \times \pi {r^2}\rho {\text{ }} - - - (3)$
From equation (2) and (3)
$\Rightarrow m = \dfrac{{2S\cos \theta }}{{\rho gr}} \times \pi {r^2}\rho$
$\Rightarrow m = \dfrac{{2S\cos \theta }}{g} \times \pi r$
$\Rightarrow m{\text{ }}\alpha {\text{ }}r{\text{ }} - - - (4)$
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. $m = 2m.$

Hence, the correct option is (B).

Note:
It should be remembered that 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. 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 $\theta = {90^ \circ }.$