Question

Water is flowing through a cylindrical pipe, of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m/s. Determine the rise in level of water in the tank in half an hour.

Answer 1

Given, internal diameter of cylindrical pipe = 2 cm
So, radius of cylindrical pipe $r=\frac{2cm}{2}=1cm=0.01m$
And,
Radius of cylindrical tank (R) = 40 cm = 0.4 m
Let level of water rise to the height of h cm.
Now, Volume of water flown out of the pipe in one second = velocity $\times$ cross sectional area of pipe = v $\times$ a = $0.4\times \pi \times {\left(0.01\right)}^2=1.256\times {10}^{-5}{m}^3$

Volume of water flown out of the pipe in half an hour = $1.256\times {10}^{-5}\times 60\times 30=0.22608m^3$

Volume of water flown into the cylindrical tank = Volume of water flown out of the pipe in half an hour

$\pi R^{2}h=0.22608m^3$

$\therefore h=\frac{0.22608}{3.14\times 0.4\times 0.4}$

$\therefore h=0.45m=45cm$

Hence, level of water rise to the height of 45 cm.