Rain water, which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius 20 cm. What will be the height of water in the cylindrical vessel if a rainfall of 1 cm has fallen?
The fallen rains are in the form of a cuboid of height 1 cm, length 6 m = 600 cm and breadth 4 m = 400 cm. Therefore, the volume of the fallen rains is
$V=600 \times 400 \times 1=240000 \mathrm{~cm}^{3}$
The fallen rains are transferred into a cylindrical vessel of internal radius r1 = 20 cm. Let, the height of the water in the cylindrical vessel is h1 cm. Then, the volume of the water in the cylinder is
$V_{1}=\pi r_{1}^{2} h_{1}=\frac{22}{7} \times(20)^{2} \times h_{1}$
Since, the volume of the water in the cylinder is same as the volume of the rainfalls, we have
$V_{1}=V$
$\Rightarrow \frac{22}{7} \times(20)^{2} \times h_{1}=240000$
$\Rightarrow \quad h_{1}=\frac{240000 \times 7}{(20)^{2} \times 22}$
$\Rightarrow \quad=190.9$
Therefore, the height of the water in the cylinder is 190.9 cm.