Question

If the surface tension of water is $0.06N/m$ , then the capillary rise in a tube of a diameter 1 mm is : $\left( {\theta = 0^\circ } \right)$
(A) $1.22cm$
(B) $2.44cm$
(C) $3.12cm$
(D) $3.86cm$

Answer 1

Hint
The height up to which the water level rises in a capillary tube is given by the formula,
$h = \dfrac{{2T\cos \theta }}{{\rho gr}}$ where we know that the density of water and acceleration due to gravity is $\rho = 1000kg/{m^3}$ and $g = 9.8m/{s^2}$ respectively. Substituting these values and the ones given in the question we can get the answer.
Formula used: To solve this problem, we use the formula,
$h = \dfrac{{2T\cos \theta }}{{\rho gr}}$
where $h$ is the height up to which the water level rises, $T$ is the tension in the surface of the water, $\rho$ is the density of water, $g$ is the acceleration due to gravity and $r$ is the radius of the capillary tube.

Complete step by step answer
We know that there is a force of attraction present between the molecules of water(cohesive force). When a capillary tube is inserted in water, then there comes into action a force of attraction between the water molecules and the glass wall of the capillary tube (adhesive force). To overcome this force of attraction, the water level rises in the capillary tube, until this force is balanced by the force of gravity. Now, this height up to which the water level rises in the capillary tube is given by the formula,
$h = \dfrac{{2T\cos \theta }}{{\rho gr}}$
In the question, we are said that the surface tension on the water is given $T = 0.06N/m$ and the diameter of the capillary tube is given as 1 mm.
So, the radius of the capillary tube is given by,
$r = \dfrac{{diameter}}{2}$
$\therefore r = \dfrac{1}{2}mm = 0.5 \times {10^{ - 3}}m$
Now, we know that the density of water is $\rho = 1000kg/{m^3}$ and the acceleration due to gravity has a value $g = 9.8m/{s^2}$ .
The value of $\theta$ given in the question is $\theta = 0^\circ$ . So, $\cos \theta = \cos 0^\circ = 1$
Now, by substituting all these values in the equation, we have
$h = \dfrac{{2T\cos \theta }}{{\rho gr}} = \dfrac{{2 \times 0.06 \times 1}}{{1000 \times 9.8 \times 0.5 \times {{10}^{ - 3}}}}$
By multiplying the values in the numerator and the denominator,
$\therefore h = \dfrac{{0.12}}{{4.9}}m$
Further solving gives us,
$h = 0.0244m$
In cm it is $h = 0.0244 \times 100cm = 2.44cm$
So, the correct answer is option (B).

Note
In the case of water, the adhesive force is more than the cohesive force, so the water rises in the glass capillary. But if a liquid had a cohesive force greater than the adhesive force then the level of the liquid depresses. The formula for the capillary rise is derived by balancing the weight of the liquid $\left( {\pi {r^2}h\rho g} \right)$ to the surface tension, $\left( {2\pi rT\cos \theta } \right)$ .