Question

Water is flowing at the rate of $2.52km/hr$, through a cylindrical pipe into a cylindrical tank, the radius of whose base is $40cm$. If the increase in the level of water in the tank, in half an hour is $3.25m$, find internal diameter of the pipe.

Answer 1

Increase in the water level in half an hour = 3.25 m = 325 cm
Radius of the water tank = 40 cm
Volume of the water that falls in the tank in half an hour = $\pi r^{2}h=\frac{22}{7}\times 40\times 40\times 325=1634286cucm$

Rate of the water flow = 2.52 km/hr
Length of water column in half an hour = $\frac{(2.52\times 30)}{60}=1.26km=126000cm$

Let the internal diameter of the cylindrical pipe be d.
Volume of water that flows through the pipe in half an hour = $\pi \times (\frac{d}{2}{)}^2\times 126000$

As we know that,
Volume of the water that flows through pipe in half an hour = Volume of water that falls in the cylindrical tank in half an hour
$⟹\frac{22}{7}\times (\frac{d}{2}{)}^2\times 126000=1634286$
$⟹\frac{22}{7}\times \frac{d^2}{4}\times 126000=1634286$
$⟹d^2\approx 16$ $⟹d=4cm$

So, the internal diameter of the pipe is 4 cm