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 per second. Determine the rise in level of water in the tank in half an hour.

Answer 1

We have,

the internal radius of the cylindrical pipe, r = 2/2 = 1 cm and

the base radius of cylindrical tank, R = 40 cm.

Also, the rate of flow, h = 0.4 m/s = 40 cm/s

Let the rise in level of water be H.

Now,

The volume of water flowing out of the cylindrical pipe in 1 sec =πr²h = π × 1 × 1 × 40 = 40π cm³

As,

Volume of water in the cylindrical tank = Volume of standing water in cylindrical pipe

⇒πR²H = 72000π

⇒R^2H = 72000`

⇒ 40 × 40 × H = 7200H

`rArr H = 72000/(40xx40)`

∴ H = 45 cm

So, the rise in level of water in the tank in half an hour is 45 cm.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.