Question

Water is flowing at the rate of 2.52 km/hr 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.

Answer 1

Solution:-

Increase in the water level in half an hour = 3.15 m = 315 cm
Radius of the water tank = 40 cm

Volume of cylinder = $\pi$$r^{2}h$

Volume of the water that falls in the tank in half an hour = $\pi$ $\times$ ${40}^2\times 315$
= $\pi$ $\times$ 40 $\times$40 $\times$315

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

Let the internal diameter of the cylindrical pipe be d.
Volume of water that flows through the pipe in half an hour = π$\times$(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
⇒ 22/7 $\times$(d/2)² $\times$126000 = 1584000
⇒ 22/7 $\times$d²/4 $\times$126000 = 1584000
⇒ d² = 16
⇒ d = $\sqrt{16}$
⇒ d = 4
So, the internal diameter of the pipe is 4 cm