Wax is coated on the inner wall of a capillary tube and the tube is then dipped in water. Then, compared to the unwaxed capillary, the angle of contact $$\theta$$ and the height $$h$$ upto which water rises change. These changes are :

First, recall the formula for the height $$ h $$ to which a liquid rises in a capillary tube:

Here, $$ S $$ is the surface tension of the liquid (water), $$ \theta $$ is the angle of contact, $$ \rho $$ is the density of water, $$ g $$ is the acceleration due to gravity, and $$ r $$ is the radius of the capillary tube. The radius $$ r $$, density $$ \rho $$, gravity $$ g $$, and surface tension $$ S $$ remain constant. Only $$ \theta $$ changes when the tube is waxed.

For an unwaxed glass capillary tube, water wets the surface, and the angle of contact $$ \theta $$ is acute (less than 90°). Typically, $$ \theta $$ is close to 0° for pure water and clean glass. Thus, $$ \cos \theta $$ is positive, and $$ h $$ is positive, meaning water rises in the tube.

When the inner wall is coated with wax, the surface becomes hydrophobic (water-repelling). Wax increases the angle of contact $$ \theta $$ because water does not wet the surface as well. For wax, $$ \theta $$ becomes obtuse (greater than 90°), often around 106° for paraffin wax. Therefore, $$ \theta $$ increases compared to the unwaxed tube.

Now, examine the effect on $$ h $$. Since $$ \theta $$ increases from less than 90° to more than 90°, $$ \cos \theta $$ changes:

• When $$ \theta < 90^\circ $$, $$ \cos \theta > 0 $$ (positive).
• When $$ \theta > 90^\circ $$, $$ \cos \theta < 0 $$ (negative).

Substituting into the formula:

For unwaxed tube: $$ h_{\text{unwaxed}} = \frac{2S \cos \theta_{\text{unwaxed}}}{\rho g r} > 0 $$ (since $$ \cos \theta_{\text{unwaxed}} > 0 $$).

For waxed tube: $$ h_{\text{waxed}} = \frac{2S \cos \theta_{\text{waxed}}}{\rho g r} < 0 $$ (since $$ \cos \theta_{\text{waxed}} < 0 $$).

A negative $$ h $$ means capillary depression occurs—water falls below the surrounding level instead of rising. The question asks for the height "up to which water rises." In the waxed tube, water does not rise; it depresses. Therefore, compared to the positive rise in the unwaxed tube, the height of rise $$ h $$ decreases (it becomes negative, so it is less than the positive value).

Thus:

• The angle of contact $$ \theta $$ increases (from acute to obtuse).
• The height $$ h $$ to which water rises decreases (from positive to negative).

Comparing with the options:

A. $$ \theta $$ increases and $$ h $$ increases → Incorrect, because $$ h $$ decreases.

B. $$ \theta $$ decreases and $$ h $$ decreases → Incorrect, because $$ \theta $$ increases.

C. $$ \theta $$ increases and $$ h $$ decreases → Correct.

D. $$ \theta $$ decreases and $$ h $$ increases → Incorrect, because $$ \theta $$ increases and $$ h $$ decreases.

Hence, the correct answer is Option C.