A cone of radius $7\text{cm}$ and height $12\text{cm}$ is completely filled with water. This water is emptied into an empty cylindrical vessel of radius $3.5\text{cm}$. What will be the height of water in this vessel?

The volume of water remains constant when transferred from the cone to the cylinder.
$$\text{Volume of cone}=\text{Volume of cylinder}$$
$$\frac{1}{3}\pi r_1^2{h}_1=\pi r_2^2{h}_2$$
Given:Radius of cone ($r_1$) = $7\text{cm}$ Height of cone ($h_1$) = $12\text{cm}$ Radius of cylinder ($r_2$) = $3.5\text{cm}=\frac{7}{2}\text{cm}$ Height of water in cylinder = $h_2$ Substitute the values into the equation:
$$\frac{1}{3}\times \pi \times 7\times 7\times 12=\pi \times \left(\frac{7}{2}\right)\times \left(\frac{7}{2}\right)\times h_2$$
Cancel $\pi$ and $7\times 7$ from both sides:
$$\frac{1}{3}\times 12=\frac{1}{4}\times h_2$$
$$4=\frac{1}{4}\times h_2$$
$$h_2=4\times 4=16\text{cm}$$
Therefore, the height of water in the vessel will be $16\text{cm}$.