Question

Capillary tubes of diameters 1, 1.5, 2 mm are dipped vertically in the same liquid. The capillary ascents of the liquid in the tube are in the ratio:
A. $2:3:4$
B. $1:1:0.5$
C. $3:4:6$
D. $4:3:2$

Answer 1

Hint: Three capillary tubes of different diameters are dipped in the same liquid. Since the liquid used is the same, the factors like the surface tension, the angle of contact, the density of the liquid and the acceleration due to gravity will remain the same.
The radius of the diameter is inversely related to the height to which the liquid rises in the capillary. Lesser the diameter, more is the height of the liquid raised, as the forces due to surface tension have more effect in a thin capillary.

Formula Used:
The height to which the liquid rises is given by:
\[h = \dfrac{{2T\cos \theta }}{{\rho rg}} \\
\Rightarrow h= \dfrac{{4T\cos \theta }}{{\rho dg}}\]
where, \[T\] is the surface tension
\[\theta \] is the angle of contact
\[\rho \] is the density of liquid
\[r\] is the radius of capillary
\[g\] is the acceleration due to gravity
\[d\] is the diameter of the capillary tube.

Complete step by step answer:
 A tube of very small bore is called capillary. If a glass capillary tube is dipped into a liquid such as water, which wets the glass, the liquid rises in the tube. However, if the glass capillary tube is dipped in a liquid such as mercury which does not wet the glass, the liquid falls in the tube (totally opposite of rising). This phenomenon of rise and fall in the capillary tube is called capillarity.

The height to which the liquid rises is given by:
\[h = \dfrac{{2T\cos \theta }}{{\rho rg}} \\
\Rightarrow h= \dfrac{{4T\cos \theta }}{{\rho dg}}\]
In the given problem, three tubes of diameters 1, 1.5, 2 are given.
\[{d_1}:{d_2}:{d_3} = 1:1.5:2\]
\[\Rightarrow{h_1}:{h_2}:{h_3} = \dfrac{{4T\cos \theta }}{{\rho {d_1}g}} :\dfrac{{4T\cos \theta }}{{\rho {d_2}g}}:\dfrac{{4T\cos \theta }}{{\rho {d_3}g}}\]
\[\Rightarrow {h_1}:{h_2}:{h_3} = \dfrac{1}{{{d_1}}} :\dfrac{1}{{{d_2}}}:\dfrac{1}{{{d_3}}} \\
\Rightarrow{h_1}:{h_2}:{h_3} = \dfrac{1}{1} :\dfrac{1}{{1.5}}:\dfrac{1}{2} \\
\therefore {h_1}:{h_2}:{h_3} = 1:0.66:0.5 \approx 1:1:0.5 \\ \]
All the terms (except the diameter terms) are cancelled because the capillaries are dipped in the same liquid. Therefore, the terms tension, angle of contact, density of liquid and acceleration due to gravity do not change.

Hence, option B is the correct answer.

Note:The rise or fall of the liquid in the capillary depends on the type and the nature of liquid used. For the liquids which wet the container, the angle of contact is less than \[{90^ \circ }\], so \[\cos \theta \] is positive and therefore height is also positive. Thus the liquid rises. And, for the liquids which do not wet the container, the angle of contact is more than \[{90^ \circ }\], so \[\cos \theta \] is negative and therefore height is also negative. Thus the liquid falls.