The lower end of a capillary tube of radius 1 mm is dipped...

The formula for the capillary depression $h$ of a liquid is given by:

$h = \dfrac{2T \cos(180^\circ - \theta)}{\rho g r}$

Substituting the given values ($T = 0.465 \text{ N/m}$, $\theta = 135^\circ$, $\rho = 13600 \text{ kg/m}^3$, $g = 9.8 \text{ m/s}^2$, $r = 10^{-3} \text{ m}$):

$h = \dfrac{2 \times 0.465 \times \cos 45^\circ}{13600 \times 9.8 \times 10^{-3}}$

$h = \dfrac{0.93 \times \dfrac{1}{\sqrt{2}}}{133.28} \approx 0.00493 \text{ m} = 4.93 \text{ mm}$

When the tube is dipped to a depth $l$ which is less than the natural depression $h$, the mercury meniscus stays at the lower end of the tube and adjusts its contact angle to a new value $\theta'$ to balance the pressure. The new depth is related to $\theta'$ by:

$l = \dfrac{2T \cos(180^\circ - \theta')}{\rho g r}$

Given that the length dipped is half of the natural depression ($l = \dfrac{h}{2}$), we get:

\dfrac{2T \cos(180^\circ - \theta')}{\rho g r} = \dfrac{1}{2} \left(\dfrac{2T \cos(180^\circ - \theta)}{\rho g r} \right)

$\cos(180^\circ - \theta') =$\dfrac{1}{2} \cos($180^\circ - 135^\circ) =$\dfrac{1}{2} \cos$45^\circ$

$\cos(180^\circ - \theta') =$\dfrac{1}{2\sqrt{2}} \approx$0.3535$

This gives $180^\circ - \theta' \approx 69.3^\circ$.

The angle made by the mercury surface with the vertical (which is the new angle of contact $\theta'$) is:

$\theta' = 180^\circ - 69.3^\circ = 110.7^\circ \approx 111^\circ$

Thus, the depression is $4.93 \text{ mm}$ and the angle is $111^\circ$.