A rectangular vessel of dimensions $20cm\times 26cm\times 11cm$ is full of water. This water is poured into a conical vessel. The top of the conical vessel has its radius 10cm. If the conical vessel is filled completely, determine the height. $\left( \pi =\dfrac{22}{7} \right)$.

Hint: Focus on the point that the volume of water for both the vessels will be equal. So, equate the volume of the two vessels to get the height of the cone.

Complete step-by-step answer:
Let us start by drawing a representative figure of both the vessels for a better visualisation.



As given in the question, the dimensions of the cuboidal vessel is $20cm\times 26cm\times 11cm$ . Therefore, the volume of water that can be stored in the cuboidal vessel is equal to the volume of the cuboidal vessel. So, using the formula: volume of cuboid = $l\times b\times h$ , we get
The volume of cuboidal vessel = $l\times b\times h=20\times 26\times 11=5720c{{m}^{3}}$
Now according to the question, the water from the cuboidal vessel is transferred to the conical vessel. Therefore the volume of the conical vessel must be equal to 5760 $c{{m}^{3}}$ in order to store the water. We know that the volume of the cone is $\dfrac{1}{3}\pi {{r}^{2}}h$.
$\therefore \dfrac{1}{3}\pi {{r}^{2}}h=5720$
Now using the value of r and $\pi $ given in the question, we get
 $\dfrac{1}{3}\times \dfrac{22}{7}\times {{10}^{2}}\times h=5720$
$\Rightarrow h=\dfrac{5720\times 7\times 3}{22\times 100}$
$\Rightarrow h=\dfrac{120120}{2200}=54.6$cm
Therefore, we can conclude that the height of the conical vessel is 54.6 cm.

Note: Make sure to convert all the dimensions to a standardized system of units,this decreases the chance of errors. Also, be sure to use the value of $\pi $ according to the questions and calculations as many times it is the value of $\pi $ which decides the complexity of the calculations in our solution. Also, we need to remember all the basic formulas for surface area and volume of the general 3-D shapes like the cone, cube, cylinder, etc.