A water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of `80 cm^(-1)` 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?

Explanation

Given , radius of tank , `r_(1) = 40 cm`
Let hight of water level in tank in half an hour =` h_(1)`
Also , given internal radius of cylindrical pipe, `r_(2) = 1cm`
and speed of water `= 80 cm//s ` i.e., in 1 water flow = 80 cm
`therefore` In 30 (min) water flow `= 80 xx 60 xx 30 = 144000 cm`
According to the question,
Volume of water in cylindrical tank = Volume of water flow from the circular pipe in half an hour
`rArr " " pir_(1)^(2) h_(1) = pi r_(2)^(2) h_(2)`
`rArr " " 40 xx 40 xx h_(1) = 1 xx 1xx 144000`
`therefore " " h_(1) = (144000)/(40 xx 40) = 90` cm in half an hour.