Question

Water is flowing at the rate of 2.52 km/h through a cylindrical pipe into a cylindrical tank, the radius of the base is 40 cm. If the increase in the level of water in the tank, in half an hour is 3.15 m, find the internal diameter of the pipe. [CBSE 2015]

Answer 1

Increase in the level of water in half an hour, h = 3.15 m = 315 cm
Radius of the water tank, r = 40 cm
Volume of water that falls in the tank in half an hour = πr²h
= π × (40)² × 315
= 5,04,000 π cm³
Rate of flow of water = 2.52 km/h
Length of water column in half an hour = 2.52 ÷ 2 = 1.26 km = 1,26,000 cm
Let the internal diameter of the cylindrical pipe be d.
Volume of the water that flows through the pipe in half an hour = $\mathrm{\pi }{\left(\frac{d}{2}\right)}^2\times 126000$
We know
Volume of the water that flows through the pipe in half an hour = Volume of water that falls in the tank in half an hour

$$\begin{gathered} \Rightarrow \mathrm{\pi }{\left(\frac{d}{2}\right)}^2\times 126000=504000\mathrm{\pi } \\ \Rightarrow {\left(\frac{d}{2}\right)}^2=4 \\ \Rightarrow d=4\mathrm{cm} \end{gathered}$$

Thus, the internal diameter of the pipe is 4 cm.