On dipping one end of a capillary in liquid and inclining the capillary at an angles 30° and 60° with the vertical, the lengths of liquid columns in it are found to be l1 and l2 respectively. The ratio of l1 and l2 is

To find the ratio of the lengths of the liquid columns inside the inclined capillary tube, let us understand how the vertical height of the liquid column relates to the length of the liquid column along the tube.

When a capillary tube is dipped in a liquid, the liquid rises to a certain vertical height $h$ above the free surface of the liquid. This vertical height is determined by the surface tension, the density of the liquid, the acceleration due to gravity, and the radius of the tube. As long as these parameters remain constant, the vertical height $h$ of the liquid column remains constant, even if the capillary tube is inclined at an angle.

Let the capillary tube be inclined at an angle $\theta$ with the vertical.
If $l$ is the length of the liquid column along the inclined capillary tube, then the vertical height $h$ is the vertical component of this length.
Using simple trigonometry, we can relate $h$ and $l$ as:
$h=lcos\theta$

Since the vertical height $h$ remains constant:
$l=\frac{h}{cos\theta }$
This implies that the length of the liquid column is inversely proportional to the cosine of the angle of inclination with the vertical:
$l\propto \frac{1}{cos\theta }$

Let $l_1$ and $l_2$ be the lengths of the liquid columns when the angles with the vertical are ${\theta }_1=30°$ and ${\theta }_2=60°$ respectively.

The ratio of the lengths $l_1$ and $l_2$ is given by:
$\frac{l_1}{l_2}=\frac{cos{\theta }_2}{cos{\theta }_1}$

Substitute the given values of the angles:
$\frac{l_1}{l_2}=\frac{cos60°}{cos30°}$

Using the standard trigonometric values, $cos60°=\frac{1}{2}$ and $cos30°=\frac{\sqrt{3}}{2}$:
$\frac{l_1}{l_2}=\frac{1/2}{\sqrt{3}/2}=\frac{1}{\sqrt{3}}$

Therefore, the ratio of $l_1$ to $l_2$ is 1:√3.