a right circular conical vessel of altitude 20 cm and base radius of 10 cm is kept with its vertex downwards if one liter of water is poured in it how high above the vertex will the water level be

The cone in this question is placed upside-down, with the vertex downwards. Observe that slicing an upside-down cone horizontally would produce a smaller cone, where:

The smaller cone shares the same vertex as the original cone,
The ratio between height [tex]h[/tex] and radius [tex]r[/tex] for the smaller cone is the same as that of the original cone.
The distance between the original vertex and the cutting plane would be the height of this smaller cone.

Thus, if the original cone is of height [tex]h_{0}[/tex] and radius [tex]r_{0}[/tex], while the smaller cone is of height [tex]h_{1}[/tex], the base radius of the smaller cone would be:

[tex]\begin{aligned}& (\text{new base radius}) \\ =\; & (\text{new height})\times \frac{(\text{new base radius})}{(\text{new height})} \\ =\; & (\text{new height}) \times \frac{(\text{original base radius})}{(\text{original height})} \\ =\; & h_{1}\times \frac{r_{0}}{h_{0}}\end{aligned}[/tex].

The volume of this smaller cone of altitude [tex]h_{1}[/tex], embedded in the larger cone of altitude [tex]h_{0}[/tex] and base radius [tex]r_{0}[/tex], would be:

[tex]\displaystyle V_{1} = \frac{1}{3}\, \pi\, \left(\frac{h_{1}\, r_{0}}{h_{0}}\right)^{2}\, h_{1} = \frac{\pi\, {r_{0}}^{2}}{3\, {h_{0}}^{2}}}\, {h_{1}}^{3}[/tex].

In this question, as long as the water does not overflow, it would be as if the water level has sliced the conical interior of this container into two halves. Water in this container would occupy the lower half, which would represent the smaller cone embedded inside this conical container. If the water level is at a distance of [tex]h_{1}[/tex] from the vertex of the conical container, volume of the conical region that water has filled would be:

[tex]\displaystyle V_{1} = \frac{\pi\, {r_{0}}^{2}}{3\, {h_{0}}^{2}}}\, {h_{1}}^{3}[/tex].

It is given that [tex]r_{0} = 10\; {\rm cm}[/tex] and [tex]h_{0} = 20\; {\rm cm}[/tex] for the conical container and that [tex]V_{1} = 1\; {\rm L} = 1 \times 10^{3}\; {\rm cm^{3}}[/tex] for the water in the container. Rearrange the equation above to find the height of the water level, [tex]h_{1}[/tex]:

[tex]\begin{aligned}h_{1} &= \sqrt[3]{\frac{3\, V_{1}\, {h_{0}}^{2}}{\pi\, {r_{0}}^{2}}} \\ &= \sqrt[3]{\frac{(3)\, (1 \times 10^{3}\; {\rm cm})\, (20\; {\rm cm})^{2}}{(\pi)\, (10\; {\rm cm})}} \\ &\approx 15.6\; {\rm cm}\end{aligned}[/tex].

(Note the cube root in the expression- volume of water in this container is proportional to the cube of the height of the water level.)