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 

\[\pi r^2 \times 80 = 80\pi  {cm}^3\]

volume of the water that flows through pipe in half an hour

\[80\pi \times 30 \times 60 = 144000\pi {cm}^3\]

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 =  \[\pi \left( R \right)^2 h = \pi \left( 40 \right)^2 h\]

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

\[\Rightarrow \pi \left( 40 \right)^2 h = 144000\pi\]

\[ \Rightarrow h = \frac{144000}{1600} = 90 cm\]

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