Question

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 tank in half an hour ?

Answer 1

The inner radius of the cylindrical pipe r =1 cm.

Rate of flow of water = 80 cm/sec

volume of the water that flows through pipe in 1sec is ${\mathrm{\pi r}}^2\times 80=80\mathrm{\pi }{\mathrm{cm}}^3$

volume of the water that flows through pipe in half an hour $80\mathrm{\pi }\times 30\times 60=144000\mathrm{\pi }{\mathrm{cm}}^3$

radius of the base of the cylindrical tank is R = 40 cm

let the water level in the cylinderical tank after half an hour be h.

volume of the raised water = $\mathrm{\pi }{\left(\mathrm{R}\right)}^2\mathrm{h}=\mathrm{\pi }{\left(40\right)}^2\mathrm{h}$

volume of the raised water in tank = volume of the water that flows through pipe

$$\begin{gathered} \Rightarrow \mathrm{\pi }{\left(40\right)}^2\mathrm{h}=144000\mathrm{\pi } \\ \Rightarrow \mathrm{h}=\frac{144000}{1600}=90\mathrm{cm} \end{gathered}$$

Thus water level will rise by 90 cm in half an hour.