Question

A capillary tube is submerged in a broad vessel filled with water such that the upper end of the tube is above the level of water in the vessel by $2cm$ . The internal radius of the capillary is $0.5mm$ . Find the radius of curvature R of a meniscus in the capillary tube. Consider the wetting to be complete, (surface tension of water is $0.075N{m^{ - 1}}$ and $g = 10m{s^{ - 2}}$)

Answer 1

Hint: In order to solve this question, we will use the general formula of the rise in height of the water in capillary tubes in terms of Radius of curvature, the density of the liquid, and surface tension, and thus we will solve for radius of curvature of the meniscus.

Formula used:
The rise in height of water in capillary tube is given by:
$h = \dfrac{{2T}}{{R\rho g}}$ where,
h is the rise in height of the water
T is the surface tension of water
R is the radius of the curvature of meniscus
$\rho = 1000kg{m^{ - 3}}$ density of the water
$g = 10m{s^{ - 2}}$ acceleration due to gravity

Complete answer:
According to the question, we have given the following parameters value
rise of water at a height of $h = 2cm = 0.02m$
surface tension of water $T = 0.075N{m^{ - 1}}$
$g = 10m{s^{ - 2}}$ acceleration due to gravity
density of water we know as $\rho = 1000kg{m^{ - 3}}$

On substituing these value to find the R in the formula $h = \dfrac{{2T}}{{R\rho g}}$ we get,
$
  0.02 = \dfrac{{2(0.075)}}{{R(1000)(10)}} \\
  R = \dfrac{{0.15}}{{0.02 \times 10000}} \\
  R = 75 \times {10^{ - 5}}m \\
 $

Hence, the radius of curvature of meniscus will be $R = 75 \times {10^{ - 5}}m$

Note: The basic unit of conversion here is used as $1cm = {10^{ - 2}}m$ and always makes all the physical quantities in the same units while solving such numerical problems in order to avoid any minor errors.