A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 m per second.
Question:

A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 m per second. In how much time will the tank be filled?

Solution:

Radius of the water $\tan k, R=\frac{1.4}{2}=0.7 \mathrm{~m}$

Height of the water tank, H = 2.1 m

$\therefore$ Capacity of the water tank $=\pi R^{2} H=\pi(0.7)^{2} \times 2.1 \mathrm{~m}^{3}$

Speed of the water flow = 2 m/s

Radius of the pipe, $r=\frac{3.5}{2}=1.75 \mathrm{~cm}=0.0175 \mathrm{~m}$

Area of the cross section of the pipe $=\pi r^{2}=\pi(0.0175)^{2} \mathrm{~m}^{2}$

Volume of the water flowing out of the pipe in one second = Area of the cross section of the pipe $\times 2 \mathrm{~m}=\pi(0.0175)^{2} \times 2 \mathrm{~m}^{3}$

Let the time taken to fill the tank be t seconds.

∴ Volume of the water flowing out of the pipe in t seconds

= Volume of the water flowing out of the pipe in one second × t

$=\pi(0.0175)^{2} \times 2 \times t \mathrm{~m}^{3}$

Now,

Volume of the water flowing out of the pipe in t seconds = Capacity of the water tank

$\therefore \pi(0.0175)^{2} \times 2 \times t=\pi(0.7)^{2} \times 2.1$

$\Rightarrow t=\frac{(0.7)^{2} \times 2.1}{(0.0175)^{2} \times 2}$

$\Rightarrow t=1680 \mathrm{~s}$

$\Rightarrow t=\frac{1680}{60}$

⇒ t = 28 minutes

Thus, the tank will be filled in 28 minutes.