**Question:**

Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 80 cms-1 in an empty cylindrical tank, the radius of whose base is 40

cm. What is the rise of water level in tank in half an hour?

**Solution:**

Given, radius of tank, r1 = 40 cm

Let height of water level in tank in half an hour = 1 cm.

Also, given internal radius of cylindrical pipe, r2 = 1 cm

and speed of water = 80 cm/s i.e., in 1 water flow = 80 cm

In 30 (min) water flow = 80x 60 x 30 = 144000 cm According to the question,

Volume of water in cylindrical tank $=$ Volume of water flow from the circular pipe

in half an hour

$\Rightarrow \quad \pi r_{1}^{2} h_{1}=\pi r_{2}^{2} h_{2}$

$\Rightarrow \quad 40 \times 40 \times h_{1}=1 \times 1 \times 144000$

$\therefore \quad h_{1}=\frac{144000}{40 \times 40}=90 \mathrm{~cm}$

Hence, the level of water in cylindrical tank rises 90 cm in half an hour.