Question

Water is flowing at the rate of 7 meters per second through a circular pipe whose internal diameter is 2 cm into a cylindrical tank, the radius of whose base is 40 cm. Determine the increase in the water level in \[\dfrac{1}{2}\] an hour.

Answer 1

Hint: Find the volume of water that flows through the pipe in half an hour and equate it to the formula of volume of the cylindrical tank, that is, \[V = \pi {r^2}h\] where ‘r’ is the radius of the base and ‘h’ is the height of the cylindrical tank.

Complete step by step answer:

It is given that the water flows out through a circular pipe, whose internal diameter is 2 cm.

The radius of the circular pipe is half of the diameter, that is, half of 2 cm.

\[{r_{pipe}} = \dfrac{1}{2} \times 2\]

\[{r_{pipe}} = 1cm\]

The area of cross-section of the pipe is given as \[\pi {r_{pipe}}^2\], hence, we have:

\[A = \pi {r_{pipe}}^2\]

\[A = \pi {1^2}\]

\[A = \pi c{m^2}...........(1)\]

The water flows at a rate of 7 m per second into the cylindrical tank.

\[7m/s = 700cm/s\]

The volume of water flowing into the cylinder per second is equal to the product of the area

of cross-section and the flow rate.

\[v = \pi \times 700\]

\[v = 700\pi c{m^3}/s\]

The volume of water that flows into the tank in half an hour is given by:

\[V = 700\pi \times \dfrac{1}{2} \times 60 \times 60\]

\[V = 1260000\pi c{m^3}............(2)\]

We know that the volume of a cylinder with radius r and height h is given as follows:

\[V = \pi {r^2}h\]

The radius of the base of the cylindrical tank is 40 cm.

\[V = \pi {(40)^2}h\]

\[V = 1600\pi h...........(3)\]

We know that equation (2) and equation (3) are equal. Then, we have:

\[1600\pi h = 1260000\pi \]

Solving for h, we have:

\[h = \dfrac{{1260000}}{{1600}}\]

\[h = 787.5m\]

Hence, the water fills up to a height of 787.5 cm.


Note: Convert all quantities into the same units before multiplying or dividing them,

otherwise, the answer will differ by powers of 10, which is wrong.