Question

A coaxial cylinder made of solid glass inside a hollow cylinder is immersed in water of surface tension ‘S’. Radius of inner and outer surface are \[{R_1}\]and \[{R_2}\]respectively. Height till which the liquid will rise in the gap between the solid and the hollow cylinder is (Density of liquid is
\[\rho \])

A. \[\dfrac{{2S}}{{{R_2}\rho g}}\]
B. \[\dfrac{{2S}}{{{R_1}\rho g}}\]
C. \[\dfrac{S}{{({R_2} - {R_1})\rho g}}\]
D. \[\dfrac{{2S}}{{({R_2} - {R_1})\rho g}}\]

Answer 1

Hint:The height to which the water will rise within the gap due to surface tension depends on the difference between the outer radius and the inner radius of the co-axial cylinder, the density of water, surface tension and the angle of contact.

Formula Used:The height to which the liquid rises due to capillarity is given by:\[h = \dfrac{{2S\cos
\theta }}{{\rho rg}}\]

Complete step by step solution:
Surface tension is the property of liquids by virtue of which its free surface behaves like a free membrane. It is given in the problem that a coaxial cylinder is made up of solid glass and is placed inside the liquid of surface tension\[S\]. The coaxial cylinder consists of two
cylinders with the inner cylinder of radius\[{R_1}\] and the outer cylinder of radius\[{R_2}\].

Therefore, the liquid in the hollow cylinder will rise within the space between inner and outer radius due to capillarity. The phenomenon of rise and fall of a liquid in the capillary tube is called capillarity.

The height to which the liquid rises due to capillarity is given by

\[h = \dfrac{{2S\cos \theta }}{{\rho rg}}\] \[ \to (1)\]

where, \[S\]is the surface tension

\[\theta \] is the angle of contact
\[\rho \] is the density of liquid
\[r\] is the radius of capillary or cylinder
\[g\] is acceleration due to gravity.

The radius \[r\]in equation (1) will be the radius of outer cylinder minus the radius of inner cylinder. Therefore, \[r = {R_2} - {R_1}\]

The angle of contact between solid glass and pure water is zero. That is \[\theta = {0^ \circ }\].
Then, \[\cos \theta = \cos {0^ \circ } = 1\]

The height of water that rises in the gap between inner and outer radius of the co-axial cylinder is
\[h = \dfrac{{2S}}{{\left( {{R_2} - {R_1}} \right)\rho g}}\]

Hence, option (D) is the correct answer.

Note:The water used should be pure. The angle of contact will only be zero when it is for pure water and clean solid glass. Impure water such as tap water may increase the height to which water will rise.

For impure water, the angle of contact will be less than \[{90^ \circ }\].