Question

Question 18
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

Given , radius of tank, $r_1=40cm$.
Let height 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 1sec water flow = 80 cm.
$\therefore$ in 30 ( min) water flow =$80\times 60\times 30=144000cm$
According to the question.
Volume of water in cylindrical tank = Volume of water flow from the circular pipe In half an hour
$\Rightarrow \pi r_1^2{h}_1=\pi r_2^2{h}_2$
$\Rightarrow 40\times 40\times h_1=1\times 1\times 144000\therefore h_1=\frac{144000}{40\times 40}=90cm$
Hence, the level of water in cylindrical tank rise 90 cm in half an hour.