Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of\[80{cm}/{\sec }\;\] in an empty cylindrical tank, the radius of whose base is 40 cm. What is the rise of water level in a tank in half an hour?

Hint: As the water flows through a cylindrical pipe, so, we will make use of the formula of volume of the cylinder. The volume of the water in a cylindrical tank equals the volume of the water that flows from the circular pipe in half an hour. So, we will equate this condition to find the value of the rise in water level in the tank in half an hour.
Formula used:
\[V=\pi {{r}^{2}}h\]

Complete answer:
From the given information, we have the data as follows.
Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of\[80{cm}/{\sec }\;\] in an empty cylindrical tank, the radius of whose base is 40 cm.
The formula that we are using to solve this question is given as follows.
The volume of a cylinder.
\[V=\pi {{r}^{2}}h\]
Where r is the radius of the cylinder and h is the height of the cylinder.
The volume of the water in a cylindrical tank equals the volume of the water that flows from the circular pipe in half an hour.
The radius of the cylindrical tank is, \[{{r}_{1}}=40\,cm\]
The radius of the circular pipe is, \[{{r}_{2}}=1\,cm\]
The height of the cylindrical tank is, \[{{h}_{1}}=?\]
The speed of water = \[80{cm}/{\sec }\;\]
The height of the circular pipe is, \[{{h}_{2}}=80\times 60\times 30=144000\,cm\]
\[\pi r_{1}^{2}{{h}_{1}}=\pi r_{2}^{2}{{h}_{2}}\]
Substitute the values in the above equation.
\[\begin{align}
  & {{40}^{2}}\times {{h}_{1}}={{1}^{2}}\times 144000 \\
 & \Rightarrow {{h}_{1}}=\dfrac{144000}{1600} \\
 & \therefore {{h}_{1}}=90\,cm \\
\end{align}\]
\[\therefore \] The rise in the level of water in the cylindrical tank is 90 cm in half an hour.

Note:
As the units of all the parameters are in cm, so, no need to change them. The condition to solve this problem was, the volume of the water in a cylindrical tank equals the volume of the water that flows from the circular pipe in half an hour.