A vertical tube with a semicircular cross section contains a...

Let the radius of the semicircular cross-section of the tube be $R$.

The surface tension force acting on any part of the wall is given by $F = T \times l \times \cos\theta$, where $T$ is the surface tension of the liquid, $l$ is the length of the line of contact, and $\theta$ is the contact angle.

Given that the contact angle $\theta = 0^{{}^{{}^{\circ}}}$, we have $\cos\theta = 1$. Thus, the force is simply proportional to the length of contact, $F = T \times l$.

For the curved part, the length of the line of contact is the arc length of the semicircle:
$l_c = \pi R$
Therefore, the surface tension force on the curved part is:
$F_c = T \times \pi R$

For the flat part, the length of the line of contact is the diameter of the semicircle:
$l_f = 2R$
Therefore, the surface tension force on the flat part is:
$F_f = T \times 2R$

The ratio of the surface tension force acting on the curved part to that on the flat part is:
$\dfrac{F_c}{F_f} = \dfrac{T \times \pi R}{T \times 2R} = \dfrac{\pi}{2}$