Water ascends to a length of 10 cm in a vertical capillary tube. If...

The vertical height of water in the capillary tube is $h = 10 \text{ cm}$.

When the capillary tube is inclined at an angle $\alpha$ with the vertical, the vertical height of the liquid column remains constant.

Let $l$ be the length of the liquid column in the inclined tube.

From the geometry of the setup, $h = l \cos \alpha$.

Given $\alpha = 45^\circ and$h = 10 \text{ cm}$.<br><br>$10 = l \cos 45^\circ$<br><br>$10 = l \left(\dfrac{1}{\sqrt{2}}\right)

$l =$10\sqrt{2} \text{ cm}